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Jorge Pullin: Okay, so our speaker today is how hey guards will speak about quantization all the volume of the simplest grain of.
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space.
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Jorge Pullin: Welcome everyone Thank you so much for being here.
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Hal Haggard: it's always a great pleasure to discuss these things with you.
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Hal Haggard: Today, I want to tell you about this topic of colonization of the volume of the simplest strain of space and it's an old topic in the Community it's something people have thought, a lot about.
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Hal Haggard: By and large.
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Carlo Rovelli: volumes very.
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Carlo Rovelli: Good me buddy.
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Hal Haggard: Can you hear me any better now.
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Abhay Vasant Ashtekar: We can hear you all perfectly fine maybe it's.
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Hal Haggard: On their end on.
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Hal Haggard: Let me know, Carlo if I need to stop at some point.
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Hal Haggard: I was just saying that this is a very old topic with many people, contributing to it by and York Lewandowski and Carlo and lease Mullen and and many others.
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Hal Haggard: Thomas team on and bruna MN, and so I in some sense need to justify why i'm returning to it.
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Hal Haggard: And I think that there's actually a very interesting thing for us to look into here, which is a relationship between preservative results and non preservative results.
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Hal Haggard: And this theme has been taken up a lot in quantum field theory in string theory in recent years and it's one that I have not seen as much in quantum gravity and that I am really interested in exploring more.
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Hal Haggard: So the work is joint with my undergraduate student here at bard College on to suntan new unto as a junior and a wonderful student a wonderful person I highly recommend you start recruiting him now.
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Hal Haggard: And I thought it would be fun to begin with a lecture the argument that i've always found very intriguing from I buy his book on non preservative canonical gravity.
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Hal Haggard: The book begins the introduction begins with the question of whether there are features of classical general relativity.
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Hal Haggard: That would indicate that non putter a bit of quantum gravity is very different from conservative quantum gravity.
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Hal Haggard: And the The argument is a very simple argument, they give lots of caveats for this argument, but I still find it a very intriguing argument.
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Hal Haggard: So the idea is to consider a massive Shell of charge, he and radius epsilon and to ask about the self energy of this Shell and asked how that self energy varies, as you very epsilon.
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Hal Haggard: So if you ignore gravity all together it's just the bare mass m zero plus the electrical self energy.
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Hal Haggard: Or if you include newtonian gravity there's also the binding energy of the mass and that corrects this this energy and both of these expressions, you can see, are divergent in the limit that epsilon goes to zero.
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Hal Haggard: But there's a fascinating thing about general relativity, which is that it's universal all forms of energy gravitate.
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Hal Haggard: And so, when we consider the binding energy of the mass we shouldn't just use the bear mass, but we should use the corrected gravitational mass.
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Hal Haggard: And so, if you put that into this formula and solve for em as a function of the radius you can do that completely explicitly it's just a quadratic equation.
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Hal Haggard: And you get the the solution on the bottom right here, and the fascinating thing about this result is that it's completely finite in the limit epsilon goes to zero.
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Hal Haggard: The The self energy just becomes the ratio of the electrical charge to the square to the newton's constant.
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Hal Haggard: So this is an indication that non perturb IV quantum gravity may have finite values for parameters that that you wouldn't notice otherwise.
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Hal Haggard: Why do I say you wouldn't notice them, otherwise well, you can also take this exact same result and reorganize it as an expansion in the newton's constant.
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Hal Haggard: So we take that same solution and we expand the solution in treating newton's constant as a small parameter.
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Hal Haggard: And what you see in this expansion is that every single term in the expansion is divergent in the limit epsilon goes to zero so so if you're taking a week gravity perspective on this self energy it looks like you have a badly divergent result.
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Hal Haggard: So I i've long taken this cautionary tale very seriously there's something very interesting about non-paternity of quantum gravity that could be very different from preservative quantum gravity.
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Hal Haggard: And that difference, you know it can show up that because of the universality of.
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Hal Haggard: gravitation and because of the non linearity of the theory those two factors should be a really key part of our quantum theory of gravity.
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Hal Haggard: But for me that was really surprised today that I want to tell you about that I think is interesting for our Community.
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Hal Haggard: Is as a new theme for me, which is that perturbed of divergence is can carry very interesting information and structure.
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Hal Haggard: That will actually inform you about the theory this topic has been studied a lot in recent years by a whole community of physicists and mathematicians and they generally know it as as resurgence so that's the name that i'll be referring to a lot in the talk.
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Hal Haggard: Okay, so, as I was saying it's a very interesting topic with, I think, a very impressively broad.
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Hal Haggard: And impactful take home for everyone and to the best of my knowledge, it hasn't been applied so much in quantum gravity so today, the talk has three parts.
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Hal Haggard: The first talk, I want the first part of the talk, I want to introduce these resurgence and preservative non-prescriptive relations for all of you.
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Hal Haggard: i'm here i'm drawing very heavily on introductions done by Christopher house adri TAO he's and Gerald done from the recent workshop in school.
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Hal Haggard: on applied research and ask them topics, this is through the Newton instant Institute in Cambridge and it's a workshop that i've been participating in.
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Hal Haggard: The second part of the talk, I want to return to the loop quantum gravity approach to the grains of quantum space.
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Hal Haggard: In particular i'll introduce the area geometry of tetrahedron their classical face space and their semi classical quantization these are old result and for much of this audience very familiar.
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Hal Haggard: Your genuine bianca and I worked on these things in in 2011 2011 yes 2011 2012 and so.
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Hal Haggard: I i'll try to go somewhat quickly through that part of the talk, but I think it's important to bring it to front of mind in order to understand.
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Hal Haggard: The results, towards the end of the talk and and please feel free to tell me to slow down if you need any background on that material.
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Hal Haggard: And then, in the final part of the talk i'll start to give you some evidence for these preservative non perturb it of relations in the quantization of the volume of the simplest grain of space a quantum tetrahedron.
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Hal Haggard: So that's the plan for the talk, as I said, the first part, I really need to give you some more details on what is a synthetic resurgence.
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Hal Haggard: So i'm going to give you.
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Hal Haggard: An introduction through to physical examples, on the one hand, you have the classical subject of rainbows and the intensity of rainbows.
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Hal Haggard: And on the other hand, i'm very interested in quantum mechanics and quantum tunneling so i'll also study the near turning point analysis of the schrodinger equation.
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Hal Haggard: and ask about what's happening on either side of this turning point, it turns out that in both these contexts, a really important topic is that of airy functions.
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Hal Haggard: And the function is plotted here on the Center of the slide you see this really interesting behavior so at the zero of X, the area function switches behavior to the right of that zero it.
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Hal Haggard: exponentially decays and to the left of that zero it oscillates and it's a surprising and and hard to understand phenomena, how does a function end up switching between these two behaviors.
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Hal Haggard: And this was relevant for stokes when he was studying so called supernumerary rainbows So in addition to the main bow that you see in this picture, you can see much fainter.
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Hal Haggard: fringes supernumerary rainbows that have smaller intensity, but they are certainly visible and stokes was interested in the amplitude and position of the supernumerary rainbows he wanted to be able to predict them very accurately.
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Hal Haggard: On the other hand, in the quantum mechanics we know exactly what this double behavior of the area function is.
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Hal Haggard: to the left of the turning point is the classically allowed region where the potential is smaller than the energy that you're considering and in the classically allowed region, the wave function oscillates.
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Hal Haggard: Well, to the right of the turning point you've entered a classically forbidden region where the energy is below the potential below the barrier.
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Hal Haggard: And you can only access that via tunneling effects and you should expect an exponential decay, so this is it's clear the area function is a very good description of the near turning point behavior of the shooting or equation.
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Hal Haggard: Okay, so let's let's go into this in some more detail let's take the schrodinger equation and expanded around a turning point.
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Hal Haggard: i'll take coordinates, such as the turning point, or at the origin and my coordinates and in this expansion, you see that the East side terms cancel out and we're left with areas equation this interesting differential equation, with the independent variable also in on the right hand side.
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Hal Haggard: Here i've kept the constant alpha explicit just so you can see the H bar dependence of this function.
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Hal Haggard: And web theory suggests that we we introduce an action in order to study this shirting or equation, so let me take this coefficient of sigh on the right hand side ticket square root that's the usual double wk be integral and.
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Hal Haggard: And and integrated up and I find an action coordinate that I can use to study this problem, so this action, coordinate it goes as two thirds X, to the three half's with the appropriate scaling.
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Hal Haggard: So we're going to take a wk beyond sites consider an exponential of this action divided by the square root of the action the derivative the action.
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Hal Haggard: and taking that on sites will will look for power series solution where I use the action coordinate as my expansion.
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Hal Haggard: So if you put this full on insights into the schrodinger equation, you find a recursion relation, for the coefficient CN and that recursion relation is easily solved in terms of oilers gamma functions.
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Hal Haggard: And this highlights a very interesting feature of this.
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Hal Haggard: This expansion, the solution, this is not a convergence series, this is not a standard power series where you expect every term to get.
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Hal Haggard: Smaller and you're converging to the answer instead it's divergent series web series and the divergence, you can see it here as the fact that these gamma functions go as factorial.
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Hal Haggard: And so I have two factorial is in the numerator and one factorial in the denominator so as n gets large this things growing factorial, which is a very fast divergent grows, so this is the character of acid tonic series.
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Hal Haggard: I wanted zero in on these coefficients because it turns out they're very interesting and surprising.
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Hal Haggard: So let's take the coefficients with the plus sign you notice I got two solutions to my recursion relation one with a plus sign one with an alternating sign.
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Hal Haggard: let's take the coefficients with the plus sign and let's just calculate the first few coefficients so C zero C one C two C three and you just get these large ratios of integers by rational numbers.
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Hal Haggard: But let's do a second thing, and this is the surprising thing to do, let's also consider these coefficients in the limit of large in in that limit, we know the the gamma functions can be written as factorial, and so one of them cancels with this n factorial and the denominator.
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Hal Haggard: And I have the.
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Hal Haggard: Divergent.
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Abhay Vasant Ashtekar: factorial growth, I was already mentioning.
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Hal Haggard: And perhaps it's natural to assume well i'm taking large N, so I should do an expansion in one over N, and if you do that, you get these coefficients i've highlighted in blue.
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Hal Haggard: It turns out, this is not the most interesting way to look at these coefficients in the large and limit there's a second way that that's much more productive for looking at them.
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Hal Haggard: let's reorganize in a in a series of factorial decay, that is let's put a one over N minus one here, such that that N minus one cancels the N minus one of the factorial.
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Hal Haggard: And then one over N minus one times N minus two for the second one that will cancel the first two terms of the factorial and organize the series that way.
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Hal Haggard: If you do that, and you pull out a factor of four thirds that four thirds is coming from the four thirds in this denominator here, you see something extremely striking, which is the exact same coefficients that I saw at lowest order are appearing at the high order expansion, so this series.
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Hal Haggard: Not only are those coefficients appearing, but they are appearing with an alternate sign you see it's minus plus and it would go on like that.
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Hal Haggard: These are precisely the SI N minus the low orders of the CN minus.
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Hal Haggard: So somehow the the series coefficients knows both about the solution with CNN plus and the solution with the CN minus and they appear at opposite ends of this expansion.
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Hal Haggard: So the immediate question is why what's going on here How is this possible, and there are many answers to that question and they're.
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Hal Haggard: very striking in different ways, but i'm going to try and give you an answer, using integral representations and thinking about integral representations.
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Hal Haggard: So it's a quick check that if you take this integral representation of the aerie function and plug it into Aries equation it indeed solves the equation.
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Hal Haggard: And so i'm going to study this integral representation, it turns out as soon as you have an integral representation it's very convenient to extend everything to the complex domain.
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Hal Haggard: So, in particular i'm going to think of the every function as a function of a complex Z and i'm going to also extend the integration parameter T to a complex domain.
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Hal Haggard: Having made those extensions you get the I can do plots like that, on the bottom left here, this is a plot in the complex T plane and.
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Hal Haggard: The colors represent the value of the real part of the exponent so blue are small values of the real part and yellow are large values of the real part.
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Hal Haggard: This is organized in such a way that the integral is exponentially decaying as you go off into a blue domain and exponentially growing as you go off into a yellow domain.
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Hal Haggard: So the blue domains are the domains of convergence of the integral they're going to be the things that we're very interested in.
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Hal Haggard: But the complexity of the problem in the interest of the problem comes when you recognize that i'm going to allow Z to be complex as well, allowing Z to be complex means that for different values of Z this picture looks totally different.
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Hal Haggard: So I i'll take the Z to have the the polar form are either the theta and the critical points are go like plus or minus route Z that's an easy check you just take the T derivative of the exponent and solve it, so we have things that go like plus or minus either the I stayed over to.
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Hal Haggard: So i'm going to study this integral and i'm going to study it along the contour gamma one, this is the definition of the area function, the Ai.
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Abhay Vasant Ashtekar: Could you just tell us again what gamma one was i'm sorry I just lost something.
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Abhay Vasant Ashtekar: yeah.
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Hal Haggard: I hadn't I hadn't said it, yet I buy so I need to say.
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Hal Haggard: It now i'm the gamma one is the contour that defines the function, so I was saying, the blue regions are the regions where the integral converges.
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Hal Haggard: And so we're going to go from one region of convergence to another region of convergence, if we went to the same region of convergence, we get a trivial integral so we're going to go from one to another in a non trivial fashion.
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Hal Haggard: And the functions just defined as the one that starts in the lower left region and ends up in the upper left region.
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Hal Haggard: So this is the contour that defines the the.
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function.
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Hal Haggard: But what I want you to notice is that that contour and the integral that results depends on the value of theta because everything in the picture depends on the value of theta.
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Hal Haggard: So the picture I have drawn for you here is for real positive Z that is for theta equals zero so when fade equals zero you get the picture i'm showing.
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Hal Haggard: And, and you get the the kind of behavior that we saw in the plot of the every function, you get an exponentially decaying behavior.
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Hal Haggard: Where does that exponential decay come from you can think of it as a saddle point domination.
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Hal Haggard: The saddle point is real and negative in this case, and that leads to an exponential decay, I have not showing you all the details of how it leads to an exponential decay it's it's a bit intricate but that's the case.
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Hal Haggard: Okay now let's change data.
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Hal Haggard: So i'm just showing you this plot again the extra waves so far looked at the behavior to the positive real access the exponential decay behavior.
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Hal Haggard: But let's continue into the complex plane of Z that is let's increase data and see what happens.
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Hal Haggard: And i'm going to increase the theta incrementally i'm not going to go straight to the negative real access i'm going to go through some values so turns out the first really interesting value is when fade equals two pi over three.
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Hal Haggard: You can see that, on the right hand side here and, as I mentioned the real part of the exponent changes, so the yellow and the blue regions have shifted and change behavior.
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Hal Haggard: Along with the the real part changing the critical points move as well, and so at two pi over three the critical points are aligned in such a way.
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Hal Haggard: That the contour gamma one that you want to go from the lower left convergent region to the upper left convergent region, no longer goes through just one saddle point.
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Hal Haggard: The steepest descent contour now actually goes through to saddle points in order to get to that region, it goes up to this point continues along a flat region to the second subtle point and then down.
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Hal Haggard: This is just the control that defined theory function that's what we mean but we see that there's now a change in behavior in the saddle point domination to saddle points are contributing.
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Hal Haggard: This is very interesting notice these two saddle points are both complex values.
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Hal Haggard: This is exactly what we've been discussing a lot in the I O que je s this this semester it's been coming up again and again, how do we think about complex saddles in the spin phone path integral.
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Hal Haggard: And this area function is showing the complex saddles matter they matter, as you very the parameters of the problem.
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Hal Haggard: Okay So here we see these two contributing let's keep going and consider well actually I wanted to show you how this change the behavior of the area function.
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Hal Haggard: So now i'm plotting the real part of the query function over the complex Z plane remember the area functions a function of Z.
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Hal Haggard: And you can see, we started out looking at the real axis, which goes kind of down into the right here and that's the exponential decay.
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Hal Haggard: Having gone all the way to fade equals two pi over three we actually have completely shifted the behavior the area function we're now climbing this pink.
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Hal Haggard: Part of the the figure and and you see an exponential growth, so the turning on of the second saddle has resulted in an exponential growth of the area function.
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Hal Haggard: Okay let's go all the way, please.
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question.
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But for for values of.
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The path.
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That is.
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New regions in the lower left and the upper left.
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Our with at the same location as.
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Hal Haggard: It makes sense to deform these parts.
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Hal Haggard: yeah that's a surprising that's a surprising feature I think it's worth getting.
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Hal Haggard: Talking I can't see you, but I think.
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Wolfgang Wieland: Ah yes.
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Hal Haggard: And that's a surprising feature Wolfgang of this particular problem, the atom tonic regions, the gamma one connects do always turn out to have the same location in this plot.
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Hal Haggard: In general, if I was doing this, if I was varying Z and I was looking at a general problem I would actually have to track my contour continuously as I varied Z in order to know what my contour should be does that make sense.
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Hal Haggard: As I very fade I would actually have to track track the contour.
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Hal Haggard: Because, as your as you're thinking, the acid toxic domains my move as I very theta.
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Hal Haggard: But in this particular problem they don't.
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Hal Haggard: it's one reason it's a nice example.
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Hal Haggard: So if we go all the way to fade equals pi so now we're going along the negative real access what you see is the picture on the right.
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Hal Haggard: And if you follow the contour as we were just discussing it turns out the way that you have to get from the lower left ask them tonic region to the lower upper left or as some tonic region.
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Hal Haggard: is to go first over the first saddle and then follow that back along the second saddle along the second contour.
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Hal Haggard: And so the aerie function has turned into a function, which is a sum of too complex exponential see the saddles are both complex values now.
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Hal Haggard: Some of too complex saddles which turns out to give you a sign function si any function that oscillates along the negative real axis.
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Hal Haggard: So the saddle point domination actually enlightens you as to why the function has these two different contributions.
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Hal Haggard: So here's a plot of the area function in blue and the to ask them tonic approximations in the two different regions, you see, at the turning point, the wk be analysis diverges that's not surprising.
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Hal Haggard: that's the normal behavior, but these two as empowered approximations are incredibly good at getting the aerie function.
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Hal Haggard: This was why stoke study them if you go back and look at stokes work he was trying to find these oscillations.
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Hal Haggard: Using a convergence series, and he was only able to get this first dip.
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Hal Haggard: After doing 19 terms and the convergence series, so he had to by hand figure out 19 terms and the convergence series Meanwhile, the first term in the ass and tonic series nails that like super super well.
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Hal Haggard: So this is the value of divergent series they're they're very good at getting quickly to to the result that you're interested in.
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Eugenio Bianchi: Okay, yes, can you can you can I clarify the relation between not so beautiful as interview series, and what we learn out when we study for standard web where we do.
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Eugenio Bianchi: The classical allowed region, the classical the forbidden region, and then in between we match and we learned that we do have to match rebury function to succeed yeah you're not doing that, so it can you can you clarify the relation with what is the matching yeah.
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Hal Haggard: So it's it's a very the picture on the left is addressing this question, to some extent it's a it's a very surprising, but true result that.
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Hal Haggard: it's exactly as you described it's a patching right so in the vicinity of a turning point, we can always approximate the potential by a linear potential.
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Hal Haggard: And this area function gives us an exact solution right an exact solution to the schrodinger equation, and so you can say well okay.
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Hal Haggard: We could try to match that exact solution to the way function that we have in the classically allowed region, but that matching well it's kind of difficult because the every function is a complicated function.
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Hal Haggard: So what do you do you treat the region to the left, ask them topically so you actually pretend like it's very far from the turning point where you're gluing.
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Hal Haggard: And, and you expand it and then that's an expansion in terms of functions that you already were working within the classically allowed region, and you can easily match coefficients.
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Hal Haggard: So the thing that surprising about it, you think when you think asked him tonic Oh, the parameter must be huge.
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Hal Haggard: But as i'm showing you in this plot is very relevant to your question this plot the atom tonic series is extremely good even very close to the point where the the.
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Hal Haggard: divergence the turning point in this case is happening, and so you actually don't have to go very far in the parameter to match effectively with the awesome topic series.
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Hal Haggard: So it is exactly that role that this is is playing.
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Hal Haggard: But there's something even more beautiful I think happening here, which is that we're seeing that the the behavior in the classically forbidden region, the behavior in the classically allowed region aren't actually independent.
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Hal Haggard: The the there's one saddle contributing in the classically forbidden region that same saddles contributing in the classically allowed region it's just that we've added on a second saddle.
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Hal Haggard: So why, why is that happening, how does this work i've kind of hopefully convinced you that it works, but the question remains why.
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Hal Haggard: And so the y is really fascinating and it helps to go back and look at the differential equation again, so this is a second order oh do.
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Wolfgang Wieland: O D.
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Hal Haggard: And we know second order ODS have to independent solutions right.
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Hal Haggard: So it must be the case that if I look at the every function along some rotated plane right I shift trying to look at a rotated area function.
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Hal Haggard: It must be the case that this solution can be expanded in the basis of solutions that we already had the aerie function and the bi airy function, which is just another contour in that plane.
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Hal Haggard: And so it must be a superposition of these two, so the reason that the second saddle started contributing is that it was always contributing.
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Hal Haggard: The the behavior of the function always had both saddles in it.
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Wolfgang Wieland: it's just that the second.
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Hal Haggard: saddle was exponentially suppressed, with respect to the dominant saddle.
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Hal Haggard: And so, this is what we call a non preservative connection formula it's an expansion that tells me the function in terms of both of the saddles it, regardless of the exponential dominance of one of them over the other.
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Hal Haggard: So that's what's always happening all the saddles are always contributing it's just that they're exponentially suppressed sometimes.
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Hal Haggard: Okay, so this is an example of a preservative non prohibitive connection the exponentially sub dominant thing is what we would call a non prohibitive effect a tunneling effect if you like.
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Hal Haggard: And the exponentially dominant thing is what we would call a classical effect right a preservative effect.
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Hal Haggard: And it turns out they're always both contributing to any analysis and it's just whether you can dig out the exponentially sub dominant piece, as to whether you can see that non preservative physics going on.
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Hal Haggard: alright.
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Hal Haggard: So Gerald Dan has this beautiful mnemonic image for this, so he is this is actually precise, in a sense, as.
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Wolfgang Wieland: well.
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Hal Haggard: The wavelengths surrounding droplets of water, so you can think of the droplets themselves as critical points critical points in the path integral, for example.
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Hal Haggard: And what you find in these resurgent analyses, is that if I expand around one of those critical points, the late terms in that expansion are actually the early wavelengths around the neighboring critical points.
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Hal Haggard: And so you can think of any expansion is being made up of late and early terms which are related critical point to critical point.
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Hal Haggard: So this is.
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Hal Haggard: going to be a great physical interest in studying the quantum mechanics of the grain of a simple grain of space.
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Hal Haggard: So i'm going to move on to that topic unless people have questions.
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All right.
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Hal Haggard: So let me just start at the classical side, this is very familiar to many of you, but we typically describe a geometry.
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Hal Haggard: In terms of links, if you like, in GR we use the metric and think about the metric is giving us the length of paths.
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Hal Haggard: or in terms of a discrete geometry, we can think about the triangle and describing a triangle by its edge links.
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Hal Haggard: But herons formula gives us a beautiful connection between the edge link description and the area description, and you can go back and forth, you can think about the area directly as your variable.
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Hal Haggard: This turns out to be an important perspective because of the classical result of Herman makovsky is that you can completely describe a tetrahedral geometry by its area vectors so vectors that are perpendicular to the faces of the.
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Wolfgang Wieland: tetrahedron and whose magnitudes or equal to the.
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Hal Haggard: Area of that face, in particular with makovsky showed is that there's a one to one correspondence between tetrahedral geometries.
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Hal Haggard: And closure relations where the four area vectors of the tetrahedron add up to zero, so this is a classical result and you could just think of it as a geometrical result.
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Hal Haggard: But what loop quantum gravity ads in that I think is very interesting is an interpretation of these area vectors as so three Lee algebra elements as angular momentum.
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Hal Haggard: When you interpret these vectors as angular momentum, you can think of them as generators of transformations.
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Hal Haggard: And, in particular the closure relation takes on this beautiful double meaning, on the one hand it generates overall rotations of the tetrahedron.
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Hal Haggard: And what we're saying is that the geometry of the tetrahedron doesn't change when you rotate it so that's why the closure is zero.
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Hal Haggard: On the other hand, it represents a sort of area flux flux of this discrete geometry, and we can think of it as as the gauss constraint of discrete geometry, so this is beautiful thing of interpreting these vectors so really algebra elements.
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Hal Haggard: Okay, but at first, you might be worried about the parameter account, so we know by counting edge links have a tetrahedron that there are six independent parameters for the shape of the tetrahedron.
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Hal Haggard: The four area vectors have 12 components and so they're clearly overkill and if we just worked with their magnitudes there's only four magnitudes and so they're insufficient.
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Hal Haggard: But closure this relation that the some of the areas equals zero gets us out of this quandary of parameter accounting.
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Hal Haggard: We seek rotational invariance that's what the closure relation was telling us, and so.
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Hal Haggard: You can, in addition to the four areas, you can add any two dot products, if you like, and you precisely describe the geometry, or you could add one dot product and one triple product.
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Hal Haggard: And these are all rotational invariance and they're perfectly good ways of describing the tetrahedron.
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Hal Haggard: So it turns out i'm going to find it very convenient to work with the magnitude of the sum of two of the area vectors.
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Hal Haggard: And with the triple product, one of the reason, where i'm particularly interested in the triple product is it turns out that this triple product is just the volume squared of the tetrahedron.
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Hal Haggard: So i'm going to use that quantity so much that I give it its own name i'm going to call it Q so Q is the volume squared of the tetrahedron.
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Hal Haggard: We can take the closure relation and just interpret it as a Victorian some if we interpreted as a Victorian some we just get for vectors tip detail the add up to zero.
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Hal Haggard: And we can describe the shape of those four vectors how those four vectors point in space by two coordinates one coordinate is the angle between the Left triangle here and the right triangle will call that angle fi.
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Hal Haggard: And the other variable is the length as I was already mentioning of the diagonal here.
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Hal Haggard: It turns out that if you keep track of those two coordinates the topology of the face base of shapes is just a sphere.
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Hal Haggard: And the fly is like a Q coordinate on that sphere and the magnitude of this diagonal is like a Z coordinate on the sphere.
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Hal Haggard: And we have a very beautiful description of the face space of tetrahedron, so this is a any point on this sphere corresponds to a unique tetrahedron.
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Hal Haggard: i've fixed to the magnitudes of these vectors in the background and the remaining two coordinates are just the fly and the magnitude of the diagonal.
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Hal Haggard: In particular, since we're interpreting these area vectors as so really algebra elements, they have a natural plus on bracket and we can compute the POs on bracket of these coordinates and it turns out that there are canonical coordinates on this face space.
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Hal Haggard: So this is the description of classical tetrahedron that i'll use.
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Hal Haggard: What you Jenny and I did in 2011 was to consider the evolution generated by the volume on this face space so consider the volume as a function on this.
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Hal Haggard: spherical face space, again I fixed the background areas, you can write it out explicitly the volume squared is proportional to the sign of the angle.
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Hal Haggard: And these deltas are the areas of these two triangles the orange triangle and the left and the pink triangle on the right, I can get the areas of those triangles via herons formula.
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Hal Haggard: And they're just in terms of the magnitude a one a two a three or four and the diagonal length, so this gives a completely explicit function on the face space.
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Abhay Vasant Ashtekar: So can ask the question I mean is this some simplification occurring or not and i'm always confused about this point that's why i'm asking for clarification is the face face really compact, or are we adding some points, or is this completely faithful pace pace or you're doing something.
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Hal Haggard: it's completely faithful there's no loss of information here at all the.
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Abhay Vasant Ashtekar: gain up information either.
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Abhay Vasant Ashtekar: You know.
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Abhay Vasant Ashtekar: we're not adding points that people make Thank you.
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Hal Haggard: And I have to be a little careful ebi there's a subtlety here the if you consider parameters such that your tetrahedron could be flat that.
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Hal Haggard: degenerate configuration, then the North Pole or the sphere actually becomes a cusp and the manifold becomes singular at that point okay so degenerate configurations are subtle special case in this space okay.
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Abhay Vasant Ashtekar: So if I leave all the degenerate configurations that the claim is that this is a complete one to one mapping that exists face faces come back and this is a faithful representation exactly Thank you.
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Hal Haggard: Having it.
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Hal Haggard: introduced canonical coordinates on this face space we can study their evolution under this volume operator.
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Hal Haggard: So i'm going to call the parameter that's conjugate to the volume or really the volume squared Lambda so Lambda is a flow parameter along the curves of constant volume.
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Hal Haggard: And I can ask how my court canonical coordinates my shape parameters and my tetrahedron evolve under that Lambda flow.
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Hal Haggard: This is what you Jenny and I studied and it turns out it's an integral system it's an amazing fact I would not have guessed that, at the outset.
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Hal Haggard: But the volume evolution of the magnitude of the diagonal squared turns out to be the square root of a quadratic polynomial.
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Hal Haggard: The quadratic polynomial has the sort of bear roots roots when Q equals zero that are given on the right here, but when you add in a non zero volume those bear roots get perturbed and you have a true.
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Wolfgang Wieland: evolution by aquatic law here square root of aquatic.
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Hal Haggard: So the solutions to this, you can find explicitly and they're given by jack kirby elliptic functions these so called SN functions.
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Hal Haggard: What does this mean it means that we can actually watch the volume evolution.
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Hal Haggard: So i've made a video here on the left is the face space we were just discussing and I put a point on that face space and allowed it to evolve along the curve of constant volume.
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Hal Haggard: On the right are just two different views of the tetrahedron that corresponds to this face space, so this is exactly the same tetrahedron just viewed from two different angles.
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Hal Haggard: And what you're seeing is an evolution of the shape of that tetrahedron that preserves the volume exactly and preserves the four face areas of the tetrahedron it.
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Hal Haggard: All you're seeing in this evolution is a change in the shape of the tetrahedron.
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Hal Haggard: alright.
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Hal Haggard: So you're genuine I use this to perform a bar sommerfeld quantization what we did was to compute the action along one of these volume contours.
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Hal Haggard: And set that action equal to N, plus a half times to page or times clunks constant and to require that the that this identity holds and solve for the V ends that saw that provides such.
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Hal Haggard: an identity, the, the result is the black dots here i'm plotting everything versus GA and number, which is the same number that's giving rise to these four.
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Wolfgang Wieland: areas.
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Hal Haggard: it's just a way to make this plot one dimensional.
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Hal Haggard: So the black solid dots are the bar sommerfeld quantization of these volume evolution contours the open circles are the result of a beautiful analysis that began with a by in Europe, many, many years ago.
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Hal Haggard: Where you quantifies this via spin networks and you find the volume quantum mechanically.
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Hal Haggard: I can do that numerically and when I do that numerically I get the open circles and you see an astonishing agreement between the bar sommerfeld values and the numerical results, so this is what you Jenny and I did some years ago.
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Hal Haggard: The first new result that I can report for all of you today is with onto with my my student and it's to actually give you explicit form for the Web wave function of.
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Hal Haggard: This exact same problem, so the wave function is in a.
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Hal Haggard: A representation that is i'm going to use the the recoupling parameter the length of the diagonal as my way function parameter so that's plotted along the horizontal here.
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Hal Haggard: And, and you can solve explicitly for the action integrals that you need to compute the wk be wave function.
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Hal Haggard: Those action integrals are elliptic functions that elliptic integrals of the third kind they're called these pies here.
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Hal Haggard: But everything here is completely explicit I can give you all of it in complete detail and, and if you put it into this formula, you find the the solid blue line here which, in the classically allowed region is in extraordinarily good agreement with the numerical values that you get.
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Abhay Vasant Ashtekar: We can do this is sort of specific way function.
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Hal Haggard: yeah this is for specific values of the volume and the for face areas.
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Abhay Vasant Ashtekar: But the way functions or specific site key.
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yeah.
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Abhay Vasant Ashtekar: we're doing.
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Abhay Vasant Ashtekar: Side sake, you have key where did that come from sorry.
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Hal Haggard: Are you asking how I got it or i'm not sure yeah.
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Abhay Vasant Ashtekar: So the question is, are you just picking up a function site Q amp T, which is given by right hand side, or does it have was another.
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Abhay Vasant Ashtekar: fundamental significance.
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Hal Haggard: i'm taking one of the values that you're genuine I found for the quantization of the volume and i'm fixing that Q and then i'm asking for an Eigen state of the volume operator.
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Abhay Vasant Ashtekar: that's what I need is an against their the one who opened it up right exactly.
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Hal Haggard: So this is in.
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Abhay Vasant Ashtekar: This is a web application and.
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Hal Haggard: i'm showing you the version that's valid in the classically allowed region we're also able to find the exponentially decaying one in the classically forbidden region.
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Hal Haggard: alright.
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Hal Haggard: So we have these results on a quantum green of space that were completely semi classical and I need to emphasize that the results both that you Jenny and I had.
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Hal Haggard: And that onto and I have our first order in each bar.
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Hal Haggard: So these results are the lowest order semi classical quantization I still think they're remarkable results because.
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Hal Haggard: You can study this thing analytically so we can actually look at the volume Eigen states and loop quantum gravity as analytic functions of all the parameters so that's fascinating but it's a lowest order in each bar approximation, and so the question is, can we go further than this.
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Hal Haggard: So what I want to tell you about now is what a the this resurgent approach to to the volume of a green of space and, in particular, I want to try and see if there are alternative non-paternity relations that we can find in this problem.
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Hal Haggard: So the first part of the talk, I told you about integral representations and about how we can understand the different saddles and an integral representation, contributing to some quantum mechanical.
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Hal Haggard: object of interest there's a second form of resurgence, which is known in their community as constructive resurgence and it's a version of resurgence, where you can do everything very explicitly you're not just thinking.
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Hal Haggard: term by term and in each bar expansion, but you're actually trying to find a full preservative non-conservative relation analytically.
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Hal Haggard: These results are constructive and beautiful for that property but they're restricted they're restricted to face spaces that are of low dimensionality.
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Hal Haggard: So i'm going to switch gears or i've been talking a lot about low dimensional systems, but i'm going to very much focus on low dimensional systems now.
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Hal Haggard: In particular i'm going to think about a face base, which can be seen as a remote surface.
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Hal Haggard: So for remote surfaces there's many, many results, and in particular for genius one remind surfaces there's a huge number of beautiful results.
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Hal Haggard: So what's a canonical example you might have in the back of your mind and thinking about this, you could consider a quadratic oscillator i'll take the symmetric quarter oscillator but you'd actually don't need to you can study more general quarter oscillators as well.
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Hal Haggard: The quarter oscillator you can think of picking some energy, and you know very well that behavior you would get an awesome story behavior in the left, well, and also the Tory behavior in the right well.
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Hal Haggard: The energy states that correspond to the two particles localized in those two have degenerate energies at the lowest order in each bar but there's a preservative sorry, a non preservative tunneling effect that actually splits that energy to that.
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Hal Haggard: That degenerate energy and gives you an exponentially small difference in energy Eigen states.
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Hal Haggard: Of this quarter well, so this is a canonical example of a remonde surface of genius one i'm going to show you why that is.
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Hal Haggard: But I wanted to point out that, actually, we have a very analogous analysis in the quantum mechanics of the volume of a grain of space.
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Hal Haggard: I told you that the evolution was the square root of a quadratic and I showed you that core deck if I take Q equals zero in this expression and plot the aquatic that results, I get the blue curve.
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Hal Haggard: Okay, on the other hand, if I allow Q, to be non zero the roots of that quarter are going to shift, but an easy way to visualize where the roots shift to is just to draw the linear.
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Hal Haggard: term that is this being subtracted off your linear in a squared I should say that linear term intersects this blue curve and four points and gives us the new route of the full Court, I see that something popped up in the chat is there a question.
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Hal Haggard: Yes, the question is, it seems that the queue operator plays the role of the hamiltonian in the face space and wondering how to define time evolution in this theory.
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Hal Haggard: So it is playing the role of the hamiltonian and the definition of time evolution is exactly what I showed on slide 22 here so time evolution is evolution in the.
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Hal Haggard: coordinate that's conjugate to Q, so the way you generate it is just via pass on brackets so define a Lambda is passed on bracket of five with Q that's an explicit formula, if you solve that formula, you know how five varies with Lambda so that's what i'm doing in this analysis.
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Okay.
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Hal Haggard: So we have this to cortex on the left of the symmetric oscillator on the right, the volume evolution.
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Simone SPEZIALE: i'll just to make sure i'm not missing something.
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Simone SPEZIALE: But now it's Jim zero right and then considering it from your description okay.
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Hal Haggard: yeah so far, nothing, nothing interesting what am I going to do we noticed that the integral representation got interesting when we complexity five variables.
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Hal Haggard: So i'm going to study complex defied classical mechanics if you've never encountered this before it's a really strange idea turns out to be immensely useful and quite beautiful.
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Hal Haggard: So if we think about the symmetric oscillator this was the momentum was the square root of a quadratic but that that's a square root, it must have singularities.
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Hal Haggard: The singularities are going to be branch points and they're given by the roots of the core deck So if I continue my variables into the complex domain I get branch cuts in fact I get for.
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Wolfgang Wieland: branch points, one for each route of the quarter quick.
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Hal Haggard: And I can connect to those branch points by branch cuts.
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Hal Haggard: We know what it means to introduce a branch cut it means you're moving to a remote surface, we have to slit along the branch cut and glue corresponding sides so here's two copies of the complex Q plane.
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Hal Haggard: I slid along the branch cuts I flip the bottom one upside down and I can glue along the branch cuts.
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Hal Haggard: And you see that I get something well it's not quite a tourist is not quite genius one in the traditional sense, because these two planes are non compact, but if we make a one point compacted vacation you see that this is actually a Taurus.
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Hal Haggard: So this is the genius one reman surface that is hiding behind the quarterback oscillator system it's in the complex defied face face.
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Hal Haggard: The same exact thing happens for our volume evolution or volume evolution has four routes, we just saw what they were they were these yellow dots.
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Hal Haggard: Those four routes act as branch points for this square root, we can take the sphere that I showed you a moment ago and slit along those branch cuts.
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Wolfgang Wieland: We make a second copy.
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Hal Haggard: We rotate those two copies and glue them along the branch cuts in the appropriate way and we obtain exactly it tourists, so the remonde surface that i'm considering is in the complex defied a five space and the complexity vacation leads to a surface that's a Taurus.
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Hal Haggard: This tourists turns out to be fascinating.
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Hal Haggard: The on the one hand, we know the topology of contours on a tourists there's an A and a B cycle or an Alpha and beta cycle i'm going to call them today.
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Hal Haggard: And in the quarter oscillator we know what the first cycle represents if I go back to the quarter picture if i'm doing an action integral between these two turning points, those are exactly the roots, I was just mentioning that's going to be like doing an integral around the a cycle.
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Hal Haggard: That integral around the cycle of pdq is exactly the bar summerfield action if I go all the way around it's the bar summer full batch and that's the thing that's going to tell me about the contest values.
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Hal Haggard: But we know the Taurus has a second contour and, naturally, independent contour which is beta the B cycle here and we can introduce a dual action and integration of the exact same quantity, the pdq remains exactly the same, but we're integrating it along a top illogically distinct cycle.
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Hal Haggard: And this, it gives rise to what i'll call for the rest of the talk and what's known in the literature as the dual action so that's why ad dual action.
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Hal Haggard: And the dual action is exactly what determines the exponentially small splitting of the degenerate energy levels of the cornick the symmetric Quebec so this thing controls the non preservatives tunneling effect that connects the two wells.
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Hal Haggard: Okay, in fact, let's turn to algebraic topology and borrow a really nice result, the number of independent cycles on an algebraic manifold is exactly equal to the number of independent one forums on that manifold.
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Hal Haggard: So, in the case of a 1d complex manifold with genus G and with P punctures pieces singularities.
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Hal Haggard: This is the number of independent cycles is just two times the genus plus two times the number of punctures minus one.
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Hal Haggard: And what do we mean by independence of one forums, we mean the independence is up to an exact form.
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Hal Haggard: So, if we take m equals one.
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Hal Haggard: Sorry that's a that's a typo that should be a G equals one a genius one case.
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Hal Haggard: And we let this one form theta be pdq.
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Hal Haggard: Then, only two G plus two P minus one of the theta and the energy derivatives of the theta that is i'm going to generate a whole tower of one forums.
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Hal Haggard: How am I going to generate the Tower i'm going to take feta and i'm going to take the energy derivative of its pre factor i'm going to get a new one form and I generate K of them.
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Hal Haggard: If you take all of them together theta and all its derivatives, they can't be independent, at some point, I have to run out of independence, because I know there are only this many that are independent So what do we get we get a linear.
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Hal Haggard: Sorry, the word is escaping me a linear relationship amongst the vedas What do I mean I mean if you take coefficients times the vedas add the whole thing up, it must be an exact form.
347
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Hal Haggard: here's where the power of having complex defied things comes in, we can take this relationship that we know just on algebraic grounds, must be there and we can do contour integrations of it.
348
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Hal Haggard: Take a closed cycle and integrate this linear dependency there, it is a long gamma well the exact form integrates to zero, the coefficients are only energy dependent, so I can pull them out.
349
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Hal Haggard: And Lo and behold, I have a relationship amongst the.
350
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Hal Haggard: action and the derivatives of the action that holds for any close cycle on the manifold.
351
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Hal Haggard: This differential equation coefficients times derivative of the action with respect to energy up to some K this differential equation that results is known as a postcard Fuchs equation.
352
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Hal Haggard: This is a famous result but I never appreciated what it meant in physics until recently.
353
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Hal Haggard: So what does it mean, how do we think about this here's the picard Fuchs equation for the quarter the symmetric work that I was just telling you about.
354
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Hal Haggard: And the idea is actually very simple, if you take an energy level, and you imagine very in that energy level a little bit because you know the potential the turning points evolved in a very systematic way.
355
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Hal Haggard: We can even find how the turning points change as we change the energy, but that means if the turning points are varying in a systematic way the actions must be varying in a very systematic way, and you can actually figure out how the action is varying as you move that curve.
356
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Hal Haggard: So what are we learning we're learning that there are not an infinite number of actions on any manifold of genius one there's only a few the independent solutions of the picard Fuchs equation Those are all the possible actions on the manifold.
357
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Hal Haggard: And it's better than that because we have a differential equation and differential equations can satisfy certain constraints.
358
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Hal Haggard: For example, RON skin relations you take independent solutions you take the determinant and you have a wrong skin condition on the solutions of this equation.
359
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Hal Haggard: And we get a connection between the classical action, a and the dual action ad that is just a preservative non preservative relation that has to hold.
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Hal Haggard: there's no way out of it, the tunneling effects encoded in ad and the spectral effects encoded in the bar sommerfeld action are related to one another there's another question in the chat.
361
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Oh it's just somebody saying they have to go.
362
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Hal Haggard: um, so this is what's called a non preservative connection formula it's a relationship between non preservative effects like tunneling and preservative effects like for some heartfelt quantization.
363
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Hal Haggard: So what is the first additional result that I can tell you all today it's a result that onto and I have.
364
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Hal Haggard: Which is to find a card Fuchs equation for this problem that we've been discussing that quantization at the green of space.
365
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Hal Haggard: So, as I very cute very that yellow line that's going through the core deck but I very in a very systematic way, and so the turning points of the evolution very in a very systematic way and it turns out there's a third order the card Fuchs equation sitting behind.
366
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Hal Haggard: So now we know that there are only three independent actions that are possible on this genius one remote surface.
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Hal Haggard: turns out you'll notice that, while it's a third order differential equation in this case there's no non derivative term, and so one solution is just a constant.
368
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Hal Haggard: This constant turns out to be very interpretive there are poles on our remote surface, but the integral around a Pole, is a constant.
369
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Hal Haggard: And so, that is one of the possible actions, the integral around a Pole, but it's just a constant in terms of the energy it doesn't vary, as you very the energy or the queue in this case.
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Hal Haggard: And so that's the the third solution here, the other two are the interesting ones they're the ones that can code preservative and non preservative effects.
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Hal Haggard: I just wanted to show you that we did some real work here here's the coefficients in that regard Fuchs equation.
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Hal Haggard: So let me try and go back to the the card physics equation has this structure there's a coefficient D a coefficient see to a coefficient see one.
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Hal Haggard: And those coefficients are given by these horribly nasty expressions.
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Hal Haggard: D is interpreted these very interesting actually D is exactly the discriminate of the quadratic it's The thing that.
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Hal Haggard: goes vanishes if, and only if two of the roots coalesce so D is very interesting I haven't found interpretations for see two and see one yet in terms of quarter can variance I kind of hope there's one, but I don't know at the moment it's a mess it's this really complicated thing.
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Hal Haggard: i'm rapidly running out of time, so i'm going to skip a proof that I really would love to present to you all, but I just I don't have time to present it.
377
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Hal Haggard: And that that the card Fuchs equation is so messy you could easily ask why do we care, even if you were able to do it it's it's not interesting.
378
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Hal Haggard: Well, the thing that you can prove is that you can do an all orders in each bar expansion of the action and every single term you know explicitly what the action is that every order in each bar.
379
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Hal Haggard: And what you can prove is that every order in each bar because of that picard Fuchs equation these integrals can be expressed as derivatives of the classical action.
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Hal Haggard: So you can relate the action at the health order and each bar to derivatives of the zero with order action.
381
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Hal Haggard: So you can find the terms that contribute any order in each bar as soon as you know, the zeros order action, but you genuine I found the zero order action.
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Hal Haggard: And so you can find the action at any order in each bar, this is striking result.
383
01:00:28,410 --> 01:00:42,600
Hal Haggard: This is not i'm not saying it's a panacea, these are hard computations figuring out what these coefficients are it's very, very complicated, but if you know the putter bit of expansion of the classical action to some order in each bar, you can find.
384
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It in terms of a zero.
385
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Hal Haggard: Okay, I summary of some more results, the quadratic that we're studying.
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Hal Haggard: Has the classically loud region which is between our two or three, but it has to classically forbidden regions between our one or two and our three or four.
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Hal Haggard: You can find the actions in each of those regions and you find a modular relation you find exactly what we were expecting that the classically forbidden actions are not independent of each other, so you can do this, all explicitly.
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Hal Haggard: And this is about the fact that this problem is actually more complicated than the ones that have been studied in the literature so far.
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Hal Haggard: So, not only do you have to do a wk be analysis of the wave function, but the coefficients that contribute to the way function.
390
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Hal Haggard: contributed every order in each bar so this problem is richer than the things that have been proven, so far, and I think that.
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Hal Haggard: In fact, some of the results i'm not 100% sure apply to this problem, yet, and so we need to close that gap, and if we are able to close that gap, it will be of great interest to the resurgence community as well, I think.
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Hal Haggard: So, in summary, I think that quantization of geometry provides a remarkable laboratory for understanding resurgent perturb active non perturbing relations.
393
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Hal Haggard: And due to the richness of its underlying quantum structure it may even require extensions of this formalism I think these are really, really fun things to study and and I hope.
394
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Hal Haggard: That they ultimately are going to contribute to our understanding of spin phone path integrals.
395
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Hal Haggard: Because the spin phone path integrals are also very complicated in terms of their saddle point structure.
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Hal Haggard: And so my real hope is that we're going to start to learn more about gravitational path integrals by thinking more deeply about resurgence, and with that i'll conclude Thank you all so much.
397
01:02:47,190 --> 01:02:50,070
Jorge Pullin: Well, questions you jr you have you ever won the race.
398
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Eugenio Bianchi: Yes, yeah how did these, these are the relation between because the question and in the Web action is really fascinating it's new to me, I wanted to ask you.
399
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Eugenio Bianchi: Can you go back to the analysis of the functions, where you describe this.
400
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Eugenio Bianchi: yeah you were saying.
401
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Eugenio Bianchi: yeah it's a solution, our second order differential equations so every solution, I can read it on a basis, and so this shows the relation between these different sub points, can you reinterpret that statement in terms of the bigger foots equation.
402
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Yes.
403
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Hal Haggard: Yes, I think you can I have not done this, though, so take that claim with a grain of salt.
404
01:03:37,920 --> 01:03:47,940
Hal Haggard: But the idea here is we're thinking of an awesome topic expansion right so we're expanding at large variable large values of the magnitude of the variable right.
405
01:03:48,510 --> 01:03:54,270
Hal Haggard: And so I could do a preservative analysis and ask every order what's happening right.
406
01:03:54,810 --> 01:04:14,130
Hal Haggard: And what I would find is that this non-paternity of connection formula would give me relations between the coefficients and the expansion on each side right but, but what we would find is that the higher order coefficients could be expressed in terms of the lower order coefficients.
407
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Hal Haggard: And the In doing so, you would actually need to use a picard Fuchs equation to figure out what the relationship was so so the the the caveat here is that I actually haven't thought about the topology of the.
408
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Hal Haggard: Complex suffocation of this linear potential problem, and so it could be that it's somehow sort of singular and the topology is not exactly that of a Taurus and you have to be a little bit more careful so um I think it should work, but I don't know 100%.
409
01:04:50,430 --> 01:04:52,170
Eugenio Bianchi: I see really interesting thing.
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Simone SPEZIALE: is, I think you answered already migration, we don't final comment, but just to make.
411
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Simone SPEZIALE: Sure, I wanted to clarify the logic in the sense that a resurgence is usually a useful because what we thought was just that perturbed by the analysis can be used to actually get number two, but the results.
412
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Simone SPEZIALE: But here, you were studying a problem which we already know the non personal, but the results, which is the volume of the.
413
01:05:19,680 --> 01:05:23,970
Simone SPEZIALE: The sorry the spectrum of the volume operator in this for violin case we know exactly.
414
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Simone SPEZIALE: So in the logic is to use these as kind of a working example to then learn how to apply these techniques in more complicated the new quantum gravity systems is that your logic.
415
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Simone SPEZIALE: or even in the case of justice volume of the tetrahedron you're hinting at something new that can be highlighted in this way.
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Hal Haggard: I think I actually have both views I, that is why I concluded on that remark is I, I agree with you that somehow.
417
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Hal Haggard: This problem is a one dimensional problem, we can do it.
418
01:05:56,190 --> 01:06:09,960
Hal Haggard: numerically you could always get more precision out of the numerical diagonal ization of the the volume operator matrix, then I could get out of this expansion I don't I don't argue with that at all.
419
01:06:10,860 --> 01:06:20,520
Hal Haggard: But but it's really different to numerically diagonal as a matrix and get the numbers than it is to understand where those numbers came from right.
420
01:06:20,940 --> 01:06:30,270
Hal Haggard: And so, for me the the interest is that these formulas are analytic there, there are things that I can like understand I can vary, I can plot things.
421
01:06:30,630 --> 01:06:43,980
Hal Haggard: I mean it took me, the reason I displayed that animation twice the this nation, it took me a long time to make this thing, but I can make this thing I can get a computer to actually plot the the evolution right.
422
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Simone SPEZIALE: But for the for volunteer volume don't, we have also some on led expressions in terms of 60 symbols.
423
01:06:50,400 --> 01:06:52,260
Simone SPEZIALE: True yeah I.
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Hal Haggard: know, I mean the the well Okay, what we have is the matrix elements as analytics.
425
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Simone SPEZIALE: team to for the eigenvalues you still have to diagnose that.
426
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Hal Haggard: Is that right that's right.
427
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Simone SPEZIALE: way okay fine okay.
428
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Hal Haggard: um so like I don't want to make I don't want to over make the claim here Similarly, I agree with you that this problem is well understood at some level but, but I think we are adding something which is that this this problem can be understood in terms of analytic formula as well.
429
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Hal Haggard: You can have a boat I may.
430
01:07:29,550 --> 01:07:38,880
Hal Haggard: My bigger point is definitely the one that you said that this is a laboratory it's a laboratory for studying resurgence in quantum gravitational path integrals more generally.
431
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Hal Haggard: And by the way, there's no hope that this constructive resurgence is going to apply to a spin phone no hope at all that this is spent film is not genius one in any sense right.
432
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Hal Haggard: But, but the the the story I started out with the story of integral representations and and their relations to path integrals that story resurgence has been demonstrated in in any finite dimensional complex and and grow and there are many results in path integrals as well.
433
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Simone SPEZIALE: Right about your last comment, I think that our results of a resurgence, also in the three dimensional case right now.
434
01:08:14,070 --> 01:08:14,430
Simone SPEZIALE: that's right.
435
01:08:15,450 --> 01:08:17,190
Simone SPEZIALE: Like or for a particle.
436
01:08:18,510 --> 01:08:19,140
Simone SPEZIALE: particle right.
437
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Hal Haggard: Even in field there is there is there a special field theories, with high degrees of symmetry.
438
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Hal Haggard: But but there's the path integral, in fact, the resurgence Community hopes that resurgence is going to give us the first mathematical definition of a path integral that we've ever had.
439
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Hal Haggard: So I don't know if that hope is I don't know how tractable that hope is but it's a hope that we might actually be able to define the path integral in terms of these kinds of properties.
440
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Simone SPEZIALE: Thank you very much.
441
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The park.
442
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Deepak Vaid: yeah hi hi, and so I guess my question has multiple parts, the first is that.
443
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Deepak Vaid: While we're looking at the aquatic potential again, I mean what was the relation with the tetrahedron.
444
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Hal Haggard: And no relation deepak I just was trying to offer you an example that wasn't tied up in quantum gravity, at the same time it's difficult to hold all of this in mind at once.
445
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Deepak Vaid: And so i'm.
446
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Hal Haggard: Trying to give you a system, you may be more familiar with and show you that all of the ideas apply in that system and that ours is actually an interesting generalization of that system okay.
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Deepak Vaid: Okay, my second question is that.
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Deepak Vaid: So, so you have a very beautiful picture for the face of a single tetrahedron.
449
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Deepak Vaid: Now it would be it does this.
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Deepak Vaid: Give us any ideas about what the if you were to increase the Valence like if you were to think of five sided ago or I inside the quality good.
451
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Deepak Vaid: What would the What would the faceless looking.
452
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Hal Haggard: um it's it's a very interesting question and it turns out that it's extremely complicated.
453
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Hal Haggard: So i've worked very hard on penta Hydra so.
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01:10:16,170 --> 01:10:22,290
Hal Haggard: Allah he drove five faces and it turns out that the face space.
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Hal Haggard: is just just much, much richer Simone and I still have an unpublished paper on some of these things that we really got to publish because it's beautiful work and it's just been sitting in the background.
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Hal Haggard: But, but the the face face is just vastly more complicated deepak so um I I studied it and one of the things I was able to Chris and I were able to show was the the volume evolution on that face space is chaotic.
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Deepak Vaid: and
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Hal Haggard: The chaos makes it just vastly more company.
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Deepak Vaid: So, so there is how simple lessons to be drawn for.
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Deepak Vaid: hire really in the pointers from the contract basis.
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Hal Haggard: I I don't I don't have any lessons.
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01:11:10,560 --> 01:11:11,400
Hal Haggard: yeah um.
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Hal Haggard: I i've discussed with Gerald done a little like what we can learn from resurgence about quantum chaos.
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Hal Haggard: And it's actually an area that the resurgence Community hasn't fully taken on, yet they have some results, but they haven't fully taken on, and I think it's it's really interesting thing to look into.
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Hal Haggard: Because you know things like the good swiller trace formula give you a relationship between cycles closed cycles in the class.
466
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Hal Haggard: And density of state and the quantum mechanics of the.
467
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Hal Haggard: chaotic problem.
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Hal Haggard: And so, this is, for me, from my mind that screaming for resurgence sort of argument but but I don't I don't know of any such argument yet.
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Deepak Vaid: Oh, and oh one last question.
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Deepak Vaid: So when we think of it angular momentum to our our geometry what happens is that.
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01:12:04,710 --> 01:12:08,790
Deepak Vaid: If you consider a spinning tetrahedron the closer constraint is no longer.
472
01:12:10,710 --> 01:12:21,990
Deepak Vaid: I mean it's no longer valid right, so you have to consider an open tetrahedron right the the faces come upon yes, so in terms of this.
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Deepak Vaid: faceless picture is there some way that you might be able to think about how that face face mind before.
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01:12:29,400 --> 01:12:32,610
Hal Haggard: yeah that's another lovely question that I wish I had a deeper answer.
475
01:12:33,690 --> 01:12:45,510
Hal Haggard: I mean you're you're kind of going immediately from for Vaillant intertwine or to a five Vaillant one is a way of the end so it's a it's a kind of another version of your previous question and.
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01:12:45,600 --> 01:12:52,560
Hal Haggard: Oh, I don't actually have a really nice way of viewing that that's complicated is my answer.
477
01:12:53,340 --> 01:12:56,250
Deepak Vaid: Well, no, I mean I would think that you know if you think about.
478
01:12:57,360 --> 01:13:04,410
Deepak Vaid: Just the classical level, you can you can increase the angular momentum of the tetrahedron starting from zero by a very small amount.
479
01:13:06,030 --> 01:13:11,760
Deepak Vaid: great and so that would cause a very small deformation of the principles in some sense yeah.
480
01:13:11,820 --> 01:13:21,570
Hal Haggard: yeah like the point, Wolfgang veal and and I thought about this point a lot at one point, Mr say because what we were thinking of was putting a Fermi on inside of the tetrahedron.
481
01:13:22,200 --> 01:13:29,610
Hal Haggard: But putting a vermaelen inside is adding a spin which is like adding a fifth face right it's exactly the context that you're thinking of.
482
01:13:30,630 --> 01:13:32,820
Hal Haggard: And it's a very small perturbation.
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01:13:34,140 --> 01:13:41,400
Hal Haggard: But we would, despite being very compelled by that pictorial Lee I don't I don't have I don't know what to say about it yet.
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01:13:41,670 --> 01:13:43,980
Hal Haggard: I would really like to to think about it more.
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Deepak Vaid: Excellent.
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Jorge Pullin: On one.
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01:13:55,020 --> 01:13:55,590
Jorge Pullin: On one.
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01:14:03,120 --> 01:14:03,270
Wolfgang Wieland: I.
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01:14:04,050 --> 01:14:04,650
Hongguang Liu: Can clean up.
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01:14:05,700 --> 01:14:06,540
Hal Haggard: Now we can hear you.
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01:14:10,140 --> 01:14:15,120
Hongguang Liu: Okay, so it's maybe a more general question about the research grant because.
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01:14:16,350 --> 01:14:19,710
Hongguang Liu: So far this combination our research grants you really, we need to see which.
493
01:14:21,360 --> 01:14:29,430
Hongguang Liu: which says apply anti relevant and which set of eyes are not relevant, so you went beyond this resurgence in the in the general totally free theory.
494
01:14:30,510 --> 01:14:44,430
Hongguang Liu: When we when we formed the integration side of brown away from the from the real integration domain, we need to, we need to know which which, like number which setup is contribute and which setup i'm still not contribute.
495
01:14:45,960 --> 01:14:47,550
Hongguang Liu: I don't know if we'll have any.
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01:14:49,050 --> 01:15:03,930
Hongguang Liu: Any any ideas from the from the resurgence, and how to distinguish which which one contribute, are we trying to not contribute or awesome contribute to the in the same so we're servants, so in the theory of resurgence, they all contribute.
497
01:15:03,960 --> 01:15:18,300
Hal Haggard: But there is a hierarchy of their exponential sub dominance, so I agree with what you said, the the idea of the current left shifts theory is to deform your initial integration contour onto the most relevant Santo points.
498
01:15:19,770 --> 01:15:28,830
Hal Haggard: And and they're there in the case of finite dimensional integrals there are some nice results about how you identify which are the most relevant.
499
01:15:29,130 --> 01:15:36,900
Hal Haggard: there's a there's a notion of neighboring saddles and how you connect to neighboring saddles the it's a very intricate topic.
500
01:15:37,830 --> 01:15:45,750
Hal Haggard: But there's an a very nice introduction to it in as part of that applied research and ask them topics workshop that I mentioned, there was a school involved.
501
01:15:46,260 --> 01:15:55,380
Hal Haggard: And the the school Christopher House gave a nice introduction to the the notion of neighboring saddles and how you figure out which ones are neighboring.
502
01:15:56,880 --> 01:16:00,120
Hal Haggard: I don't you know you're thinking about a very complicated.
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01:16:01,290 --> 01:16:02,700
Hal Haggard: Situation Hong Kong that.
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01:16:03,060 --> 01:16:09,270
Hal Haggard: I don't know whether their their techniques are going to be applicable yet, but I think it's worth investigating.
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01:16:10,020 --> 01:16:18,750
Hongguang Liu: Okay Okay, thank you and also yeah because because of this also concerned inflating the most informed and save even more complicated for adults, because they.
506
01:16:19,350 --> 01:16:30,810
Hongguang Liu: That sounds like that are multi valued said ours appears in be and why because he always hotly where they contribute on not on calls they contribute I think in the resurgence.
507
01:16:31,380 --> 01:16:38,400
Hongguang Liu: Because I noticed some paper by for example down the stairs and also with multi value, instead of actually I important in the.
508
01:16:39,510 --> 01:16:46,350
Hongguang Liu: us a definition of the of the old departing to Google, so they should contribute, but some other people who may not be doing this.
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01:16:47,760 --> 01:17:02,520
Hal Haggard: yeah yeah I mean this is a very active area and lots is happening all at once, so I don't have any specific comments on that, but, but I think if you found the done papers and they had already that's a really good place to start okay.
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01:17:04,320 --> 01:17:04,830
Jorge Pullin: Well, can.
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01:17:07,020 --> 01:17:09,150
Hal Haggard: I hired alive of getting work done.
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01:17:09,660 --> 01:17:14,100
Wolfgang Wieland: Thanks thanks for that really nice talk so my question is.
513
01:17:15,900 --> 01:17:18,360
A bit of regarding.
514
01:17:19,800 --> 01:17:38,160
problems that had so from the you mentioned at some point that so once it gives you the interface, so one faith based integral gives you the four sommerfeld approximation for the eigenvalues and the other one something like the degeneracy.
515
01:17:39,900 --> 01:17:47,220
Do can be used these to get a handle on better days of volume volume gap.
516
01:17:48,240 --> 01:17:48,630
Whether.
517
01:17:49,770 --> 01:17:50,190
We get.
518
01:17:53,670 --> 01:18:01,980
Wolfgang Wieland: In order days in area gap on gravity can these methods for us to understand whether there is a warrior forum you.
519
01:18:03,540 --> 01:18:11,730
Hal Haggard: yeah that's a fascinating question Wolfgang I actually haven't thought about that question yet i'm.
520
01:18:12,930 --> 01:18:31,080
Hal Haggard: The i've always found the question of volume gap, a little bit subtle because because it's related to what you're mentioning we we know the matrix elements as Simone and I were discussing and for any finite values of Jay those are going to give us finite.
521
01:18:32,490 --> 01:18:42,420
Hal Haggard: volume and so it's really a question of order of limits and what you're thinking about right, so we have the emirati parameter, which controls the area gap.
522
01:18:42,810 --> 01:18:53,220
Hal Haggard: And then we have to kind of ask well are we scaling both the angular momentum and the emirati parameter, or what kind of a limit, are we investigating.
523
01:18:53,490 --> 01:18:55,170
Hal Haggard: And, and I find the question of the.
524
01:18:55,170 --> 01:19:05,220
Hal Haggard: area gap to be subtle because of we've been noticing that's a been another theme of the last few seminars that you can't take the large angle, or the large area limit.
525
01:19:05,550 --> 01:19:17,280
Hal Haggard: naive Lee you're going to ruin send me classics if you do that, you have to take it carefully and take into account the area gap, at the same time, so so this question and I i'm worried that we haven't.
526
01:19:18,450 --> 01:19:27,330
Hal Haggard: posed a well post question with the volume gap yet and and I want to, I want to think more deeply about the right regime for asking that question.
527
01:19:29,250 --> 01:19:30,300
Absolutely related.
528
01:19:31,710 --> 01:19:43,800
acne technical question so in this analysis, the or so, to get these two dimensional space space, the four areas of the areas of the four sides of the tetrahedron.
529
01:19:45,990 --> 01:19:49,020
kept constant right that's right.
530
01:19:52,500 --> 01:19:54,420
So, this would enter somehow is that.
531
01:19:55,800 --> 01:20:00,630
Hal Haggard: Exactly that's that's what i'm fixing properties, the boundary geometry.
532
01:20:05,400 --> 01:20:06,360
yeah thanks for the.
533
01:20:10,500 --> 01:20:11,760
Jorge Pullin: Okay, any other questions.
534
01:20:18,210 --> 01:20:27,480
Jorge Pullin: If they're not let me point out that in the chat Europe has preliminary announcement for the upcoming topics workshop February 14 to 18th.
535
01:20:30,390 --> 01:20:31,290
Hal Haggard: Thanks everybody.
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01:20:32,160 --> 01:20:32,910
Jorge Pullin: Thank you bye bye.