 OK, thank you. So I'm Takuyo Kuda from the University of Tokyo. And the organizers asked me, actually, not just boundary, but also extended operators in supersymmetric localization with extended operators in supersymmetric quantum field theories. So I think, OK, by extended, what is meant is that the operators are not necessarily local. So they can have some non-trivial dimensions. And the examples include line operators, such as Wilson-Poufft line operators. And there are also surface operators. And then you can boundaries, interfaces, also known as domain walls. And we can also have coupled systems of QFTs of different dimensionalities. I think I can include them as examples of extended operators in quantum field theories. And we can do various things. And we can, indeed, do supersymmetric localizations with these extended operators. And also, they play significant roles in the 2D, 4D, or AGT correspondence, or related 3D, 3D correspondence, and so on. And these objects are also important in other contexts, such as conformable strap, holography, and so on. And maybe most significantly, especially the line operators or loop operators were first introduced as order parameters of gauge theories. So the behaviors of their expectation values or correlation functions as functions of some parameters can be used to classify possible phases of quantum field theories. And so today, I will focus on the boundaries and interface, the monthly boundaries in two dimensions. So this is the topic of today. And at least my current plan is to talk about line operators, Wilson and Toothed operators in four dimensions tomorrow. That's my plan. And OK, so today's topic, this is about in two dimensions. So today's topic is closely related to what Benini discussed last week. And OK, the topic of tomorrow, the topic will be about four dimensions, so it will be closely related to what Peter discussed last week. OK. And I do believe that the examples I will discuss are important examples. But I also want to emphasize and illustrate the general principles that can be learned from the examples I will concretely discuss. So my examples will be interesting and important, but I also want to emphasize the general aspects. OK. So usually, I write down the plan of my presentation. So for today's lecture, here is the plan. So I first, I'm already giving an overview of my lectures. And then I will discuss the so-called B-type boundary conditions. And then I'll explain that something called matrix factorization will be necessary. And then I will actually perform supersymmetric localization and explain how to compute the so-called hemisphere function function. OK. And then I will tell you that the hemisphere function has a meaning as the central charge of D-brain when the two-dimensional theory is used as a word sheet theory of type two string theory. And also, as an application of the hemisphere function, I plan to discuss the interface. OK. And of course, the questions are welcome. So please feel free to stop me and ask questions anytime. OK. So I want to introduce B-type boundary conditions. And in order to do that, I need to explain the setup. So this is basically the same as what Benny discussed. So we are going to consider two-dimensional N equals 2 comma 2 theory on the two-spheres or, well, or hemisphere. All right. Actually, I want to give you references. So as a general reference, elementary pedagogical reference, I give you section 39 over text book called MiraSymmetry, written by Holly and other people. Well, you can Google MiraSymmetry, Holly, and then you can find the PDF from Clay Mathematical Institute. That's true. Yeah, but this is the book, yes. OK, maybe you can put the clay. And then you'll find the PDF. Or you can get it from Inspire. So this is the text book. And the actual content of my lecture today will be based on my paper on the hemisphere function function. And actually, on the hemisphere function function, there are three papers. One is by Sugistan Terashima. Terashima was here last week. Another, Honda and myself. And yet another one by Holly and Romo. These are all from August 2013. So basically, these papers were posted on the archive simultaneously. And of course, naturally, I will be mostly using the conventions of my own paper. Yes. And our convention is very similar to Doroud Benigny. And sorry, Doroud, Reflog, Gomis, and Riva are using. Yeah, OK. So the setup is a two-dimensional n equals 2,2. And because we are on a curved manifold, we want to consider some background, supergravity background. And OK, I'm interested in a particular version of two-dimensional supergravity called U1B. And this is a dimensional reduction of four-dimensional n equals 1 supergravity, the new MIMO supergravity, which Fesucha discussed last week. OK. So in particular, vector r symmetry. Yeah, so this supergravity can be coupled to a theory with vector r symmetry. And I want to consider the following metric. So let's see. So F, so theta and phi are the usual polar coordinates on the truth here. And F behaves like this for theta close to 0. And OK, because we are interested in essentially Hems here, let's say, I want to restrict to, well, actually, OK. Yeah, OK. So this is the behavior for theta near 0. And OK, on the truth here, theta takes value between 0 and phi. And phi takes value between 0 and 2 phi, as usual. OK. And sometimes I will use the field bind. So and I will be using the hat to denote the frame index. So this is the index for the field bind. So E1 hat is F times d theta. And E2 hat is L sine theta d phi. OK, well, if F is simply L, then this corresponds to the round metric of radius L. And you may also take L squared cosine theta cosine square theta plus L tilde squared sine squared theta, where both L and L tilde are some parameters, positive parameters. And this is also a good example of F. I need to experiment a bit. And OK, we also have U1R symmetry gauge field. So VR, so this is one form, 1 minus L over F of theta d phi like this. And if you remember, I think from Benin's talk, this supergravity, this version of supergravity also contains something I think he denoted by H. So this is a gravity photon field strength. And this is proportional to 1 over F. OK, am I clear? Is everything clear? No question? OK. Do you know what the partition function is going to depend on? I mean, for example, in the 4D case, we were explaining that it depends only on the complex structure, et cetera. So do you know if the similar analysis is specific to D and what the conclusion is? Well, I know the result of the localization calculation. Let's see. Yeah, it depends on the context. But for example, for the Henshaw function in the conformal case, there is an analysis using the superwire anomaly by Bacchus and Plankner. And they showed that it depends hormonically on the Kera module. And here I introduced the deformations of L and L theta. So do you know if they exhaust all the further form deformations of S2 sheets? So S2 partition function, there is no analog of B for S2 partition function. Yeah, so here I introduced the deformation, but the partition function actually is independent of the deformation. So it's actually enough to compute the partition function on the round sphere. I see. I'm asking you that if you take a more general metric, et cetera, are you sure that this is the only answer you get? Or if you take a crazy complex structure, et cetera, is there a possibility of getting another answer? For the truth, I don't think there are many options. Sorry, say that again. You say when theta goes to 0, F, theta approaches N. Yes? So then it becomes 0. At this one? Yeah, you said a singular point. Yeah, so right. So in it, F goes, this part becomes 0 at theta equals 0, and that's necessary for this gauge field to be smooth, because this is angular variable. Yeah, I think I'm going actually too slowly, so OK. Now, I should now write down the generalized, generalized chilling spinner. Can you read? So this is the condition for supersymmetry to be preserved. Y is two-component spinner, and I'm going to define the gamma matrices soon. So there are two kinds of spinors, epsilon and epsilon bar. Minus 1 over 2f, gamma mu, gamma 3, epsilon bar. Epsilon and epsilon bar have opposite R charges. And my convention for gamma matrices is the theta hat corresponds to gamma 1 hat. This is the first sigma power matrix. Gamma phi hat is gamma 2 hat minus ii, and gamma 3 just denotes the chirality matrix in two dimensions. Yes? OK. And on the round sphere, we have SU 2 slash 1 symmetry. But the deformation, and also, when we put a boundary along the equator, theta equals pi over 2, the symmetry SU 2 slash 1, it gets broken to SU 1 slash 1. So that's the symmetry we'll be interested in. It'll be interesting. How do I do this? So I will be considering very explicit spinors, super symmetry parameters. So I'm just writing down the solutions to the general Killing-Spinner equations. So I will be considering exponential minus i over 2 theta gamma 2 hat. So this is a matrix, and it acts on exponential of i over 2 times phi 0, and the epsilon bar equals i over 2 theta gamma 2 hat 0. It is minus i over 2 phi. Yeah, so these serve the general Killing-Spinner equations, which I wrote down. And if you compute the super symmetry transformations, and then you find that the super symmetry squares to bosonic symmetries, and the bosonic symmetries involve some Killing vector field. So it gives a rotation of the sphere. And the rotation is such that the boundary at theta equals pi over 2 is preserved. So I will be putting the boundary at theta equals pi over 2. And in this limit, the spinors, the super symmetry parameters becomes both basically 1 1 up to r symmetry transformation. And this is because gamma 2 hat is 2 by 2 matrix. And if we set x1 to be L tilde times theta minus pi over 2, and x2 is L times phi, then locally, near the boundary, the system looks like this. So there is a boundary at x1 equals 0, or theta is pi over 2. And the remaining part is, so the region on the left is the bulk of the hemisphere. And if you, I haven't told you the rule, but if you compute the super charge, so in terms of the spinors, they look like this. And in terms of super charges, it looks like, in the convention of the text book, media symmetry textbook, it looks like this. And this such super symmetry is known as B type super symmetry. So yes. But this is a simple thing. And it may be possible to, OK. I think it's an interesting question, whether one can actually put a boundary somewhere else. And at least some interfaces can be put at places other than the equator. I know that, for example, the Wilson loop can be placed away from the equator. So I think it's an interesting question, whether one can put a boundary somewhere else. And for example, do super symmetric localization. But I don't know for sure whether one can do localization for the boundary at a more general location. So this is a choice. And this choice is nice, because the super symmetry locally looks like a B type. And I think Benny told you that near the north and the south poles of the two sphere, the super symmetry looks like a type. So the theory looks like a model. And but now, if you consider a boundary along the equator, then the super symmetry there is a B type. So if you consider, for me, the aim of this talk is to compute the Hermitian function. And so this Hermitian function, then Hermitian function gives an interesting, the Hermitian function gives an interesting coupling between A model observables at the north pole and B type observables at the equator. So let me write it equation. So just by analysis of super symmetry, it looks like the Hermitian function is of this form. So something that depends on the boundary or boundary condition. So it's a D-brain preserving B type super symmetry. So it's called B-brain. And on the north pole, we have A model. So that means that if you do not insert any operator, then it corresponds to the identity operator. The algebra of operators of topologically twisted A model is called the AC ring. So the Hermitian function appears to be computing a coupling like this, coupling a pairing like this. And I think I will come back to this later. So my goal is to compute this function function. OK, so far, I have only explained the background. And I want to tell you about the field content. So we have a vector multiplied for gauge group G. And this includes the gauge of A mu, two real scalars, sigma 1 and sigma 2. And there are gauge in lambda alpha and lambda alpha bar. So alpha takes 1 and values 1 and 2. And there is a real auxiliary field D. And then we also consider the Kairou mutiplet. And it's in representation, in representation R of G. And the fields are phi, psi, alpha, and F. So these mutiplet have already appeared in the lectures last week, I think many times. Yes. So if I put that equal to E by 2, yes, I would get these skinners, yes. But if I put gamma 2 multiplied by E, I would get real exponents with 3. I mean, how exponents disappear? So you're asking about the computation from here to here? Yeah, OK, so gamma 2 hat. Because gamma is complex matrix, and phi is also complex parameter. So if you apply complex matrix to complex parameter, I would get something real. Yes, and you're asking how to get rid of this? Yeah. Yes, this is by arithmetic gauge transformation, yeah. So we'll be interested in vector and Kairou mutiplet. So this is actually relatively complicated. OK, so now I want to say a little bit about supersymmetry transformations, but I'm not going to write all of the transformations. So for example, for the vector mutiplet, we have delta of a mu going to like minus i over 2, epsilon bar gamma mu lambda plus lambda bar gamma mu epsilon. And my convention is something like this. So for example, if I write epsilon bar gamma mu lambda, then this means that epsilon bar alpha with upper index, now gamma matrix has lower left, upper right, lower right index for the spinner. And for the spinners, the indices are raised and lowered by charge conjugation matrix. Of course, these are details, but I gave some exercises to check the supersymmetry of boundary conditions. So you are going to use some transformations like this. OK, and then there are the transformations. For the Kairou mutiplet, we have delta phi. So the transformation of the complex scalar is epsilon times psi, where the contraction rule is basically gamma removed. And gamma psi is i times gamma mu epsilon d mu phi plus i epsilon sigma phi plus gamma 3 epsilon sigma 2 phi plus. Well, actually, so there is some convention dependence on the sign of the arches. And I'm choosing, I think, OK, some convention, which is actually the opposite from the paper, I think. And then, OK, there is some formula for delta f like this. OK. Excuse me, are these transformations adjusted like those on the sphere? Yes, yes, so this is exactly the transformation on the sphere, that's right. So yeah, so the hemisphere is just half of the sphere. So we are using the same background and the same transformations. Yeah, and I'm considering the boundary in two dimensions, but the story can be repeated in other dimensions, like three and four dimensions. And the people have indeed studied localization on such spaces. People have considered hemispheres in four dimensions definitely. And in three dimensions, I think people have considered S1 times two-dimensional hemisphere. So say that again. Do we have the two, both of you choose the symmetry on the hemisphere or the part of it? Part of it, part of it, that's right. Yeah, so by beta, I mean really a subalgebra of full n equals 2 comma 2. And in particular, beta subalgebra has two supercharges. And it's also possible to consider a type and a type and a b type are related by a mirror symmetry. And from four dimensions, these setups arise by the dimensional reduction of either new minima of gravity, new minima of supergravity, or the minima of supergravity. OK, now I want to write down the supersymmetry boundary conditions of the b type. So for the vector multiplied, I will consider the following. So at theta equals pi, pi over 2. So we are considering the field here. So we require that field certifies the following. So here, a1 really means, OK, I can also write a theta. And by boundary condition, we require that the field strength is 0 at the boundary. So you see that a1 is 0, but I don't do anything on a2 or a5. Well, except that, OK, essentially this is a Neumann boundary condition on a5. So this actually preserves gauge symmetry along the boundary. So this is the boundary condition I want to consider. But it's also possible to consider a boundary condition that breaks symmetry, but I'm not going to do that today. OK, so this is for the vector multiplied. Now I'm going to describe boundary conditions on the chiral mass plate. The main boundary condition on the chiral mass plate is a Neumann boundary condition on the chiral mass plate, which says that d1, pi equals 0, yi gamma 3 psi equals d1, yi gamma psi equals 0 and f equals 0. But it's also possible to consider an alternative boundary condition, which says that phi is constant, OK. I choose a constant to be 0, so phi is 0, yi gamma psi equals d1 yi gamma 3 psi equals 0, d1 hat e to the minus i phi f plus i d1 phi equals 0. OK, the appearance of e to the minus i phi may be annoying, but it can be removed by, again, our asymmetry gauge transformation. So to some extent, this is a matter of convention. So if you go to a different gauge, this disappears, and then the boundary condition looks more natural. Yeah, so one of the exercises for this lecture is to check that these boundary conditions, actually, in flat space, the flat space version of these boundary conditions preserves beta supersymmetry. So that's one of the exercises. OK. Are these boundary conditions elliptic? I think that was consistent with the equation of motion. And Schumer was reading this as to the other issue. I see. OK, so the way it's written, these are not elliptic, I think, because I think imposing these differential boundary conditions on the fermions is, I think, too strong if you try to formulate variational principle. Yeah, but for supersymmetry localization, this is enough. And that's what I'm doing. Yes. So somehow, it's hard to do the perturbation theory at all for the passing difference, but you can never rest localize to find the equation. Yeah, so I think in this case, you can actually just remove the, well, I think for the normal, I think you can just remove this part. Yeah. The thing that, so for example, if you have this equation, then you can do supersymmetry transformation, and I think you find this. Right, so yeah, so if you want to do perturbation theory, then you need to be more careful about the boundary condition. OK. Now, OK, so these are the boundary conditions. OK, any more questions? OK. So these are the boundary conditions, and I'm still in the process of defining the theory. So defining the theory on the manifold with boundary, yes. I don't have what? I don't. I get AC ring? Yeah. Yeah, so the question is the following. So I wrote that the Hemi-Seer-Valscombe function equals the overlap between the B-brain state and the identity operator in the AC ring. And but she pointed out that I only have a gate theory, and I don't have supersymmetry, so how does it make sense? So that's the question, right? Yeah, so did I? Well, what I meant was that was a rough equality, and actually, I probably should have put a question mark whether that equation is true or not. And yeah, I was going to mention it later, but I can tell you now. So we are considering a gate theory, and in some cases, so with the right matter content and charge assignment, this gate theory can flow to super conformal field theory. And then let's see. Historically, once you get the formula for the Hemi-Seer-Valscombe function, from no example, it was apparent that Hemi-Seer-Valscombe function could compute something called the central charge, which can also be written as the overlap between the B-brain boundary state and the identity element in the AC ring or the Ramonamon ground state. So when we wrote these papers, it was a conjecture that the Hemi-Seer-Valscombe function computes such an overlap. And more recently, I've already mentioned, there is a work by Bacchus and Plankner, which uses the super-wire anomaly to show that this conjecture is actually true. So it's not trivial. It's not trivial that the calculation in gate theory actually computes the overlap between the boundary state and the identity operator. But it can be shown independently. Yes? To be a state? Right. OK, so the question is, why is a Hemi-Seer-Valscombe function is a number rather than a state? So I think what it meant is that the path integral on a manifold with boundary can indeed represent a wave function. But in general, in quantum field theory, so it's good because I need to emphasize the general principle, in quantum field theory, we need to distinguish two kinds of boundaries. So one, I forgot the general name, but OK. One type of boundary condition supports a state, actually a normalizable state in the Hilbert space. So some type of boundary on the manifold represents a state. On the other hand, there are boundaries that are really boundaries, so these are not an intermediate state, but rather it corresponds to the end of the world. So some boundaries in quantum field theory are end of the world. And here we are, OK, actually I'm not completely sure if there is a clear distinction, but that's a general story. And here I'm considering the boundary, that is the end of the world. And in that case, an example is a d-brain in string theory. I mean, the boundary in the sense of the world sheet. So if you have a d-brain boundary condition, then you can construct a state, which is a boundary state. So it's a state, but it's not normalizable, OK? So if you have a genuine boundary in quantum field theory, then you can formally construct a state, but it's not normalizable. So if you compute a normal square, it actually diverges. So in that sense, I do not regard the Hermitian partition function as a wave function. It's more like a partition function. OK, any other question? OK. So I want to define a theory on spacetime with boundary. And in order to do that, of course, I need to specify the Lagrangian. Lagrangian and actions. I'm going to write action. Actually, yes, also. And there are two types of actions. So one is a physical action, and the other one is a localization action. And the theory is specified by the physical action. And it takes the form. So the vector-multiplet part, the chiral-multiplet part, superpotential term, topological term, and the FI parameter. OK. So basically, I think I'm going to omit giving these expressions, because they were essentially given in Benjamin's talk, I think. But there is one point. Yes? OK, good. So the only important part is the superpotential term, which looks like a FI PIW, where W is a superpotential. So it depends on phi. It looks like this. I think the remaining terms, I think most of you know. And the important thing is that if you compute the SUZI variation of the physical action, then all of it vanishes except the superpotential term. So you can, OK, yes, also. Essentially, the remaining terms can be written as SUZI exact term plus boundary term. And the boundary terms vanish by the boundary conditions. That is true for vector and chiral. OK, the state term actually needs to be super-symmetrized. And if you do that, it's super-symmetric. And the FI term is also, well, usually it's OK. On the truth here also, we use the super-symmetric version. Well, and it's super-symmetric action on the truth here. Yes, so anyway, what's important is that the superpotential, the SUZI variation of the superpotential term is not automatically 0. And this looks like epsilon gamma mu psi i 0 i w plus conjugate. So it becomes a boundary term. And this means, what does this mean? This means the following. This implies the following. So the Hamilton function function is supposed to be given by the path integral over field configurations subject to p-type boundary conditions. And the theory is defined by the physical action. And we might put localization action also. But the problem is that this does not represent a super-symmetric system because the variation does not vanish. So we need to do something about it. So this is a well-known problem. And this term is called the Werner term. And this Werner term needs to be cancelled. And in order to cancel it, we include the following term. Trace in some vector space v, path order of exponential of some version of script A, script A phi. So these are some version of Wilson loop. But this Wilson loop, so this is called the boundary interaction. And this looks like the usual super-symmetric Wilson loop, A phi plus i sigma 2 in some representation for this vector space v. And this vector space v is called the Champ-Payton. So this is familiar from string theory. Champ-Payton vector space. And it carries some representation of the gauge group, flavor group, and r-symmetry group. Let's see. And then there is a contribution corresponding to r-charge. If there is flavor symmetry, there is also twisted mass. So this is the, OK, I think I forgot to mention. Twisted mass is the scalar component for the vector-multiply, the background vector-multiply for the flavor symmetry. Well, this was explained in Beninist lectures. And now there are some terms that depend on what I call q and its conjugate q bar. And here, q is some operator that depends on phi. And it's a linear operator that maps v to v, Champ-Payton space to itself. So what I want to do is to, yeah. So the role of q is that if you compute the supersymmetry variation of the integrant, of the path integral, then the variation of the boundary interaction should cancel the random term. That's the role of the q and the boundary interaction. Let's see. What's next? No. It's actually an interesting computation. And you can, it's an interesting computation. Also, it's relatively non-trivial. And if you do it, it's actually rather long computation. And for details, you can look at the paper by Herbst, Forry, and Page from, I think, 2008, 2007 or 2008. So anyway, the upshot is that the supersymmetry variation of the superpotential term, it is in minus superpotential terms, times trace in v of p of exponential of i d phi a phi. So this is 0 if q of phi squares to the superpotential by the identity operator v. Is the statement clear? OK. Yeah, so this is the most important property of this operator q, where it's also called the Tachyon profile, where it's also called the matrix factorization. And for this, it's got to make sense. q has to obey some constraints. And OK, let me write it this way. So suppose that g is an element of the gauge group. And the flavor symmetry group, which I wrote by GF. And suppose that rho is a representation of a gauge group and the flavor group on the champeton space v. Then we need that it satisfies the following. So this is necessary for this to be symmetry. And we also need that for r symmetry, so there is a representation of r symmetry on the champeton space v. And then q has to satisfy this. By capital R, I mean the action of r symmetry on the Kairou-Mathsburg field. The sine convention here is a little strange, but this is a standard choice. So these equations need to be satisfied. And in the exercise, I give at least one example. And OK, I didn't explicitly ask you to do this, but you can also try to check. With some appropriate choice of representations, the example I give in the exercise actually satisfies this. And also I want to explain why this is called matrix factorization. So if the champeton space has a structure, so if it's z to graded, and actually it has to be z to graded, then maybe this is part of the constraint. So q has to be odd, so this is one requirement. And then we can, OK, using this convention, we can write q as block diagonal matrix with non-zero entries A and non-zero of diagonal entries A and B. And then q square equals B can be written as AB equals BA equals W times identity. So you see that, yeah, so A and B indeed give a factorization of this hormone function W, which is usually polynomial. OK, so I see. OK, so I explained the matrix factorization. Now I want to perform a localization calculation. And basically I'm going to recycle what Benini did on the two sphere. So, OK, first of all, I'm going to use the same localization action. So I'm going to omit a lot of things. I can omit a lot of things because Benini did most of the work. Yeah, so we use the same localization action. And but actually we are going to specialize. And here I'm going to consider Coulomb branch localization. Higgs branch localization should also be interesting, but it has not been done yet. So now for the vector model, remember that we require that the field strength is 0 on the boundary. So this gives a strong constraint on the localization locus. In particular, it means that, OK, so the field strength is 0 on the localization locus. Sigma 1 is 0. And then what remains is the zero model, sigma 2. So the constant model of the sigma 2. So we get the path integral reduces to a finite dimensional integral over sigma 2. And for the color market, we get f equals 0. I think f is also 0. And of course, the fermions are 0. Then we can evaluate the physical action on the localization locus. And we find that e to the minus physical action is given by e to the, OK, t times twice of sigma, OK. So I'm writing the result as if the gauge group is 1. I'm writing the result for the gauge group u n for gauge group u n. And t here, t is the complex wide FI parameter, which, OK. I think I could write 2 pi, so Benin was using xi. So I think in his convention, t would look like 2 pi xi minus i theta, theta, or theta is the topological theta angle. OK. And the boundary interaction, we also need to evaluate the boundary interaction on the localization locus. And if you do that, you find that this is given by e to the minus 2 pi i times sigma plus m, here by sigma, I really mean minus i l times sigma 2. And here m is minus l times sigma 2 times the flavor part. So this is the classical part, OK. I'm going to discuss the one loop part, one loop determinant. So the general principle is the following. So without boundary, people have already computed the one loop determinant. And the localization locus is also a specialization of the Cia case. And now, when computing the one loop determinant, we only need to contribute. We only need to keep contributions from the most. That obeys the boundary condition. So that's what we're going to do, OK. And, OK, for the Kailor multiplet with Neumann boundary condition, so this can be written as a product of weights in the matter representation and the infinite product, which is divergent. And, OK, as Benini did, we might do data function regularization and get like this. And if you have a flavor symmetry and a twisted mass, we also need to include them. But the way it enters is the same as sigma, OK. Dirichlet boundary condition. This may be a bit interesting. Again, we have a product of weights. And there is an infinite product. Now, if you do data function regularization, you get something, OK, one of gamma function, one of gamma function. But I claim that this is wrong. This is wrong, OK. So I claim that data, so this is how to read, right? Data function regularization is, OK, I say, not always correct. More precisely, OK, data function regularization does not always automatically give the correct result. And now, I'm going to give you the correct result and explain how to justify or derive it from first principle. You were correct. One loop determinant for the Dirichlet boundary condition includes extra factors. Minus 2 pi i, so very specific factors. Pi i times w times sigma, I think I need to include minus i over q times q over 2 divided by 1 minus w sigma minus i q over 2. So then why do we need to include those factors, right? So that's a crucial question, right? And there are two explanations. So one explanation which we ordinarily had in a paper was that there is a duality between Dirichlet boundary condition and Neumann boundary condition plus some boundary conditions, boundary interactions. So there is something called Atiyabot Shapiro construction of diesel brains. And there is some known duality between the boundary conditions. And in order for the one loop determinant to be consistent with such duality, we need to include these factors. But by now, I have not published it, but there is a first principle derivation of this, which is the Paris-Villard's regularization. The computation for the hemisphere has not been done, but Paris-Villard itself, I wrote a paper one year ago. So you can take a look if you're interested. So if you apply Paris-Villard's, then the ratio between Dirichlet and Neumann one loop determinant turns out to be 1 minus e to the e to the 2 pi i times w sigma plus minus i i times q over 2. And the Paris-Villard's, this preserves gauge and the flavor symmetry. So if you demand that flavor symmetry, gauge symmetry is preserved, gauge symmetry and r symmetry is preserved, then Paris-Villard's regularization gives a gauge invariant, gauge invariant of the manifestly supersymmetric regularization. And this ratio is unambiguous. It's not ambiguous. So this is a figure, and it's consistent with these expressions. Yes? So if you connect to Paris-Villard's, you already have the Paris-Villard's after a grander level before localization, or you have the Paris-Villard's afterwards. But if you add the Paris-Villard's from the outset and supersymmetric, it's not consistent. At which stage you introduce the Paris-Villard's later? Well, it's a general story, I think. So we should start from the beginning, by including the Paris-Villard ghost. But then, is it consistent with the supercharger you're using? Yes. OK. There is a supersymmetric version of Paris-Villard's. That's what I'm saying. Yes. Yes. What's the point of what's going with the data function of the definition? Yeah, so in principle, there is no reason that the data function regularization should work. OK? What I'm saying is not, again, published. But in general, you can relate the Paris-Villard's regularization, heat corner regularization, and the data function regularization by using the heat corner method. And one can check how these regularizations are compared with each other. But usually, what people have done in localization with data function regularization is not very systematic. And so I would say that, in general, there is no prior reason that the data function regularization should work. So its validity should be checked by other methods, like reality. And the relation between the data function and the Paris-Villard is clear for bosons. But for fermions and spinars, the situation is more careless. More care is necessary. And especially, in the presence of boundary, the situation is more subtle. Oh. How you got sure which one is correct and which one is wrong? Yeah, so I mean, Paris-Villard's regularization preserves the gage symmetries. So if you are interested in results that are compatible with gage symmetries, this is the correct result. But actually, OK, the no-man Dirichlet boundary condition themselves, actually, the boundary condition themselves break some other symmetries, like charge conjugation symmetry. And if you want charge conjugation symmetry, if you give more priority to charge conjugation symmetry, then you should combine some regularization method with proper counter terms. So in general, two different regularizations are related by choice of counter terms. And regularization and counter terms, their combination is the invariant thing. The physical meaning thing is the combination of regularization, UV regularization, and counter terms. And the correct choice should be made according to which symmetry you want to preserve. And if you want to preserve gage symmetry, this is the correct result. Yeah, OK. I think I want to give an expression for the vector model plate and then stop for today. So for the vector model plate, there is a product over positive roots. And again, there is an infinite product, so this alpha. Alpha is a positive root like this. And OK, for this, this functional regularization gives a consistent result. And it is like this. Yeah, so I finished giving one loop determinant. And yeah, I want to stop for today.