 be a smooth four-manifold with topological condition that B1 and B2 plus are 0, 1 for B2 plus. B2 plus, this came about because the U-plane integral or the integral of the Coulomb branch of the Ashley 2 Gates theory is only non-venecing if B2 plus is smaller or equal to 2, 1. It seems hard to evaluate it for B2 plus as 0. So we'll focus on the B2 plus is one case. The B1 is 0 is more to simplify the discussion, to avoid zero modes for one of the fermionic fields. Then one of the Incudians was a lattice L, which corresponded to the second cosmology over the integers. And we mod out the torsion. We have a quadratic form or a bilinear form on this lattice just from the intersection or integration of the wedge particle of two forms over the four-manifold. So this is B. And we can extend it over the reals. Actually, later we will also use complex arguments. If it has a quadratic form, then q, qk is just simply Bkk or we will write it as k squared. Then we have the following piece of information. j is the period point. Let j be the period point. But in other words, it is the unique self-dual 2 form in the forward light cone. Since B2 plus is equal to 1, we have one positive direction. So there's also therefore only one direction which is self-dual in front of h2mr. So we can say it's in the forward light cone or it is effective that it has positive inner products with curves in m. And we normalize it to be of norm 1. qj is equal to questions about that. Feel free to ask yesterday after the talk to a few questions about this notion. But we have we work on a manifold of B2 plus as 1. And for that reason, there is a unique self-dual 2 form which spans this direction in h2mr. And we call this one the period point. And it depends on the metric because j is star j. The self-duality condition depends on the metric for the watch time. Then using this period point, we can project elements of k in h, or let me say, k in l. You can project a k in l to a positive and a negative component. k plus is Bj and then log times j. So this is a times r. And similarly for k minus, this is then k minus k k plus. All right, then we set up this u-plane integral as follows. So I denoted us phi mu j, and then we could include an observable o. And we arrived at the following expression. It's d tau, where it's d tau bar, nu tilde tau times a sum over fluxes or sum over the lattice, psi mu j tau bar, and then this observable o. So let me just recall the different elements, nu tilde of tau was d tau, then a u times to the power of the Euler number of m. These are the curvature couplings. And then there is a term bu to the power of the signature. Other lectures were using here the E of m, but I used k. And then the sum over fluxes psi j tau tau bar reads the sum over k in the lattice, possibly shifted by a half lattice element. Actually, the whole thing is multiplied by 1 divided by the square root of y. y is the imaginary part of tau. Then we get a sine minus 1 to the b k. I will say a bit more about all the ingredients I haven't discussed before in a minute. Let me just finish this expression. The bilinear form b kj, then we get q, which couples to the negative definite components of k, and q bar couples to the positive definite components of k. Is there a question? Can you even know where k is k? Yeah, very, very good. I'll get there in a second. First, I'll say something about mu. Mu lies in l divided by 2, a half lattice element. Then the capital K, for a complex manifold, one can take the canonical class. More generally, it is an integral lift of the second Stevel Whitney class, w2m to the lattice l. y is the imaginary part of tau. I already mentioned. And just recall the topological observables we discussed yesterday. We can take for O the point class, which corresponds to two times this observable u, which was a trace of phi squared. We can also exponentiate it e to the 2pu, where then p would be a fugacity for this point class, the point observable. And so another class of observables was i minus x. This minus sign will maybe become apparent a bit later. But this is a surface observable. And so x is a two-cycle in the manifold n. Questions? OK, so our task today is to evaluate this u-plane integral. And so let me give a little intermezzo about modularity and modular forms before we're turning to this integral. We'll leave the integral for the moment. So just to introduce some notation and some modular objects, let me start by recalling the modular group as of 2z, which is simply the group of integer matrices, a, b, c, and d in z, such that the determinant is equal to 1. And then a modular form is a function f from the upper half plane h to c, such that f of a tau plus b divided by c tau plus d is equal to c tau plus d to the power w. And then it gives back the function by itself. And this w is the weight. So this is the classic definition of a modular form. In a sense, it's quite surprising that there are even examples, especially holomorphic ones, because it's a strong condition on the functions. Good examples exist. And so typically, the examples have an even weight w if you want to satisfy this equation on the nose. If you allow complex phases, you can get some more interesting examples. For example, the datacent eta function, eta has weight, w is 1 half, and it's defined as follows. Theta of tau is q to the 1 over 24, n is 1 to infinity, q to the n. And it has weight, w is 1 half. It transforms into azzol to z, but it will pick up some extra phases compared to this form. All right, to get more interesting examples, we should also look at congruent subgroups. We will just mention one of them. It's noted by gamma upper 0n. And this is the subgroup of azzol to z, which is just the elements of azzol to z. But with b is equal to 0 modulo n. So in particular, n is 4 will be important for us. Maybe for completeness, let me mention one other one, namely gamma n, which is a subgroup of this one, in fact, cd in azzol to z, such that abcd is equal to the identity matrix modulo n. So since it's a subgroup of the full-modular group, we have more examples. And in particular, the Jacobi-Theta series are modular forms for gamma 2, 1 half modulo forms gamma 2. And I'll just recall the definition that the 2 tau is equal to the sum over r in the half integers q to the r squared over 2. Theta 3 is basically the same thing, except that we take a sum over integers n in z, qn squared over 2, and theta 4 is equal to the same thing as theta 3. But then we have a minus 1 inserted, minus 1 to the n, q to the n squared over 2. And yeah, one thing you see is that modular forms have a Fourier expansion just because you can have 1, 1, 0. And then 1 here, that transformation implies that they have a Fourier expansion. And often, the coefficients have interesting arithmetic information. OK, so these are some of the modular forms which are useful for the following. And as the title suggests, you also need some modular forms. This is a generalization of the theory of modular forms, which I will just explain by giving two examples. So this subject of modular forms, for those who don't know, that goes back to the 20s by the Indian mathematician Robin Hoodian was looking at some q-series which were different from the classical modular forms. And there were some open ends which finally got resolved in the 2000s by Sviggars and Sagyay. So one of the type of q-series people discover is the following function f of tau, which reads as follows. It's minus 1 divided by theta 4 of tau. And then a sum over integers n and z. And we get a bit more complicated expression, minus 1 to the n, q to the 1 half n squared minus 1 divided by 8. And the denominator, 1 minus q to the n minus 1 half. One can expand this quite easily using a computer. One get the following series, q to the 3 over 8. And then 1 plus 3 times q, 1 half plus 7 q, et cetera. And in fact, it's one of the q-series which appears in umbral moonshine, a paper by Chang, Duncan, and Hart. OK, so this is a more complicated function. And it does not transform as a modular form. So yeah, the main statement is it does not transform as a modular form. But we can quite easily modify it to a function which does transform as a modular form. So f tau, however, the following function does be considered a so-called completion f hat of tau. f hat of tau and tau bar. This is a non-holomorphic function. And on the right-hand side, we get the holomorphic part f of tau plus an additional term, which reads as follows minus i over 2. And then an integral from minus tau bar to i infinity. Then the dedicate eta function to the power 3 and the square root of minus i w plus tau dw. So we add this so-called period integral to it. Although it's not quite an integral over a period. And the integrand is the dedicate eta function we had on the board a minute ago. And then there is an additional factor. And one can show that if you do a modular transformation, that this function spits out a similar integral. It transforms back to this function. But it transforms with a shift. It spits out an integral similar to this one, which is then actually a period integral. It goes from, say, 0 to i infinity if you do the tau to minus 1 over tau transformation. So then the claim is that this holomorphic part kind of absorbs that shift. And then this whole thing transforms as a modular. So this function f hat does transform 1 half model of form for gamma 0 form. So yeah, I'm not going to derive this. You have to believe me. And if you take the tau bar derivative of this f hat tau tau bar, then we see that we get no contribution from f. But we just act on the period integral. And we get minus i divided by 2 square root of 2y times the complex conjugate of tau cubed. Let me give one more example, which is related to this one function, which is denoted in the literature for specific reasons as 8s and then super 2 of tau. And it reads, we can express it in terms of f tau. In fact, 24 times f tau. And then there is a term, which is actually a model of form theta 2 to the 4 tau plus, sorry, this 2 comes outside, theta 3 to the 4 tau, divided by eta tau cubed. And if you make a series expansion, this reads as 2 cubed to the minus 1 over 8 and minus 1 plus 45 cubed plus 231 cubed squared, et cetera. And people may recognize this as the famous model of form, which gives the dimensions of the Mace Monsang group. So this appeared in the context of Mace Monsang following the work of Eguchi, Oguri, and Tashikawa. And then I think this formula appeared first in the paper by the Bokar, Merti, and Saki. Sorry? Is it a character of some theory? Yeah, there's a character of a two-dimensional theory. Yeah, that's right. In the sigma model, we have target space k3. But it won't be very important in this talk, but it is indeed the case. So again, this is a mock model of form. And it does not transform as itself as a model of form by itself. But we need to add a non-hologomorphic term. We can read off the non-hologomorphic term by this one. If we add 24 times that non-hologomorphic term, then it transformed actually as a model of form for s out to z. So now this extra term makes it transform as a model of form for s out to z. So maybe one last thing, too, of this intermission about modularity is that since we have this kind of symmetry group of the functions, we don't have to consider the functions on the full upper-half plane. We only have to consider them on a fundamental domain. Domain, so if we consider h, modulo, the full modular group, the classical fundamental domain people take is symmetric around the vertical axis. It's the keyhole fundamental domain, which has boundaries at minus a half and a half. And then this is part of the unit circle. And in fact, the two boundaries are identified under the modular group. And also these two arcs are identified under modular group, just that the only kind of open end is at the top. And for the other congruent subgroup h, mod gamma 04, it turns out we get six copies of this keyhole fundamental domain. And like this, so we get, again, the keyhole fundamental domain. But we get another five images. These ones are the horizontal copies like this. And then there are two inversions, one going down to tau is equal to 0, and one going down here to tau is equal to 2. So if we go back to cyberquitten theory, then the cyberquitten solution basically gives these quantities like u and the adu in terms of modular functions. We have for u of tau, you saw yesterday, it starts with q to the minus 1 over 4. But we can completely give a full expansion in tau. And this reads then 1 half theta 2 to the 4 plus theta 3 to the 4 divided by theta 2 squared theta 3 squared. And this combination is invariant under gamma 04. And just one other quantity that we use is this quantity, the adu. It appeared as one of these topological couplings. And this one reads 1 1 half times theta 2, 2 of tau times theta 3 of tau. OK, so this is very strong that we have now in all, we know u explicitly as function of tau. And now if we go back to the u plane integral, now I can specify the integration domain because you might have noticed that when I wrote down the u plane integral that I didn't specify on the domain. So we get since this order parameter u is invariant under gamma 04, we know now that we should integrate over this fundamental domain. So we are integrating over 8's, gamma 04, and then d tau, which is d tau bar, and then this quantity mu tilde tau times psi mu j. And for this integral to be well defined, it should actually be invariant under the choice of fundamental domain. So if you would take a different choice of fundamental domain, one should be able to show that the integrand remains the same and one can in fact do this using these expressions and model transformations of this psi at the possibility of the observable. And I could also mention how nu tilde looks like, nu tilde in terms of model functions. It actually is quite simple. It is minus i over 8 times theta 4 to the 13 minus b2 over eta to the 9. OK, so it is quite useful if you want to demonstrate that this is in fact a well-defined integral independent of the fundamental domain. It is a useful expression. Do I need questions? All right, so this integral was evaluated by more written for small values of b2. Say b2 is 1 and b2 is 2. And also one way to do it is you can start from an so-called empty chamber and then use wall crossing to get to the chamber you are interested in. This quantity. So this is the contribution. To mention that, let me draw the picture of the u-plane, just the plane where u lives, which has these three special points, i infinity and lambda u is equal to minus 1 and plus 1. So if you compute the full path integral, you get a function support from these strong coupling singularities. And for the manifolds with b2 plus is 1, you get a contribution from the rest of the u-plane. So we are calculating the contribution from the coulomb branch to the full path integral of the theory. But it only contributes for b2 plus is 1 and 0. So yeah, so thanks for asking the question. We see that these points u is minus 1 and u is plus 1 and that corresponds to these two cusps at the bottom of the real axis. And the weak coupling is at tau 2 i infinity. So in fact, this integral is an improper integral because it does go all the way to i infinity. And also if you look at the measure, it is also non-compact if you go down to the real axis. So what we ought to do is to cut off this, cut off the domain at some finite plus large y. And then let us also, and then at the other end, we cut it off at 0.1 over y, imaginary part of 1 over y. So then we can determine the integral on this compact domain. Let me call it h mod comma 0, 4. And then with y as a subscript. And if we are lucky, we can take the limit of y to infinity without any issues. Yeah, so the two sides are identified. Yeah, I'm not quite sure at the bottom. I imagine that this one and this one and this one are, no, I imagine that this is identified and this and this. And then we can go over the, in the middle. So it is, yeah, you can think of it as a sphere, which then has kind of three cusps and the other questions. OK, so to continue working with a generic, so maybe let me first include this cutting of the domain. I'll put the y here and then the limit of y to infinity. OK. And now in order to continue with the evaluation, let me say something about the classification of the lettuces, classification of lettuces with p2 plus is 1. So for the let, if l is old, this is a major, yeah, a major topic for definite lettuces, but it appears quite straightforward if, in fact, if you have p2 plus is equal to 1. Since if l is old, we can decompose it as follows. It's just, we can bring it to the diagonal form. The lettuces l is unimodular for any form manifold and we can bring it to a diagonal form since that it takes this form of one positive direction and b2 minus one negative directions. And if l is even, then l also takes a simple form. It is this two-dimensional self-dual lettuce. So this has intersection form 0, 1, 1, 0, plus n times the negative e8 lettuce, b2 minus 2 over 8. Some number times the negative e8 lettuce. OK, so we'll use this decomposition of the lettuce. I will focus now for l odd just to make things not too technical and for the mistake within the time. Now I pick a convenient metric which satisfies this or respects this decomposition. So I take for this period point j that j is equal to 1 and then I take b2 minus one zeroes. And in this case, our sum over fluxes psi mu j in fact factorizes. So we can write it as some function f little f mu 1 tau, tau bar, which has all the non-holographic dependence, times a theta function or a theta series which sums over this and over the negative definite part of the lettuce. So this is a theta series l minus mu minus. And this, let me spell out this function f mu 1. So mu is just the first component of this conjugated class mu, our tau bar. It complicates i to the pi i mu divided by 2 square root of 2i times the sum over k in l plus mu minus 1 to the k minus mu. k and then q bar k squared over 2. And then this second theta series, theta l minus mu minus is more classical. It's simply a sub over k minus in l minus plus mu minus 1 to the b, restricted to the negative definite lettuce k minus q minus k minus squared divided by. So if we pick a convenient metric with respect to the standard form of the diagonal form for the odd lettuce, then our sum over flux psi factorizes it to the two of these functions. And if you want to pick another metric, you can do so by using the Waclasing formula, say starting formula. OK, so if you evaluate this function f mu, one gets the following. tau bar is equal to 0 if mu is an integer, basically because of this term at the bottom. Whereas if mu is a half integer, we recover the cubic eta, the dedicated eta function to the power free. This should be just set. I hope that that resolves the question. This is just one component of the lettuce. Does that help? And I have omitted the one subscript from all the mu's in this line, so it's just a number here. All right, so maybe given this is answer, people may see where this is going. We can now, all the non-olomorphic dependence of the integrant is in this function. And we saw when we discussed the model of forms, we had an anti-derivative of eta tau cubed. So we can write the integrant as a derivative to tau bar and devaluate using stokes theorem mu j. Let me say for an observable e to the power 2p u. Find us the limit of y to infinity. Just need a little bit more space. So this is equal to the limit of y to infinity, the integral of h mod gamma 0 4, gamma 0 y. Now we have here d tau, which is d tau bar. And we have nu tilde tau. And then in this series psi, we can write it as the d tau bar of f hat, tau tau bar times this series over the negative definite lattice tau, which is just the function of tau. OK, so we can then apply a stokes theorem. It's the limit of y to infinity of the integral over d tau and the boundary of this domain y. Then we get nu tilde tau f hat of tau. So since we go to the boundary of, well, let me just tau tau bar, theta l minus nu minus tau. And here I still wanted to include e to the 2p, the point observable e to the 2p u. Same thing here. So if I continue with this integral, the integral of d tau over the boundary here, if we do first the cos of infinity, it's the integral from minus half to three and a half. So this will just pick up the constant term in tau, or actually four times the constant term because the interval has length four. So we get that this is equal to the limit of y to infinity four times the constant term of mu tilde tau tau. And now let me write the tau bar in terms of tau and capital Y because that's where the cusp at infinity is. So I write as tau minus 2 times i y and then theta l minus mu minus tau to give the constant term of this one which I've noted by the term q to the zero. And there are two more contributions from the other two cusps, which let me just abbreviate it here as sending tau to minus one over tau. And for the other one it is sending tau to two times tau minus one and all the q to the zero. And now as a final step if you take this, you see that the only capital Y dependence in a sense is sitting here in this non-holomorphic part of this model of form. In fact, that part is sub-leading. If you take this limit y to infinity keeping tau fixed then that period integral disappears and you are just left with the holomorphic part. So we can take this limit y to infinity and just write and just take the holomorphic terms in the bracket. So we get mu tilde tau times f tau times theta l minus mu minus tau e to the 2 pu, sorry, the constant term of this. And then we have the contributions from the other two cusps, 2 minus 1 over tau to the 0 plus tau minus 1 divided by tau. So in fact, if we take a simple example of say p2 which says p2 is equal to 2 1, then one reproduces the Donsen invariance determined by Lothar. And in fact I think the expression is actually quite similar to his expression. Although he completed it I think using Warcrossing starting from the hitchable surface and then blowing down. We also see it as actually quite a universal expression because in fact we had a power here, nu tilde to the power 13 minus b2 theta 4. And in fact we get theta, this one can be expressed as a power of theta 4. In fact the theta 4 of this one is actually cancelling the ones of this one. And then it appears actually to be quite a universal answer for the U-plane integral for manifolds with an odd intersection form but otherwise leaving b2 arbitrary. Again in this way evaluated for arbitrary smooth manifolds with p2 plus is equal to 1, b1 is equal to 0. Are there any questions? So maybe let me make a couple of comments if you want to include the surface observables. I want to include the surface observables e to the i minus x plus i plus x. So I am including a, this is the q closed surface observables which couples to the negative, to the anti-self dual part of the connection if you go to the IR. And somehow if you want to connect to expressions in the literature of mock model of forms it is most natural to include also a q-exact observable, q-exact observable which couples to the positive definite part. And then the U-plane integral takes basically what this amounts to. If you include this is that the sum over flux is psi mu j tau tau bar is replaced by, will include also elliptic variables. So we get psi mu j tau tau bar and the rho and the rho bar where rho is an element of l tensor c and it is explicitly written as x divided by 2 pi times du dA. So this is determined in terms of this 2-cycle x and it is multiplied by this combination du dA which makes it transform effectively with a weight minus 1 which is then a Jacobi variable. And then one can also basically run the same game but then using Opel-Lech sums rather than these mock-modeler forms. One thing I should add is that this q-exact operator was a bit subtle. We run into a problem with calculating the one-point function of this observable using the regularization I was using in this talk and we had to develop a new regularization including counter terms to cancel the infinities which showed up and all that can be done. So it is a good observable to include, especially if you want to match to the literature on Opel-Lech sums in the modular form literature. So we have seen that we can very explicitly evaluate this u-plane integral now for pure accuracy to Gates theory and we hope this helped maybe to also evaluate similar integrals for theories of other more complicated Coulomb branches. Thank you for your attention.