[MUSIC] In the proof of this theorem we get more. Proof. In the proof, we analyze the modular behavior of this Jacobi form. The corresponding function, which we'll have to study in the pullback of this function for that equal to zero, but the technique of the slash action, which were developed in the last section, gives us better result than we formulateing the theorem. So, I would like to analyze the modular behavior of this function, with respect to the slash KM action for arbitrary element of the modular group S sub 2 zed plus 2. [SOUND] To do this, I use the following. Multiplication in the Jacobi. Is equal to M times M to the power minus 1. Times x0 times M. So we got here the conjugation with the respect to M and the function. Pi is invariant with respect. To the section. Therefore. We have the following general property. Is equal to pi. K M. m to the power minus 1 is 0. This result we have for any Mn S a2 z. Now, I would like to take an arbitrary M in the group gamma x. In this case, M times x minus x belong to zed to the square. We can use the following. Calculation. M inverse times x0 = [M to the power -1 X,0] [X,0] to the power -1] times [X,0] X to the power minus 1 is equal to minus x 0. And then, we get. M inverse x minus x. Then the central extension, all curves point to minus the determinant M inverse, x, x. We have minus, because we see minus here. Times x0. But this vector belongs to zed to the square because M is in gamma x. We can continue our calculation, and we get an element from the integral, Heisenberg group times the rational center element. I can write this determinant as (X,MX), because the determinant of M is equal to 1 times x0. Therefore, we can calculate. Our function. In this particular case. And we get e to the power 2vim determinant (MX,X) We have the column in order to avoid this minus times 5 K M X 0. It means that we proved the modular property, not only for the force pull back, but for the Jacobi form. We have proved for arbitrary M in gamma x We have the following modular behavior. Of the Jacobi form after the rational Heisenberg transformation. This additional factor We denote by xm is a character of gamma x group. Why? Because we have the slash action here, after that the second index. So if I put now z = 0, we obtain the modular equation for the function phi x. So, let me denote this equation like G. X. We can specialize the modular equation J x for z equal to 0. The we get then the function of x is modular With respect is modular certainly, with a character, if the modular form was a character. E to the power 2 pi M, the determinant M, X, X Pi x for any M in gamma x. To finish the prove we have to analyse the fully expansion of this modular form [SOUND] As definition, this is 2 pi i m ( q to the square tau + pq) times phi tau q plus p and we can use the formula for the Fourier expansion, Of Jacobi modular form. 2 pi i (n tau + l Zed). But Zed is equal to q tau plus 1. So we have to add this term here. We have this summation with respect all indexes slash for mn minus is greater or equal to 0. Let's calculate these fourier expansion. We have the sum. First of all, let's try to find a factor which doesn't depend on n, on n, we, with e to the power, maybe we can put it before the summation. e to the power two pi i m p q, that is this term. Times the sum a n l then we can add this term. E to the power 2 pi i l p. Then i to the power 2 pi i Then m times q to the square plus l q plus n times tau. I denote, the power, by this power by n. The summation with respect to all n and m, or and l and l, such that 4 and m, minus l to the square, equatable to 0. But N, Is always greater to 0, because the discrimen of this quadratic polynomial is equal to l squared minus 4 m n is always less or equal to zero. Therefore, function phi x is holomorphic. At infinity. To thing is to prove, I have to analyze the Fourier expansion of the function Phi in all cusps. We have to analyze the free expansion of function x in all casps, for any M in SL2 Zed. [SOUND] But we can do this using [SOUND] the functional equation Jx. And Jx will describe the modular behavior of our translated Jacobi. Translated by its Jacobi form. And then we can use this to calculate tho Foley expansion. This is equal to the pull back, Of this Jacobi. But we can calculate the Fourier expansion of this function using the same argument. M inverse x, this is Another vector in Q to the square. So the similar calculation, Show that Tau, is equal, some Fourier coefficient, e to the power N tau, where N is greater equal to 0. So, the function, phi x in tau is holomorphic. In all cusps. Of the modular group gamma x. And our theorem is proved. Remark The condition, then fourier coefficient of polymorphic Jacobi form are equal to 0 unless, 4 nm minus l to the square is greater or equal to 0 is equivalent, To the fact that, For any x in Q to the square the function phi x tau is holomorphic, At infinity. So today we have two description, or two interpretation of this condition to be holomorphic for Jacobi form. This interpretation Is given in terms of the pull backs. Operational Jacobi form and the second interpretation was given in terms of Siegel modular forms. Our Jacobi form of weight in index m is holomorphic. If the corresponding function on Siegel upper half plane is holomorphic. So, the two points of view on Jacobi form gives us two additional interpretation of this condition for Jacobi form to be holomorphic at infinity. [SOUND] [MUSIC]