[MUSIC] The next subsection of this chapter is automorphic correction. This is a central point of this lecture. And I would like to give you today the second explanation of this rather complicated Jacobi automorphic vector which we have in the modular functional equation of any Jacobi form of weight k and index m, phi is a Jacobi, for example, weak Jacobi form of weight k and index m. We can see that as a case of one variable. So as a case of Jacobi modular form of. So this vector, this complicated automorphic vector. We gave one explanation. Our first explanation was the following. Instead of modular form on the product of upper half-plane and C, we consider modular form on the Siegel upper half plane. Now, tau and omega are in H1 and imaginary part of this matrix Z is positive. So instead of our function, we can see there's a function phi in three variable 2 pi i m omega. Then the modular group, the Jacobi modular group, is a parabolic subgroup of Sp2 Z of of the Siegel modular group of genus two. And this function which we denote phi tilde m in z. Is gamma J modular form. On H2. This is the first explanation of Jacobi modular equations. But now, I would like to explain to give another interpretation, only on the modular equation. This modular equation. Of Jacobi form of weight k and index m. And to do this, I introduce the new function which I call the automorphic correction of Jacobi form. By definition, the capital phi with index m in tau, z is equal to the e to the power minus 8 p to the square m G2 in tau z to the square phi tau z. So we add a new vector, a quasimodular vector, so we can call this correction quasimodular correction. But why this additional vector is very, very useful. Because the automorphic correction capital Phi now Is invariant, is a modular form of weight k with respect to group [INAUDIBLE] z. More exactly, this function satisfies the following functional equation, c tau plus d to the power k, capital Phi m2 z. It means that with this additional quasimodular vector, we cancel the Jacobi automorphic vector. You can prove this by yourself and have no difficulties. It's simply, you have to write the functional equation for the automorphic correction. And we got this additional fractional linear term in this place. So the functional equation, this is easy exercise. We have had so many formulas in our lecture today, so I leave this formula for you. You'll have no difficulties to prove it. You have to use only the modular property, the quasimodular property of the quasimodular Eisenstein series G2. [INAUDIBLE] this additional term, cancel the exponential automorphic vector in the modular equation of Jacobi form. But now let's analyze the corollary of this vector. So now I can write down the Taylor expansion of this automorphic correction. If I write down the Taylor expansion in z. Then all Taylor coefficients [INAUDIBLE] modular form of weight k plus n with respect to full modular group. This fact follows directly from this functional equation. Now I can write down the Taylor expansion of the original Jacobi form. So phi is our original weak Jacobi form of weight k and index m. And we write down its Taylor expansion using the previous Taylor expansion. So we have e to the power now plus 8 p to the square m G2(2) z to the square times the Taylor expansion of the automorphic correction. And we see that all Taylor Coefficients of the original Jacobi form are quasimodular. So we proved the theorem for [INAUDIBLE] again. That for any n, fn(2) is quasimodular. Though the automorphic correction is a very useful fact, because we can control the Taylor expansion. Now some corollary from this theorem. So let me formulate the first corollary. Phi is a weak Jacobi form of weight k and index m. Then we know. That the order of 0 of this function is smaller or equal to 2m. We proved this theorem in the beginning of our lectures. So, this is the theorem about the number of zeros of any Jacobi form in the fundamental parallelogram. So, and now we see that the first two m Taylor coefficients of any weak Jacobi form determine this form uniquely. But we can analyze the first two m Taylor coefficients of our Jacobi form. We can analyze the first 2m Taylor coefficient of its automorphic correction. So we have the following inequality for the dimension of the space of weak Jacobi form, smaller or equal to the sum of the dimension of the space M k plus e the summation from 0 to 2m. In principle, we have a better estimation, because let's assume that k is even. k is even. Then our Jacobi form is even function. And in this case, we have a better estimation, because we know that all weight will be even. In fact, we have to add 2 in here. So it gives us the estimation of the dimension of the space of weak Jacobi form. [MUSIC]