[MUSIC] Let me analyze this rather complicated formula once more. So we use the property, it's a general property of Fourier coefficient of all Jacobi forms. Let me fix it once more. Fourier coefficient depends only on the hyperbolic nome of its index and the class of the vector in the lattice in the dual modular latus. So in the left-hand side and the right-hand side, we have the same Fourier coefficient with the different indices. So we can modify h, fix element h by any vector v in L. Using this fact, we rewrite the Fourier expansion, first of all, using the summation over all classes, In the discriminate group. There are exactly determinate to fail classes. Then we modify the hyperbolic nome of Fourier coefficient because the hyperbolic nome of the index is exactly 2 times n. Then we fixed a summation with respect to vector v. And we get our splitting principle, we get this function depends only on, wait, sorry. We have to add here the variable tau. This function depends only on tau and this is exactly our modular form. Fh in tau. Maybe I write the formula for the Fourier expansion of this function, Once more. So this is summation of all hyperbolic nomes. N in general is rational, but N + h square / 2 is integral e to the power 2 pi r N tau. This is the new function. We have this function for any class in the lattice modular L. So we proved. Let me formulate now this result. Theorem, For any weak Jacobi form of weight k and index, the lattice L phi in tau z = the sum of this finite sum contains determinant of L terms, fh tau times tau Lh (tau,zeta ). And the Fourier expansion of these two functions are the following. This is Jacobi theta series with characteristic h. We simply that this translation to any vector V is the latest N and this is Fouier expansion of the function F(h). Now I would like to analyze the Fourier expansion or function of f(h). Fourier expansion have this form. So I would like to analyze a (N + h square / 2h). I would like to understand for each indices this coefficient is not equal to 0. First of all, let's assume that phi is holomorphic Jacobi form. Then a(n,l) Is not 0 only if the hyperbolic nome of the index and this is equal to 2N because the hyperbolic nomes is 2N is non-negative. It means then in our function O and R are non-negative, so let's fix it here. N is greater to, equal to 0, if phi is holomorphic. It means that this function is holomorphic at infinity. Fh tau is holomorphic, At phi infinity for any h. Certainly even if Imodify our condition a little if we take for class form. Then we'll have better estimation, and our function fh tau vanishes at infinity. Now I would like to analyze the property of Fourier coefficient for weak Jacobi form. To do this, I would like to introduce the notion of the nome of the class modular a. Let's consider an element in this finite group. Then by definition, h square is the minimum, Scalar square of all element in h. And I would like to take the following invariant of the latest a. By definition, this is the maximum, Nome of the classes, Of this final group. This is max over h of the minimal square of vector in h Now using this notion, I would like to analyze Fourier expansion. So analyze the same coefficient a (N + h square / 2, h). If our Jacobi form is weak, We know then this Jacobi form is a non-vanishing coefficient only if n is greater or equal to 0. But according to our definition, n is equal to N + L square / 2. In other words, a (N + h2 / 2, h) is not equal to 0. It follows then, N is greater or equal to -h square / 2 or in terms of our new notation, mu(L). That's formulated as this fact, this proposition. Phi is a weak form, then a(n, l) is not equal to 0. Follow then, 2 n- l to the square. I am sorry. In this inequality, I forgot certainly the coefficient one-half, because here we have h square / 2. And now here, we have -M(h). So this is a property of the weak Jacobi forms. And now let's consider running them both. Let's take a weak Jacobi form of factors z and r. Rate k and Index M, which means weak form of weight k for the latest 2M where 2M is the latest of when1 times mu square = 2m. Mu 2m, easy to see, we have to analyze all classes of the dual lattice. So we have to analyze x square / 2 m. And this is clear when mu Is greater or equal to -M / two because the maximum of nome we have exactly if x here is equal to M. Maybe I can write it more carefully but I hope you can prove it by yourself. And now what we can create Fourier coefficient If we write down the Fourier coefficient in the terms of the lattice. So we have here x times, U square, but U square is 2m is not equal to 0, then 2m- 2n, I'm sorry, -x2 / 2m is greater or equal -m / 2. Therefore if we use the usual notation of the Eichler-Zagier modular form, this greater equal to -m to the square. And this is true If the Fourier coefficient of weak Jacobi form is not 0. This fact was proved in the Eichler Zagier book, Jacobi Form on the page 105. You see that our proposition gives us the proof of this classic effect. [SOUND]