[SOUND] The Free Expansion all the of i Has the following home. Has the sum, over n. Then, the sum of vector in but the three coefficients depends on ln + lu. I add here point to show that we have direct sum here. Octagonal sum. And we have to take summation, over o and q in this one-dimensional lapis such that two n minus, Lm to the square is greater or equal to lu to the square. This condition means that our Jacobi form is holomorphic at infinity. This sum is corresponding for the coefficient. This is for the coefficient of the pullback. I add here the index u( n, l, m). And now, let's try to analyze this condition more carefully. It might happen then for any no banishment for the coefficient, L U is not equal to 0. This is true in general but it might help In this case, Our new modular form Will be So you see the principle, we have this possibility to construct Jacobi cast form starting from very simple Jacobi forms in many variables. Now, I would like to consider the first example of this effect, lets call this question as the problem. To formulate it, i need one more definition. It's better to say. I would like to define the order of Jacobi form in several variables at infinity. The order, Of a Jacobi form at infinity is by definition. The minimum, Hyperbolic, no, Of the indices of all non-zero fourier coefficient of our Jacobi form. In particular, this order is non-negative, If we have a holomorphic Jacobi form and strictly positive if we have a Jacobi cusp. Our problem is the following, I would like to estimate The order at infinity of the delta product for 0 Dm. This is our question. But certainly I would like to take I'm sorry not this function but its pullback. On the octagonal complement of some vector U Dm. More exactly, I consider the following vector U. 2(b1, ..., bm) where the greatest common divisor of the integral coordinate (b1, bm) = 1, and the sum of the coordinate is odd. In this case, this vector Is a primitive Vector. In Dm, we cannot divide U by 2 because Dm contains only vector with even some of it's quarter. Now, I would like to estimate the order at infinity, but we need casterolls for the extension of Jacobi the series. By definition, the Jacobi Theta series of this route axis is equal to the product of m Jacobi, Theta series. So and we have the following, Fourier expansion of this product. First of all, in all Jacobi theta series we have the chromatic character. This is a character so we can write down the product of this character in the following way. N1, nm, r integral. Then we have e to the power 2 pi i. I write it for the expansion, more or less. Standard 4, n1 to the square. So nm to the square over 8. +n1 over 2 times z1+..+nm over 2 times zm. This is the fourier expansion of the product of m Jacobi theta series. Now, I would like to change the notation a little. [SOUND] Let us continue. I would like to denote. And E over 2 le. Then, we can rewrite this sum as pi i (l1 to the square and so on lm to the square) tau. I'm sorry, I forgot the variable to the expansion let me add if we are able. So now I can rewrite this for the expansion in terms of the new, in this is le we get the sum l1 and so on, lm are half integral. More exactly, the element of the jewel latus epsilon. Instead of the Chroneca character, we have the product two L 1- this is an integer- so 2, l, m. Then it was a power pi i. Then, l, the vector l, times tau + 2(l L- this is a vector L one so LM. You can compare this formula -- Let me add our notation. This is -- -- Jacobi Theta series of the root lapis Dm to z. Please compare this formula with the formula for the Jacobi -- -- Theta series of the unimodular axis E eight. You see that we have the same summation under exponent back in the case of the latest D eight, we have additional character. [SOUND] In front of the exponent. So and now to solve the problem which we put, we have To analyze the fully expansion of the two back of the power function on the octonganal compliment of the letter U. -- to understand this pullback, we have to analyze this factor- how we can derive it? It would be two times -- (lm + lu) u over U2 scalar squared, zed m + zed u. And you will have to analyze it for zed = 0. But now, this is really the crucial point of our calculation. Let's try to analyze. This kind of product of L and U. This is equal to two times L one, B one plus so on 2 times lmbm. But 2 times li is always equal to 1 module 2 if not, the correspondent character is equal to 0. Then the sum is congruent to one module two because the sum of B is one module two. So, we see that l, u is always nonzero. And now we can formulate the solution of our problem. [SOUND] [MUSIC]