[MUSIC] Proof. First, I would like to calculate the Fourier expansion of theta product. We take k theta product. We have calculate the Fourier expansion of two theta functions. But now, we'll have K to the function and the formula is quite similar. This is a product. Of symbol. And now, I can write in a pulling notation. N is a vector with the component from n1 to n k. This is a vector in the Z to the power k. Then we have, q to the power, n1 to the square over 8 +n2 to the square over 8 and so on. We can write this as the scalar product of n. And N over 8 then we have R to the power A1 times N1 + A2 times N2 and so on till MK over 2R to the power N. The vector A over 2. In the index of this Jacobi form is, maybe I write it here, index m = A1 to the square, and so on. A k to the square over 2. And this is the scalar square of the vector a. Now, we have to calculate the hyperbolic norm of the index of the non trivial Fourier coefficient. This is our Fourier coefficient. So I can denote this number as N over 2, this index as l over 2. Sorry, this is not n over 2, but certainly n over 8. Is this in? We have to check. Here as a norm. Capital N over 8, l over2 is = 4 times N over 8 times the index M- N 2 to the square. And we can calculate these quadratic form in m. In the coordinate of the vector m, explicitly. Will be 1 over 4, the sum neaj-nj a i/g) to the square. The summation of all i, strictly smaller than j. This calculation reflects the fact. Then the hyperbolic norm of the index of the Fourier coefficient is non negative. But now our idea is to find, Its minimal. Value for odd. This is very important restriction. Coordinate n1, n2 and so on. N k because all coordinate are owed u 2 symbol. And now, I can make this calculation for theta quark. The function theta a now. This is theta. A times theta b times theta a plus b. So the key is =3. And you could check that in this case first of all we have the coefficient 1 over 4 but also we have 3 factor k is equal to 3. And then, you can calculate our norm in this case explicitly. And we get the following function. Please, you have to check this. Then this =( a + 2b )n1- 2a plus b times m2, -a -b times m3 to the square +2 times a to the square +ab, plus B to the square. N1 + N2, -N3 to the square. And you see, now we can estimate the minimum value of the hyperbolic norm of the indices of non trivial Fourier coefficient. The first term here is positive so this is greater or equal than 2 times a to the square plus a b plus b to the square. Because, this sum, certainly is odd number. And we see, then, the corresponding product not only certainly this is a form, but not only we have this very important estimation on the minimal value of the index of hyperbolic norm. And now, I can prove. Then the beta quark is [INAUDIBLE]. The first non trivial Fourier coefficient of theta function that is q to the power 1 over 24. So if you calculate the inverse function what we get Get q to the power -1 over 24 plus something, where we have only non negative power of 2. So you see that. The first Fourier coefficient for us have the following form. So we have to check. The following. Fourier coefficients. So Fourier expansion. F N over 2, l over 2 times, sorry, again. I make these small mistakes. This N over 8. And here, we have q to the power n over 8 minus 1 over 24, rl over 2. So this is exactly the best part of the Fourier expansion. And we have to check that the index of the corresponding Fourier coefficient has good hyperbolic norm. So let me calculate it. The norm of this coefficient. Is equal to the norm, N, 8, L over 2- 4 times 1 over 24 times the index of our form. The index this is A to the squared +b to the squared +A plus b to the squared over 2. But according to the result. We, good in the previous slide. We see that the norm is greater or equal to the index over 12. Greater equal to a to the square, plus d to the square, +a, +b to the square, over 12. And now we have a to the square, +b to the square, +a +b to the square, over. 12, which is greater or = 0. So, our theta block, is polymorphic. Moreover, under what condition, we could have. Inequality. We have inequality. If and only if N1 +N2 -N3 is = + -1. Only under this condition, we could have zero in the first term, and exactly this. Coefficient in the second. But then You see we could have an equality on down to the questions and n3 is equal to n1 + n2 + -1. You analyze, analyze the corresponding no, and you can get this to check that this inequality is possible if and only if A- B. Is divisible by 3, modular 3. You can see this immediately, if you put the formula for n3 here. Let's check this. [FOREIGN] >> [FOREIGN] >> [FOREIGN] >> [SOUND] [MUSIC]