[MUSIC] So, let us fix the function which we'd like to consider. First of all, we see that. 3e + 3k = 0 modular 24, which is equivalent to the fact that e plus k is divisible by 8. Moreover, the sum of the coefficient ak is even. It means that we would like to consider the following function. Eta to the power 3e times theta. a1, so on, times theta. ak and the sum of these indexes is even. And I would like to consider the Jacobi modular forms was trivial character of the minimal possible weight and index for the corresponding combination. It means that let us consider the case when e + k = 8. Let me construct a table, first of all. e then k. We can start by 8. k in this case is equal to 0. What function do we have of the minimum possible weight? So this is eta to the power 24. Certainly this is Ramanujan delta function, the modular form of weight 12 with respect to the full modular group. This is our starting function. And now we get different elliptization of this function, in some sense. If e =7, k = 1. So, what product do we have? Eta to the power 21. Now I would like to get the minimal index. If we put here the factor theta, then we still have an untrivial character with respect to the Heisenberg group, because the sum a1+...+ak is even. So I cannot put here eta. But I could put here eta 2. Then 2 is even. What function do we get? We get the Jacobi form of weight, 11 and index 4 over 2, 2. We consider this function. Moreover, we know [COUGH] that this is the only function. Certainly this is a cusp form, but this is the only function of weight 11 and index 2. You see this idea of theta and eta blocks gives us very good example. Let us consider our study. Equal to 6, 2, k equal to 2. Eta to the power 18. Now we can take theta to the square because 1 plus 1 is even. And we get. The first Jacobi cusp form of weight 10 and index 1. And we know we proved this in the last lecture. Then the space of Jacobi cusp form of this weight and index is one dimension. So e equal to 5, k equal to 3. Then we have eta to the power 15. Times a + 1, a1 + 2 + a3 is even. So the minimal combination has a falling form. Theta 2, theta 1 to the square. What weight do we have? The weight is 15 + 1, 16 + 2, 18, so we have weight 18 over 29, then we calculate the index, 4 + 2, 6 over 2, 3. And certainly this is a cusp form. This a cusp form of weight 9 and index 3. So the exercise which we would like to formulate to analyze. As a space. This is a simple question, let us continue. e is equal to 4. k is also 4. Eta. To the power 12. Then the product of 4 theta series, we simply can put eta to the power 4. 4 times 1 is 4, is even. So we get. Jacobi cusp form of weight, 8, 16 over 2 and index 2. This is the only cusp form of weight 8 and index 2. We have prove this in the last lecture. Now e equal to 3, k equal to 5, eta to the power 9. So, we have to add five terms, theta a1 till theta a5. The sum is even, so we put theta 2 times theta to the power of 4. This function has the minimal index. So, let us calculate the weight. The weight is seven. The index, 4 plus 4 over 2 is 4. So this is a function, and please try to analyze the corresponding space of Jacobi cusp form. I can put here two question. Internalize this space. Cusp 4. And this space. For example, do we have a Jacobi cusp form of weight 6 and index 4? It means this function which we constructed is the first cusp form of index 4 or not. Now, we have to finish our table. e equal to 2, k equal to 6. Eta. To the power 6. 6 term we can put eta to the power 6, and we get a function which we considered before of weight 6 and index 3. This is the only cusp form. Of this weight and index. Now. E equal to 1, k equal to 7, eta3. Now we have seven terms so we can end the sum of correspondent a is even. So the minimal index we get if I put theta 2 times eta to the power 6. So the weight is now 3 plus 14 plus 610, the weight is 5 Index 6 plus 4 over 2 over 5. So now we found a Jacobi cusp form of this weight and this index is the same question we can put when doing mathematics. Let's try to find or to analyze this space. But this is not the last case. e is equal to 0, this is the last case. And k = 8. Certainly, the function with the minimal index has this form, theta to the power 8. This is Jacobi form of weight 4 and index 4. And this is the first Jacobi modular form in our list which is not a cusp form. Please try to prove the following fact, that the dimension of this space is equal to 2. Can we construct Jacobi cusp form of weight 4 using the same procedure? Because in the first eight lines start in chromes a very classical modular form of Ramanujan. All these functions are cusp form. Can we construct cusp form of weight 4? This is our next question. [MUSIC]
