[MUSIC] I can give you the following exercise. Let's consider the generator of the space of Jacobi cusp form. Of weight 10 and index 1. This is the Eichler–Zagier case. Then, this function. Is equal theta function to the power 18 times the square. Jacobi theta series. Then, H10 phi 10, 1 is equal to the function phi tilde 12, 1. Which is a cusp form of weight 12 and then this one, where. The 0s value of this function is equal, up to the constant, to the Ramanujan Delta function, in tau. In particular, 1 over c phi tilde is a function phi 12, 1 such that this particular weight is just a function. Please find this [INAUDIBLE]. Our question is to find. C. The second question. Let us take function E 4, 1, and t, z which is of the Jacobi series for the latest E8. For the root vector V. VE is 88, V to the square = 2. Please try to calculate H4 E4 1. This is the Jacobi form for 6 contain index 1. Please try to go calculate it's 0s way. But now, I would like to analyze an obligation of the result which we get more exactly. I would like to use the second function and to prove the following theorem, theorem. You can find this theorem in Aigner, Ziegler book. This is a theorem about algebraic instruction of the of the weak Jacobi forms of even weight. By definition, this is a grated ring of the weak Jacobi form of even weight, arbitrary index, K in z MZ0. And the theorem claims that this graded ring has two generators of the graded ring of s equal to z modular forms. The first generator this is a function phi -1. The second phi- 1. The proof. You can prove this theorem by induction. And I give you only the main idea and you can check without any problem the basis of this induction. But now, I would like to give you the main idea of this induction. Let us consider an arbitrary function. Of weight 2 k and index m. Then, if zero's value. Is a modular form of weight 2k. Have to check to change the sprint 2k. In one variable. It may be zero but in any case this is a modular form of weight 2 K. With respect to the full modular group. Now, we can see this following function, phi tilde, is equal to phi minus, this is our function, F 2k Tau times 1 over 12. Phi, 0, I'm sorry, here. Let us start this phi 0 1, tau z to the power m was a co efficient 1 over 12 to the power of m. What function do we have? This function for that =0, is =0 identically. Why? Because, This is the 0. Our function phi(0,1) at 0 is a modular form of weight 0. It's a constant. But this constant is =12. Because we have correlated the first zeros constant ten of of our coefficient. So this is a constant 12, for that equal to zero. It follows then. The tailor expansion of this function. Starts from z to the square. F tilde and z to the power n is = 2 Because the weight of this function is even and we have only even power of that in the. But we know that the function phi -1, Has zeros exactly of order 2. And this is the only zero of this Jacobi form, because the Jacobi form of weight one has exactly two zeros in any fundamental domain. Therefore, the quotient Is Jacobi form of weight. 2k + 2 and index m -1 and this is weak Jacobi form because we know a part of the Taylor expansion of this function. So this is the standard recursion induction argument. And now, you can use induction on m. Certainly, you have to change or you have to check. As a basis of this induction by I leave this as a formula and a proof of the basis of this induction for you. So you see that within the differential operator. We can construct the generators of the graded ring of the Jacobi form. And using our result, you can try to generalize this theorem of Eichler–Zagier to the case with Jacobi form, in many variables. So generalization. Of the two. Let's put the same question about. The weak, Maybe I write it as a term. Graded ring of Jacobi forms. 4 the latest a. And graded ring certainly of weak. This is very important. So let me add this. Weak Jacobi forms because as you can see. This is a graded ring of all Jacobi form. This ring usually has infinite number of generators. For example, I don't like to prove the theorem but for many root system you can find the graded ring of weak Jacobi form if you add some condition of symmetries. For example, let us consider the Jacobi form for the root system Dn. going to construct the weakJacobi form of weight -N for this like this by definition this is product theta zed 1 over eta q in eta. Till theta In (zn) over eta cube (eta). So this is a weak Jacobi form of weight -n for the lattice Dn. And this function is also symmetric. Symmetric in a sense than any permutation. Of that one Z n give us the same function. I don't like to formulize as this definition. But, this is more was clear in this case. And if we can see the symmetric function. Then, now, you can use the differential operator For this lattice Dn. And you get, A Jacobi form of weight- n + 2 for this lattice. It's again, symmetric with Jacobi form. But it's weight is- N+2. Please, find it's Q zero for coefficient Using this argument and this procedure. You can really try to analyze the greater dream. Of weak Jacobi form of arbitrary weight for the latest Dn with arbitrary index m. So here, I also put star. Since the gradient ring Jk,Dn star weak And all this function, we can see the on the symmetric function. So please to analyze this graded ring. It's possible to find all generators and this is very very good exercise which you can make, use into the. We consider it. And this question is very close to the modern mathematic because the same question for the latest E8 is. Almost open. Now, I can add another exercise. I can come back to the previous case, sorry. And I can add the following exercise. Try to find that the graded ring of all weak Jacobi form of Eichler–Zagier is a quadratic extension Of these gradeed ring and if it was a generated function, you have to take the Jacobi form of weight -1 and index 2. This is a function theta into z over N cubed theta. Moreover, please find its quadratic equation for this function in terms. Of phi 0,1 and phi -2,1. [MUSIC] [SOUND].
