Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

It is well known that characters of affine Lie algebras have certain modular properties. For instance, the linear span of all irreducible characters at a given level must be invariant under a certain action of $SL(2,\mathbb Z)$. In the case of affine $E_8$ there is only one irreducible level $1$ representation, the basic representation $V(\Lambda_0)$, and the (specialized and normalized) character can be written as $$\chi_{V(\Lambda_0)}(q)=\frac{E_4(q)}{\eta(q)^8}.$$ The RHS can be achieved as a sum of characters of another affine algebra. Affine $so(16)$ has $4$ level one representations. Besides the basic representation another one of these is $V(\Lambda_4)$, where $\Lambda_4$ denotes the fundamental weight whose finite part is the highest weight for one of the half spin representations. Using specialized and normalized characters again we have $$\chi_{V(\Lambda_0)}(q)+\chi_{V(\Lambda_4)}(q)=\frac{E_4(q)}{\eta(q)^8}.$$ I am interested in which elements of $\mathbb Z [E_4,E_6,\Delta]/(E_4^3-E_6^2-1728\Delta)$ also can show up here. It's not hard to use the above to get $\frac{E_4(q)^n}{\eta(q)^{8n}}$, so a good starting spot I'm wondering about is:

Question:is there an affine Lie algebra and a finite set of virtual representation $V_1,...,V_n$ such that $$\chi_{V_1}(q)+...+\chi_{V_n}(q)=\frac{E_6(q)}{\eta (q)^{12}}$$

The need for virtual representations is certainly necessary since the RHS will have some negative coefficients. I suspect the answer is no, because I'm guessing the whole thing is tied to even unimodular lattices and the second way above of getting $\frac{E_4(q)}{\eta(q)^8}$ comes from the connection between $E_8$ and $SO(16)$. So if not, is it possible to achieve this by some other infinite dimensional algebras whose characters have modular properties, e.g. generalized Kac-Moody algebras, vertex operator algebras, etc...

share|improve this question

1 Answer 1

up vote 5 down vote accepted

Since you allow virtual characters you should definitely expect such a thing (due to the general philosophy of writing down Eisenstein series as linear combinations of theta series after Siegel, Weil and others).

Here is an explicit construction. Take the simplest affine Kac-Moody Lie algebra, namely $A_1^{(1)}$, and take the level to be $1$. Then there are (essentially) two integrable highest weight $A_1^{(1)}$-modules of this level. Let's denote them by $V_1$ and $V_2$ for simplicity.

As usual,let $$\theta_{00}(q)=\sum_{n \in \mathbb{Z}} q^{n^2}, \ \ \theta_{10}(q)=\sum_{n \in \mathbb{Z}} q^{(n+1/2)^2}.$$ Then the corresponding (homogeneous) characters are $\chi(V_1)(q)=\frac{\theta_{00}}{\eta}$ and $\chi(V_2)(q)=\frac{\theta_{10}}{\eta}$.

You can easily show that

$$E_6=-33 \theta_{00}^4 \theta_{10}^8+ \theta_{00}^{12}+\theta_{10}^{12}-33 \theta_{00}^8 \theta_{10}^4$$

Now $\frac{E_6}{\eta^{12}}$ is just a linear combination of level $12$ integrable $A_1^{(1)}$-modules (view each summand as a tensor product of 12 level one modules).

One more thing. Your quotient reminds me of Serre's paper "Sur la lacunarite des puissances de $\eta$", on the lacunarity of even powers of the $\eta$-function. In the case of $\eta^{14}$, he uses a nice identity

$$\frac{E_6}{\eta^{12}}=\frac{\varphi_{K,c_+}+\varphi_{K,c_-}}{\eta^{14}},$$ where $\varphi_{K,c_{\pm}}$ are certain CM modular forms of weight $7$ (the field is $K=\mathbb{Q}(\sqrt{-3})$ and $c_\pm$ are Hecke characters). I wonder if the right-hand side can be linked to anything in representation theory.

share|improve this answer
    
Very nice, thank you! Do you know of a good introduction to this Siegel-Weil way of expressing Eisenstein series in terms of theta functions. I have only heard about this in passing before and would like to have a better understanding of it. –  charris Jan 9 '12 at 18:26
    
My understanding is that "M. Eichler, Uber die Darstellbarkeit von Modulformen durch Thetareihen, J. Reine Angew. Math. 195 (1956), 156-171." would be the right place to start. –  Antun Milas Jan 10 '12 at 16:49

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.