Getting certain modular functions from characters - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T02:13:31Z http://mathoverflow.net/feeds/question/74961 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/74961/getting-certain-modular-functions-from-characters Getting certain modular functions from characters charris 2011-09-09T00:52:15Z 2012-01-07T23:54:39Z <p>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:</p> <p><strong>Question</strong>:is there an affine Lie algebra and a finite set of virtual representation $V_1,...,V_n$ such that $$\chi_{V_1}(q)+...+\chi_{V_n}(q)=\frac{E_6(q)}{\eta (q)^{12}}$$</p> <p>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...</p> http://mathoverflow.net/questions/74961/getting-certain-modular-functions-from-characters/85165#85165 Answer by Antun Milas for Getting certain modular functions from characters Antun Milas 2012-01-07T23:24:28Z 2012-01-07T23:54:39Z <p>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).</p> <p>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. </p> <p>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 $\chi(V_2)(q)=\frac{\theta_{10}}{\eta}$.</p> <p>You can easily show that </p> <p>$$E_6=-33 \theta_{00}^4 \theta_{10}^8+ \theta_{00}^{12}+\theta_{10}^{12}-33 \theta_{00}^8 \theta_{10}^4$$</p> <p>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).</p> <p>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 </p> <p>$$\frac{E_6}{\eta^{12}}=\frac{\varphi_{K,c_+}+\varphi_{K,c_-}}{\eta^{14}},$$ 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. </p>