Getting certain modular functions from characters - MathOverflow most recent 30 from http://mathoverflow.net2013-06-19T02:13:31Zhttp://mathoverflow.net/feeds/question/74961http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/74961/getting-certain-modular-functions-from-charactersGetting certain modular functions from characterscharris2011-09-09T00:52:15Z2012-01-07T23:54:39Z
<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#85165Answer by Antun Milas for Getting certain modular functions from charactersAntun Milas2012-01-07T23:24:28Z2012-01-07T23:54:39Z<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>