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...