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When $\mathfrak g$ is a complex, simple, simply laced Lie algebra of rank r then the (specialized) character of the basic representation for the corresponding affine Lie algebra $\hat {\mathfrak g}$ is given by $\chi_{\hat {\mathfrak g}}(q)=\frac{\Theta _{\mathfrak g}(q)}{\eta (q)^r}$, where $\Theta _{\mathfrak g} (q)$ is the theta function for the root lattice of $\mathfrak g$. It is well known that $\Theta _{E_8} (q)$ is equal to the normalized Eisenstein series of weight 4, $E_4 (q)$. We also have another basic result that $E_4(q)=\frac{1}{2}(\theta_2 (0,q)^8 +\theta_3 (0,q)^8 + \theta _4 (0,q)^8)$, where $\theta_2(x,q)$, $\theta_3(x,q)$, $\theta_4(x,q)$ are the three classical even Jacobi theta functions. Being odd, the other Jacobi theta function $\theta_1(x,q)$ vanishes at $x=0$. My question is in regards to the unspecialized form of the character; that is, I'm wondering if evaluating the character on a non-zero element of the Cartan subalgebra produces something of the form $$\chi _{\hat {E_8}}(x_1,...,x_8,q)=\frac{\frac{1}{2}(\prod_{i=1}^8\theta_2 (x_i,q) + \prod_{i=1}^8\theta_3(x_i,q) + \prod_{i=1}^8\theta _4 (x_i,q)+\alpha \prod_{i=1}^8\theta_1 (x_i,q))}{\eta (q)^8}$$ for some $\alpha$? Moreover, in the case $\alpha = 0$, does it make sense that Taylor expanding the numerator of this expression and writing it in terms of elementary symmetric polynomials in $x_1^2,...,x_8^2$ one should find that everything below degree 8 can be written as quasimodular forms times only powers of $p_1(x_1,...,x_8)=x_1^2+...+x_8^2$? This is what I got when I plugged it into Mathematica and in fact the coefficients are similar to the Eisenstein-Jacobi series $E_{4,1}(z,q)$. What would probably be very relevant is if there's anything regarding Jacobi forms in several variables that has been studied that is similar to the "development coefficients" which are well studied in Eichler and Zagier "The Theory of Jacobi Forms".

I've tried looking through a good bit of the literature regarding both of these matters, but couldn't find what I was looking for. If anyone could point me in the right direction, I would be greatly appreciative.

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up vote 3 down vote accepted

Your (related?) question brought me here. I'm not sure if I'm following you right, but I guess your first question asks: is there {$h_1,...,h_8$},a basis of the Cartan subalgebra, such that $${\rm tr}_{L(\Lambda_0)} q^{L(0)-1/3} x_1^{h_1} \cdots x_8^{h_8}$$ equals the right hand side for some $\alpha$? (Here $L(\Lambda_0)$ is your basic representation). I assume you are flexible on the choice of a basis of the Cartan, right? For instance, your formula obviously doesn't hold if you take (for some purposes natural) $h_i=\omega_i$ (fundamental co-weights).

Let me ignore the denominator which doesn't change. Start from the usual Euclidean $\mathbb{Z}^8$-lattice with orthonormal basis $\epsilon_i$. Then, $D_8$ lattice, as you know, consists of all $v \in \mathbb{Z}^8$ such that the sum of coordinates of $v$ is even. To get the $E_8$-lattice you also include all $v \in (\mathbb{Z}+\frac{1}{2})^8$ such that (again) the sum of coordinates of $v$ is even. Set $$\theta_3(q^{1/2},x)=\sum_{n \in \mathbb{Z}} q^{n^2/2} x^{n}$$ $$\theta_4(q^{1/2},x)=\sum_{n \in \mathbb{Z}}(-1)^n q^{n^2/2} x^n $$ $$\theta_2(q^{1/2},x)= \sum_{n \in \mathbb{Z}} q^{(n+1/2)^2/2} x^{n+1/2}$$ $$\theta_1(q^{1/2},x)=\sum_{n \in \mathbb{Z}} (-1)^n q^{(n+1/2)^2/2} x^{n+1/2} $$

So you will be summing over the $D_8$-lattice and another "shifted" $D_8$.

For the $D_8$ part, you get $$\prod_{i=1}^8 \theta_3(q^{1/2},x_i)+\prod_{i=1}^8 \theta_4(q^{1/2},x_i)$$ multiplied by $\frac{1}{2}$.The shifted part contributes with $$\prod_{i=1}^8 \theta_2(q^{1/2},x_i)+\prod_{i=1}^8 \theta_1(q^{1/2},x_i)$$ multiplied by $\frac{1}{2}$. The resulting formula looks very similar to yours (with $h_i=\epsilon_i$).

This construction is essentially decomposition $L(\Lambda_0)=L_{D_8}(\Lambda_0) \oplus L_{D_8}(\Lambda_8)$ into irreducible (standard) $D^{(1)}_8$-modules (I hope $\Lambda_8$ is correct here). All this is very natural if you are familiar with the spinor construction of ${E_8}^{(1)}$ (see this monograph ).

Just in case you prefer $${\rm tr}_{L(\Lambda_0)} q^{L(0)-1/3} x_1^{\omega_1} \cdots x_8^{\omega_8},$$ simply express $\omega_i$ as a linear combination of $\epsilon_j$'s, and rearrange $x_j$'s accordingly.

I hope this helps. My apologies if I said something wrong.

n.b. $x_i$ should be $e^{2 \pi i x_i}$ to be consistent with " $\theta_1$ vanishes at $x_i=0$".

EDIT: (Too long for a comment.) Regarding your second question and the fact that up to degree $8$ in $x_i$'s there is only contribution from the power sum $p_2(x)=x_1^2+ \cdots + x_8^2$.

Actually something similar happens if you consider one-point functions on the torus so it might be related. For a holomorphic rational CFT $V$ (resp. VOA) of central charge $c$, meaning that there is a unique irreducible sector (resp. module), the vector space of one-point functions (essentially traces of degree zero operators coming from vertex operators)
$$\langle Z(v,\tau), v \in V \rangle$$ is a fairly nice space of modular forms with a mild pole at infinity. Basically after you multiply with $\eta^{c}(\tau)$ you expect to get all holomorphic modular form. For the ($c=24$) moonshine module this was shown by Dong and Mason here , but I think it must also hold for $c=8k$ (shown maybe in this more recent paper.) . Now back to $V=L_{E_8}(\Lambda_0)$, which is holomorphic with $c=8$. If you accept what I said, each $\eta^8(\tau) Z(v,\tau)$ is holomorphic (possibly zero) for $\Gamma(1)$ of weight $deg(v)$. So if you pick $v$ of degree $2,4,6$ you will be generating one-dimensional spaces spanned by $\frac{E_6(\tau)}{\eta^8}$, $\frac{E_8(\tau)}{\eta^8}$ and $\frac{E_{10}(\tau)}{\eta^8}$. However in degree $8$ you have a contribution from the cusp form $\Delta$, so the dimension jumps by one. If I remember correctly, up to degree $8$ you can use conformal vector to get those, but higher you also have to construct some primary vectors coming from spherical harmonics to get a nonzero $Z(v,\tau)$, and then you apply a result of Waldspurger . So I agree with you, the Weyl invariant of degree $8$ (for $E_8$) is most likely relevant in your case.

Dong and Mason (and their collaborators/students) wrote several papers on this subject, so it might be useful to read some of their work. For instance, this paper , might be also useful for your purposes.

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Thanks! For the second part, even with $\alpha=1$ Taylor expanding gives multiples of $(x_1^2+...+x_1^2)^n$ until degree $8$ in the $x_i's$. Could this have to do with the exponents of $E_8$ being $2,8,...$? – charris Jan 10 '12 at 21:46
@charris: I edited my answer based on your comment. – Antun Milas Jan 11 '12 at 14:29

Gritsenko has studied Jacobi forms in several variables in a series of papers: search for his papers on mathscinet with "Jacobi" in the title.

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Great, thank you for the reference. I will look into it. – charris Mar 29 '11 at 16:35

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