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Let $\Phi_+$ be the set of all positive roots for a Kac-Moody algebra. Denote by $\alpha_i$ the simple root associated with node $i$ by for $i \in \{1, \ldots, n-1\}$ and by $\beta$ the simple root associated with $n$.

The Dynkin diagram for $\tilde{E}_8$ is \begin{align} \circ - \circ - & \circ - \circ - \circ - \circ - \circ - \circ \\ & \ | \\ & \ \bullet \end{align} where $\bullet$ corresponds to the simple root $\beta$. The degree of a root is the coefficient of the root at $\beta$. Has the number of degree $d$ real roots in $\Phi_+$, $d \in \mathbb{Z}_{\ge 1}$, been computed for the root system $\tilde{E}_8$? Are there some references about this? Thank you very much.

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I don't believe there is a reference for this, for this follows immediately from the description of real roots in affine root systems. Namely, by Proposition 6.3(a) in "Infinite dimensional Lie algebras" by V. Kac the real roots of $\Delta=\mathsf{E}_8^{(1)}$ are of the following form. Number the simple roots as $\alpha_0,\alpha_1,\ldots,\alpha_8$, where $\mathring\Delta=\langle \alpha_1,\ldots,\alpha_8 \rangle$ is the subsystem of type $\mathsf{E}_8$, so $\alpha_0$ is the affine root. Then $$ \Delta^{\mathrm{re}} = \{ \alpha+n\delta \mid \alpha\in\mathring\Delta,\ n\in\mathbb{Z} \}, $$ where $\delta=\sum_{i=0}^8 c_i\alpha_i$, and the coefficients $c_i$ are as on the following diagram:

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In what follows all roots are assumed to be real.

Now it's a matter of a simple calculation. The degree $0$ roots form a subsystem of type $\mathsf{A}_8$, so there are $36$ positive roots among them. The number of degree $1$, $2$ and $3$ roots in $\mathring\Delta$ is, respectively, $56$, $28$ and $8$. The degree $1$ roots in $\Delta$ are either in $\mathring\Delta$ or of the form $\alpha+\delta$, where $\alpha\in\mathring\Delta$ has degree $-2$, so there are $56+28=84$ of them; the same works for degree $2$ roots.

Degree $3$ roots are

  • either in $\mathring\Delta$ (and form an $\mathsf{A}_7$ subsystem);
  • or of the form $\alpha+\delta$, where $\alpha\in\mathring\Delta$ is of degree $0$;
  • or of the form $\alpha+2\delta$, where $\alpha\in\mathring\Delta$ is of degree $-3$.

Thus in total there are $56+8+8=72$ roots of degree $3$.

Since there are no roots of degree $\geqslant3$ in $\mathsf{E}_8$, one concludes that $$ \#\left\{ \text{positive real roots of degree $d$ in $\mathsf{E}_8^{(1)}$} \right\} = \begin{cases} 36, & \text{if}\ d=0, \\ 72, & \text{if}\ d \equiv 0 \pmod{3},\ d\neq 0, \\ 84 & \text{otherwise.} \end{cases}$$ The formula for the number of all real roots of a given degree (both positive and negative) is even more simple, namely, $72$ for $d\equiv0\pmod3$ and $84$ otherwise.

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  • $\begingroup$ The method is good, but I think you have some typos. First of all, $E_8$ has $120$ positive roots, but $36+56+28+8 = 128$. I think $36$ should be $28$, because the degree $0$ part of $E_8$ is $A_7$, not $A_8$. Also, $\alpha$ ranges over all $E_8$ roots, not the positive ones, and the OP asked to count real $\tilde{E}_8$ roots, not positive ones. I think the right numbers are $72$ for $d \equiv 0 \bmod 3$ and $84$ otherwise. If you want to restrict to positive $\tilde{E}_8$ roots, then I think your final answer is right. $\endgroup$ – David E Speyer Jul 19 '19 at 15:19
  • $\begingroup$ @DavidESpeyer But we look at the degree $0$ part of $\mathsf{E}_8^{(1)}$, not $\mathsf{E}_8$. Also, here I only work with the real roots, so omit the word "real" for brevity (I shall edit). And the author asks for real positive roots of degree $d\geqslant1$ (the latter implies positivity). $\endgroup$ – Andrei Smolensky Jul 19 '19 at 15:30
  • $\begingroup$ You are right that he says "real roots in $\tilde{Phi}_+$" and I misread him, so you are answering the right question. I still think $36$ should be $28$. $\endgroup$ – David E Speyer Jul 19 '19 at 15:36
  • $\begingroup$ Beside the $28$ degree $0$ positive roots in $\mathsf{E}_8$ there are also $8$ degree $0$ roots that involve $\alpha_0$ with coefficient $1$. $\endgroup$ – Andrei Smolensky Jul 19 '19 at 16:07
  • $\begingroup$ Right, the final count is right. Oh, I see. You write "The degree $0$ roots form a subsystem of type $\mathsf{A}_8$, so there are $36$ positive roots among them. ", but you mean the degree $0$ roots in $\tilde{E}_8$, not the degree $0$ roots in $\Delta^0$, of which there are $28$ positive ones and $56$ in total. You do not explicitly say how many degree $0$ roots there are in $\Delta^0$, although you are need that for the degree $3$ computation ($72 = 8+56+8$). I now see that it is possible to parse all your sentences as right (and I definitely agree that the final count is right). $\endgroup$ – David E Speyer Jul 19 '19 at 16:18

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