Which roots are fixed by simple reflections of the Weyl Group?

Let $$\Phi$$ be a root system of a semisimple Lie Algebra, and $$W$$ it's Weyl group. Let $$\Delta = \{ \alpha_1, \dots, \alpha_l \}$$ be a root basis, and let $$w_i \in W$$ be the simple reflection corresponding to $$\alpha_i \in \Delta$$

Is there any way to compute the number $$m$$ of roots $$\alpha \in \Phi$$ such that $$w_i(\alpha) = \alpha$$ for a fixed $$i$$?

I know that simple roots perpendicular to $$\alpha_i$$ are fixed. By looking at Dynkin Diagrams, we see that there are either $$l-2, l-3,$$ or $$l-4$$ such simple roots. For each fixed simple root, it's negative is also fixed, so immediately we have found $$2l-4, 2l-6$$ or $$2l-8$$ fixed roots.

However, is there a way to compute the exact number $$m$$? Failing that, is there a way to compute $$m \pmod 4$$?

I am asking because I am interested in comparing $$n = |\Phi|$$ and $$m \pmod 4$$. I am aware they must both be even numbers, and cannot be congruent modulo $$4$$, but I am trying to see why exactly that is.

Any help would be appreciated, thank you!

• It seems that starting from the third paragraph, you assume your Lie algebra to be simple / $\Phi$ to be irreducible. -- In general, the roots whose number you are looking for are exactly those which are perpendicular to $\alpha_i$.Now it seems to me that for any root $\alpha$, the roots perpendicular to it form a sub-root system; and if $\alpha$ happens to be one of your $\alpha_i$, I would hope (but am not sure) that that root system can be read of the Dynkin diagram with all vertices neigbouring $\alpha_i$ (and corresponding edges) erased. – Torsten Schoeneberg Jan 5 at 5:26
• Thank you for this response. I suppose I mistakenly thought that the root system is irreducible when the Lie algebra is semi simple as opposed to simple. I will think about this a bit more – user366818 Jan 5 at 14:26
• @TorstenSchoeneberg sorry i forgot to tag you earlier – user366818 Jan 5 at 16:43

Let $$\Phi$$ be a root system with inner product $$\langle, \rangle$$ which w.l.o.g. is chosen as invariant under the automorphism group $$A(\Phi)$$. As remarked in a comment, the question is equivalent to: for a given root $$\alpha$$, of what type is the root system $$\alpha^\perp := \lbrace \gamma \in \Phi: \gamma \perp \alpha\rbrace$$?

Remark first that we can obviously reduce to the case that $$\Phi$$ is irreducible; also, $$A(R)$$ (even the Weyl group) acts transitively on roots of the same length, which implies that in a simply laced root system, all $$\alpha^\perp$$ are of the same type, whereas in the other cases, there are at most two types, namely $$\alpha^\perp$$ for $$\alpha$$ being an arbitrary short resp. long root.

Now how to compute this resp. these two? It's not quite as straightforward as I thought in that comment, but the following works: After choosing a set of simple roots $$\alpha_1, ..., \alpha_l$$, one root whose angle to all positive roots can be read from easily available information is the highest root $$\beta$$ resp. its negative, because people have put it in the extended Dynkin diagrams (in this image, $$-\beta$$ corresponds to the green vertex). One sees that in all cases except $$A_{n\ge 2}$$, $$-\beta$$ is connected to only one of the simple roots, say that one is $$\alpha_k$$ (meaning $$-\beta$$, and hence $$\beta$$, is orthogonal to all other simple roots $$\alpha_{j\neq k}$$!). But then we have, for an arbitrary root $$\sum_{i=1}^l n_i \alpha_i$$, that

$$0 = \langle \beta, \sum_{i=1}^l n_i \alpha_i \rangle= n_k\langle \beta, \alpha_k\rangle \Leftrightarrow n_k=0$$ or in other words, the root system $$\beta^\perp$$ is the one we get from the Dynkin diagram of $$\Phi$$ by erasing $$\alpha_k$$, i.e. the vertex which connects to $$-\beta$$ in the extended Dynkin diagram (and all edges which are linked to it). For the type $$A_{n \ge 2}$$, an analogous argument (using that the $$n_i$$ are either all nonnegative or nonpositive) shows that one has to erase the first and the last vertex. Now also note that if there are two root lengths, the highest root $$\beta$$ is always a long root.

For the short roots in types $$B, C, F$$ and $$G$$, use the following trick: going to the dual root system preserves orthogonality, but switches long and short roots. Since the dual of $$B_n$$ is $$C_n$$, while $$F_4$$ and $$G_2$$ are self-dual, we get:

• For any root $$\alpha$$ in $$\Phi$$ of type $$A_n$$, the root system $$\alpha^\perp$$ is of type $$A_{n-2}$$ for $$n \ge 3$$, and non-existent for $$n=1,2$$.
• For any long root $$\alpha$$ in $$\Phi$$ of type $$B_n$$, the root system $$\alpha^\perp$$ is of type $$A_{1}\times B_{n-2}$$ for $$n \ge 3$$, and of type $$A_1$$ for $$n=2$$.
• For any long root $$\alpha$$ in $$\Phi$$ of type $$C_n$$, the root system $$\alpha^\perp$$ is of type $$C_{n-1}$$ for $$n \ge 2$$.
• For any short root $$\alpha$$ in $$\Phi$$ of type $$B_n$$, the root system $$\alpha^\perp$$ is of type $$C_{n-1}$$ for $$n \ge 2$$.
• For any short root $$\alpha$$ in $$\Phi$$ of type $$C_n$$, the root system $$\alpha^\perp$$ is of type $$A_{1}\times B_{n-2}$$ for $$n \ge 3$$, and of type $$A_1$$ for $$n=2$$.
• For any root $$\alpha$$ in $$\Phi$$ of type $$D_n$$, the root system $$\alpha^\perp$$ is of type $$A_{1} \times D_{n-2}$$ for $$n \ge 4$$ (where $$D_2 =A_1 \times A_1$$).
• For any root $$\alpha$$ in $$\Phi$$ of type $$E_6$$, the root system $$\alpha^\perp$$ is of type $$A_{5}$$.
• For any root $$\alpha$$ in $$\Phi$$ of type $$E_7$$, the root system $$\alpha^\perp$$ is of type $$\require{enclose} \enclose{horizontalstrike}{E_6} D_6$$.
• For any root $$\alpha$$ in $$\Phi$$ of type $$E_8$$, the root system $$\alpha^\perp$$ is of type $$E_7$$.
• For any root $$\alpha$$ in $$\Phi$$ of type $$F_4$$, the root system $$\alpha^\perp$$ is of type $$C_3$$.
• For any root $$\alpha$$ in $$\Phi$$ of type $$G_2$$, the root system $$\alpha^\perp$$ is of type $$A_1$$.

A visual inspection of the root systems of rank $$\le 3$$ (see e.g. all of rank 2, $$A_3$$, $$B_3$$) gives results consistent with the above.

I leave it to you to compute or look up the numbers $$m = \lvert \alpha^\perp \rvert$$ that you are interested in.

• I wonder if there is anything wrong or unclear with this answer? – Torsten Schoeneberg Jun 4 at 17:06