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It is well-known that the normalizer of the diagonal torus $\mathbb{D}_{n}$ inside the group $GL_n$ is generated by the permutation matrices. What are the generators of the this normalizer (of $\mathbb{D}_{2n}$) inside the symplectic group $Sp_{2n}$?

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I did this awhile ago, but didn't write it down. However, I can tell you how to calculate the generators at least modulo the diagonal torus. It might take you awhile.

Let $z = \textrm{antidiag}(1,-1,1,-1, ... , 1, -1)$. And let $$G = \textrm{Sp}_{2n} = \{ x \in G : x^t z x = z \}$$

The standard maximal torus of $G$ is $D = \textrm{diag}(t_1, ... , t_n, t_n^{-1}, ... , t_1^{-1})$. If you take for granted that $G$ is semisimple, you can at least find a set of generators for $N_G(D)$ modulo $D$, which isn't quite what you want, but close.

Let $X(D)$ be the free abelian group of rational characters of $D$, which has basis $e_1, ... , e_n$ (where $e_i[\textrm{diag}(t_1, ... , t_n, t_n^{-1}, ... , t_1^{-1}] = t_i$). Let

$$\Delta = \{e_1 - e_2, ... , e_{n-1} - e_n, 2e_n \}$$

Let $\alpha \in \Delta$. Modulo $D$, there is a unique element $w_{\alpha}$ in $N_G(D)$ which does not lie in $D$ and which commutes with the elements in $(\textrm{Ker } \alpha)^0)$, the connected component of the kernel of $\alpha$. In other words, $[N_{Z_G((\textrm{Ker } \alpha)^0)}(D) : D] = 2$. For all the $\alpha$ except $2e_n$, $\textrm{Ker } \alpha = (\textrm{Ker } \alpha)^0$.

For example, $w_{e_1-e_2}$ is going to be something like

$$\begin{pmatrix} 0 & -1 \\ 1 & 0 \\ & & 1 \\ & & & \ddots \\ & & & & 1 \\ & & & & & 0 & -1 \\ & & & & & 1 & 0 \end{pmatrix}$$

although you may need to tweak the signs $\pm 1$ in a few places to make it actually lie in $G$.

Anyway, the general theory of root systems tells you that $N_G(D)/D$ is generated by the images of these elements $w_{\alpha} : \alpha \in \Delta$.

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