Intersection of identity components

Question

Let $e_{1}$ and $e_{2}$ be involutions in the algebraic group $G=\operatorname{PGL}_{n}(\mathbb{C})$. Do we have $$C_{G}(\langle e_{1},e_{2}\rangle)^{\circ} = C_{G}(e_{1})^{\circ}\cap C_{G}(e_{2})^{\circ}.$$ It's obvious $C_{G}(\langle e_{1},e_{2}\rangle)^{\circ} \leqslant C_{G}(e_{1})^{\circ}\cap C_{G}(e_{2})^{\circ}.$ If yes, can it be generalised to two elementary abelian $2$-subgroups $E_{1}, E_{2}$.

Here $H^{\circ}$ denotes the identity component of the group $H$.

I can only see this through some examples. For instance, in $G=\operatorname{PGL}_{4}(\mathbb{C})$, let $$ x=\left[ \begin{array}{cc} -1 & 0 & 0 & 0\\ 0 & 1& 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & 1 \end{array} \right],f= \left[ \begin{array}{cc} 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{array} \right].$$ Then $$C_{G}(x) = \left[ \begin{array}{cc} a_{1} & 0 & a_{3} & 0\\ 0 & b_{2} & 0 & b_{4}\\ c_{1} & 0 & c_{3} & 0\\ 0 & d_{2} & 0 & d_{4} \end{array} \right] \sqcup f\left[ \begin{array}{cc} a_{1} & 0 & a_{3} & 0\\ 0 & b_{2} & 0 & b_{4}\\ c_{1} & 0 & c_{3} & 0\\ 0 & d_{2} & 0 & d_{4} \end{array} \right].$$

Let $$ y=\left[ \begin{array}{cc} 1 & 0 & 0 & 0\\ 0 & -1& 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array} \right]. \mbox{So } C_{G}(y) = \left[ \begin{array}{cc} u_{1} & 0 & u_{3} & u_{4}\\ 0 & v_{2} & 0 & 0\\ s_{1} & 0 & s_{3} & s_{4}\\ t_{1} & 0 & t_{3} & t_{4} \end{array} \right].$$ So $C_{G}(\langle x,y\rangle)^{\circ} = \left[ \begin{array}{cc} * & 0 & * & 0\\ 0 & * & 0 & 0\\ * & 0 & * & 0\\ 0 & 0 & 0 & * \end{array} \right]\cong A_{1}T_{2}$ which is, indeed, the intersection of the identity components of the two centralisers.

Answer 1

$\DeclareMathOperator\Cent{C}\newcommand\oG{\overline G}\newcommand\oe{\overline e}$Put $\oG = \operatorname{PGL}_n(\mathbb C)$. I hope you will permit me to denote the involutions by $\oe_i$ instead of $e_i$.

At first, it matters only that we are dealing with semisimple elements (which is implied by being an involution in characteristic $\ne 2$). It doesn't even matter at first that we are dealing with $\oG = \operatorname{PGL}_n(\mathbb C)$; I will put $G = \operatorname{GL}_n(\mathbb C)$, but there is a general theory of $z$-extensions, of which every connected, reductive group $\oG$ admits one, which are to $\oG$ as $\operatorname{GL}_n(\mathbb C)$ is to $\operatorname{PGL}_n(\mathbb C)$.

If $t$ is any semisimple element of $G$, then $\Cent_G(t)$ is connected, $\Cent_G(t) = \Cent_G(t)^\circ \to \Cent_{\oG}(\overline t)^\circ$ is a surjection, and $\Cent_G(t)$ is the full pre-image in $G$ of $\Cent_{\oG}(\overline t)^\circ$, where $\overline t$ is the image of $t$ in $\oG$. In particular, if $e_i$ is a lift of $\oe_i$, then $\Cent_{\oG}(\oe_1)^\circ \cap \Cent_{\oG}(\oe_2)^\circ$ is the image in $\oG$ of $\Cent_G(e_1) \cap \Cent_G(e_2) = \Cent_G(e_1, e_2)$. Thus, the question is whether $\Cent_G(e_1, e_2)$ is connected. If so, then $\Cent_{\oG}(\oe_1)^\circ \cap \Cent_{\oG}(\oe_2)^\circ$ is the image $\Cent_{\oG}(\oe_1, \oe_2)^\circ$ in $\oG$ of $\Cent_G(e_1, e_2) = \Cent_G(e_1, e_2)^\circ$.

In the full generality that I have considered so far (where $G$ has connected centre and simply connected derived group, and the $e_i$ can be any semisimple elements), this need not be true. Even in your setting, where $\oG = \operatorname{PGL}_n(\mathbb C)$ and the $\oe_i$ are involutions, I do not know whether $C_G(e_1, e_2)$ is always connected. However, you have indicated that you are interested in the case where $\oe_1$ and $\oe_2$ belong to a common maximal torus $\overline T$ (which is the same as $\oe_2$ belonging to $\Cent_{\oG}(\oe_1)^\circ$). I now specialise to that case, without requiring that the $\oe_i$ are involutions. Let $T$ be the pre-image in $G$ of $\overline T$. Then $T$ is a maximal torus in $G$.

In general (not just for $G = \operatorname{GL}_n$), the map from $\operatorname{stab}_W(e_1, e_2)/\langle s_\alpha : \alpha(e_1) = \alpha(e_2) = 1\rangle$ to $\Cent_G(e_1, e_2)/\Cent_G(e_1, e_2)^\circ$ is an isomorphism, where $W$ is the Weyl group of $T$ in $G$. Now, finally specialising to $G = \operatorname{GL}_n$ and so identifying $W$ with $\operatorname S_n$, and conjugating $T$ if necessary so that it is the diagonal torus of $G$, an explicit computation shows that $\operatorname{stab}_W(e_1, e_2)$ is the product of permutation groups that respect the decomposition of $\{1, \dotsc, n\}$ into maximal subsets $I$ such that all diagonal entries of $e_1$ with entries in $I$ are equal, and all diagonal entries of $e_2$ with entries in $I$ are equal; and that this is just $\langle s_\alpha : \alpha(e_1) = \alpha(e_2) = 1\rangle$. Thus, $\Cent_G(e_1, e_2)$ is connected, so your desired conclusion follows.