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.



Intersection of identity components
$\DeclareMathOperator\Cent{C}\newcommand\oG{\overline G}\newcommand\oe{\overline e}$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](https://mathoverflow.net/questions/433237/intersection-of-identity-components#comment1115408_433237) 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.


