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Here is a theorem from representation theory:

$G$ is a reductive algebraic group in characteristic not equal $p$ and $E$ is an elementary abelian subgroup of $G$.

Suppose $G$ is complex and view the Lie algebra of G as a $\mathbb{C}E$-module with character $\chi_L$. Then dim $C_G(E)=(1/|E|)\sum_{x\in E}\chi_L(x)$.

I am being silly here.

I considered $E$ generated by \begin{pmatrix} -1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix} in $PGL_4(\mathbb{C})$. The centraliser apprently has dimension 9. But the Lie algebra of $PGL_n(\mathbb{C})$ is $\mathfrak{sl}(n,\mathbb{C})$ of traceless matrices. So by the formula, we would get dimension $0$...

How do I get the dimension right using the formula?

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    $\begingroup$ Why would it be $0$? The underlying vector space of the representation might consist of traceless matrices, but that does not mean any operator on that space has trace $0$ (as operator). I mean, for $x=Id$ do we not get $\chi_L(x)=15$ (the dimension of the Lie algebra)? So I suspect for the non-trivial $x$ in $E$ (which is represented by the matrix you write down), we have $\chi_L(x)=3$, which looks plausible ... $\endgroup$
    – Torsten Schoeneberg
    Commented May 31, 2022 at 19:30
  • $\begingroup$ Yes, this makes sense to me partly. How do I get the representation $\chi$ exactly? Still stuck in this silly thinking, I guess. $\endgroup$
    – scsnm
    Commented May 31, 2022 at 21:04
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    $\begingroup$ I would assume it's the adjoint? I.e. a matrix $A$ representing an element of $PGL_4$ acts on $\mathfrak{sl}_4$ via $\rho(A): X \mapsto A^{-1} X A$. Then $\chi(A)$ is the trace of $\rho(A)$. $\endgroup$
    – Torsten Schoeneberg
    Commented May 31, 2022 at 21:07
  • $\begingroup$ Do you have it? I think I do. I could add this as answer later if you want, or you can answer yourself (to not let it go unanswered). $\endgroup$
    – Torsten Schoeneberg
    Commented Jun 1, 2022 at 15:02
  • $\begingroup$ Thank you so much for following up! If you could add it as an answer, I'd appreciate it! $\endgroup$
    – scsnm
    Commented Jun 1, 2022 at 22:15

1 Answer 1

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So we have $G=PGL_4(\mathbb C)$, accordingly the Lie algebra is $V= \mathfrak{sl}_4(\mathbb C)$, the traceless $4\times 4$-matrices. Note that the representation of $G$ on $V$ is the adjoint, i.e. an element $x\in G$ (and by restriction, $x\in E$), represented by a matrix $A \in GL_4(\mathbb C)$, acts on $X \in V$ via

$$\rho(x): X \mapsto A^{-1}XA.$$

Now $\chi_L(x)$ is defined as the trace of this $\rho(x) \in End_{\mathbb C}(V)$. So the fact that the underlying vector space of the representation consists of traceless matrices is kind of irrelevant, we are looking at traces of operators on that ($15$-dimensional) space $V$.

Now since your group $E$ is of order $2$, consisting of the identity $id$ and the non-trivial element $b$ represented by your matrix $B=\pmatrix{-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1}$, everyting is quickly computed: Of course $\chi_L(id)=\dim(V)=15$, whereas $\rho(b)$ is seen to operate with eigenvalue $-1$ on the six-dimensional space $$\pmatrix{0&*&*&*\\*&0&0&0\\*&0&0&0\\*&0&0&0}$$ but with eigenvalue $1$ on the nine-dimensional $$\pmatrix{*&0&0&0\\0&*&*&*\\0&*&*&*\\0&*&*&*} \subset \mathfrak{sl}_4(\mathbb C)$$.

So $\chi_L(b) = 9\cdot 1+6\cdot(-1)=3$, and the formula gives out the answer you want.

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  • $\begingroup$ Sorry that I got a bit confused again. When seeing the operation of $\rho (b)$, a basis of 15 elements are chosen. These elements are the 16 $E_{i,j}$ excluding $E_{1,1}$. Is it correct? Thx $\endgroup$
    – scsnm
    Commented Jun 18, 2022 at 7:44
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    $\begingroup$ No, for $i=j$ the $E_{i,i}$ are not even contained in $\mathfrak{sl}_4$ (they each have trace $1$!). A pretty common basis for $\mathfrak{sl}_n$ is made up of the off-diagonal elementary $E_{i,j}$, $i \neq j$ on the one hand, and for the diagonal on the other hand, one often takes $E_{11}-E_{22}, E_{22}-E_{33}, ..., E_{n-1,n-1}-E_{nn}$. So e.g. that $1$-eigenspace has dimension $9$ coming from $6$ off-diagonal and $3$ diagonal basis matrices. $\endgroup$
    – Torsten Schoeneberg
    Commented Jun 18, 2022 at 16:58

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