Let $G$ be a connected split reductive group over a field $k$ of characteristic $p$. Let $\mathfrak{g}:=T_e(G)$ denote its Lie algebra. Let $T$ be a maximal split torus and $W$ the Weyl group (of the root system).

I appreciate references for the following statements.

  1. I believe if the size of $k$ is not too small, then $N_{G(k)}(T(k))/T(k) = W$, where $N$ denotes the normaliser. What is the precise condition on $k$ for this to hold?

  2. I believe it is almost always true that $T_e([G,G])=[\mathfrak{g}, \mathfrak{g}]$. Again what is the precise statement here? I.e. how big $p$ would have to be for this to hold?

  3. Suppose $k$ is finite and $G$ is semisimple and simply connected. Then, I believe that $G(k)$ is almost always perfect; i.e., $[G(k), G(k)]=G(k)$. Can one make this precise?

  • $\begingroup$ Assertion 1. is true with no condition on $k$, see Milne's book 'Algebraic groups' (CUP), Prop. 21.1. $\endgroup$ Jun 6, 2022 at 6:59
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    $\begingroup$ @Matthieu Romagny I don't think that's true, since if $k$ has two elements and $T$ is the diagonal torus in $G=\mathrm{GL}_2$, we have $T(k)=\{1\}$ so $N_{G(k)}(T(k))=G(k)$. Note that Milne is talking about $N_G(T)(k)$, which is not always isomorphic to $N_{G(k)}(T(k))$. $\endgroup$ Jun 6, 2022 at 8:07
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    $\begingroup$ 1 is not true as stated here in full generality, e.g. if |k|=2 and G is not a torus, or if |k|=3 and G is Sp_2n. Milne 21.1 is about the scheme-theoretic Weyl group and in this question we're talking about normalisers within the group of k-points. My guess is that 1 is true if |k|>3. $\endgroup$ Jun 6, 2022 at 8:08
  • $\begingroup$ Ahah thank you Alexander and Peter for clarifying! Indeed I read too fast. $\endgroup$ Jun 6, 2022 at 9:36
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    $\begingroup$ For Q3, if |k|>3 then G(k) is perfect under your hypotheses, for in this case G(k) is generated by its root SL_2(k)'s (by the Steinberg presentation), and each SL_2(k) is perfect. $\endgroup$ Jun 7, 2022 at 7:13

1 Answer 1


Let me address 1. Assume that for all roots $\alpha$, there exists $t\in T(k)$ with $\alpha(t)\neq 1$. I believe this implies that $(N_{G(k)}(T(k)))/T(k)=W$. This condition is true whenever $|k|>4$, as for each $\alpha$ there is a corresponding $SL_2\to G$ and the restriction of $\alpha$ to the image of the torus in $SL_2$ is of the form $\alpha(u)=u^2$.

Consider the adjoint representation of $G$. Since $G$ is reductive, the kernel of the adjoint representation is the centre of $G$. So we lose no relevant information by only considering the adjoint representation. Let $g\in N_{G(k)}(T(k))$. Let $\alpha$ be a positive root and let $t\in T(k)$ be such that $\alpha(t)\neq 1$. Let $s=g^{-1}tg$. Note that $s\in T(k)$.

Think of $g$ as a linear operator from $\mathfrak{g}$ to itself. Writing $\mathfrak{g}=\mathfrak{h}\oplus (\oplus_{\alpha} g_{\alpha})$, let $A$ be the component of the matrix representing $g$ that sends $g_{\alpha}$ to $\mathfrak{h}$. Let $v\in \mathfrak{g}_{\alpha}$ be nonzero. Then $Av=sAv=Atv=\alpha(t)Av$ which implies that $A=0$.

This argument and a similar one in the other direction shows that the summands $\mathfrak{h}$ and $(\oplus_{\alpha} g_{\alpha})$ are $g$-invariant. Therefore $g\in N_G(\mathfrak{h})$ (the algebraic group).

Here I want to claim that $N_G(\mathfrak{h})=N_G(T)$. But Lie algebras scare me in positive characteristic and I'm not seeing it right now, hopefully someone will come along in the comments and put me out of my misery. Certainly there is an inclusion $N_G(T)\subset N_G(\mathfrak{h})$ and I recall hearing somewhere that $N_G(T)$ was a maximal Zariski-closed subgroup of $G$ which would do the job, alas I don't know a reference or proof for this.

Assuming that last part is correct, it finishes the job. And if we want to go in the other direction, I think the minimal Levi corresponding to any $\alpha$ for which no $t$ exists with $\alpha(t)\neq 1$ will lie in the normaliser but I didn't dot my i's and cross my t's when thinking about it.

Hopefully we have now an answer whenever $k\neq \mathbb{F}_3$ and a criterion that can be checked on each group for $k=\mathbb{F}_3$.

  • $\begingroup$ $N_G(T) = N_G(\mathfrak h)$ fails when $p = 2$ and $G = \operatorname{SL}_2$, even though (for $\lvert k\rvert > 2$) your no-root-vanishing condition is satisfied. But actually it's easier: $C_G(T(k))^\circ$ (note rational points on $T$, but not on $G$) is a connected subgroup of $G$ containing $T$, so it is generated by $T$ and the root groups corresponding to roots vanishing on $T(k)$, of which by assumption there aren't any; so $C_G(T(k))^\circ$, which is clearly normalised by $N_G(T(k))$, equals $T$. $\endgroup$
    – LSpice
    Mar 10 at 15:17

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