I felt rather confused when I was taking a course on linear algebraic groups, because I found that I could not explain explicitly some subtle differences between properties of algebraic groups and properties the groups of their rational points. (I wasn't sure whether the three famous text books (Borel, Humphreys, Springer) dealt with such questions or not (maybe implicitly?), because so far I didn't find expected solution in them.)

My questions are as follows:

(1) Let $G$ be a linear algebraic group (LAG) over an algebraically closed field $k$. Is there any characterisation (equivalent conditions) for $G$ being reductive or semisimple, by dealing with similar properties of the group of $k$-points $G(k)$?

For example, if for $G(k)$, as an abstract group, the only solvable unipotent normal subgroup (now we can't talk about closedness and connectedness) is $\{1\}$, can we deduce that $G$ is reductive over $k$?

(2) Let $G$ be a reductive group over an algebraically closed field $k$. Suppose $K/k$ is an extension and $K$ is also algebraically closed. Recall the Weyl group is constructed by first taking a maximal torus $T$ of $G$ (since $k$ is algebraically closed, $T$ is split), then compute $W(k):=(N_G(T)/Z_G(T))(k)$. (Am I right here? or is there anything imprecise?)

Then is it obvious that the $W(k)$ is isomorphic to $W(K)$? (Similarly, is it obvious that $G/k$ is reductive (resp. semisimple) iff $G/K$ (as a $K$-group) is reductive (resp. semisimple)?)

(3) Suppose $G$ is a reductive group defined over a subfield $F$ of $k$ ($k$ is algebraically closed), $G_{ss}:=[G,G]$ the derived semisimple subgroup and $Z(G)$ the center of $G$, then we know that $Z(G)(k)\times G_{ss}(k)\rightarrow G(k)$ (usual group multiplication) is surjective. But my professor reminded me that it's an important fact that on the level of $F$-rational points, $Z(G)(F)\times G_{ss}(F)\rightarrow G(F)$ needn't be surjective. Is there any explicit handy example for this phenomenon?

I tried to study the three famous texts, but so far I didn't find what I need in these texts (since I'm a beginner, I guess they might be discussed in a rather implicit way but I didn't realize that). Can any expert give me some instructions on these questions? Thanks a lot in advance!


1 Answer 1


You should probably in the future, ask such multi-part questions separately.

(1) I think this is not super well-posed, as I am not sure what 'unipotent' means. Do you mean that for all representations $G(k)\to \mathrm{GL}_n(k)$ that the element is sent to a unipotent element of $\mathrm{GL}_n(k)$? This notion is a priori stronger than being unipotent (which requires this only for algebraic such representations). Can you clarify what you mean here?

(2) Note that $\mathrm{Aut}(T)$, the automorphism sheaf of $T$ as a $k$-group scheme, is representable by $\underline{\mathrm{GL}_n(\mathbf{Z})}$. Therefore, as $N_G(T)/Z_G(T)$ embeds into $\mathrm{Aut}(T)$, we see that $N_G(T)/Z_G(T)$ must be an etale (and thus constant as we're over an algebraically closed field) group scheme. In particular, for any extension $K/k$ one has that $(N_G(T)/Z_G(T))(k)=(N_G(T)/Z_G(T))(K)$ and moreover, one can commute the operation of 'taking quotient' and 'taking $k$-points' here (and likewise for $K$, if $K$ is algebraically closed).

For your second question, I assume you mean that if $G$ is ana algebraic group over $k$, then $G$ is reductive if and only if $G_K$ is reductive. This is true as long as $K/k$ is separable. The reason is that $R_u(G)_K=R_u(G_K)$ (e.g. see [Milne, Proposition 19.9]) and evidently $R_u(G)$ is zero if and only if $R_u(G)_K$ is zero, as $\mathrm{Spec}(K)\to \mathrm{Spec}(k)$ is faithfully flat. The same argument applies for the semi-simple case with $R_u(G)$ replaced by $R(G)$.

(3) Let $\mu:=Z(G)\cap G^\mathrm{der}$, so then $Z(G)\times G^\mathrm{der}\to G$ is a surjection of algebraic groups. For any extension $K$ of $F$ one then has an exact sequence

$$1\to \mu(K)\to Z(G)(K)\times G^\mathrm{der}(K)\to G(K)\to H^1(K,\mu)$$

where the last term is the $K$-Galois cohomology with coefficients in $\mu$. This gives us a good hint of where to look: we need to find $\mu$ and $F$ with $H^1(F,\mu)\ne 0$.

Let's take $F=\mathbb{Q}$, $k=\mathbb{C}$, and $G=\mathrm{GL}_{n,F}$. Then, observe that $\mu=\mu_{n,F}$ and $H^1(F,\mu)=F^\times/(F^\times)^n$. As this is rarely zero, this seems like a good place to look. And, indeed, it's easy to see what the issue is. Here $Z(G)$ is the diagonal matrices in $\mathrm{GL}_{n,F}$ and $G^\mathrm{der}=\mathrm{SL}_{n,F}$. If $g\in \mathrm{GL}_n(F)$ is in the image of $Z(G)(F)\times \mathrm{SL}_n(F)$ then we can write $g=zh$ with $z\in Z(G)(F)$ and $h\in \mathrm{SL}_n(F)$. But, observe then that $\det(g)=\det(z)\det(h)=\det(z)$. But, if $z=\mathrm{diag}(r,\ldots,r)$ then $\det(z)=r^n$ and so $\det(g)=r^n$. So, $g$ will not be in the image as soon as $\det(g)$ is not an $n^\text{th}$-power in $F$. I am sure you can find such an element.

  • $\begingroup$ I'm very sorry that I forgot to reply to you... And I'm also very sorry to pose several questions under one title (I'll pay attention to this next time!). Your explanation is very helpful and heuristic! Thanks a lot! But I don't know what is the definition of the "automorphism sheaf". Could you give some further details? $\endgroup$
    – youknowwho
    Commented Nov 22, 2022 at 13:32
  • $\begingroup$ I have not yet learnt the language of group schemes. My professor told me that if I want to deal with problems related to automorphic forms, then reading Borel-Humphreys-Springer is much more important than using scheme-theoretic viewpoint, because the language of varieties is more explicit. (For example, "rational points" are just solutions with coordinates rational----nothing to do with the residue field as in the scheme language.) $\endgroup$
    – youknowwho
    Commented Nov 22, 2022 at 13:35
  • $\begingroup$ From your answer I'm doubt whether the invariance of the absolute Weyl group can be understood directly and intuitively from the variety language? Well, but I should admit some aspects (such as $R_u(G)_K=R_u(G_K)$) is transparent from the framework of schemes. $\endgroup$
    – youknowwho
    Commented Nov 22, 2022 at 13:38

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