11
$\begingroup$

Let $G$ be an almost simple algebraic group defined over a field $K$. Then we know that, for $H = G/Z(G)$, the set of rational points $H(\overline{K})$ is a simple group (when considered with the group operation inherited from $G$). Must the set of rational points $H(K)$ also be a simple group?

$\endgroup$
11
  • 5
    $\begingroup$ No, e.g., for $G=\mathrm{PSL}_2=\mathrm{PGL}_2$ one gets the abelianization $K^*/{K^*}^2$. $\endgroup$
    – YCor
    Feb 27, 2023 at 16:28
  • $\begingroup$ Not sure I understand this. What equals $K^*/{K^*}^2$? $\endgroup$ Feb 27, 2023 at 16:36
  • 5
    $\begingroup$ The abelianization of $\mathrm{PGL}_2(K)$ is $K^*/{K^*}^2$. And, as an algebraic group, $\mathrm{PGL}_2=\mathrm{SL}_2/Z(\mathrm{SL}_2)$. $\endgroup$
    – YCor
    Feb 27, 2023 at 16:40
  • 3
    $\begingroup$ I think Steinberg shows that, for a simply connected group over a finite field, there are only a very few small cases where this fails. However, "somewhere in Steinberg" is vast literature to search. $\endgroup$
    – LSpice
    Feb 27, 2023 at 21:23
  • 1
    $\begingroup$ @LSpice yes, sorry I indeed meant "non-simply-connected (semisimple) groups" $\endgroup$
    – YCor
    Mar 1, 2023 at 23:06

4 Answers 4

13
$\begingroup$

If I've understood the question right, there are some low rank examples. Here's one. Let $G$ be $Sp_4$, the symplectic group of $4\times 4$ matrices, defined over ${\mathbb F}_2$. Then $Sp_4({\mathbb F}_{2^n})$ is simple for $n>1$, but $Sp_4({\mathbb F}_2)$ is isomorphic to the symmetric group $S_6$, which is not simple.

$\endgroup$
3
  • 4
    $\begingroup$ I suppose the point of these low rank examples is that even if you try to deYCor your question by stipulating that G is simply connected, and only quotient out the centre after taking K-rational points, then these low rank examples will persist. $\endgroup$ Feb 27, 2023 at 23:25
  • $\begingroup$ Just to have it said, these are not just low-rank examples, but small-field (not even small-characteristic) examples; for fields of order $4$ or more, this issue does not arise (although we do still usually have to consider, not the rational points of the full adjoint group, but only the image of the simply connected cover). $\endgroup$
    – LSpice
    Mar 1, 2023 at 15:24
  • 1
    $\begingroup$ Exactly. The entire list is finite. $\endgroup$ Mar 1, 2023 at 18:26
12
$\begingroup$

Steinberg - Variations on a theme of Chevalley, Section 5.6, shows that, if $K$ is a field, $G$ is a split, simply connected, simple group and $G_\text{ad}$ is its adjoint quotient, then the image of $G(K)$ in $G_\text{ad}(K)$ is a simple abstract group unless $G$ is of type $\mathsf A_1$ ($G = \operatorname{SL}_2$), $\mathsf B_2 = \mathsf C_2$ ($G = \operatorname{Spin}_5 = \operatorname{Sp}_4$—essentially @DaveBenson's example), or $\mathsf G_2$ over a field of characteristic $2$, or $\mathsf A_1$ over a field of characteristic $3$.

$\endgroup$
7
  • $\begingroup$ Wow, even including $G_2$ !?!?! :) $\endgroup$ Feb 27, 2023 at 21:36
  • $\begingroup$ Oops, I originally wrote "adjoint" (despite getting it right in my comment), but @‍YCor's comment reminded me that that hardly ever works even in type $\mathsf A_1$. Fortunately, my error did not affect the type $\mathsf G_2$ that astonishes @paulgarrett. 😄 $\endgroup$
    – LSpice
    Feb 27, 2023 at 22:34
  • 1
    $\begingroup$ Aha. "Ree groups". Keywords, forsooth! :) Thanks! :) $\endgroup$ Feb 27, 2023 at 22:55
  • 1
    $\begingroup$ $G_2({\mathbb F}_2)$ has $U_3({\mathbb F}_3)$ as a subgroup of index two. $\endgroup$ Feb 28, 2023 at 0:36
  • 1
    $\begingroup$ I think it's an example of the fact that there are only so many small simple groups, and some of them happen to coincide. Another example is that $U_4({\mathbb F}_2)$, $PSp_4({\mathbb F_3})$ and the derived group of $W(E_6)$ are all isomorphic. $\endgroup$ Feb 28, 2023 at 9:24
12
$\begingroup$

This is a very classical topic in algebraic group theory. If $k$ is a field and $G$ is an quasi-simple algebraic group defined over $k$ which is $k$-isotropic, then the group $G^+\subseteq G(k)$ generated by the unipotent elements in the $k$-parabolic subgroups is 'in most cases' simple modulo its center. (An algebraic group is quasi-simple if it has no proper connected normal subgroups defined over $k$, and $k$-isotropic if it has a $k$-split torus.)

The precise result, due to J. Tits, is as follows [J. Tits, Algebraic and abstract simple groups, Ann. Math. 80 (2), 1964]. If $G$ is a quasi-simple $k$-isotropic algebraic group defined over a field $k$ with at least 4 elements, then $G^+/Z(G^+)$ is simple. (Tits explains in the article also the exceptional cases over fields with 2 or 3 elements. Exceptions occur only in $k$-rank 1 or 2).

This reduces the question to the quotient $G(k)/G^+$. The computation of this quotient is known as the Kneser-Tits problem. If $G$ is simply connected, then the quotient is in many cases trivial. This quotient is also called the Whitehead group $W(G,k)=G(k)/G^+$. If $G$ is simply connected and split or quasi-split, then $W(G,k)=1$ [Tits, Groupes de Whitehead de groupes algebriques sur un corps, Sem. Bourbaki vol. 1976/77]. The Whitehead group is also known to be trivial for certain fields (such as $\mathbb R$), or for certain types of groups.

In the example $G=SL_n$ over any field $k$ with enough elements we have $G(k)=G^+$ (the typical unipotents being the conjugates of the upper triangular matrices with 1 on the diagonal) and indeed $SL_n(k)/Z(SL_n(k))$ is simple (and $SL_n$ is simply connected as an algebraic group). Thus $W(SL_n,k)=1$.

If $G$ is not simply connected, the answer is more complicated. The adjoint group of $SL_n$ is the quasi-simple group $G=PGL_n$, which is not simply connected. In this case $PGL_n(k)^+=PSL_n(k)=SL_n(k)/Z(SL_n(k))$. The quotient $PGL_n(k)/PSL_n(k)=G(k)/G^+$ is isomorphic to $k^*/(k^*)^n$.

[Note: the map $SL_n(k)\to PGL_n(k)$ is not surjective on the $k$-rational points. The terminology I use here differs from YCor's notation above, but I think it is in accordance with many books on algebraic groups, such as Borel or Milne.]

So for simply connected $k$-isotropic quasi-simple algebraic groups, Tits' Theorem reduces the question to the computation of the Whitehead group.

If the group $G$ is not $k$-isotropic, then $G(k)$ may have many nontrivial normal subgroups. Consider for example the special orthogonal group $G=SO_n$ for the standard bilinear form. For $n\geq 5$, this group is quasi-simple. For $k=\mathbb R$, the compact Lie group $G(k)$ is simple modulo its center. But if $k$ is a non-archimedean real closed field (eg. the field of nonstandard reals $^*\mathbb R$), then the matrices infinitesimally close to 1 generate a normal subgroup in $G(k)$.

$\endgroup$
1
  • 2
    $\begingroup$ In Tits' article, he considers also the exceptions over small fields. As observed in the other answers, they occur only for certain groups of low rank, of types $A_1$, $A_2$, $B_2$, and $G_2$. $\endgroup$
    – Linus
    Mar 1, 2023 at 7:29
4
$\begingroup$

One should mention the spinor norm as a source of obstructions. Let $k$ be a field of characteristic $\neq 2$ and let $V$ be a vector space equipped with a non-degenerate symmetric bilinear form $\langle \ , \ \rangle$. Let $O_V$ be the orthogonal group of $V$, let $SO_V$ be the determinant $1$ subgroup of $O_V$, let $PO_V$ be the quotient $O_V/\pm \text{Id}$ and let $PSO_V$ be $SO_V$ if $\dim V \equiv 1 \bmod 2$ and $SO_V/\pm \text{Id}$ if $\dim V \equiv 0 \bmod 2$.

The group $SO_V$ is quasi-simple for $\dim V \neq 4$, and $PSO_V$ is the adjoint quotient.

For $v \in V$ with $\langle v,v \rangle \neq 0$, the reflection over $v$ is the linear map in $O_V$ given by $$r_v(w) = w - 2 \frac{\langle v,w \rangle}{\langle v,v \rangle} v.$$ The spinor norm is a character $O_V(k) \to k^{\ast}/(k^{\ast})^2$ determined by the property that $r_v \mapsto \langle v, v \rangle$. We can restrict the spinor norm to $SO_V(k)$.

If $\dim V \equiv 1 \bmod 2$, then $PSO_V(k) = SO_V(k)$, so the restriction of the spinor norm gives a character of $PSO_V(k)$.

If $\dim V \equiv 0 \bmod 2$, then let $D$ be the discriminant of the quadratic form $\langle \ , \rangle$. I compute that the spinor norm of $- \text{Id}$ is $D$, so the spinor norm passes to a map $PSO_V(k) \to k^{\ast}/\langle D, (k^{\ast})^2 \rangle$.

This will very often give nontrivial quotients of $SO_V(k)$ and $PSO_V(k)$. I believe, however, that it doesn't give any interesting quotients of $\text{Spin}_V(k)$: I think that the spinor norm is trivial on the image of $\text{Spin}_V(k)$.

$\endgroup$
1
  • $\begingroup$ Indeed, your last statement fits into Steinberg's / Chevalley's result mentioned in my answer that the image in the adjoint group of the simply connected group is almost always simple. $\endgroup$
    – LSpice
    Feb 28, 2023 at 19:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.