Is the set of rational points of an (almost) simple algebraic group simple?

Question

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?

Answer 1

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.

Answer 2

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 $3$.

Answer 3

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 $G(k)$.

Answer 4

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 $\text{Spin}_V(k)$.