7
$\begingroup$

I have seen a theorem attributed to Chevalley, to the effect that a sub-semi-group in an algebraic group is Zariski dense if and only if the subgroup it generates is Zariski dense. Would anyone happen to have a reference?

$\endgroup$
1
  • 6
    $\begingroup$ Presumably the algebraic group $G$ is over an algebraically closed field $k$ and irreducible (to avoid silliness), so it is finite type (not just locally so). The Zariski-closure of a sub-semigroup is a sub-semigroup, so you want that if $M$ is a Zariski-closed subsemigroup and $M(k)$ generates $G(k)$ then $M=G$. It suffices to show $\dim M = \dim G$. By irreducibility of $G$, it suffices to show every element of $G(k)$ is a "word" in a bounded number of letters from $M(k)$. Now see Prop. 2.2 in Ch. I of Borel's textbook on algebraic groups. Chevalley's constructible image theorem is the crux. $\endgroup$
    – nfdc23
    Commented Aug 7, 2016 at 1:44

3 Answers 3

16
$\begingroup$

Here is one proof. It suffices to show a Zariski closed submonoid M of an algebraic group G is an algebraic group. Let $x$ be in $M$. The left translation by $x$ is an injective polynomial morphism from M to M and hence surjective by Ax-Grothendieck. So $xy=1$ for some $y$ in $M$. Since $G$ is a group $y$ is the inverse of $x$ in $G$.

You should assume here the field is algebraically closed of course.

Edited much later. For clarity, let me put the argument I gave in the comments also in the answer as it is easier. If $M$ is a closed submonoid of an algebraic group $G$, then $M$ is a subgroup. Let $x\in M$. Then $\lambda_x\colon G\to G$ given by $\lambda_x(g) = xg$ is an isomorphism as a map of algebraic varieties. Thus $\lambda_x(M)=xM$ is a closed subset of $G$. Iterating this, we obtain a chain $M\supseteq xM\supseteq x^2M\supseteq \cdots $ which must stabilize since $G$ is a Noetherian space. Thus $x^kM=x^{k+1}M$ for some $m$. Since $G$ is a group, this implies $M=xM$ and so there exists $y\in M$ with $xy=1$. It follows that $M$ is closed under inversion, hence a subgroup.

$\endgroup$
8
  • 1
    $\begingroup$ A very cool argument (though probably not the one Chevalley had in mind :)) $\endgroup$
    – Igor Rivin
    Commented Aug 7, 2016 at 3:01
  • 1
    $\begingroup$ I think I learned it from either Putcha or Renner's book on algebraic monoids or from one of Brion's papers. I think there is also in one of those books a direct proof that the inverse of a matrix is in the Zariski closure of its powers. $\endgroup$ Commented Aug 7, 2016 at 3:08
  • 1
    $\begingroup$ One interesting point is that this seems morally close to the "strong approximation" proof that I had in mind for cases of interest to me (Zariski dense-> modular projections are onto, but that last condition is obviously the same for groups as for semi-groups. Which is somehow very much Ax-Grothendeickian. $\endgroup$
    – Igor Rivin
    Commented Aug 7, 2016 at 3:23
  • 2
    $\begingroup$ You can even avoid Ax-Grothendieck by observing that the translation (being an automorphism of $G$ as a variety) induces an isomorphism of $M$ with a closed subset of $G$ (hence of $M$). In other words, it is a closed immersion of $M$ into itself! The rest is an exercise (especially since $M$ is reduced) via dimension arguments. $\endgroup$ Commented Aug 8, 2016 at 15:05
  • 2
    $\begingroup$ @Laurent Moret-Bailly, I think you can just use Noetherian condition on the descending chain of closed subsets xM > x^2M>... to get stabilization and then use cancellation. $\endgroup$ Commented Aug 8, 2016 at 20:25
6
$\begingroup$

The whole point is that for every $g\in\mathrm{GL}_n$, $g^{-1}$ belongs to the Zariski closure of $\{g^k:k\ge 1\}$. This is an elementary consequence of noetherianity.

Embedding into $\mathrm{SL}_{n+1}$ if necessary, we can suppose $g\in\mathrm{SL}_n$. Let $I_m$ be the set of polynomials (in $n^2$ variables) that vanish on $g^k$ for all $k\ge m$. Then $I_m\subset I_{m+1}$ for all $m$. Define $T.P(x)=P(gx)$: so $T$ is an automorphism of the ring of polynomials. Then $I_{m+1}=\{P:\forall k\ge m:P(g^{k+1})=0\}=\{P:T.P\in I_m\}=T^{-1}I_m$. Also by noetherianity, for $m$ large enough we have $I_m=I_{m+1}$. So for any $m$, choosing $m'$ large enough, we get $I_m=T^{m'}I_{m+m'}=T^{m'}I_{m+m'+1}=I_{m+1}$. In particular $I_{-1}=I_0=I_1$, which implies that every polynomial vanishing on $g^k$ for all $k\ge 1$ vanishes on $g^{-1}$.

[It looks like I didn't use being in $\mathrm{SL}_n$, but I did: I used that a regular function on $\mathrm{SL}_n$ is the restriction of a polynomial on the space of matrices. This is not true with $\mathrm{GL}_n$.]


Edit: here's a slightly more abstract reformulation of the above argument.

Lemma. Let $(X,\le)$ be a noetherian poset and $f$ an automorphism of $X$, and $x\in X$ such that $f(x)\ge x$. Then $f(x)=x$.

Proof: the bi-infinite sequence $(f^n(x))_{n\in\mathbf{Z}}$ is increasing (allowing equality). It easily follows that it is either strictly increasing, or constant. For a noetherian poset, the first possibility is excluded. $\Box$

We apply this to $X$ the poset of ideals of $k[G]$ and, for $g\in G$, $x=x_0$ is the ideal of $u\in k[G]$ vanishing on $\{g^n:n\ge 0\}$, and $F(P)=P(g^{-1}x)$, and $f$ is the automorphism of the poset of ideals of $k[G]$ induced by $F$. (So $f^n(x)$ is the set of ideals vanishing on $\{g^j:j\ge n\}$.)

$\endgroup$
1
$\begingroup$

I also had this question: that is, I needed a reference for this fact. Of course, doing an internet search now leads to this exact MathOverflow post! So I looked further and found an almost reasonable reference (copied from MathSciNet):

Solomon, Louis(1-WI) - An introduction to reductive monoids. Semigroups, formal languages and groups (York, 1993), 295–352. NATO Adv. Sci. Inst. Ser. C: Math. Phys. Sci., 466 Kluwer Academic Publishers Group, Dordrecht, 1995 ISBN:0-7923-3540-6

On page 297, line -17, Solomon says that the Zariski closure of a monoid is again a monoid. However, it seems to me that the reference he gives (in Footnote 5) is broken (namely, reference [25] has no Lemma 2.1). On page 298, line 2, he says that the Zariski closure, intersected with $\mathrm{GL}(n)$, is a group. This time the references (in Footnote 6) look good to me.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .