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I would like to show that any Zariski-closed subsemigroup of $SL_n(\mathbb{C})$ is a group. If I understand correctly, this is consequence 1.2.A of .

Is there a more elementary proof? For $SL_2(\mathbb{C})$, the result is quite easy to show directly, or using the Hilbert basis theorem, .

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Note that the linked paper relies on the older Chevalley viewpoint about algebraic groups and algebraic geometry over arbitrary fields, which Hochschild also followed much later on. This doesn't work well in prime characteristic, so eventually the framework used by Chevalley and others shifted (and schemes came in). But for a field like $\mathbb{C}$ none of that really affects the question here. – Jim Humphreys Nov 22 '10 at 13:57
P.S. Tag added. – Jim Humphreys Nov 22 '10 at 17:52
What exactly is the Chevalley viewpoint? Considering algebraic groups as varieties rather than schemes? Or working with algebraic groups through their Hopf algebras? – darij grinberg Nov 22 '10 at 17:53
@darij: It's a question of using older language about varieties, with the ground field being arbitrary (but infinite). The function algebras certainly play a leading role, especially in Hochschild's Springer graduate text following Chevalley's 1955 book. Much of this works well enough in char 0, but otherwise gets out of control when you construct quotients, etc. Chevalley's version of algebraic geometry was a step toward what is now standard (and may have helped speed the transition). My education however started out with Weil's book ;-( – Jim Humphreys Nov 22 '10 at 18:43
up vote 23 down vote accepted

It is quite elementary. Let $S$ be the semi-group in question. Then for any $g \in S$, the set $g^kS$ for $k=1,2, \dots$ is a decreasing sequence of closed sets, hence it has to stabilize. So, $g^kS=g^{k+1}S$ implies that $gS=S$. Hence $S$ is closed with respect to taking inverse, and therefore is a group.

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Thanks! Do you know where I might find a reference for this argument? – Colin McQuillan Nov 22 '10 at 13:02
@Colin: Unfortunately, I don't know a reference for this. – Keivan Karai Nov 22 '10 at 13:19
I'm also curious about whether this is written down somewhere, since I have a vague recollection of seeing the argument before. Anyway, the result is in a way analogous to the fact that a subsemigroup of a finite group is a subgroup. – Jim Humphreys Nov 22 '10 at 13:51
P.S. My "vague recollection" is probably based on Exercise 7.5 in my 1975 book on algebraic groups: A closed subset of an algebraic group which contains the identity and is closed under taking products is a subgroup. Actually, the identity element comes along for free. (As the context of my Section 7 suggests, the exercise is inspired essentially by Chevalley's 1955 book but was perhaps stated explicitly later on.) – Jim Humphreys Nov 22 '10 at 15:51
Dear Pete: au contraire. The argument adapts to work scheme-theoretically, and then has an important application to scheme-theoretic normalizers (useful for reductive gps): for finite type gp $G$ over a field $k$ (any char.) and a smoth closed subscheme $V$ in $G$, the functor $N_G(V)$ assigning to any $k$-alg. $A$ the set of $g \in G(A)$ such that $g V_A g^{-1} \subseteq V_A$ is rep'td by a closed subsemigp (usually not smooth even if $G$ is, when char($k) > 0$). Want $N_G(V)$ to act on $V$ by automorphisms (i.e., stable by inversion)! See Def. A.1.9ff. in "Pseudo-reductive gps". – BCnrd Nov 22 '10 at 16:10

I think it's a matter of basic linear algebra.

Generally, let $k$ be a field, and $L$ be some finite-dimensional $k$-algebra. (In our case, $k=\mathbb C$ and $L=\mathrm{M}_n\left(k\right)$.) If $A\in L$ is invertible, then $A^{-1}$ lies in the Zariski closure of the set $\left\lbrace 1,A,A^2,A^3,...\right\rbrace$.

Proof. Let $k\left[L\right]$ denote the algebra of all polynomial functions from $L$ to $k$ (where a "polynomial function" means a function that can be written as a polynomial in the coordinates). (If $k$ is infinite, this is isomorphic to the non-naive algebra of coordinate functions, i. e. the symmetric algebra $\mathrm{S}\left(L^{\ast}\right)$, but we don't care about this isomorphy and therefore we don't need $k$ to be infinite.)

Let $P\in k\left[L\right]$ be a polynomial such that $P\left(A^i\right)=0$ for every $i\in\mathbb N$. We must then prove that $P\left(A^{-1}\right)=0$ as well.

Define a $k$-algebra $U$ by $U=\bigoplus\limits_{i=0}^N L^{\otimes i}$ as a vector space, but with the multiplication being inherited from $L$ on each summand. So, as a vector space $U$ is a "cropped" tensor algebra over $L$, but as an algebra it is a direct product!

Let $N=\deg P$. Then the polynomial $P:L\to k$ can be written as $P=p\circ s$, where $U=\bigoplus\limits_{i=0}^N L^{\otimes i}$, where $s:L\to U$ is the canonical map given by

$s\left(B\right)=1\oplus B\oplus \left(B\otimes B\right)\oplus \left(B\otimes B\otimes B\right)\oplus ...\oplus B^{\otimes N}$,

and $p:U\to k$ is some $k$-linear map. (In fact, this follows from the properties of the tensor algebra, because here we are NOT using the algebra structure on our $U$, but we are only using the vector space structure on $U$, and as I said, as a vector space $U$ is just the tensor algebra of $L$ "cropped" at $N$, which is enough for linearlizing polynomial maps of degree $\leq N$.)

Now consider the element $s\left(A\right)\in U$. This element $s\left(A\right)$ is invertible (since $A$ is invertible, so that $A^{\otimes i}$ is invertible for every $i$, and since the multiplication on $U=\bigoplus\limits_{i=0}^N L^{\otimes i}$ is componentwise), and the algebra $U$ is finite-dimensional (although its dimension is usually quite large). Thus, $s\left(A\right)^{-1}$ lies in the $k$-linear span of the set $\left\lbrace 1,s\left(A\right),\left(s\left(A\right)\right)^2,\left(s\left(A\right)\right)^3,...\right\rbrace$ (because if $u$ is an invertible element of some finite-dimensional $k$-algebra, then $u^{-1}$ lies in the $k$-linear span of the set $\left\lbrace 1,u,u^2,u^3,...\right\rbrace$; this is easily proven using the fact that any element of a finite-dimensional $k$-algebra is algebraic over $k$). Since $s$ is a multiplicative map, we have $\left(s\left(A\right)\right)^i=s\left(A^i\right)$ for all $i$, so that this becomes: The element $s\left(A^{-1}\right)$ lies in the $k$-linear span of the set $\left\lbrace s\left(1\right),s\left(A\right),s\left(A^2\right),s\left(A^3\right),...\right\rbrace$. Since $p$ is a linear map, we can apply $p$ here and obtain: The element $p\left(s\left(A^{-1}\right)\right)$ lies in the $k$-linear span of the set $\left\lbrace p\left(s\left(1\right)\right),p\left(s\left(A\right)\right),p\left(s\left(A^2\right)\right),p\left(s\left(A^3\right)\right),...\right\rbrace$. Now $p\circ s=P$, so this becomes: The element $P\left(A^{-1}\right)$ lies in the $k$-linear span of the set $\left\lbrace P\left(1\right),P\left(A\right),P\left(A^2\right),P\left(A^3\right),...\right\rbrace$. So when $P\left(A^i\right)=0$ for all $i\in\mathbb N$, then $P\left(A^{-1}\right)=0$, qed.

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Let $G \subset SL_n(\mathbb C)$ be a Zariski closed subsemigroup. The map $$\alpha(x,y) := (x,xy)$$ defines an injective self-map of $G \times G$ (see as algebraic varieties over $\mathbb C$). By the Ax-Grothendieck theorem, this map is bijective and hence an isomorphism. It is now a standard argument to construct the inverse map for $G$ out of the inverse of $\alpha$.

Is there a way of not using the Ax-Grothendieck theorem or anything like this?

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