Zariski-closed subsemigroups of SL_n(C) are groups

2010-11-22T11:33:30Z

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 http://www.heldermann-verlag.de/jlt/jlt03/BOSLAT.PDF .

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, .

Answer by Andreas Thom for Zariski-closed subsemigroups of SL_n(C) are groups

2010-11-22T12:30:22Z

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?

Answer by Keivan Karai for Zariski-closed subsemigroups of SL_n(C) are groups

2010-11-22T12:50:27Z

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.

Answer by darij grinberg for Zariski-closed subsemigroups of SL_n(C) are groups

2010-11-22T12:51:44Z

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.

Zariski-closed subsemigroups of SL_n(C) are groups - MathOverflow

Zariski-closed subsemigroups of SL_n(C) are groups

Answer by Andreas Thom for Zariski-closed subsemigroups of SL_n(C) are groups

Answer by Keivan Karai for Zariski-closed subsemigroups of SL_n(C) are groups

Answer by darij grinberg for Zariski-closed subsemigroups of SL_n(C) are groups