2
$\begingroup$

In (some version of) the proof of the fact that any affine algebraic group is a linear algebraic group, there is an important step as follows (for example in Borel's book "Linear Algebraic Groups", Prop 1.10):

Let $G$ be the affine algebraic group, $e_1,e_2,\ldots$ be a basis of its regular function ring (possibly infinite). Then for any regular function $f$, we can write $f(gh)=\sum_{i=1}^M v_i(g)e_i(h)$. (Borel wrote this as tensors but the idea is the same.) Thus the orbit of $f$ under the right action of $G$ lies inside the finite dimensional space $V$ spanned by the $v_i$'s, and the orbit gives a finite dimensional subspace $W$ which is stable under the action of $G$. (Then we proceed with this finite dimensional invariant $W$ into some $GL_N$)

My question is: Is there an example that $V\ne W$? Or can we prove $V=W$ under some conditions? I only check the affine line case and could not beat $V=W$.

Thank you!

$\endgroup$
1
  • $\begingroup$ Added a tag 'algebraic-groups'. $\endgroup$ Oct 16, 2014 at 19:23

2 Answers 2

1
$\begingroup$

Let me make sure I understand the question: $V$ is spanned by all functions of the form $g \mapsto f(gh)$, $W$ is spanned by the functions $g \mapsto v_i(g)$, it is clear from the given formula that $W \subseteq V$ and you are asking whether $V=W$. Also, I am assuming you are dealing with reduced algebraic groups over a algebraically closed field $k$, since otherwise what you mean by an orbit is a bit confusing. So I can think of elements of $\mathcal{O}(G)$ as functions on the $k$ points of $G$.

Yes. Since the $e_i$ are a basis for $\mathcal{O}_G$, in particular $e_1$, $e_2$, ..., $e_m$ are linearly independent. So we can find $h_1$, $h_2$, ..., $h_m$ in $G(k)$ so that the matrix $\left( e_i(h_j) \right)$ is invertible.

Inverting the equations $$f(g h_j) = \sum v_i(g) e_i(h_j)$$ we see that $g \mapsto v_i(g)$ is in the span of functions of the form $g \mapsto f(gh)$, as desired.

$\endgroup$
1
  • $\begingroup$ Thank you. Actually I wanted to define $V$ as the space spanned by $g\mapsto v_i$'s, and $W$ spanned by $g\mapsto f(gh)$, but I see what you mean. And it is a little surprising to me that most textbooks just claim $W\subset V$ instead of $W=V$. $\endgroup$
    – Ted Mao
    Oct 16, 2014 at 1:44
1
$\begingroup$

I don't think the question has a yes or no answer, since the formulation is too loose to make sense. In particular, the second paragraph isn't close enough to Borel's formulation. The ideas here are elementary but tricky to write down correctly in terms of indices. (For example, it's unclear what the index $M$ in the question depends on.)

This argument goes back to the 1956-58 Chevalley seminar, which started with a rough formulation of algebraic geometry over an algebraically closed field (later adapted to the scheme setting in SGA3, etc.). Essentially the problem is to show that an affine algebraic group is isomorphic to a closed subgroup of some matrix group over the field. (Borel and others also treat fields of definition.) All you are given here is a finitely generated algebra of functions on the given group $G$, from which you have to extract a highly non-unique finite dimensional subspace leading to a faithful representation of the group.

The original arguments can be seen online in the early expose 4 (section 4.1), which was presented by Grothendieck here. Note that Cartier edited a typeset and mildly corrected version of the text, published in 2005 by Springer.

In his Columbia lectures Linear Algebraic Groups, written up by Bass and published in 1969 by W.A. Benjamin, Borel treated this topic in Proposition 1.10 (as mentioned in the question). This follows a somewhat ad hoc updating of the foundations used by Chevalley. Here Borel formalizes the arguments a bit differently, making more explicit the Hopf algebra comultiplication in the algebra of functions on $G$. In the argument, both left and right translation of functions under $G$ come into play, which makes the notation somewhat heavy. (All this material got included, with chapters added, in the Springer GTM second edition published in 1991.)

My book, also called Linear Algebraic Groups, was published as a Springer GTM in 1975, and here I covered the linearization theorem in Theorem 8.6. My arguments are close to Borel's but the notation differs somewhat. Later Tonny Springer published his own book with Birkhauser in 1981, followed by a second edition in 1998. His Theorem 2.3.7 covers similar ground.

Though these treatments differ in style and sometimes in notation, the basic ideas are the same: find a finite generating set of functions for the function algebra which serve as matrix coefficient functions for a suitable faithful representation of $G$. In spite of the non-uniqueness of choice, such matrix coefficients can be shown to exist spanning a finite dimensional subspace of the functions which is suitably invariant under the group action. What complicates matters notationally is the need to work with two variables at times, which the comultiplication formalizes in a natural way. This use of tensor products is suppressed in the formulation of the question here.

$\endgroup$

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.