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!



A question from the proof of affine algebraic group is a linear
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.


