# Why is this group action a morphism of varieties?

I am examining a proof that any affine connected algebraic group is a closed subgroup of some $GL_n$, and I am stuck on a fine point.

Let $G$ be an affine algebraic group in the sense that it is a group and also an affine variety over an algebraically closed field $k$, and such that the multiplication map $G\times G\rightarrow G$ and inverse map $G\rightarrow G$ are morphisms of varieties. Suppose that it is also connected.

$G$'s action on itself by (right or left) multiplication induces an action of $G$ on its coordinate ring $k[G]$. I have been convinced by the proof I am studying that there exists a finite-dimensional $k$-linear subspace $W\subset k[G]$ that is $G$-invariant and on which $G$ acts faithfully: essentially, if $f_1,\dots,f_m$ are $k$-algebra generators for $k[G]$, then all the $g(f_i)$'s live in a finite-dimensional subspace of $k[G]$ and this is $W$. This gives us an inclusion

$$\varphi:G\hookrightarrow GL(W)$$

so that $G$ is realized as a subgroup of $GL(W)\cong GL_n$ for $n=\dim W$. The argument I am studying now asserts that $\varphi$ is a morphism (with the variety structure on $GL(W)$ given by its identification with $GL_n\subset k^{n\times n}\cong \mathbb{A}_k^{n^2}$) and deduces that the image of $\varphi$ is a closed subvariety of $GL(W)$. I believe the argument except that I am having trouble explaining to myself why $\varphi$ has to be a morphism.

Why is $\varphi$ a morphism?

Here are my (very few) thoughts so far: selecting a basis $w_1,\dots,w_n$ for $W$, we can describe the $g$-action on $W$ in terms of its action on the basis:

$$g(w_i)=\sum_j c_{ij}(g)w_j$$

Then $g\mapsto (c_{ij}(g))$ is an explicit description of $\varphi$. $c_{ij}:G\rightarrow k$ is the pullback of the $ij$th coordinate function on $GL(W)$ under $\varphi$. I need to know that this is a regular function on $G$. It must be implied by the fact that the multiplication map $G\times G\rightarrow G$ is a morphism.

This would be sufficient. Technically I also need to know that the pullback of $1/\delta$ is regular, where $\delta$ is the determinant function on $GL(W)$ (this is because the coordinate ring of $GL(W)$ is generated by the coordinate functions and $1/\delta$). However, it follows from $c_{ij}$ being regular for each $i,j$, because the pullback of $1/\delta$ is $1/\det (c_{ij}(g))$. If the $c_{ij}$ are all regular, then $\det (c_{ij})$ is regular, and nonvanishing, so $1/\det (c_{ij})$ is regular.

That's as far as I've gotten. Thanks in advance.

Given any finite dimensional vector space $V$ with basis $\{e_1, \ldots, e_n\}$ and an algebraic action $\phi\colon G \times V \to V$ the induced map $\phi\colon G \to \operatorname{GL}_n(k)$ is a morphism of varieties.
As you said you need only show that the composition $c_{ij} = x_{ij}\phi$ of $\phi$ with the coordinate function $x_{ij}$ giving the $(i, j)^\text{th}$ entry of the matrix is regular. But this is true because $c_{ij}$ is a composition of regular functions: $$\begin{matrix}G & \xrightarrow{g \mapsto (g, e_j)} & G \times V & \xrightarrow{\quad\phi\quad} & V & \xrightarrow{\quad x_i\quad} & k\end{matrix}$$ where $x_i$ is dual to $e_i$.