What is a faithful representation?

Question

In the notes I am reading, it states: A representation fo an affine group scheme $G$ is a morphism $r: G \rightarrow GL_V$. It is faithful if it is injective.

In this notes they are defining schemes using their functors of points, so here a morphism is a natural transformation, which I understand as equivalent to assigning a map $G(R) \rightarrow GL_V(R)$ for all $k$ algebras $R$ so that given any morphism $R \rightarrow R'$ the diagram (which I don't know how to create on Latex) commutes.

What does it mean for this to be injective? Any explanation would be appreciated. Thank you.

Answer 1

In any category you have a notion of monomorphism, which is often taken as a generalization of the notion of injectivity (in $\mathbf{Set}$ and many usual categories such as $\mathbf{Grp}$, the monomorphisms are precisely the injective maps)

Here "injective" is probably meant as "a monomorphism $G\to GL_V$".

Now I don't know enough algebraic geometry, but there is probably an interpretation of "monomorphism" in terms of the maps $G(R)\to GL_V(R)$, for algebras $R$. For instance if this map is injective for all $R$, then $G\to GL_V$ is a monomorphism; I don't know about the converse, hopefully someone with more algebraic geometry knowledge can fill that gap.

If you look in the category of affine schemes, which is dual to the category of rings, then a map $\hom(A,-)\to \hom(B,-)$ is a monomorphism if and only if the associated $B\to A$ is an epimorphism. But if it is an epimorphism, given $R$, $\hom(A,R)\to \hom(B,R)$ is clearly injective; so in the category of affine schemes, monomorphisms are precisely the maps that are injective on each algebra; however I don't know what happens if you go to schemes.

Answer 2

Let $G$ and $H$ be finite type algebraic groups over a field $k$. Then, the following are equivalent for a group map $f:G\to H$:

1. The kernel $\ker f$ is trivial as a scheme (i.e. $(\ker f)(R)=\ker f(R)=1$ for all $R$ i.e. $G(R)\to H(R)$ is injective for all $R$ i.e. $\ker(f)=\mathrm{Spec}(k)$).
2. The map $f$ is a closed embedding.
3. The map $f(\overline{k}):G(\overline{k})\to H(\overline{k})$ is injective and the map on Lie algebras $(df)_e:\mathrm{Lie}(G)\to\mathrm{Lie}(H)$ is injective.

If you assume that $\mathrm{char}(k)=0$ you can even remove the condition on Lie algebras in 3. To see this, note that 1. evidently implies that $f(\overline{k}):G(\overline{k})\to H(\overline{k})$ is injective. Conversely, if $f(\overline{k})$ is injective then $(\ker f)(\overline{k})$ is trivial. In characteristic $0$ the scheme $\ker f$ is automatically reduced, and since it's reduced and finite type the claim that $(\ker f)(\overline{k})$ is trivial implies that $\ker f=\mathrm{Spec}(k)$.

In positive characteristic you cannot remove the Lie algebra condition 3. as the example of the trivial map $\mu_p\to\mathrm{Spec}(\mathbb{F}_p)$ shows. That said, even in positive characteristic you can remove the Lie algebra condition if you assume that $f$ is smooth since then the above argument I mentioned for characteristic $0$ still works since $f$ smooth implies that $\ker f$ is smooth.

Being 'injective' for a map of algebraic groups finite type over $k$ means that one of these three equivalent conditions holds. So a faithful representation $f:G\to\mathrm{GL}(V)$ means that one of the above three conditions holds. I usually think of the condition $\ker f=1$, but it's useful to know it's a closed embedding, and it's useful to know that you need only check that it's injective on $\overline{k}$-points in positive characteristic.