# Are linear algebraic groups always finite-dimensional?

Recall that a complex linear algebraic group $G$ is an affine algebraic group (i.e. an affine group variety over $\mathbb C$). It's a well-known fact that there is always a closed embedding $$G \hookrightarrow \operatorname{GL}(n, \mathbb C).$$ My concern is that this seems to imply that $G$ is finite-dimensional as a variety, just because $\operatorname{GL}(n, \mathbb C)$ is for any particular integer $n$ (it has codimension 1 in $\mathbb A^{n^2}$, hence dimension $n^2-1$). This doesn't mesh well with my intuition, because I vaguely know that if $R$ is a non-Noetherian $\mathbb C$-algebra, then $\operatorname{Spec} R$ should be infinite-dimensional as a variety. Unfortunately I don't know any of these examples well-enough to try to construct a group operation on one of them to figure out what's happening.

I'd appreciate it if someone could explain why it should be true that any $G$ should be finite-dimensional; for example if having the group law places some restriction on the size of $G$ that I'm not seeing. It's also possible that my claim is wrong or doesn't make sense in some way; if so I'd also be happy if someone could point out the mistake. Thanks.

EDIT: To clarify on Qiaochu's comment: a linear algebraic group $G$ for me is an $k$-variety $G$ equipped with the structure of a group such that both the group multiplication $G \times G \to G$ and the inversion $G \to G$ are morphisms of varieties. But I think the issue was actually that I don't know the definition of variety: these need to have finite type. Thanks all.

• There's no such thing as an infinite dimensional variety. Sure, it's a scheme, but schemes and varieties are not the same thing. I'm curious about how you define linear algebraic groups though if the definition doesn't yield something tautologically finite dimensional. Mar 15, 2016 at 23:21
• @MattSamuel You could define something which is affine, and a group scheme, but not actually finite type. Something like $\mathbf{G}_{m,\mathbb{Q}(t)}/\mathbb{Q}$. As a more reasonable example, affine group schemes which are not finite type come up quite a bit in the study of Tannakian categories. There the 'fundamental group' of the category is an affine group scheme which is a pro-variety. In other words, it's an inverse limit of algebraic groups. Mar 16, 2016 at 0:28

For the sake of being explicit, here's a simple example of an infinite-dimensional affine group scheme, even a reduced one. You might call it infinite-dimensional affine space $\mathbb{A}^{\infty}$. Its functor of points takes a commutative $k$-algebra $R$ ($k$ the underlying field) and assigns

$$R \mapsto R^{\infty} = \prod_{i=1}^{\infty} R$$

with group operation given by pointwise addition. As a Hopf algebra,

$$\mathbb{A}^{\infty} = \text{Spec } k[x_1, x_2, \dots ]$$

with comultiplication given by extending

$$\Delta x_i = x_i \otimes 1 + 1 \otimes x_i$$

(which should look familiar if you've ever written down the Hopf algebra of functions on affine space $\mathbb{A}^n$). You can think of $\mathbb{A}^{\infty}$ as a cofiltered limit of the finite-dimensional affine spaces, and this picture can be generalized to arbitrary affine group schemes.

Matt Samuel's comment did it for me: I'd misunderstood the definition of "affine variety". Originally I thought this was just a reduced scheme over $\mathbb C$ except that we only consider the closed points (i.e. look at $\operatorname{MSpec} R$ for some reduced $\mathbb C$-algebra $R$). Now I see that the various true definitions all require some finiteness condition on the variety (e.g. from a scheme point of view, we want it to have finite type over $\mathbb C$). Thanks all.