Distinguish smooth affine algebraic group by its underlying variety

If $$G_1, G_2$$ are two connected affine smooth algebraic groups over an algebraically closed field $$k$$, and $$G_1 \cong G_2$$ as algebraic varieties, must they be isomorphic as algebraic groups?

Although connected unipotent groups of same dimension all have same underlying varieties (i.e the affine space), I suspect that with suitable reductive assumptions (e.g one of them is reductive + other conditions), $$G_1$$ and $$G_2$$ must be isomorphic as algebraic groups.

There is a classification of connected reductive groups, but do the topological invariants (e.g fundamental group + dimension + cohomology) determine the group?

What if $$k$$ is only assumed to be a perfect field?

Example: the dimension $$1$$ case and semisimple rank $$1$$ case are easy by classification. For instance, $$SL_2$$ and $$PGL_2$$ are different because they have different fundamental groups.

Edit: another way of saying the question is: how to characterize those $$G$$ on which there exists only one / finitely many smooth (reductive) algebraic group structures? If one is reductive, must the other be?

• Speculation: the connected component of the automorphism group (imagine it exists...) will give you back the group. – Piotr Achinger Dec 23 '19 at 20:07
• @PiotrAchinger, for reductive groups, this recovers the adjoint quotient; for example, it can't tell $\operatorname{SL}_2$ from $\operatorname{PGL}_2$. (But it doesn't seem to address the question of recovering the algebraic structure from the purely geometric structure, right?) – LSpice Dec 23 '19 at 20:33
• A silly example: you can’t distinguish direct and semi-direct products, e.g., $\operatorname{PGL}_2 \ltimes \operatorname{PGL}_2$, with the trivial action and the conjugation action (or are those isomorphic?). – LSpice Dec 23 '19 at 22:41
• @LSpice for any group the semidirect product with conjugation is isomorphic to the product. Proof: exercise – Ben Wieland Dec 25 '19 at 17:35

$$G_1:=GL(2)(\cong SL(2)\rtimes \mathbb G_m)$$ is isomorphic as a variety to $$G_2:=SL(2) \times \mathbb G_m$$ via the map $$A\mapsto \Big(\big(\begin{smallmatrix}\det(A)^{-1} & 0 \\ 0 & 1 \end{smallmatrix}\big)A , \det(A)\Big).$$ We will show that $$G_1$$ and $$G_2$$ (which are both reductive) are not isomorphic as algebraic groups over fields of characteristic 0. This is not immediate, as for $$k$$ a perfect field of characteristic 2 there is an isomorphism of abstract groups $$\psi: GL(2)(k) \to (SL(2) \times \mathbb G_m)(k)$$ given by $$\psi(A) = (\det(A)^{-\frac{1}2}A, \sqrt{\det(A)})$$.

So suppose $$\varphi: G_1 \to G_2$$ is an isomorphism.

First proof: Note that $$Z(G_1)=\mathbb G_m$$ and $$Z(G_2)=C_2 \times \mathbb G_m$$. Then $$\varphi$$ induces an isomorphism $$Z(G_1) \cong Z(G_2)$$, which is absurd since over a field with characteristic different from 2 the equation $$x^2=1$$ has 2 solutions in $$\mathbb G_m$$ and 4 solutions in $$C_2 \times \mathbb G_m$$.

Second proof: As $$\mathbb G_m$$ is diagonalizable and $$\varphi^{-1}$$ preserves this property, so is $$\varphi^{-1}(\mathbb G_m)$$ and hence is contained in a maximal torus, which are all conjugate. So $$\psi:= \varphi^{-1}|_{\mathbb G_m}$$ gives rise to a so-called cocharacter $$\psi \in Hom(\mathbb G_m, T)$$ for $$T$$ the maximal torus of diagonal matrices of $$GL_2$$. By standard facts about cocharacters, $$\psi$$ has the form $$z\to (z^a, z^b)$$ for some integers $$a, b$$. Since the domain of $$\psi$$ is central in $$G_2$$, so must be $$\text{im } \psi$$, which forces $$a=b$$. Since $$\psi$$ is injective on $$\bar{k}$$-rational points, we have $$a=\pm 1$$. Then $$\psi(-1)=(-1, -1)$$.

On the other hand, $$\varphi^{-1}$$ affords a 2-dimensional algebraic representation of $$SL_2$$, so $$\varphi^{-1}(SL_2)=SL_2$$. This finally gives a contradiction as the images of $$\varphi^{-1} (SL_2)$$ and $$\varphi^{-1}(\mathbb G_m)$$ are not disjoint.

• One can still wonder about semisimple groups …. – LSpice Dec 23 '19 at 22:36
• $\psi(A) = (\det(A)^{-1/2}A, \sqrt{\det(A)})$ – reuns Dec 26 '19 at 11:32

Let's consider the last version of the question over $$\mathbb C$$. We will prove two facts: I, II.

I. If two complex algebraic groups $$G_1$$ and $$G_2$$ are diffeomorphic, then they are either both reductive or both non reductive.

II. For any reductive complex algebraic group there exist at most finite number of diffeomorphic to it complex algebraic reductive groups.

Obvious remark. Of course, two complex Lie groups that the same as algebraic varieties are diffeomorphic.

Proof of I. This follows from the following claim.

Claim. A complex algebraic $$n$$-dimensional group is reductive if and only if it is homotopy equivalent to an (orientable) real compact manifold of dimension $$n$$.

Clearly, this shows that a non-reductive group can not be even diffeomorphic to a reductive one.

Let's see how to prove this claim. We will use two alternative characterisations of reductive groups.

First, a complex algebraic group is reductive if and only if it has a compact real form, see for example here:

http://www.math.uchicago.edu/~mbergeron/ComplexReductive.pdf

If follows from this, that any complex reductive group is homotopy equivalent to a compact manifold of half dimension, its real compact form. See, for example, "Other characterisations" here: https://en.wikipedia.org/wiki/Reductive_group . For example $$(\mathbb C^*)^n$$ is homotopy equivalent to $$(S^1)^n$$. In particular $$H_n(G,\mathbb Z)$$ of a complex reductive group $$G$$ of dimension $$n$$ is $$\mathbb Z$$.

Second. The mid-dimensional homology group vanishes for non-reductive groups, even more, non-reductive groups are homotopy equivalent to compact manifolds of dimensions less than the half. To prove this one uses the following definition of reductive groups: A non-reductive group has a nontrivial normal unipotent subgroup. See here https://en.wikipedia.org/wiki/Reductive_group. Now, if we quotient a group by a normal unipotent subgroup, the homotopy type doesn't change. So we can quotient until we get a reductive group.

Proof of II. If is easy to see from the proof of I, that if $$G_1$$ is diffeomorphic to $$G_2$$, then their (half-dimensional) compact forms are homotopy equivalent. So we just need to know that in each dimension there exists only finite number of connected compact Lie groups. This is indeed well known, see for example the references in the following mathoverflow answer: Classification of (compact) Lie groups

• These aren’t reductive groups (as the question suggested were sought), but fattenings up of unipotent groups by the same reductive group, so the (counter)example doesn’t seem to address the spirit of the question. – LSpice Dec 23 '19 at 22:36
• LSpice, I changed the answer so that it addresses the question in the positive direction – Dmitri Panov Dec 25 '19 at 13:14
• Why does $G_1$ diffeomorphic to $G_2$ imply that their compact real forms are diffeomorphic? All I see how to show from your argument is that their compact real forms are homotopy equivalent. – John Pardon Dec 25 '19 at 22:42
• You are right, I should have said they are homotopy equivalent. I wonder though if they will in the end be diffeomorphic because they have a metric with a non-negatively definite curvature operator. Anyway, I don't know how to prove that they are diffeomorphic, I'll change the answer (but of course for the argument one just needs to know that the thing is homotopy equivalent to a compact lie group of half dimension). – Dmitri Panov Dec 25 '19 at 23:28
• @DmitriPanov, isn't there also a Iwasawa decomposition $G = K\cdot A \cdot N$ for complex reductive algebraic groups? But $K\times A \times N$ is not reductive as soon as $N$ is nontrivial, so how can one reconcile this with claim I? – Guntram Dec 26 '19 at 8:31