# Algebraic Groups Connected and Reduced?

Lastly I was a bit surprised about a statement regarding the difference of group schemes to algebraic groups at wiki

https://en.wikipedia.org/wiki/Group_scheme

Let me quote it: "... Group schemes arise naturally as symmetries of schemes, and they generalize algebraic groups, in the sense that all algebraic groups have group scheme structure, but group schemes are not necessarily connected, smooth, or defined over a field...."

My question is simply why is this provides a distinguishing criterion? Are algebraic groups allways connected? Up to now I never heard that without some extra assumtions (e.g. irreducibility) all algebraic groups are connected. Or do I oversee here a detail?

Futher question: What about beeing reduced? Should algebraic groups always be reduced as varieties? If yes, where does this in their definition flows in?

• You're going to find different definitions in every textbook. Algebraic groups are usually defined as groups that are also varieties, and it's common for a variety to always be irreducible (thus connected) and reduced. I'm a little surprised to see "smooth" on the list. I think "not always defined over a field" is the most important distinction listed, though I'd say "not always finite type over a field" instead, since even group schemes over fields can be quite far from being algebraic groups. – Slade Dec 9 '18 at 0:11
• ...and a variety is for you a scheme over $k$ such that it is integral (so irred and reduced) and the structure morphism $X \to Spec(k)$ is separated and of finite typ? Sorry, for this definition question, but variety is another candidate which is in defined in quite every ag book in a way far away from uniqueness... – KarlPeter Dec 9 '18 at 0:25
• My point was that almost everybody agrees "algebraic group" = "group object in category of algebraic varieties", but there's some room for disagreement when defining varieties. The definition you gave is pretty close to my preferred definition, though I might add "geometrically integral" (i.e. the base change to the algebraic closure is integral). – Slade Dec 9 '18 at 1:25
• @Slade Is there any particular reason that you think smoothness is surprising? It's guaranteed in characteristic $0$ if your group is finite type, and in characteristic $p$ the only time I really run into non-smooth groups are dealing with things like connected finite flat group schemes (e.g. $\mu_p$) whose study has a 'different feel' than the study of algebraic groups (which usually means linear algebraic groups or, perhaps, abelian varieties but those are smooth anyways). – Alex Youcis Apr 3 at 9:48

## 1 Answer

As Slade points out in the comments, this is largely a matter of definition. The basic assumptions that most people make is that your group $$G$$ is connected and smooth (the latter actually being guaranteed in characteristic $$0$$ for finite type groups). This automatically implies that $$G$$ is geometrically integral. To see that it's geometrically connected you can see the discussion of such matters here and irreducibility follows from the whole classical "smooth implies local rings are domains" and then "connected plus local rings are domains implies irreducible".

That said, there's actually a fairly convincing argument that one should consider non-connected groups in the sense that really simple groups can quickly give rise to disconnected groups via simple operations.

For example, consider the following well-known theorem:

Theorem(Steinberg): Let $$G$$ be a (connected) reductive algebraic group over $$k=\overline{k}$$. Then, the property that for every semisimple $$s\in G(k)$$ satisfies that $$C_G(s)$$ (its centralizer) is connected is equivalent to the claim that its derived subgroup $$G^\mathrm{der}$$ is simply connected.

So, for example, if you are dealing with adjoint groups like $$\mathrm{PGL}_n$$, then taking centralizers of semisimple elements needn't give you connected groups! As a simple example, convince yourself that if $$s:=\begin{pmatrix}-1 & 0\\ 0 & 1\end{pmatrix}$$ then $$\pi_0(C_{\mathrm{PGL}_2}(s))=2$$.

EDIT: Here's a reference for the claim that in characteristic 0 group schemes locally of finite type are smooth. The point is that it suffices to show that $$G$$ is reduced since then generic smoothness gives you a dense open locus of smooth points and then you can use the homogeneity of $$G$$ to show that all closed points of $$G$$ are regular, so that $$G$$ is smooth. The reducedness then comes from a trick/theorem of Cartier.

• Thank you for the answer. Following question: Could you give a reference/sketch of the proof that if group scheme $G$ is given over a field $k$ of characteristic $0$ is of finite type then it must be smooth? – KarlPeter Apr 3 at 19:40