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Before asking the question I should say that I don't know much about algebraic groups and I'm not sure if the question has the right level for MO. If not, please let me know and I will delete the question that emerged when I was trying to understand some theorems in a paper on the vanishing range of the cohomology of finite groups of Lie type.

Call an algebraic group (over an algebraically closed field) simple if it is simple as an abstract group and almost simple if it has finite center. Every almost simple algebraic group is of one of the following types:

$$A_n, B_n, C_n, D_n, E_6, E_7, E_8, F_4, G_2$$

Question: What are the simply connected simple algebraic groups (over an alg. closed field with no assumption on the characteristic) and of which type are they ?

Counter-examples: $SL_{n+1}$ is simply conntected almost simple of type $A_n$, but not simple while $PSL_{n+1}$ is simple but not simply connected of the same type.

Added: The mathematical problem is solved by Jay's answer. But I'm still confused about the terminology: In contrast to finite groups, where each simple group has a definite name, simple algebraic groups doesn't seem to have a name on their own (apart from the classical types). For example, by comments of Yves Cornulier and Jim Humphreys, for each type there is a unique simply connected simple alg. group. In my opinion it would be reasonable to give them a special name, for example $E_6^{sc}$. However the typical wording in the literature is like "let $G$ be a simply connected, connected complex algebraic group of type $E_6$".

Is there a particular reason for this convention ?

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@Demin: The question is "simple" (so to speak) but takes extra thought in prime characteristic. Since people tend to think about the notion of simply connected in topological terms for Lie groups, it's always a good idea to add a reminder about the meaning in the algebraic setting of Chevalley's seminar on classification. – Jim Humphreys Feb 26 '13 at 19:32
@Demin: for each given type there is a simply connected algebraic group of this type, unique up to isomorphism. – YCor Feb 26 '13 at 20:21
You will sometimes see such naming conventions used, especially for finite reductive groups. For instance people sometimes write $E_6^{sc}(q)$. I would imagine the reason for the way it's written is the following. When proving certain theorems about connected reductive algebraic groups you reduce to the case where G is simple and either adjoint or simply connected. You will then often analyse each type individually. So one would start with "let G be a simple simply connected algebraic group" and then proceed to "assume that G is of type A_n" for instance. – Jay Taylor Mar 1 '13 at 6:48
up vote 11 down vote accepted

If $G$ is an adjoint algebraic group then it is always simple as an abstract group, (EDIT: This is because any proper normal subgroup of a simple algebraic group must be finite and lie in the centre). In general assume $G$ is connected reductive algebraic group and let $\pi : G \to G_{ad}$ be an adjoint quotient of $G$. Then $G$ is simple as an abstract group if the kernel of $\pi$ is trivial because $\pi$ is then an isomorphism of abstract groups. If $\pi$ does not have a trivial kernel then it is clear that $G$ is not simple as an abstract group.

For an example: in characteristic 2 the symplectic group $Sp_{2n}$ is simple as an abstract group as it is isomorphic (as an abstract group) to the corresponding adjoint group. However these two groups are not isomorphic as algebraic group (only isogenous).

EDIT: I suppose this didn't really explicitly answer your question. Explicitly we have a simple simply connected group is simple as an abstract group if and only if it is on the following list:

  • $G_2$, $F_4$ or $E_8$ any characteristic.
  • $B_n$, $C_n$, $D_n$ (n>2) or $E_7$ in characteristic 2.
  • $E_6$ in characteristic 3.
  • $A_n$ if $n+1$ is a power of the characteristic.
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@Jay: As you observe, the issue of abstract simplicity mainly arises in prime characteristic. But you have to be more careful about type $A_n$, where the simply connected group is $\mathrm{SL}_{n+1}$. Its center consists of scalar matrices corresponding to the $(n+1)$th roots of unity in the algebraically closed field. – Jim Humphreys Feb 26 '13 at 19:02
@Jim Thank you, in my haste I had forgotten (as always) that $SL_{n+1}$ is of type $A_n$. I have corrected my answer accordingly. – Jay Taylor Feb 26 '13 at 19:28
Jay, thank you very much for your answer. According to Yves comment, in each type of your list there is exactly one simply connected group. (1) Is this group also simple ? (2) Do you happen to know the simply connected groups of the listed types ? I believe for the classical types these are $\text{SL}_{n+1}$ $(A_n)$, $\text{SO}_{2n+1}$ $(B_n)$, $\text{Sp}_{2n}$ $(C_n)$, $\text{SO}_{2n}$ $(D_n)$. Is this right ? – Demin Hu Feb 26 '13 at 21:48
@Demin: You need to do more background reading in Lie theory. The 1956-58 classification for algebraic groups by Chevalley over an arbitrary algebraically closed field is essentially the same as the Lie group list of Cartan-Killing. So there is just one simply connected group of each simple type. In your list, SO must be replaced by Spin. The exceptional types are harder to construct as linear groups, but three types are both simply connected and adjoint. – Jim Humphreys Feb 26 '13 at 22:41
@Demin As Jim says the simply connected groups of types $B_n$ and $D_n$ are $Spin_{2n+1}$ and $Spin_{2n}$ respectively (they are double covers of the corresponding special orthogonal groups). You might find section 1.11 in Carter's book "Finite Groups of Lie Type" a useful read as he reviews the classification of simple algebraic groups. The simply connected groups you list are simple algebraic groups and are simple as an abstract group under the conditions I gave in my answer. If this answers your question it would be good to accept the answer so people know the question is answered with – Jay Taylor Feb 27 '13 at 8:31

The almost simple algebraic groups $G$ are classified by their root datum (defined here): in short, this is a quadruple $(X,X^{\ast},R,R^{\ast})$ with $X,X^{\ast}$ dual finitely generated free abelian groups and $R\subset X, R^{\ast}\subset X^{\ast}$ finite subsets (satisfying some conditions). Think of $X$ as the character group of some maximal torus in $G$ and $X$ a collection of roots (in particular, $R$ will define a root system once we tensor $X$ with $\mathbb{R}$). Then, $R^{\ast}\subset X^{\ast}$ are the corresponding co-objects (cocharacters, coroots). This datum characterises almost simple algebraic groups in the following sense: the type of $G$ is determined by the root system determined by $R\subset X$ (type ABCDEFG, as you've mentioned) and the extra information supplied by knowing the co-objects determines the weight lattice of this root system. The simply connected groups are those groups for which the weight lattice of the root system of $G$ is equal to $X$; this is the same as those groups of each type with the `largest' (finite) centre. The simple groups are those groups for which $\mathbb{Z}R = X$, that is, when the root lattice equals the character lattice. Hence, the simple simply connected groups are those $G$ whose weight lattice is equal to its root lattice.

Some nice, introductory notes are available here. Also, see Chevalley's Collected Works here.

I suppose this does not completely answer your question (I haven't given a list of the simply connected simple groups) but hopefully this will give you an idea of the general framework.

EDIT: I appear to have been too slow! As I was off looking for the required list @Jay Taylor has given a complete answer.

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George, thanks for your good explanation and the references. – Demin Hu Feb 26 '13 at 20:45

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