# Non-isomorphic complex Lie groups with the same exceptional Lie algebra for $\mathfrak{g_2,f_4,e_6,e_7,e_8}$?

An exceptional complex Lie algebra is a simple Lie algebra whose Dynkin diagram is of exceptional (nonclassical) type. There are exactly five such Lie algebras: $$\mathfrak{g}_{2}$$, $${\mathfrak {f}}_{4}$$,$${\mathfrak {e}}_{6}$$, $${\mathfrak {e}}_{7}$$, $${\mathfrak {e}}_{8}$$; their respective dimensions are 14, 52, 78, 133, 248.

Usually, given a complex Lie algebra,

there could be non-isomorphic connected complex Lie groups with the same given Lie algebra.

For example, the $${\rm SO}(N)$$ and $${\rm Spin}(N)$$ can have the same Lie algebra $${\mathfrak {so}}_{n}$$, but they are non-isomorphic Lie groups because $${\rm SO}(N)={\rm Spin}(N)/(\mathbb{Z}/2\mathbb{Z})$$ has smaller center than $${\rm Spin}(N)$$. In particular, $${\rm SO}(5)$$ and $${\rm Sp}(2)\simeq {\rm Spin}(5)$$ are non-isomorphic Lie groups with isomorphic Lie algebras $${\mathfrak {so}}_{5}\simeq{\mathfrak{sp}}_2$$.

Questions:

1. It is commonly said that the Lie groups with given Lie algebra $$\mathfrak{g}_{2}$$, $${\mathfrak {f}}_{4}$$,$${\mathfrak {e}}_{6}$$, $${\mathfrak {e}}_{7}$$, $${\mathfrak {e}}_{8}$$ are $$G_2$$, $$F_4$$, $$E_6$$, $$E_7$$, $$E_8$$. However, do we have non-isomorphic Lie groups with the same exceptional Lie algebra $$\mathfrak{g}_{2}$$, $${\mathfrak {f}}_{4}$$,$${\mathfrak {e}}_{6}$$, $${\mathfrak {e}}_{7}$$, $${\mathfrak {e}}_{8}$$?

2. What are the centers $$Z(G)$$ of these Lie groups $$G$$? For $$G_2$$, $$F_4$$, $$E_6$$, $$E_7$$, $$E_8$$ and possibly others with the same given Lie algebra?

3. What are the homotopy groups $$\pi_d(G)$$ of these Lie groups $$G$$ for lower dimensions? say $$d=0,1,3,4,5,...$$?

We already know that $$\pi_2(G)=0$$ for any Lie group.

• These are pretty basic questions which are probably a better fit for math stack exchange. But certainly there are multiple different Lie groups which have the same exceptional Lie algebra as their Lie algebra: see e.g. en.wikipedia.org/wiki/List_of_simple_Lie_groups#List for a starting point. Aug 22, 2020 at 22:12
• Not only is this problem subject to explicit enumeration as @TimCampion says; there's not much enumeration to do: $\mathsf E_8$, $\mathsf F_4$, and $\mathsf G_2$ are all simply connected and adjoint (= centreless, for semisimple groups), so they are the unique (linear) Lie groups with their Lie algebras (and can be realised as the automorphism groups of their Lie algebras, if you've already got those). Aug 22, 2020 at 22:55
• For $\mathsf E_6$ and $\mathsf E_7$ the centres of the simply connected groups (equivalently, the fundamental groups of their adjoint quotients) are cyclic of prime order (3 and 2, respectively), so you have only the automorphism groups of the Lie algebras and their simply connected covers. Aug 22, 2020 at 22:56
• If "aut. Lie alg." counts as explicit but "s.c. cover" doesn't, then the $\mathsf E_6$ and $\mathsf E_7$ sitting inside $\mathsf E_8$ (as derived groups of Levi subgroups) are both simply connected. I think this is how Frank Adams constructs them in the lovely book @Mare references. If I recall correctly, he gets $\mathsf F_4$ and $\mathsf G_2$ by folding (paper by Stembridge, also lovely). Aug 22, 2020 at 22:58
• I guess in light of all this discussion I agree that it would be nice for someone to post an answer cataloging all the possible groups. Aug 23, 2020 at 0:03

I prefer to use the language of algebraic groups. All algebraic groups and Lie algebras are defined over $$\Bbb C$$.

1. Let $${\mathfrak g}$$ be a semisimple Lie algebra. Consider the automorphism group $${\rm Aut\,}{\mathfrak g}$$, its identity component $$G^{\rm ad}:=({\rm Aut\,}{\mathfrak g})^0$$, and the group of outer automorphisms $${\rm Out\,} {\mathfrak g}:=({\rm Aut\,} {\mathfrak g})/({\rm Aut\,} {\mathfrak g})^0$$. We say that $$G^{\rm ad}$$ is the adjoint group (or the group of adjoint type) with Lie algebra $${\mathfrak g}$$. Note that $$Z(G^{\rm ad})=\{1\}$$.

2. Starting with a semisimple Lie algebra $${\mathfrak g}$$, one can construct the simply connected group $$G^{\rm sc}$$ with Lie algebra $${\mathfrak g}$$; see Steinberg, Lectures on Chevalley groups, AMS, 2016. Note that $$\pi_1(G^{\rm sc})=\{1\}$$. This algebraic group $$G^{\rm sc}$$ has the following universal property: for any algebraic group $$H$$ with Lie algebra $${\mathfrak h}$$ and for any homomorphism of Lie algebras $$\varphi_{\rm Lie}\colon {\mathfrak g}\to{\mathfrak h}$$, there exists a unique homomorphism of algebraic group $$\varphi\colon G^{\rm sc}\to H$$ inducing $$\varphi_{\rm Lie}$$.

3. For any connected algebraic group $$G$$ with Lie algebra $${\mathfrak g}$$, there exists a canonical surjective homomorphism $$\rho\colon G^{\rm sc}\to G$$ inducing the identity isomorphism on $${\mathfrak g}$$; see above. We have $$\pi_1(G^{\rm sc})=\{1\},\quad \pi_1(G)={\rm ker}\,\rho.$$ On the other hand, we have a canonical surjective homomorphism $${\rm Ad}\colon G\to G^{\rm ad}\subseteq {\rm Aut\,} {\mathfrak g}$$ with kernel $$Z(G)$$. Write $$C=Z(G^{\rm sc})=\pi_1(G^{\rm ad}).$$ The homomorphism $${\rm Ad}\colon G\to G^{\rm ad}$$ induces a homomorphism $$i\colon \pi_1(G)\to\pi_1(G^{\rm ad})=C.$$ Moreover, the homomorphism $$\rho\colon G^{\rm sc}\to G$$ induces a homomorphism $$j\colon C=Z(G^{\rm sc})\to Z(G).$$ In this way we obtain a short exact sequence $$1\to\pi_1(G)\overset{i}{\longrightarrow} C\overset{j}{\longrightarrow} Z(G)\to 1.$$

Conversely, for each subgroup $$F\subseteq C$$ one can associate a connected semisimple group $$G_F:=G^{\rm sc}/F$$ with Lie algebra $${\mathfrak g}$$, with fundamental group $$\pi_1(G_F)=F$$, and with center $$Z(G_F)=C/F$$. In this way we obtain a canonical bijection between the set of subgroups of $$C$$ up to conjugation by $${\rm Out\,} {\mathfrak g}$$ and the set of isomorphism classes of connected semisimple algebraic groups with Lie algebra $${\mathfrak g}$$. It is known that $${\rm Out\,} {\mathfrak g}$$ is canonically isomorphic to $${\rm Aut\,} {\rm Dyn}({\mathfrak g})$$, where $${\rm Dyn}({\mathfrak g})$$ is the canonical Dynkin diagram of $${\mathfrak g}$$.

4. Let us return to our exceptional simple Lie algebras. The group $$C=C({\mathfrak g})$$ can be found, for instance, in tables in the book by Bourbaki "Lie Groups and Lie Algebras, Chapters 4-6", or in the book by Onishchik and Vinberg "Lie Groups and Algebraic Groups", Springer-Verlag, 1990.

For $${\mathfrak g}_2$$, $${\mathfrak f}_4$$, and $${\mathfrak e}_8$$ we have $$C({\mathfrak g})=\{1\}$$. Thus there is only one (up to isomorphism) algebraic group $$G^{\rm sc}({\mathfrak g})=G^{\rm ad}({\mathfrak g})$$ with Lie algebra $${\mathfrak g}$$.

For $${\mathfrak g}={\mathfrak e}_6$$ we have $$C({\mathfrak g})\simeq {\Bbb Z}/3{\Bbb Z}$$. This group has no nontrivial subgroups. Thus there are exactly two connected algebraic groups (up to isomorphism) $$E_6^{\rm sc}$$ and $$E_6^{\rm ad}$$ with Lie algebra $${\mathfrak e}_6$$. We have $$Z(E_6^{\rm sc})=\pi_1(E_6^{\rm ad})\simeq{\Bbb Z}/3{\Bbb Z}.$$

For $${\mathfrak g}={\mathfrak e}_7$$ we have $$C({\mathfrak g})\simeq {\Bbb Z}/2{\Bbb Z}$$. This group has no nontrivial subgroups. Thus there are exactly two connected algebraic groups (up to isomorphism) $$E_7^{\rm sc}$$ and $$E_7^{\rm ad}$$ with Lie algebra $${\mathfrak e}_7$$. We have $$Z(E_7^{\rm sc})=\pi_1(E_7^{\rm ad})\simeq{\Bbb Z}/2{\Bbb Z}.$$

5. The real forms of a connected algebraic group of an exceptional type correspond bijectively to the real forms of (or real structures on) its Lie algebra. My favorite way to classify those is via Kac diagrams. See Table 7 in the book by Onishchik and Vinberg. The number of real forms is 2 for $${\mathfrak g}_2$$, 3 for $${\mathfrak f}_4$$, 3 for $${\mathfrak e}_8$$, 4 for $${\mathfrak e}_7$$, 5 for $${\mathfrak e}_6$$. These real forms are listed also in Table V in Chapter X of Helgason's book "Differential Geometry, Lie Groups, and Symmetric Spaces" (Helgason lists all non-compact forms). Helgason classifies real forms using the original method of Kac with infinite dimensional Lie algebras. Onishchik and Vinbeg use another method, which gives exactly the same answer (the same Kac diagrams).

• I think it may be important to note what is paid for by using the language of algebraic groups (which I also prefer). For example, "there is only one algebraic group with Lie algebra $\mathfrak g$" (where $\mathfrak g$ is $\mathfrak e_8$, $\mathfrak f_4$, or $\mathfrak g_2$) is true only to the extent that we speak of groups over alg. closed field; it's not true over $\mathbb R$ (as you say later). Also, the alg. groups that we are calling connected and simply connected need not be either when regarded as real Lie groups (by taking real points). Finally, we miss non-linear groups this way! Aug 23, 2020 at 16:08
• In particular, if a semisimple algebraic $\Bbb R$- group $\bf G$ is a connected (over $\Bbb C$), and if it either compact or simply connected (simply connected over $\Bbb C$), then the group of $\Bbb R$-points $\bf G(\Bbb R)$ is connected. Aug 23, 2020 at 16:22
• One thing is wrong with this answer: the exceptional Lie algebra $\mathfrak{e}_6$ has a non-trivial outer automorphism. Aug 23, 2020 at 21:20
• @SamHopkins: And you wrote that this question would be more suitable for Math StackExchange.... Aug 24, 2020 at 9:52
• @MikhailBorovoi: Haha, indeed. Well I knew that in principle this was all worked out by Cartan 100 years ago, but there are many moving parts to say the least! Aug 24, 2020 at 12:45

I have just dumped the comments in an answer, and hope someone will make a better answer than this. (EDIT: Someone has! See @MikhailBorovoi's answer.) I'll be happy to delete this one—or you can just edit this one, which is CW to avoid reputation (since I'm just compiling comments).

One thing that hasn't been mentioned in the comments yet is Question 3. Asking about $$\pi_0$$ of a group based on its Lie algebra is in some sense meaningless; the Lie algebra of a group only sees its identity component, so one may make $$\pi_0$$ as bad as one allows a discrete Lie group to be (countable?) without changing the Lie algebra. For $$\mathsf E_8$$, $$\mathsf F_4$$, and $$\mathsf G_2$$, $$\pi_1$$ is trivial. For $$\mathsf E_6$$ and $$\mathsf E_7$$, the adjoint form (arising as the automorphism group of the Lie algebra) has fundamental group that is cyclic of prime order (3 and 2, respectively). However, all of these statements are about the complex groups, hence for compact forms, and I know that other real forms may have different fundamental groups, but don't know how they differ. Hopefully someone will fill this in.

As you mention, $$\pi_2$$ is trivial, and @AndréHenriques gives a reference in the comments pointing to Borel's paper An application of Morse theory to the topology of Lie groups proving that $$\pi_3$$ is infinite cyclic for simple groups, but I'm not sure if that's for algebraically simple groups, or for abstractly simple groups (i.e., is a finite centre allowed?). Again, hopefully someone will edit this answer, or add their own.

@SamHopkins said:

… certainly there are multiple different Lie groups which have the same exceptional Lie algebra as their Lie algebra: see e.g. http://en.wikipedia.org/wiki/List_of_simple_Lie_groups#List for a starting point.

@TimCampion said:

Of course, as questions (2) and (3) seem to hint at, the general rule for finding Lie groups $$G'$$ with the same Lie algebra as $$G$$ is (1) find the universal cover $$\tilde G$$, (2) find the center $$Z(\tilde G)$$, (3) enumerate all discrete subgroups $$\Gamma \subseteq Z(\tilde G)$$, (4) take $$G' = \tilde G/\Gamma$$ for each $$\Gamma$$, and (5) look for more direct descriptions. How straightforward is it to hunt down all these steps for all these groups in the literature (I'm no expert)? If nothing else, recording sources for each step here would make this information easier to find on the internet."

@Mare said:

@TimCampion At least to me it seems to be quite difficult to find much information on the exceptional Lie groups in the standard textbooks. There is the textbook "Lectures on Exceptional Lie Groups" by Adams which might contain most information but it seems to be out of print in my country at the moment. It would be interesting to see whether there is a modern textbook on Lie groups that contains also detailed descriptions and properties for the exceptional Lie groups and their Lie algebras. I have nearly 10 books on Lie groups but none has full detailed information on all exceptional cases."

I then blathered on for a while (1 2 3 4), but note that first I forgot to think about real forms, and even when I remembered I got it wrong by over-analogy to the $$p$$-adic case—see important corrections (1 2) of my wrong statements about non-existence of non-split real forms:

Not only is this problem subject to explicit enumeration as @TimCampion says; there's not much enumeration to do: $$\mathsf E_8$$, $$\mathsf F_4$$, and $$\mathsf G_2$$ are all simply connected and adjoint (= centreless, for semisimple groups), so they are the unique (linear) Lie groups with their Lie algebras (and can be realised as the automorphism groups of their Lie algebras, if you've already got those). For $$\mathsf E_6$$ and $$\mathsf E_7$$ the centres of the simply connected groups (equivalently, the fundamental groups of their adjoint quotients) are cyclic of prime order (3 and 2, respectively), so you have only the automorphism groups of the Lie algebras and their simply connected covers. If "automorphism group of Lie algeba" counts as explicit but "simply connected cover" doesn't, then the $$\mathsf E_6$$ and $$\mathsf E_7$$ sitting inside $$\mathsf E_8$$ (as derived groups of Levi subgroups) are both simply connected. I think this is how Frank Adams constructs them in the lovely book @Mare references. If I recall correctly, he gets $$\mathsf F_4$$ and $$\mathsf G_2$$ by folding (paper by Stembridge, also lovely). I thought we had an MO question about how to lift foldings of root systems to groups, but I can't find it (although …); but there is a nice question Beautiful descriptions of exceptional groups. (EDIT: And, oops, I forgot about real forms, of which $$\mathsf E_6$$ and $$\mathsf E_7$$ have non-split ones but $$\mathsf E_8$$, $$\mathsf F_4$$, and $$\mathsf G_2$$ because both s.c. and adjoint, do not. Further EDIT while compiling: This is wrong, as @VítTuček and @SamHopkins pointed out for $$\mathsf F_4$$ and $$\mathsf E_8$$, respectively. In fact I was, impressively, wrong on every count; there's also a non-split form for $$\mathsf G_2$$. I should have known that this uniqueness statement was wrong even before it was pointed out, because every real (linear) Lie group has at least compact form as well as a split one.)

@SamHopkins said:

Relevant old MO question: Lie algebras to classify Lie groups.

@VítTuček said:

@LSpice What do you mean that $$\mathsf F_4$$ does not have a non-split real form? There are three non-isomorphic real forms of type $$\mathsf F_4$$. See e.g. https://mathoverflow.net/a/96477/6818.

@SamHopkins said:

$$\mathsf E_8$$ also has 3 real forms according to the list I linked.

• I appreciate your effort and voted up, BUT this is not an answer! Something more precise and direct counts an answer... Aug 23, 2020 at 14:14
• I agree that it is not a very good answer, which is why I hope someone will make a better one; but it seems to me that it literally answers your Question 1 and Question 2. I did not want to editorialise, so included pretty much everything mathematical; but feel free to edit to something more streamlined. Aug 23, 2020 at 14:29
• I will post an answer very soon. I am typing it. Aug 23, 2020 at 14:35