Let $G_1$ and $G_2$ be compact connected (not necessarily semi-simple) Lie groups. Assume that the underlying smooth manifolds of $G_1$ and $G_2$ are diffeomorphic and that the underlying abstract groups are isomorphic. Is it true that $G_1$ and $G_2$ are isomorphic as topological groups? I believe that by Scheerer's theorem we know that universal covers are isomorphic.

P.S.: it is true that there has been a lot of variations of this question on this site but I believe this exact question has not been answered yet:

EDIT: Let us assume that we have proved that if two connected compact Lie groups are abstractly isomorphic, they are isomorphic (this is achieved in YCor's answer). I think the following argument extends this to general compact Lie groups.

Assume we have two compact Lie groups $G_1$, $G_2$ that are abstractly isomorphic. Note that in a compact Lie group, every subgroup of finite index is open (and open subgroups are closed), so every subgroup of finite index is a union of some connected components. Therefore, the identity component of a compact Lie group can be characterized as the subgroup of maximum finite index (which happens to be normal) and the group of connected components can be characterized as the maximum finite quotient. Therefore, from the abstract isomorphism class of $G_1$ and $G_2$ we recover extensions of Lie groups $$ 0\rightarrow (G_1)^0\rightarrow G_1\rightarrow \pi_0 G_1\rightarrow 0, $$ $$ 0\rightarrow (G_2)^0\rightarrow G_2\rightarrow \pi_0 G_2\rightarrow 0, $$ which are equivalent when considered as extensions of abstract groups (strictly speaking, we only know that there is an abstract isomorphism $G_1\rightarrow G_2$ and this by itself does not mean that the extensions are equivalent; but the above algebraic characterization of the identity component implies that any isomorphism $G_1\rightarrow G_2$ must send $(G_1)^0$ to $(G_2)^0$). We want to prove that these extensions are equivalent as extensions of Lie groups.

If we prove that for a connected compact Lie group $N$ and a finite Lie group $H$, the forgetful map from the set of equivalence classes of extensions of Lie groups to the set of equivalence classes of extensions of abstract groups $\mathrm{Ext}_{Lie}(H, N)\rightarrow \mathrm{Ext}_{Grp}(H, N)$ is an injection, we win. An extension of Lie groups $$ 0\rightarrow N\rightarrow G\rightarrow H\rightarrow 0 $$ defines a morphism of abstract groups $s:H\rightarrow \mathrm{OutAut}(N)$ called characteristic homomorphism (here $\mathrm{OutAut}(N)$ is the quotient of the group of smooth automorphisms of $N$ by the group of inner automorphisms, considered as the subgroup of $\mathrm{AbOutAut}(N)$, the group of abstract outer automorphisms of $N$). If two extensions are equivalent as extensions of abstract groups, they define the same characteristic homomorphisms so we have partitions $$ \mathrm{Ext}_{Lie}(H, N)=\bigcup_{s\in \mathrm{Hom}(H, \mathrm{OutAut}(N))} \mathrm{Ext}_{Lie}(H, N)_s, \qquad \mathrm{Ext}_{Grp}(H, N)=\bigcup_{s\in \mathrm{Hom}(H, \mathrm{AbOutAut}(N))} \mathrm{Ext}_{Grp}(H, N)_s $$ compatible with the forgetful map.

Now let $Z(N)$ denote the center of $N$ considered as an $H$-module via characteristic homomorphism. Theorem 18.1.13 (c) of "Structure and geometry of Lie groups" says that if $\mathrm{Ext}_{Lie}(H, N)_s$ is non-empty, then there is a bijection $H^2(H, Z(N))\rightarrow \mathrm{Ext}_{Lie}(H, N)_s$. Theorem 8.8. of MacLane's "Homology" says that if $\mathrm{Ext}_{Grp}(H, N)_s$ is non-empty, then there is a bijection $H^2(H, Z(N))\rightarrow \mathrm{Ext}_{Grp}(H, N)_s$. An inspection of the two proofs shows that these bijections are in fact compatible with the map $\mathrm{Ext}_{Lie}(H, N)_s\rightarrow\mathrm{Ext}_{Grp}(H, N)_s$, i.e. the 3 maps form a commutative triangle. Hence the map $\mathrm{Ext}_{Lie}(H, N)_s\rightarrow\mathrm{Ext}_{Grp}(H, N)_s$, being a composition of two bijections, is a bijection. Therefore, the forgetful map $\mathrm{Ext}_{Lie}(H, N)\rightarrow\mathrm{Ext}_{Grp}(H, N)$ is an injection (and it fails to be a surjection precisely because $\mathrm{OutAut}(N)\subsetneq \mathrm{AbOutAut}(N)$ for general compact connected Lie group).

  • $\begingroup$ Thanks for the edit. What you write "OutAut" is almost universally written "Out". $\endgroup$
    – YCor
    Mar 9, 2019 at 12:56

2 Answers 2


Yes. It's even true under the weaker assumption that $G_1,G_2$ are isomorphic abstract groups.

First the semisimple case: if two compact semisimple Lie groups are abstractly isomorphic, then they are isomorphic topological groups; indeed every abstract isomorphism is continuous (E. Cartan and van der Waerden).

Next; the abelian case: an abelian compact Lie group is topologically isomorphic to $(\mathbf{R}/\mathbf{Z})^d$ for some $d$, and the subgroup of elements of order dividing $2$ is isomorphic to $(\mathbf{Z}/2\mathbf{Z})^d$, so the abstract isomorphism class retrieves $d$. Yet unlike in the semisimple case, there are plenty of discontinuous group automorphisms.

If $G$ is an arbitrary compact connected Lie group, then we have $G=SZ$, where $S=[G,G]$ is its derived subgroup, $Z$ is the center. In particular, the abstract isomorphism class of $G$ retrieves the topological group isomorphism class of both $S$ and $Z$, and also the isomorphism class of the finite group $S\cap Z$.

Still this is not enough. We need to prove more precise results in the semisimple and abelian cases, involving a finite subgroup.

We consider pairs $(G,H)$, $G$ being topological group and $H$ a subgroup. A abstract isomorphism $(G,H)\to (G',H')$ is an group isomorphism $f:G\to G'$ mapping $H$ onto $H'$; it is called a topological isomorphism if $f$ is a homeomorphism.

If $G,G'$ are semisimple connected compact Lie groups, then any abstract isomorphism $(G,H)\to (G',H')$ is topological: this follows from automatic continuity mentioned above.

Now let $(G,H),(G',H')$ be abstractly isomorphic pairs, with $G,G'$ abelian compact connected Lie groups, and $H,H'$ finite subgroups; I claim that they're topologically isomorphic. By the above abelian case, we can suppose that $G=G'=(\mathbf{R}/\mathbf{Z})^d$. So $H,H'\subset(\mathbf{Q}/\mathbf{Z})^d$; hence there is an automorphism of $(\mathbf{Q}/\mathbf{Z})^d$ mapping $H$ onto $H'$. That is, an element $f$ of $\mathrm{GL}_d(\widehat{\mathbf{Z}})$ with $f(H)=H'$. It is not hard (for $d\ge 1$, which we can suppose) to check that the determinant map is surjective on the stabilizer of $H$. Hence, we can choose $f\in\mathrm{SL}_d(\widehat{\mathbf{Z}})$. Since $f\in\mathrm{SL}_d(\mathbf{Z})$ is dense in $\mathrm{SL}_d(\widehat{\mathbf{Z}})$, and since the set of $f$ with the above property is open, we can choose $f'\in\mathrm{SL}_d(\mathbf{Z})$ such that $f$ coincides with $f'$ on $H$. Hence, $f$ extends to a topological automorphism of $(\mathbf{R}/\mathbf{Z})^d$.

Let's now conclude. Let $f:G\to G'$ be an abstract isomorphism between connected compact Lie groups. Write $G=SZ$ and $G'=S'Z'$ as above, and $H=S\cap Z$, $H'=S'\cap Z$. Then $f(S)=S'$ and $f$ induces a topological isomorphism $S\to S'$.

Also $f$ induces an abstract isomorphism $(Z,H)\to (Z',H')$. By the above, there exists a topological isomorphism $(Z,H)\to (Z',H)$ coinciding with $f$ on $H$. Then define a homomorphism $u:S\times Z\to G'$, $(s,z)\mapsto f(s)g(z)$; it is continuous. Then, since $g=f$ on $H$, $u$ factors through a continuous homomorphism $G\to G'$, mapping $sz$ to $f(s)g(z)$ for $s\in S$, $z\in Z$, which is readily seen to be a group isomorphism, and we are done.

  • $\begingroup$ YCor, what if we drop "connected" assumption? A compact abelian Lie group is isomorphic to a direct product of a torus and a finite abelian group (the short exact sequence has to split). For some $p$, our finite abelian group does not have $p$-torsion, so your trick should still recover the dimension. Then the component group can be recovered as the maximal finite quotient (i.e. in a compact Lie group, finite index sugroup is open, open subgroups are closed, so any finite index subgroup is a union of connected components). $\endgroup$
    – rori
    Mar 6, 2019 at 16:43
  • $\begingroup$ At first sight it sounds plausible that the result (abs. isomorphic implies isomorphic) extends to arbitrary compact Lie groups. By automatic continuity, it holds when the connected component is semisimple. In general, it sounds some further work is necessary. $\endgroup$
    – YCor
    Mar 6, 2019 at 16:59

A Theorem of van der Waerden says that every abstract homomorphism of a compact group into a unitary group is continuous. For a reference see http://dx.doi.org/10.1007/BF01188647. This implies that your two compact groups which are isomorphic as abstract groups are isomorphic as continuous groups and hence as Lie groups.

[edited]This is only for compact semi-simple groups and not arbitrary compact groups (as @YCor has pointed out).

  • 9
    $\begingroup$ This is false for the circle group, which has discontinuous automorphisms (as an abstract group it's isomorphic to $(\mathbf{Q}/\mathbf{Z})\times\mathbf{Q}^{(c)}$. Van der Waerden's theorem holds for semisimple compact connected Lie groups. $\endgroup$
    – YCor
    Mar 6, 2019 at 15:20

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