Suppose $A,B$ are smooth manifolds and there exists a binary operation on the product manifold $A \times B$ making it into a Lie group.

  1. Does this guarantee that there exist binary operations on both $A$ and $B$ making them each into Lie groups?
  2. If the answer is yes, can it be done in such a way that the product group $A \times B$ is equal to the original Lie group?
  3. If the answer to that is yes, is it necessary?

I tried briefly to find a counterexample browsing a table of Lie groups but was unsuccessful.

  • 4
    $\begingroup$ I'm not sure about the first part, but it is definitely possible for $G$ to be diffeomorphic to $A \times B$ with $A$ and $B$ groups but not have $G = A \times B$ as groups. For example: $GL_n(\mathbb{C})$ is diffeomorphic to $SL_n(\mathbb{C}) \times \mathbb{C}^\times$ as a manifold, but it is not isomorphic to $SL_n(\mathbb{C}) \times \mathbb{C}^\times$ as a group. $\endgroup$
    – Nate
    Aug 30, 2020 at 18:25
  • 6
    $\begingroup$ According to this MSE answer, $SO(8)$ is diffeomorphic to $SO(7)\times S^7$, but $S^7$ has no Lie group structure. Maybe somebody can find a reference for a proof of that diffeomorphism? $\endgroup$
    – Jack Lee
    Aug 30, 2020 at 19:17
  • 2
    $\begingroup$ I took the liberty of numbering the questions. I'm not entirely sure what you mean by "is it necessary" in question 3; can you clarify? $\endgroup$ Aug 30, 2020 at 19:23
  • 2
    $\begingroup$ One famous example is given by an exotic $R^4$: It is a smooth $4$-manifold $W$ which is not diffeomorphic to $R^4$ but homeomorphic to $R^4$. However, it is known that $W\times R$ is diffeomorphic to $R^5$ since there are no exotic $R^5$'s. $\endgroup$ Aug 30, 2020 at 19:27
  • $\begingroup$ By "equal" The OP meant "diffeomorphic"? $\endgroup$
    – C.F.G
    Aug 30, 2020 at 19:48

2 Answers 2

  1. First, the example mentioned by Jack Lee: The group $SO(8)$ acts transitively on the unit sphere $S^7$ with point-stabilizers isomorphic to $SO(7)$. This gives $SO(8)$ structure of a principal $SO(7)$-bundle over $S^7$. This bundle is nothing but the orthonormal frame bundle of $S^7$. Since $S^7$ is parallelizable, its orthonormal frame bundle is trivial, hence, $SO(8)$ is diffeomorphic to $S^7\times SO(7)$. However, $S^7$ (as any sphere apart from $S^1$ and $S^3$) is not homeomorphic to any Lie group.

  2. A more difficult example is an exotic $R^4$: It is a smooth 4-dimensional manifold $W$ homeomorphic to $R^4$ but not diffeomorphic to it (there is actually continuum of diffeomorphism classes of exotic $R^4$'s). On the other hand, $W\times R$ is homeomorphic to $R^5$, hence, diffeomorphic to $R^5$ since there are no exotic $R^n$'s for $n\ne 4$. Now, $R^5$, of course, has structure of a Lie group. But if an $n$-dimensional Lie group is contractible, it has to be diffeomorphic to $R^n$. I am quite sure the same works in one dimension lower, when one uses $W$ equal to the Whitehead manifold.


Question 2 has well-known counterexamples. Generally, it's known that every connected Lie group $G$ is diffeomorphic to a product $K \times \mathbb{R}^n$ where $K$ is its maximal compact subgroup, but $G$ is generally not isomorphic to a product of Lie groups diffeomorphic to $K$ and $\mathbb{R}^n$. A nice small example is $G = SL_2(\mathbb{R})$, whose maximal compact is $SO(2)$, and hence which is diffeomorphic to a product

$$SL_2(\mathbb{R}) \cong S^1 \times \mathbb{R}^2.$$

(this can be established quite explicitly, e.g. using Iwasawa decomposition). However, $SL_2(\mathbb{R})$ has a simple Lie algebra $\mathfrak{sl}_2(\mathbb{R})$, so it isn't isomorphic to any nontrivial product of positive-dimensional Lie groups.


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