# If $A \times B$ is a Lie group, are both $A$ and $B$ Lie groups?

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.

• 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.
– Nate
Aug 30, 2020 at 18:25
• 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? Aug 30, 2020 at 19:17
• 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? Aug 30, 2020 at 19:23
• 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. Aug 30, 2020 at 19:27
• By "equal" The OP meant "diffeomorphic"? Aug 30, 2020 at 19:48

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.