Elementary (but perhaps non-trivial) question on Lie groups

[Paul]

Let two Lie groups have underlying group structures which are
isomorphic and underlying smooth manifolds which are homeomorphic.
Are these two Lie groups necessarily isomorphic?

This question was asked by a classmate of mine at grad school about 22
years ago. The prof replied that he didn't know but that he didn't
think it was an interesting question. He (the prof) suspected that
the answer is that yes, the two Lie groups are isomorphic.

Paul Epstein

[G. A. Edgar]

In article
<7952acdf-3eca-49df...@hb4g2000vbb.googlegroups.com>,

You must at least add "connected" --- otherwise take the product of the
line by two non-isomorphic finite groups of the same order.

--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/

[Paul]

On Feb 2, 1:45 pm, "G. A. Edgar" <ed...@math.ohio-state.edu.invalid>
wrote:
> In article
> <7952acdf-3eca-49df-a9cb-0dd31109b...@hb4g2000vbb.googlegroups.com>,

I agree that the connected case is more interesting.
However, I'm probably making a mistake but I don't see how your remark
leads automatically to a family of counterexamples.

For example, suppose G1 and G2 are groups with the same finite order
but G1 is abelian and G2 is nonabelian.

Then, when considering the group structures alone, the product of the
line with G1 is abelian but the product of the line with G2 is
nonabelian.

So we don't get a counter-example.

Paul Epstein

[G. A. Edgar]

In article
<d9aa4964-d55d-4f99...@j14g2000vba.googlegroups.com>,

OK, my first idea was bad. How about this:
Let R be the reals with the usual topology and let Q be the rationals
with the discrete topology. Then RxQ and RxQxQ are homeomorphic as
topological spaces, isomorphic as groups, but not isomorphic as
topological groups.