# Smooth Manifolds and Lie Group Action

I have started to read about Lie group action on smooth manifolds.

A question popped up in my mind and I am not sure it's silly or not.

Diff (M) = space of all smooth diffeomorphisms is a large group and usually it's not a Lie group which acts transitively on M.

Is it possible to extract out a Lie group from Diff (M) which still acts transitively?

I am sure it's not true. Can you please provide me an insight as to why?

What would be the consequences? For example, would it have any effect on sectional curvature if the above group contains its isometric group?

## 2 Answers

As AreaMan's answer is sufficient for the isometry group part of the OP's curiosity, I'll just mention the differential topological part.

I think you may know of the notion of homogeneous space, a (differentiable) manifold $M$ with transitive Lie group action $G$ as diffeomorphisms. And it is well-known that such manifolds can be represented as a quotient $G/H$ by a Lie subgroup $H$. So you have asked the existence of a manifold which is not of such form. This MO posting gives one(in fact, infinitely many).

• Thank you very much. I need to learn a lot to understand that MO post.... thanks again. – Illuminata Mar 26 '16 at 8:10
• Do you understand why $(S^1 \times S^2) \# (S^1 \times S^2)$ retracts to a punctured $S^1 \times S^2$? I never learned too well the "calculus" of these topological manipulations. Excellent link, btw. Thank you. – Lorenzo Mar 26 '16 at 17:47
• @AreaMan // Sorry for delay in comment; I was busy recently. As any connected sum $M\# M$of two copies of the same manifold is a double of a punctured manifold $M\setminus D^n$, the natural retract is the identification of the corresponding two points which are the same in $M\setminus D^n$. – cjackal Mar 30 '16 at 4:10

(This is too long to be a comment, too naive to be an answer.)

If $M$ is a Riemannian Manifold, then the group of isometries is a Lie group. This is an old (and apparently difficult) theorem of Myers-Steenrod.

https://en.wikipedia.org/wiki/Myers–Steenrod_theorem

Generally this will not act transitively, if your metric is bumpy or something.

(I think) it's not always possible to put a metric on a manifold for which it becomes a homogeneous space.

Here is a "proof:"

I googled around a bit and found this (a paper by Mostow!): http://www.jstor.org/stable/1969997?seq=1#page_scan_tab_contents

In particular, if a homogeneous space has solvable fundamental group, then this theorem puts a bound on the rank in terms of its dimension. (He defines what he means by rank.)

But now, because (apparently), fundamental groups of 4-manifolds can be pretty much anything you want (they need to be finitely generated), you can probably cook up a counter example.

But you would have to actually read that paper somewhat more carefully than I just did in order to be sure. (I can't read past the first page right now, his definition of homogeneous is on page 3. Maybe it is something completely different!) Let me know!

https://mathoverflow.net/questions/15411/finite-generated-group-realized-as-fundamental-group-of-manifolds

• Thanks ...take two disjoint open set . Put one a flat metric and another one a metric which has constant positive scaler curvature. Extend it to a metric on the whole of M . Then it's isometry group can not be transitive . Right ?? – Illuminata Mar 26 '16 at 8:02
• What about the actual question .. I didn't say it got to be an isometry group. Do you think it's possible to extract out a lie group from Diff (M) ? – Illuminata Mar 26 '16 at 8:04
• Illuminata// A Lie subgroup of $Diff(M)$ which acts transitively on $M$ gives a homogeneous space structure on $M$, so my post is the (negative) answer for this. – cjackal Mar 26 '16 at 13:18