Mapping Class Group and Path-Component of Id.

[Bacle]

Hi, everyone:

Given a smooth, orientable manifold X, we turn Aut(X)
the collection of all self-diffeos. of X into
a topological space, by using the compact-open
topology. Aut(X) is also a group under composition.

The mapping class group M(X) of X is defined as the
quotient:

M(X):= Aut(X)/Aut_id(X)


where Aut_id(X) is the path-component of the
identity map --which coincides with the isotopy
class of IdX in the compact-open topology.
(group operation is composition, of course)


My question:
In order for M(X) to be a group, we must have
Aut_id(X) be a normal subgroup of X. How do we know
that Aut_id(X) is normal in X?. I think we need for
X to be a topological group or something, but I
am not sure.

Thanks For any Help.

 

[quasar987]

Eh? You need H:=Aut_id(X) to be normal in G:=Aut(X). I.e. gHg^{-1} < H for all g in G. And this is obvious since g o h o g^{-1} ~ g o id o g^{-1} = id.

 

[mathwonk]

since multiplication by a is continuous, if S is a connected set then so is aS. and so is aSa^-1. This seems to do it.

 

[Bacle]

Eh? You need H:=Aut_id(X) to be normal in G:=Aut(X).

Right. My carelessness

I.e. gHg^{-1} < H for all g in G. And this is obvious since g o h o g^{-1} ~ g o id o g^{-1} = id.

I agree, Q, but what I am confused about is that I was told that Aut_idX is normal
in Aut(X) only when AutX is a Lie group or a topological group. And I don't see how
we need this assumption in your argument.

Maybe Wonk's reply: using the continuity of multiplication, which is part of the def. of Lie groups
or topological groups.

 

[mathwonk]

Since 1 is in S, then also a.1.a^-1 = 1 is in aSa^-1. So aSa^-1, is a connected set containing 1. Thus aSa^-1 is a connected set that contains 1, hence it lies entirely in S. so S is normal.

 

[Bacle]

Yes, but what I meant to say is that , AFAIK, we need the condition of
Aut(X) being a topological group for the result to hold, and I did not see
where that assumption was being used in Quasar's post.