# Mapping Class Group and Path-Component of Id.

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.

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quasar987
Homework Helper
Gold Member
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
Homework Helper
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.

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.

Last edited:
mathwonk