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If we consider the group of automorphisms of an orientable surface, then the subgroup that contains the orientation-preserving automorphisms will be of index two. Why is that?

Any explanation will be appreciated.

Thank you.

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2 Answers 2

up vote 6 down vote accepted

If $S$ is connected and you have chosen an orientation , you get an orientation morphism $$or:Aut(S)\to Or(S)=\lbrace +1,-1\rbrace$$ If you know that $S$ admits of an orientation reversing automorphism (this is the key point!), that morphism will be surjective and you can conclude thanks to the exact sequence of groups
$$1\to Aut^+(S) \to Aut(S)\to\lbrace +1,-1\rbrace \to 1 $$

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Very clear! Thanks. –  yaa09d Feb 3 '12 at 2:01

I'm assuming by "surface" you mean a (real?) manifold.

Anyway, let $G\cong \text{Aut}(S)$ be the group automorphisms on your surface $S$. Let $H$ be the subgroup of orientation preserving automorphisms (you should show actually show that this is indeed a [normal] subgroup).

The order of the quotient group $G/H$ is the index of $H$ in $G$ and consists of elements of the form $\{\overline{\text{id}_H},\bar{\phi} \}$ where $\bar{\phi}$ is the equivalence class of orientation reversing automorphisms of $S$. (Note: the identity map $\text{id}_H$ is orientation preserving.)

The product $\bar{\phi}\cdot\bar{\phi}$ is the coset of the automorphism $\phi\circ\phi$, which has to be orientation preserving since [on any suiably chosen neighborhood of $S$] $\text{det}(\phi\circ\phi)=\text{det}(\phi)\text{det}(\phi)=(-1)(-1)=1$. So the quotient $G/H$ is indeed of order 2, hence the index of $H$ is 2.

I've been a bit lax about neighborhoods and charts (assuming we're talking about manifolds) and this is something you should work out, but what I've said is more or less the idea of what's going on.

Hope that helps.

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Thanks for your help! –  yaa09d Feb 3 '12 at 2:02

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