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Hi, everyone.

Assume $S$ is a genus at least 2 orientable closed surface. And there is a simplical complex defined on $S$ called Curve complex.

It is well known that any automorphism of surface $S$ acts on curve complex $\mathcal {C}(S)$ isometrically.

Now I want to know that

For any periodic and irreducible automorphism $f: S\rightarrow S$, is there a connected subcomplex $W\subset \mathcal {C}(S)$ with infinite diameter such that $f(W)=W$?

Staylor constructed such a $W$. Now there is a handlebody $H$ such that $\partial H=S$. Let $f$ be as above, $W$ be the disk complex of $H$.

Now I wonder:

Is it still possible that $f(W)=W$?

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4  
What kind of subcomplex? The orbit of a vertex under f is such a thing. –  Richard Kent Nov 20 '12 at 2:49
    
@R.Kent I have edited it again. –  yanqing Nov 20 '12 at 5:37
6  
Now I don't see why you can't just take W to be the whole curve complex... –  HJRW Nov 20 '12 at 14:51
    
In my deleted answer my eyes somehow leapt right over "periodic" and landed on "irreducible". –  Lee Mosher Nov 20 '12 at 21:47
1  
What is the subcomplex $W$ for? What properties do you want it to have? –  Sam Nead Nov 20 '12 at 23:29

2 Answers 2

No, this is false. The point is that if you take an irreducible periodic automorphism $f$, there is no disjointly embedded collection of curves of $\mathcal{C}(S)$ which is left invariant by $f$ up to isotopy.

Now suppose that the action of $f$ on $\mathcal{C}(S)$ has a fixed point. Then this fixed point lies in a unique minimal simplex of $\mathcal{C}(S)$ which is preserved by $f$ (one may describe this as the intersection over all simplices containing the fixed point, since the action is simplicial). But then the vertices of this simplex are permuted by $f$, and therefore $f$ is not irreducible, a contradiction.

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@Agol Since I required that $f$ is not reducible, $f$ can not fix any vertex on $\mathcal {C}(S)$. –  yanqing Nov 20 '12 at 5:52
    
@Yanqing: usually when you say "f fixes a subspace", you mean that it fixes each point of the subspace pointwise. I took this interpretation of your question (where I figured you meant a subspace, not a subcomplex, since vertices can't be fixed), so in particular, that there is some fixed point. So I think you need to consider carefully what you mean by "fixed subcomplex". –  Ian Agol Nov 20 '12 at 6:00
    
@Agol: I have edited it again. –  yanqing Nov 20 '12 at 6:20

HW seems to have the best response. If, however, you are looking for a proper subcomplex try the following:

First choose a hyperbolic metric on $S$ so that $f$ can be realized as an isometry. Then pick your favorite curve on $S$ and consider the set of curves in the curve complex with length less than or equal to your curve. Call this set $X$. Since $X$ has finite diameter, some neighborhood $N(X)$ will be connected. $N(X)$ is invariant under $f$ by construction. Now there are various ways to enlarge $N(X)$ to have infinite diameter, preserving invariance. For example, take $W$ to be $N(X) \cup \cup_i f^i \gamma$, where $\gamma$ is an infinite geodesic ray beginning at a point of $N(X)$.

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