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Let $f:X\to Y$ be a non-constant morphism of smooth projective connected curves over $\mathbf{C}$ (or compact connected Riemann surfaces).

Suppose that $X=Y$ and that $f$ is not the identity.

Why are the fixed points of $f$ isolated?

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    If they weren't isolated, they would have an accumulation point, so by the identity theorem $f$ would be the identity.2012-02-04
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    Alternately, a smooth connected projective curve over $\mathbb{C}$ is irreducible, and if $f$ is not the identity then its set of fixed points is a proper subvariety, so a finite set of points.2012-02-04
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    So what if I replace $X$ by a surface (a two-dimensional variety)? Then, a proper subvariety could be an entire curve so they don't have to be isolated. I don't see how the argument in the complex case breaks down. The identity theorem still holds to give that $f$ is the identity..right?2012-02-04
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    Actually, come to think about it, to see that the fixed points form a proper subvariety one uses the fibre product interpretation, right? The scheme of fixed points is $X_f\times_X X$, where the first $X$ goes with $f$ and the second with the identity.2012-02-04
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    By surface what do you mean exactly? (the term "Riemann surface" refers to a complex *curve*, a complex variety of complex dimension *one*) There are tons of smooth connected projective complex surfaces (varieties of complex dimension two) with automorphisms with non-isolated sets of fixed points!2012-02-04
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    In your context it is much more simple to view the set of fixed points as the intersection of the graph of $f$ with the diagonal $\Delta\subseteq X\times X$.2012-02-04
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    @Ali: all you have to do is observe that $f(x) = x$ is an algebraic condition on $x$.2012-02-04

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