Connected components of fixed point set in a connected manifold The following questions occurred to me while reading this paper (p. 573).
Let $M$ be a closed, connected, oriented Riemannian manifold of even dimension. Let $T$ be an isometry of $M$ (orientation-preserving if needed). Let $F\subseteq M$ be the fixed point submanifold of $T$.
Question 1: Does $F$ have only finitely many connected components?
Question 2: Must each component of $F$ have even dimension?
Both of these properties of $F$ are claimed in the paper. For the first claim, I guess one would need to know that the connected components are isolated from each other. So I'm curious, for example, why can't the fixed point set consist of an infinite number of (non-isolated) discrete points?
If there is a reference that explains this, that would also be nice.
 A: For your first question, let $X\subseteq M$ denote the fixed point set of $f$, and assume for a contradiction that $X$ contains infinitely many components.  Label the first countably many of them $F_1, F_2,..$.
Now, create a sequence $x_i$ with $x_i\in F_i$.  Because $M$ is compact, some subsequence of the $x_i$ converges to some $x\in M$.  By abuse of notation, We'll assume the subsequence is the original subsequence and we'll write $\lim x_i = x$.
We first claim that $x\in X$. This is a simple consequence of continuity if $f$:  $$f(x) = f(\lim x_i) = \lim f(x_i) = \lim x_i = x.$$
Now, let $U\subseteq M$ be a normal neighborhood of $x$, so that there is a unique minimal geodesic connecting $x$ to any point in $U$.  Because $\lim x_i = x$, there are infinitely many $x_i\in U$.  Suppose $x_j\in U$ is one of these.
Connecting $x$ to $x_j$ by a minimal geodesic $\gamma$, we find that $f\circ \gamma$ is a minimal geodesic connecting $x$ to $x_j$.  By uniqueness, $f\circ \gamma = \gamma$, so $\gamma(t)$ is in $X$ for all $t$.  Having a path from $x$ to $x_j$ in $X$, we deduce $x\in F_j$.
However, we can repeat this argument for a different $x_k\in U$, we find that $x\in F_j \cap F_k$ for $j\neq k$.  This contradiction implies that there must have only been finitely many components.
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For the second question, suppose $f:M\rightarrow M$ is an isometry which preserves orientation.  Let $F\subseteq M$ be a connected component of the fixed point set of $f$.  I'm assuming you already know that $F$ is an embedded submanifold.  We'll prove that $F$ has even codimension in $M$.  That is, $\dim M - \dim F$ is an even number.  (For your application, $M$ itself is even dimensional, which them implies $F$ is).
Let $p\in F$.  The tangent space $T_p M$ has an orthgonal splitting $T_p M = T_p F \oplus V$ for some complementary subspace $V$.  The goal is to show that $V$ has even dimension.
To that end, consider the differential $d_p f$.  Because $d_p f$ obviously acts as the identity on $T_p F$, it follows that $d_p f$ preserves $V$ in the sense that $d_pf$ maps $V$ to itself by some isometry.  Also, because $d_p f$ preserves the orientation and acts as the identity on $T_p F$, it must act via an orientation preserving isometry of $V$.
In addition, $d_p f$ cannot fix any  non-zero $v\in V$, for otherwise the points $\exp(tv)$ are fixed by $f$, so $F$ would not have been a connected component of the fixed point set.
Because $d_p f:V\rightarrow V$ is an isometry, all its (complex) eigenvalues have modulus one.  Those with non-zero imaginary part come in complex conjugate pairs.  And by the previous paragraph, $1$ cannot be an eigenvalue of $d_p f:V\rightarrow V$.  It follows that that number of $-1$-eigenvalues has the same parity as $\dim V$.  In particular, $\det(d_p f:V\rightarrow V) = (-1)^{\dim V}$.  Because $f$ preserves orientation, this determinant is positive, so $\dim V$ must be even.
