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I want to answer this question: Let $M$ be a smooth manifold, and let $f: M \rightarrow M$ be a diffeomorphism. Let $\mathrm{Fix}_f$ be the fixed points of $f$, and suppose that $x \in \mathrm{Fix}_f$ is not isolated. Show that $df_x$ has an eigenvalue of 1.

I can show that $\mathrm{Fix}_f$ is closed (may or may not be relevant). I can also show that IF there is a curve $\gamma(t)\subset \mathrm{Fix}_f$ with $\gamma(t_0)=x$ such that $\gamma'(t_0)$ is nonzero, then $\gamma'(t_0)$ is an eigenvector of $df_x$ with eigenvalue 1.

But what if $x$ is a limit point of a discrete subset of $\mathrm{Fix}_f$? Could this ever happen?

user31800
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    Have you tried weakening your assumption by supposing that $\gamma(t)\cap \text{Fix}_f$ merely accumulates to $x$ as $t\to t_0$? – Kevin Carlson Jul 17 '15 at 23:53
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    Incidentally, this can indeed occur, although the diffeomorphism can't be analytic. Something like $(x,y)\mapsto(x,y+e^{-1/x^2}\sin(1/x))$ should give an example. – Kevin Carlson Jul 18 '15 at 00:07

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Since you said you would like a solution with less analysis, here's an argument using intersection theory: In $M \times M$, let $\Delta$ (the diagonal) and $Z$ denote the graphs of $\operatorname{id}_M$ and $f$, respectively: \begin{align*} \Delta &= \{(x,y) \in M \times M : y=x\}\\ Z &= \{(x,y) \in M \times M : y=f(x)\}. \end{align*} We can identify $\operatorname{Fix}_f$ with $\Delta \cap Z$. Both $\Delta$ and $X$ are (properly) embedded half-dimensional submanifolds of $M \times M$, so any points where they intersect transversely must be isolated. Thus if $x$ is a non-isolated point of $\operatorname{Fix}_f$, then $\Delta$ and $Z$ are not transverse at $(x,x)$: \begin{equation*} T_{(x,x)} \Delta + T_{(x,x)} Z \subsetneq T_{(x,x)} (M \times M) \end{equation*} A dimension count shows that $T_{(x,x)} \Delta \cap T_{(x,x)} Z\neq \{0\}$. Since tangent vectors to $\Delta$ and $Z$ are of the form $(v,v)$ and $(v,df_x v)$, respectively, for $v \in T_x M$, the intersection $T_{(x,x)} \Delta \cap T_{(x,x)}Z$ consists of pairs $(v,df_x v)=(v,v)$. We conclude that $df_x$ has 1 as an eigenvalue.

Kyle
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My friend helped me solve this problem. His solution uses more analysis than I would like, but here it is. Choose a chart for $M$ on an open set $U$ identifying $x$ with the origin, and then choose $V \subset U$ so that $f(V) \subseteq U$. Now we may view $f$ as a diffeomorphism in a neighborhood of the origin in $\mathbb{R}^n$ that has 0 as a non-isolated fixed point.

By (analytic) definition, the pushforward $df_0$ satisfies $$ \lim_{||h|| \rightarrow 0} \frac{f(h)-df_0 h}{||h||} = 0. $$ Let $(x_i)$ be a sequence of fixed points of $f$ converging to 0. Then the sequence $(x_i/||x_i||)$ is contained in a compact set, and hence has a convergent subsequence $(x_{i_k}/||x_{i_k}||)$. Manipulating the above equation yields $$ \lim_{k \rightarrow \infty} \frac{x_{i_k}}{||x_{i_k}||} = df_0(\lim_{k \rightarrow \infty} \frac{x_{i_k}}{||x_{i_k}||}). $$

user31800
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