Fixed points of diffeomorphisms: eigenvalues of the pushforward 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?

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 
$$
\lim_{k \rightarrow \infty} \frac{x_{i_k}}{||x_{i_k}||} = df_0(\lim_{k \rightarrow \infty} \frac{x_{i_k}}{||x_{i_k}||}).
$$
