# Problem of the Week #251 - Jun 13, 2017

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#### Euge

##### MHB Global Moderator
Staff member
Here is this week's POTW:

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Let $f : M \to M$ be a self-map of a smooth manifold $M$. Prove that the graph of $f$ is transversal to the diagonal of $M$ if and only if the fixed points of $M$ are nondegenerate, i.e., for all fixed points $p$, $+1$ is not an eigenvalue of $df_p$.

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#### Euge

##### MHB Global Moderator
Staff member
No one answered this week's problem. You can read my solution below.

Suppose the graph of $f$, $G(f)$ is transversal to the diagonal $\Delta(M)$ of $M$. Let $p$ be a fixed point of $M$. As $df_p - \text{id}$ is a linear operator on the finite dimensional vector space $T_pM$, it suffices to show that $df_p - \text{id}$ is onto. Fix $v\in T_pM$. Since $(p,p)\in G(f)\cap \Delta(M)$, the transversality condition yields $$T_{(p,p)}(M\times M) = T_{(p,p)}(G(f)) + T_{(p,p)}(\Delta(M))$$ or $$T_pM\times T_pM = G(df_p) + \Delta(T_pM)$$
Hence $(0,v) = (u,df_p(u)) + (w,w)$ for some $u,w\in T_pM$. It follows that $v = df_p(u) - u = (df_p - \text{id})(u)$, and consequently $df_p - \text{id}$ is onto, as desired.

Conversely, supposed the fixed points of $M$ are nondegenerate. If $p$ is a fixed point and $v,w\in T_pM$, there exists an $a\in T_pM$ such that $v - w = a - df_p(a)$. Since $v - a = w - df_p(a)$, then $(v,w) = (a,df_p(a)) + (v - a, w - df_p(a)) \in G(df_p) + \Delta(T_pM)$. Thus, $T_pM\times T_pM = G(df_p) + \Delta(T_pM)$ for fixed points $p$, which implies $G(f)$ intersects $\Delta(M)$ transversally.

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