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Let $X$ be a connected $n$-dimensional manifold and $f:X\to X$ a differentiable function satisfying: $f\circ f =f$. Prove that for all $p\in X$ that $rk_pf\leq rk_{f(p)}f$ and subsequently that $rk \;f$ is constant along $f[X]$. Can anyone give me a hint? I have no idea where to begin

2 Answers2

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OK. Here is one hint:

  1. $Rank (S \circ T) \le \min(Rank(S), rank(T))$, where $S$ And $T$ are linear transformations, and the codomain of $T$ is the domain of $S$. Apply this to the case $T = Df_p$ and $S = Df_{f(p)}$.

For the second part ("subsequently that ..."), I don't see how to use the first result in any way. If we let $A = f(X)$, then evidently we have $f | A = id_{A}$, but that doesn't mean that $f$ has constant rank on $A$.

John Hughes
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  • Those are two hints! – Mariano Suárez-Álvarez Nov 29 '14 at 15:36
  • Yeah...I started writing three, and decided the third one was bad. I've now changed "Three" to "two". :) – John Hughes Nov 29 '14 at 15:37
  • How did you arrive at that first hint? Also note that $f$ must be differentiable yet not necessarily linear. –  Nov 29 '14 at 15:54
  • While $f$ is not linear, $df_p$, for a fixed $p$, is a linear map from $T_p X$ to $T_{f(p)} X$; the rank of $f$ at $p$ is defined to be the rank of this linear map. So I arrived at the first hint by saying "I need to know something about the rank of the composite function $f \circ f$; that's the rank of the composition of two linear xforms...and that's no greater than the min of their ranks, because the dim of the image of a linear transform is never more than the dim of the domain. – John Hughes Nov 29 '14 at 17:16
  • The exercise explicitly stated that the connectedness of $X$ was required for the second part, I don't see how that is used here –  Nov 29 '14 at 18:56
  • I didn't write a solution to the exercise; I gave hints, as you asked, to get you started. In particular, I did nothing about the claim that the rank is constant on $X$ when $X$ is assumed connected. – John Hughes Nov 29 '14 at 21:34
  • Thank you for the hints, I would like to point out a problem with the second hint though. Given is that $f(f(x)=f(x)$ not that $f(f(x))=x$ –  Dec 01 '14 at 19:06
  • You're right...I misread that bit. I've edited, and now I'm down to one hint. :( – John Hughes Dec 01 '14 at 20:46
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John Hughes has already provided a wonderful hint for the fact that $rk_p f \leq rk_{f(p)} f$, so I'll focus on the other part of the statement.

First, a counterexample if $X$ is not connected is provided by letting $X$ denote two disjoint copies of $\mathbb{R}^2$ and letting $f$, on the first copy, project to the $x$ axis, while $f$, on the second copy, it is just the identity.

Now, some hints for the case where $X$ is connected. First, since $f$ is continuous, $f(X)$ is also connected.

Now, since the rank is an integer bounded above by $\dim X$, it follows that there is some $p\in X$ achieving the maximum rank. The fact that $rk_p f \leq rk_{f(p)} f$ now implies that $q = f(p)$ also achieves maximal rank.

Set $Z = \{s\in f(X): rk_s f = rk_q f\}$. Now, $q\in Z$ so $Z$ is non-empty. Prove that $Z$ is both open and closed. Since $f(X)$ is connected, this will imply $Z = f(X)$, so $f$ has constant (maximum) rank on $f(X)$.

Sketch for "closed": The condition $rk_s f = rk_q f$ is defined by the vanishing of determinants (polynomials) of submatrices of $d_s f$, so is closed.

Sketch for "open": Having smaller-than-maximum rank is given by the vanishing of more determinants of submatrices, so having smaller-than-maximum rank is closed, so having maximum rank is an open condition.

Jason DeVito - on hiatus
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  • I thought that something like this would do it, but couldn't quite get there. As usual, by posting an answer, I've managed to learn something! – John Hughes Dec 02 '14 at 04:21