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


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.

  • $\begingroup$ 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! $\endgroup$ – John Hughes Dec 2 '14 at 4:21

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 $A$.

  • $\begingroup$ Those are two hints! $\endgroup$ – Mariano Suárez-Álvarez Nov 29 '14 at 15:36
  • $\begingroup$ Yeah...I started writing three, and decided the third one was bad. I've now changed "Three" to "two". :) $\endgroup$ – John Hughes Nov 29 '14 at 15:37
  • $\begingroup$ How did you arrive at that first hint? Also note that $f$ must be differentiable yet not necessarily linear. $\endgroup$ – D. Vente Nov 29 '14 at 15:54
  • $\begingroup$ 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. $\endgroup$ – John Hughes Nov 29 '14 at 17:16
  • $\begingroup$ The exercise explicitly stated that the connectedness of $X$ was required for the second part, I don't see how that is used here $\endgroup$ – D. Vente Nov 29 '14 at 18:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.