I am reading Introduction to differential topology, T.H Bröcker & K. Janich and I could not solve the following exercise:

Let $(E,\pi, X) $ be a vector bundle over a connected space $X$, let $f: E \rightarrow E$ be a vector bundle homomorphism and $f\circ f=f$. show that $f$ has constant rank.

I tried to find an open neighborhood $U$ s.t on this neighborhood $f$ has constant rank $k$. Then I would like to use connectedness of $X$ to extend this rank everywhere. But obviously this is not working since I have an counterexamples on the $\mathbb{R}\times\mathbb{R}^{2}$ and also I haven't used the fact that $f\circ f=f$.

If you can help me I will appreciate. Thanks in advance.

  • 5
    $\begingroup$ "constant rant" in title is pretty funny $\endgroup$
    – BrianO
    Oct 13 '15 at 20:05
  • $\begingroup$ upps !! Thank you for your remark :) $\endgroup$ Oct 14 '15 at 3:38

Let $n$ be the rank of $E$, let $I\subset M_n(\mathbb{R})$ be the set of idempotent matrices, and let $I_k\subset I$ be the set of idempotent matrices of rank $k$ for each $k$ such that $0\leq k\leq n$. By looking at what $f$ looks like in local trivializations of $E$, you can see that it suffices to prove that for each $k$, $I_k$ is clopen as a subset of $I$.

How do you prove that $I_k$ is clopen in $I$? There are several ways you could do this, but the following is probably the easiest. Every rank $k$ idempotent matrix has characteristic polynomial $x^{n-k}(x-1)^k$. The map sending a matrix to the coefficients of its characteristic polynomial is a continuous map $M_n(\mathbb{R})\to\mathbb{R}^n$. It follows that each $I_k$ is closed in $I$, and thus also open, since $I\setminus I_k=\bigcup_{j\neq k} I_j$ is a finite union of closed sets and hence also closed.

  • $\begingroup$ First of all thanks a lot for your answer. But I still have some points to ask: As far as I could see, you are defining a map $\phi$from $X$ to $\mathbb{R}^{n}$ given by $\phi(x,f_{x})$ equals to the coefficients of the characteristic polynomial. However continuity of this map is not clear for me since the map is not from $M_{n}(\mathbb{R})$ to $\mathbb{R}^{n}$. Even if I write this function using local trivializations then $f$ will looks like $f(x,v)=(x,A_{x}v)$ where $A_{x}\in I_{k}$ $\endgroup$ Oct 15 '15 at 3:35
  • $\begingroup$ In a local trivialization, you can identify $f$ with the function $x\mapsto A_x$, which is a map $X\to M_n(\mathbb{R})$ (which actually lands in $I\subset M_n(\mathbb{R})$). Composing it with the "coefficients of the characteristic polynomial" map gives you a map $X\to \mathbb{R}^n$. Note that a priori this map is only locally defined (on an open subset of $X$ where you have chosen a local trivialization), but it is actually globally defined because a different choice of local trivialization will conjugate $A_x$ by some invertible matrix, which does not change the characteristic polynomial. $\endgroup$ Oct 15 '15 at 3:47
  • $\begingroup$ Why does the map $X \to M_n(\mathbb{R})$ such that $ x \mapsto A_x $ needs to be continuous? I think we need to use continuity of this map to then use the fact that $X$ is connected. $\endgroup$
    – user392559
    Sep 30 '19 at 0:31
  • $\begingroup$ @user392559: That is trivial: identifying $E$ with $X\times\mathbb{R}^n$, then the $i$th column of $A_x$ is just the second coordinate of $f(x,e_i)$ where $e_i$ is the $i$th standard basis vector. $\endgroup$ Sep 30 '19 at 0:53

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.