# idempotent bundle homomorphism has constant rank

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.

• "constant rant" in title is pretty funny Oct 13 '15 at 20:05
• upps !! Thank you for your remark :) 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.
• 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 $A_{x}\in I_{k}$ Oct 15 '15 at 3:35
• 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. Oct 15 '15 at 3:47
• 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. Sep 30 '19 at 0:31
• @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. Sep 30 '19 at 0:53