# Can the constant rank theorem for smooth manifolds be generalized to nonconstant rank?

The constant rank theorem says that if $f\colon M→N$ is a smooth map whose rank equals some fixed $k≥0$ at any point of $M$, then, locally with respect to $M$ and $N$, the map $f$ assumes the easiest possible form: $f(x_1,…,x_k,…,x_m)=(x_1,…,x_k,0,…,0)$.

I am interested in a more general situation: the rank of $f$ is at most $k$ at any point of $M$ and can actually drop below $k$ at some points. The classification of local forms for such maps seems to be too difficult. However, in my case one can replace the map f with any map homotopic to it via a homotopy $M×[0,1]→N$ that is itself a smooth map of rank at most $k$. Thus, one can inquire about the possible local forms of smooth maps of rank at most $k$ up to smooth homotopy of rank at most $k$. I am mostly interested in the case $k < \operatorname{dim} M$.

Such a problem could presumably be solved using the techniques of catastrophe theory, but my knowledge of this area is virtually nonexistent.

Is it possible to write down a finite list of local forms of smooth maps of rank at most $k$ considered up to a smooth homotopy of rank at most $k$?

This is motivated by the following question, which arises in connection to holonomy and parallel transport: given some $k<m$, can any smooth map $S^m→\mathbb{R}^n$ of rank at most $k$ be extended to a smooth map $D^{m+1}→\mathbb{R}^n$ of rank at most $k$?

• Have you looked at the book "Stable Mappings and their Singularities" by Golubitsky and Guillemin? If you are allowed to alter the map up to homotopy then you are proably asking about local forms of stable maps, which is exaclty the subject of singularity theory. Apr 29, 2017 at 17:40
• @MarkGrant: It seems to me that when k < dim M ≤ dim N, maps of rank k are never stable: a small perturbation (using a homotopy of rank at most k) will change such a map into a map with very different image. For instance, the constant rank map is clearly not stable: a small perturbation (given by a homotopy of rank at most 1) turns it into a nonconstant map. So it's not obvious to me how one could apply the theory in the Golubitsky-Guillemin book in this context. May 4, 2017 at 9:19
• Ah, I see, I missed the subtlety that the homotopy itself has rank at most $k$. May 4, 2017 at 9:30

Your motivating question has a negative answer: Consider the inclusion $\iota:S^m\to\mathbb{R}^{m+1}$, which has rank at most $m$. If there were a smooth extension $f:D^{m+1}\to\mathbb{R}^{m+1}$ of rank at most $m$ everywhere, then all the points of $D^{m+1}$ would be critical points for $f$ and hence, by Sard's Theorem, the set of critical values, i.e., the entire image $f(D^{m+1})$, would be a set of measure zero in $\mathbb{R}^{m+1}$. In particular, it would have to miss some point $p$ in the interior of $D^{m+1}$. However, since $p$ is in the interior, the inclusion $\iota:S^m\to\mathbb{R}^{m+1}\setminus\{p\}$ is not null homotopic, and hence there cannot be any extension $f:D^{m+1}\to \mathbb{R}^{m+1}\setminus\{p\}$.
If you want a map of rank less than $m$ (which is the dimension of $S^{m}$), just take the Hopf map $\pi:S^3\to S^2\subset\mathbb{R}^3$, which has rank $2<3$ everywhere. Then, by the same Sard's Theorem argument as above, coupled with the fact that $\pi$ generates $\pi_3(S^2)\simeq\mathbb{Z}$, it follows that $\pi$ cannot be extended to a smooth map $f:D^4\to\mathbb{R}^3$ of rank at most $2$ everywhere.