# Constant rank mappings and submersions

Just to be clear: the objects we're talking about here are smooth ($$C^\infty$$) manifolds (without a boundary) and submersion is defined as a map between manifolds, which has constant rank that is equal to the dimension of the codomain.

While trying to do an exercise I kept stumbling upon the idea that "submersion is locally a projection and therefore an open map". This idea is not the problem. The "problem" is the theorem that states the following:

Let $$M$$ and $$N$$ be smooth manifolds, $$dim M=m$$, $$dim N=n$$, and let $$f:M\to N$$ be a smooth mapping of a constant rank $$r$$. For every $$p\in M$$ there is a (smooth) chart $$(U, \varphi)$$ at $$p$$ and chart $$(V, \psi)$$ at $$f(p)$$, such that $$f(U)\subset V$$ and such that $$f$$ has coordinate representation $$\psi \circ f \circ \varphi ^{-1} (x_1,\ldots ,x_r,x_{r+1}, \ldots ,x_m)=(x_1,\ldots ,x_r,0,\ldots ,0)$$

Doesn't this mean that every constant rank mapping, not only submersions, is locally a projection? It seems to me that I've completely misunderstood the idea of "locally being a projection", because that should be something that's very characteristic of submersions. Also, if every constant rank mapping is locally a projection, that would mean there are no constant rank mappings from compact manifolds to euclidean space.

So, my question is: what do people mean when they say "submersion is locally a projection"?

• It's a trivial difference. In one instance, "projection" refers to a map $(x_1, \ldots, x_m) \mapsto (x_1, \ldots, x_r)$, and in the second instance "projection" refers to a map $(x_1, \ldots, x_m) \mapsto (x_1, \ldots, x_r, 0, \ldots, 0)$. The difference isn't very interesting. In one case the projection is onto $\mathbb{R}^r$ and in the second, the projection is onto the subset $\mathbb{R}^r \subset \mathbb{R}^n$. – Jake Mirra Nov 5 '20 at 4:48
• @JakeMirra Oh, I see. But the mapping in the case $r\neq n$ is still open? Because it seems like it is and that would mean there are no constant rank mappings from compact manifolds to euclidean space. Is that even true? – blue Nov 5 '20 at 4:59
• No, if $r<n$, the mapping is no longer open. And of course there are constant rank mappings to $\Bbb R^n$ when the rank is less. See @Alekos's example. – Ted Shifrin Nov 5 '20 at 6:17

## 1 Answer

This local form $$(x_1,\ldots, x_r,x_{r+1},\ldots, x_m)\to (x_1,\ldots, x_r,0,\ldots,0)$$ is written slightly misleadingly. If we examine a few special cases, we can see that these maps are not all "projections" in the sense which you intend. If $$m\ge n=r$$, we have that the map is of the form $$(x_1,\ldots, x_m)\mapsto(x_1,\ldots, x_r)$$ and is indeed locally a bona fide projection. If $$m\ge n>r$$, then the map looks like $$(x_1,\ldots, x_m)\mapsto (x_1,\ldots,x_r)\mapsto (x_1,\ldots, x_r,0,\ldots,0)$$ and hence a composition of a projection and then an inclusion. In the case where $$r=m\le n$$, this map becomes $$(x_1,\ldots, x_m)\mapsto(x_1,\ldots, x_m,0,\ldots, 0)$$ which is an inclusion. If $$r, we get $$(x_1,\ldots, x_m)\mapsto (x_1,\ldots, x_m,0,\ldots,0)\mapsto (x_1,\ldots, x_r,0,\ldots, 0)$$ which is composition of an inclusion and a projection.

The moral of the story is that there are a few varied behaviors. For an example, take the inclusion of $$S^2\hookrightarrow \Bbb{R}^3$$. This is a constant rank $$2$$ map, so the theorem tells us that locally it looks like $$(x_1,x_2)\mapsto (x_1,x_2,0)$$. I.e. locally it is a standard inclusion. It is an example of an inclusion of a compact manifold into Euclidean space, and does not contradict the theorem.