@TedShifrin what's the geometric intuition behind the constant rank theorem? Or, rather, what's the intuition behind a smooth map which has chart on which it's linear?

@arrow: The point is that any submersion looks (in appropriate coordinates) like a projection, any immersion looks like a linear inclusion, and any map of constant rank looks like a composition of the two. Basically, the inverse function theorem tells you that you if you have a map R^n\to R^n of maximal rank, you can change coords to make it the identity.

@TedShifrin that's the explanation I'm able to find online, but I still don't really have any intuition for the significance of having a chart on which a smooth map is linear

It's the case for any map of constant rank.

In fact, you can specify the linear map to be a $r\times r$ identity.

That's the content of the theorem, but what does it mean geometrically?

being of constant rank about a point just says the dimension of the image of the derivative is constant, but what about having a "linear chart"?

To me that's totally geometric.

I don't doubt that :)

What do you mean by geometric?

Choosing the right coordinates to understand a function is sort of the whole point of manifolds, I suppose.

When you read that a smooth map has a chart on which it is linear, what picture do you have in mind?

The point is that all reasonable maps have this property. Maps that are not of constant rank are quite difficult to understand. That's a whole field of geometry/topology/algebraic geometry called singularity theory.

Immersions give you (at least locally) submanifolds. Submersions give you (with some conditions, locally) fiber bundle structure.

These are very geometric ideas.

Yes I'm okay with immersions and submersions

I'm just worried that I don't have any intuition for what constant rank means

But constant rank can be written as a combination (composition) of those.

I understand, but such composites can be harder to imagine..

Maybe I'm looking too hard at the wrong thing

If $f\colon \Bbb R^n\to\Bbb R^m$ has rank $r$ on an open set, then basically I project $\Bbb R^n$ to an appropriate $\Bbb R^r$ and then include into $\Bbb R^m$.

Or you can rig it so that you are projecting onto an $\Bbb R^r$ inside $\Bbb R^m$.

Yes, that's exactly the content of the constant rank theorem. So when you read that a smooth map has a chart on which it's linear you don't have a particular picture in mind for how the restriction of the smooth map looks like between the manifolds?

I guess that's really the same thing, but I'm thinking of projecting onto the image of $df_x$ for some fixed $x$, by changing coordinates in $\Bbb R^m$.

I visualize it specifically as saying the image of the map is an $r$-dimensional submanifold.

Ah, okay!

Is that OK?

The problems come when rank drops somewhere ... :(

Rank can drop? Isn't it lower semicontinuous?

Again, this is all local (so topological things like for immersed vs. embedded submanifolds can bug you).

Of course rank can drop :) Start at the low point and then it can go up from there.

haha sorry about that.

Think about $f(x)=x^2$ or $x^3$ for the simplest.

Semicontinuity always confuses me :P

As far as immersed vs embedded I think the usual figure 8 vs a heart example is very helpful

BTW, there are lots more people to talk geometry/topology with in the usual chatroom.

May I bug you with one last thing for now?

Well, don't forget the dense "line" on the torus, too.

I feel uncomfortable asking my silly questions there

Don't be silly.

I was helping someone with basic 8th grade geometry yesterday.

Yes, but my condition is much more frowned upon because I learned abstract nonsense way before getting familiar with examples and diving into the basics

Well, true, I am not a fan of abstract crap.

I remember :) And so if a student came to you speaking of sheaves and cohomology but did not know some elementary examples, perhaps you'd be suspicious

No, not suspicious. Lots of people are educated that way, sadly.

Anyhow, your question is just unwinding some of the stuff we just said. Geometrically, it's using the representation of a manifold locally as a graph over a tangent space (if you think of a surface in $\Bbb R^3$, for example), pushing down to the tangent space (perhaps with a bit more change of coordinates), doing the linear map, and then using the fact that the image is an $r$-dimensional submanifold which is likewise a graph.

I think the best thing to do is work out examples.

Interestingly, in most of my career, I've used immersions and submersions, but rarely the rank theorem.

It shows up in some beautiful exercises, but one can always avoid it :P

Maybe I should give you one because it will help with your intuition here. The exercise is to prove that if you have a $C^1$ (or better) retraction $X\to Y\subset X$, then $Y$ is always a submanifold of $X$.

This is false for $C^0$.

You can always restrict charts and make the sets balls.

So it's really the same formulation except we make sure our charts are balls and then use diffeomorphisms to tangent spaces

Ooh, what a pretty exercise

It stumped me for days when I was a first-year grad student. And the proof I came up with didn't use the rank theorem. But it's cool for you to look at :)

Perhaps I'll solve it within a decade :D

Thank you very much for your time! I'm off to sleep!

If you know to use the Rank Theorem, it's not so bad :P

Night.