Topological dimension of the image of a smooth curve in a manifold

[Rick_D]

Here is the situation I am concerned with -

Consider a smooth curve $g:[0,1] \to M$ where $M$ is a topological manifold (I'd be happy to assume $M$ smooth/finite dimensional if that helps). Let $Im(g)$ be the image of $[0,1]$ under the map $g$. Give $Im(g)$ the subspace topology induced by $M$.


The question is this --- as a topological space, does $Im(g)$ always have (topological) dimension $\leq 1$ ?

Note that it IS important not to restrict $g$ to be either injective or an immersion - then the result is straightforward. The tricky thing is that $g$ may not be constant rank and also may not be a submanifold of $M$ (or even a manifold at all).

Also, a reference to a proof is fine, I don't really need to know HOW to prove it, I just need to be certain that it is true. It certainly seems intuitively obvious...

I've seen a number of statements (without reference or proof) that $Im(g) [/[STRIKE][/STRIKE]itex] must have zero Lebesque measure. I don't immediately see how this would answer the above question, so any references or proofs on this front would be useful as well.<br /> <br /> Thanks!$

 

[quasar987]

There are the so-called space-filling curves that are continuous surjective maps [0,1]-->[0,1]^n for any n. In particular, making the n-cube into a torus, you get a continuous curve [0,1]-->T^n whose image has dimension n since it is the whole torus.

Note that such a curve cannot be injective. Otherwise, it would be a bijection from a compact space into a Hausdorff space, hence a homeomorphism, which would contradict invariance of dimension.

No differentiable space-filling curve can exist, but there are ones that are smooth almost everywhere.

These things are discussed in the Dover book "Counter-examples in topology"

 

[lavinia]

Here is the situation I am concerned with -

Consider a smooth curve $g:[0,1] \to M$ where $M$ is a topological manifold (I'd be happy to assume $M$ smooth/finite dimensional if that helps). Let $Im(g)$ be the image of $[0,1]$ under the map $g$. Give $Im(g)$ the subspace topology induced by $M$.


The question is this --- as a topological space, does $Im(g)$ always have (topological) dimension $\leq 1$ ?

Note that it IS important not to restrict $g$ to be either injective or an immersion - then the result is straightforward. The tricky thing is that $g$ may not be constant rank and also may not be a submanifold of $M$ (or even a manifold at all).

Also, a reference to a proof is fine, I don't really need to know HOW to prove it, I just need to be certain that it is true. It certainly seems intuitively obvious...

I've seen a number of statements (without reference or proof) that $Im(g) [/[STRIKE][/STRIKE]itex] must have zero Lebesque measure. I don't immediately see how this would answer the above question, so any references or proofs on this front would be useful as well.<br /> <br /> Thanks!$

$<br /> I think this is your answer.<br /> <br /> For smooth curves around any point where the derivative is not zero there is a neighborhood of the point mapped diffeomorphically onto its image. this i think is the Inverse Function Theorem so the image around such a point is a 1 dimensional manifold.<br /> <br /> Near a point where the derivative is zero this is still true if the point is isolated - not hard to prove. If there is an interval where the derivative is zero then this interval is mapped to a single point. So locally the image looks like a one dimensional manifold or a point.<br /> <br /> <br /> Since your interval is compact there can be no points of accumululation that are not in the image. This in any neighborhood of a point where the derivative is not zero the image is an embedded 1 dimensional manifold.$

 

[Bacle2]

Sard's theorem may help for when the curve is embedded in R^m :

http://en.wikipedia.org/wiki/Sard's_theorem

 

[lavinia]

Sard's theorem may help for when the curve is embedded in R^m :

http://en.wikipedia.org/wiki/Sard's_theorem

I think the Inverse function theorem works because one of the projections onto one of the coordinate axes must have non-zero derivative.

 

[Bacle2]

I think the Inverse function theorem works because one of the projections onto one of the coordinate axes must have non-zero derivative.

Do you mean when M is embedded in R^n ? I don't know if the OP is assuming this.

And, my bad: Sard's theorem tells us something about the measure of the image,(since, in the 1xm Jacobian, every point in I=[0,1] is necessarily a critical point) but there is no connection in this respect between measure and topological dimension; e.g., in R^n, any (measurable)subset of dimension less than n will have n-dimensional Lebesgue measure 0, so that does not narrow it down much, definitely not enough.

 

[lavinia]

Do you mean when M is embedded in R^n ? I don't know if the OP is assuming this.

And, my bad: Sard's theorem tells us something about the measure of the image,(since, in the 1xm Jacobian, every point in I=[0,1] is necessarily a critical point) but there is no connection in this respect between measure and topological dimension; e.g., in R^n, any (measurable)subset of dimension less than n will have n-dimensional Lebesgue measure 0, so that does not narrow it down much, definitely not enough.

I mean that id the derivative is not zero at some point then the derivative of the composition of the curve with one of the projections must also be non-zero. Smoothness then allows the Inverse Function theorem to guarantee a local diffeomorphism. This seems OK.

 

[Bacle2]

A small variant of the problem: if we're looking to preserve Hausdorff dimension--I assume by topological dimension the OP meant Lebesgue covering dimension-- then we can use that continuously-differentiable maps on a compact metric space, i.e., [0,1] , are Lipschitz, so that there is a bound to the scaling of the diameter d of the sets in the cover, and it follows that the Hausdorff dimension is preserved.

 

[Bacle2]

Luckily, a topologist/geometer friend of mine helped me to this result:

http://books.google.com/books?id=Lt...ge&q=Szpilrajn and lebesgue dimension&f=false

In which Szpilrajn (sp?) showed that inductive (topological) dimension is less than the Haudorff dimension.

So, for an alternative argument: we start with I=[0,1]. Then we use the fact that

a C^oo (C^1 is enough, AFAIK) map is Lipschitz, so that f preserves the Hausdorff

dimension , and then use Szpilrajn's inequality. It then follows, if I did not miss any thing,

that the image curve has topological dimension <=1.

 

[Rick_D]

Thanks!

That's definitely what I was looking for - the books I have on dimension theory only talk about the inductive and covering dimensions, and I just found out about the Hausdorff dimension. This definitely does the trick - thanks again.