Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Everyone:

Given smooth manifolds $M,N$ ($m$- and $n$- manifolds respectively) Sard's theorem says that for $f:M \to N$ in $C^k$ ; $k \geq 1$, the image of the set of critical points of $f$ in $M$ --points in $M$ where the Jacobian has rank $< m$ --has measure zero in N:

http://en.wikipedia.org/wiki/Sard%27s_theorem

Now, I know what it means for a set to have measure zero in the case of $M, N$ being Euclidean spaces, but I am not clear on how we define measures on manifolds; I know we may pull back different types of objects from $\mathbb R^n$ into a manifold, like $k$-forms, etc., but I don't know if/how one pullsback a measure from $\mathbb R^n$, since I am not even clear on what kind of object a measure is. Is it just a $0$-form, i.e., just a function? If so, can we define a measure globally on a manifold, or do we just pullback separately for each chart?

share|improve this question
3  
Well, to define measure zero you don't really need a measure. You can just say that a set has measure zero if it has measure zero in all charts. This answer of Willie Wong seems to address your more specific question. –  t.b. Nov 5 '11 at 20:35
    
Thanks, but I am not even sure of how to define a measure on a single given chart, because I don't even know how a measure on a manifold nor on a given chart are defined. –  J.K Nov 5 '11 at 20:37
    
In the case of an oriented Riemannian manifold, we usually integrate against a volume form, which is a global generator for the top exterior power of the cotangent bundle. But we may define measures in more general settings. Have you studied any measure theory? –  Zhen Lin Nov 5 '11 at 20:37
1  
@J.K: You said you know how to define measure zero on subsets of $\mathbb{R}^n$. Now call a set in $M$ "of measure zero" if its image in every chart has measure zero in $\mathbb{R}^n$, that's not entirely satisfactory but sufficient for most purposes. Please have a look at Willie's answer I linked to but I'm sure somebody will provide an answer to your actual question. –  t.b. Nov 5 '11 at 20:41
    
@t.b: thanks; I will read Willie's answer, but, just for now, let me follow up with Zhen Lin: yes I know some measure theory; generalized measure spaces and sigma algebras. I guess I would/could then define a sigma algebra on my manifold, maybe the sigma-algebra generated by the open sets in the manifold (with its standard metric topology given by the interinsic metric). But then what? Do I use some sort of pullback metric by the chart map? –  J.K Nov 5 '11 at 21:01

2 Answers 2

Having measure zero is something that can be defined without actually defining a measure on the manifold. A subset $A$ of $N$ can be defined to have measure zero if for every chart $\phi:U\to\mathbb R^n$ of $N$, $\phi(U\cap A)$ has measure zero in $\mathbb R^n$.

You would want this to be something that is invariant under diffeomorphism. That is the case because smooth maps between subsets of $\mathbb R^n$ preserve the property of having Lebesgue measure $0$.

share|improve this answer

Select a point p belongs to M. U be the neighborhood of d. And a open set U in M. In M there are many points other than p ,say pi' points . and defined sigma algebra on U, which contains all open sets say Ui's and defined measure on that sigma algebra to 0 to infinity.those points are punctured point of m there measure is zero.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.