Register to reply 
Tangent vectors as derivations 
Share this thread: 
#1
May909, 10:39 AM

P: 43

I've encountered a definition of the tangent vector via the notion of derivations on the manifold and I have some problems with it. I would actually like to show that every derivation can be expressed as a directional derivative, but I'm not ver successful in doing so.
I have this definition of the derivation: A derivation X in the point p on the manifold M is a linear functional on [tex]C^{\infty}(M)[/tex] such that the Jacobi rule holds: [tex]X(fg) = X(f)g(p)+f(p)X(g)[/tex], where [tex]f,g \in C^{\infty}(M)[/tex]. (The tangent space in p is then the set of all derivations in p.) Now we know that each directional derivative, i.e. each operator [tex]v_i\frac{\partial }{\partial x_i}[/tex] (in some local coordinates [tex]x_i[/tex]) is a derivation. Thus there is a map [tex]h: DD \rightarrow D [/tex] between the set of directional derivatives (DD) and the set of the derivations (D) and it is easy to see that this map is linear. Now I'd like to show that this map is bijective, i.e. that every derivation can be uniquely expressed by a directional derivative. I'm able to show the injectivity of this map, by showing that the kernel of h is {0}. To do this, I take any directional derivative [tex]V=v_i\frac{\partial }{\partial x_i}[/tex] (in some coordinate chart U, with local coordinates [tex]x_i[/tex]) and plug in a function [tex]f_j=\chi x_j[/tex], where [tex]\chi[/tex] is a smooth bump function around the point p with support in the chart U. This gives us that all [tex]v_j=0[/tex] and thus V=0. But how do I show surjectivity? Note that the the functions are defined globally on the whole manifold, not locally (i.e. not via germs). How can I do this? 


#2
May909, 10:54 AM

Emeritus
Sci Advisor
PF Gold
P: 16,099




#3
May909, 12:25 PM

P: 43

Can you be more specific?



#4
May909, 01:01 PM

Emeritus
Sci Advisor
PF Gold
P: 16,099

Tangent vectors as derivations
I'm not sure if I can without starting to do the problem for you!
You said that your problem is that things are defined globally, so you circumvent that by working locally: solve the problem on a coordinate chart rather than the whole manifold. Now, typically one of three things happen: (1) It's easy to assemble solutions on coordinate charts into a solution on the whole manifold. (2) The fact that you can assemble solutions on coordinate charts into a solution on the whole manifold is a deep theorem. (3) There is a homology group (or similar object) that tells you that cannot produce a solution on the whole manifold unless certain conditions are met. 


#5
May909, 04:40 PM

P: 43

I'm not really sure what you mean by "solution".
I can prove the surjectivity in the R^n case. If I defined the derivation to be acting on the germs of [tex]C^{\infty}(M)[/tex] functions, I'd be done. But since I defined the derivation to be defined on the whole [tex]C^{\infty}(M)[/tex], it's not clear to me, if this property will extend to the whole manifold (i.e. if it will still hold true if I take functions defined globally). (also, the manifold isn't assumed to be paracompact) 


#6
May909, 06:55 PM

Emeritus
Sci Advisor
PF Gold
P: 16,099

Whoops, my bad, localglobal isn't the general method you want to use, it's excision  you want to prove that what happens away from P doesn't matter. Can't you use your bump function to show the values of your functions outside of your chart are irrelevant to the derivation?



#7
May909, 07:37 PM

P: 491

You are acting on germs of [tex]C^{\infty}(M) [/tex] functions! Think of it this way: if two functions agree on a neighborhood of a point p, all derivatives (including the first one) of their difference vanish at p.



#8
May909, 09:26 PM

Emeritus
Sci Advisor
PF Gold
P: 16,099




#9
May909, 10:33 PM

P: 491

I thought he figured out how to prove it otherwise. Never mind, here's a shot at it.
Fix a coordinate chart at p, and assume by restriction if necessary that the coordinate chart is a ball in R^n centered at 0. For x near p, we have [tex] f(x)f(0) = \int_0^1 \frac{d}{dt} f(tx) \text{ dt} = \int_0^1 \sum \frac{\partial f}{\partial x_i} (tx)(x_i) \text{ dt} [/tex] So if [tex]f \in C^\infty(p), \text{ } f(x) = f(0) + \sum x_i g_i(x), \text{ where } g_i(0) = \frac{\partial f}{\partial x_i} (0)[/tex]. Since constants annihilate derivations, we have [tex] D f(x) = D(f(0)) + \sum D[x_i g_i(x)] = 0 + \sum D(x_i) g_i(0) + \sum (0)D(g_i(x)) = \sum g_i(0)D(x_i) [/tex] But that's a directional derivative, so we're done. 


#10
May1009, 03:43 AM

P: 43

Zhentil: My definition of derivation is not via germs of functions. The derivation acts on functions which are defined globally. Thus the proof for a coordinate chart is not sufficient. 


#11
May1009, 05:12 AM

Emeritus
Sci Advisor
PF Gold
P: 9,225

The set of derivations at p has an obvious vector space structure, and the partial derivative operators at p clearly span a subspace of that vector space. The dimension of the subspace is the same as the dimension of the manifold. So if you can prove that the dimension of the tangent space at p (i.e. the vector space of derivations at p) isn't greater than the dimension of the manifold, you're done.



#12
May1009, 08:32 AM

Emeritus
Sci Advisor
PF Gold
P: 16,099

Or... (just checking) is the problem that you haven't realized you can take a bump function defined on your chart and turn it into a global function by defining it to be zero everywhere outside of your chart? 


Register to reply 
Related Discussions  
Tangent vectors to a manifold  Differential Geometry  32  
Why are tangent vectors like this?  Differential Geometry  4  
Tangent spaces and derivations  Differential Geometry  14  
Tangent vectors  Differential Geometry  21  
Tangent vectors  General Math  1 