
#1
Apr2806, 10:47 AM

P: 867

I thought I knew what a pullback map was until I read my notes again, and now I'm not so sure. So I have a few questions to ask.
Firstly, if I have two manifolds [itex]M[/itex] and [itex]N[/itex] with different coordinate systems and possibly different dimensions, then I can construct a map [tex]\phi\,:\,M\rightarrow N[/itex] and a function [tex]f\,:\,N\rightarrow\mathbb{R}[/tex] We can easily compose [itex]\phi[/itex] with [itex]f[/itex] to construct a new map which appears to pull the function back through N to be a function from M to N. The new map [tex]\phi_*\,:\,M \rightarrow\mathbb{R}[/tex] is called the pullback of [itex]f[/itex] by [itex]\phi[/itex]. Now suppose that we have another function [itex]g\,:\,M\rightarrow\mathbb{R}[/itex]. Can we create a function on N that consists of g and [itex]\phi[/itex]? The answer is no, and we need some help. My first question is: Is the pushforward map, [itex]\phi^*[/itex], a map between the tangent spaces of M and N? Whereas the pullback map is a map between the manifolds themselves. If the pushforward map is between the tangent spaces then we must only be able to "pushforward" a tangent vector at a point p. So we can't say what [itex]\phi^*(f)[/itex] is, instead we must say what [itex](\phi^*(V))(f)[/itex] is? Is this correct? So unlike pulling back functions we push forward vector fields and say that the action of pushing forward a vector field on a function is the action of the vector field on pulling back the function. This is kind of confusing And for my second question: Can you pullback a vector? Can you pullback a dualvector (one form)? Ooh, that is interesting... What about a mixed tensor? 



#2
Apr2806, 03:37 PM

Sci Advisor
HW Helper
P: 9,421

the confusion is based on the phrnomenon thata map may not be one to one, so several points of M may map to the same point of N.
hence you cannot push forward vector fields, only individual vectors. i.e. if i have a field of vectors on M, and I choose a point y on N, at which I would like to place a vector, I do not know which point x of M to choose from which to push a vector forward to N. if the map were an isomorphism, then there would always be exactly one point of M lying over my point of N, and I could look at that one point for a vector which I could then easily uch forward to N. a field of covectors is easily pulled back however from N to M. if a i have a field of covectors on N, and I choose a point x of M, then there is only one image point y of x in N. so I choose the covector lying at y and pull it back to x. 



#3
Apr2806, 04:10 PM

Emeritus
Sci Advisor
PF Gold
P: 16,101

Are you sure you have your notation right? If I have a map [itex]\phi : M \rightarrow N[/itex], then I'm used to seeing the notation [itex]\phi_* : TM \rightarrow TN[/itex] for the map on the tangent spaces in the forward direction, and [itex]\phi^* : T^*N \rightarrow T^*M[/itex] for the map on the cotangent spaces in the backward direction.
And in general, I see superscript stars for things that go backward, and subscript stars for things that go forwards. I would have expected to see the pullback of your map [itex]f : N \rightarrow \mathbf{R}[/itex] written as [itex]f^* : M \rightarrow \mathbf{R}[/itex]. Or maybe all that category theory has addled my brain. To be even more lowbrow, think in terms of matrices. What sort of action can an mxn matrix have on column vectors of size m? Column vectors of size n? Row vectors of size m? Row vectors of size n? 



#4
Apr2806, 08:35 PM

P: 867

Some geometry questions 



#5
Apr2806, 08:44 PM

P: 867

[tex]\phi_*\,:\,T_pM \rightarrow T_{\phi(p)}N[/tex] But, isn't [itex]\phi_*[/itex] just the composition of two maps: [itex]\phi\,:\,M\rightarrow N[/itex] and [itex]f\,:\,N\rightarrow \mathbb{R}[/itex]. So perhaps we (or I am) are getting confused with the naturally induced differentiable map and the composition map [itex]\phi_* f = f\circ\phi[/itex]  because even though I neglected to put the f in my equation in my first post, I am talking about [itex]\phi_* f = f\circ\phi[/itex] in this thread and not the differential map. 



#6
Apr2806, 09:35 PM

Mentor
P: 6,037

This is true by definition, i.e., by the definition of "function". Suppose for (the same) x, f(x) = y1 and f(x) = y2. Since f is a function, you can say what about y1 and y2? Regards, George 



#7
Apr2906, 01:09 AM

P: 867

If f is a function and f(x) = y_1 and f(x) = y_2 then y_1 = y_2 because f can only take on one value of y for each value of x.
I see, so it is quite straightforward. 



#8
Apr2906, 11:54 AM

Sci Advisor
HW Helper
P: 9,421

hurkyl's advice to forget the manifold is good to take the first step, of seeing what individual objects can be pushed or pulled.
but it makes the second step invisible, of seeing which fields of objects, as opposed to individual objects, can be pushed or pulled. Manifolds, or at least open sets of points, are crucial to seeing why you cannot push forward vector fields, as observed above. 


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