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Need help understanding the pushforward

by Identity
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Identity
#1
Feb5-12, 12:08 PM
P: 150
In my notes, the following two functions are defined:

Suppose [itex]M^m[/itex] and [itex]N^n[/itex] are smooth manifolds, [itex]F:M \to N[/itex] is smooth and [itex]p \in M[/itex]. We define:
[tex]F^*:C^\infty (F(p)) \to C^\infty (p)\ ,\ F^*(f) = f \circ F[/tex]
[tex]F_{*p}: T_pM \to T_{F(p)}N\ ,\ [F_{*p}(X)](f) = X(F^*f) = X(f \circ F)[/tex]

I understand the first function, [itex]F^*[/itex]; it maps [itex]f[/itex], a function on [itex]C^\infty(F(p))[/itex], to [itex]f \circ F[/itex], a function on [itex]C^\infty(p)[/itex].

However, I don't understand the second one, [itex]F_{*p}[/itex]. Since [itex]X(f) \in T_pM[/itex], it follows that [itex]f \in C^\infty (p)[/itex]. But then how is
[tex][F_{*p}(X)](f) = X(F^*f)[/tex]
defined? After all, in the definition of [itex]F_{*p}(X)[/itex], [itex]f[/itex] is a function on [itex] C^\infty (p)[/itex], not [itex]C^\infty(F(p))[/itex], so how can we evaluate [itex]F^*f[/itex]?
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morphism
#2
Feb5-12, 12:26 PM
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X(f) isn't in TpM -- X is.

[itex]F_\ast[/itex] takes X to [itex]F_\ast(X)[/itex]. The question now is, what is [itex]F_\ast(X)[/itex]? We want it to be an element of TF(p)N, i.e., it should be a point derivation at F(p) on N, i.e., you need to know how to evaluate Fp(X) at smooth germ f at F(p).
quasar987
#3
Feb5-12, 12:30 PM
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So F*p takes a vector X at p (derivation on the germs of smooth function at p) and sends it to a vector F*pX at F(p) (derivation on the germs of smooth function at F(p)). So what is this vector F*pX then? How does it act on a germ f at F(p)? This is what the formula is telling you: it says (F*pX)(f) is just X(F*f). And this makes sense, since F*f =f o F is indeed a germ of functions at p.

Identity
#4
Feb5-12, 12:44 PM
P: 150
Need help understanding the pushforward

Sorry, I've never heard of the term 'germ' before, can you explain please?
joebohr
#5
Feb5-12, 01:04 PM
P: 57
Quote Quote by Identity View Post
Sorry, I've never heard of the term 'germ' before, can you explain please?
A germ is essentially just a local topological structure.
lavinia
#6
Feb5-12, 02:03 PM
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Quote Quote by Identity View Post
In my notes, the following two functions are defined:

Suppose [itex]M^m[/itex] and [itex]N^n[/itex] are smooth manifolds, [itex]F:M \to N[/itex] is smooth and [itex]p \in M[/itex]. We define:
[tex]F^*:C^\infty (F(p)) \to C^\infty (p)\ ,\ F^*(f) = f \circ F[/tex]
[tex]F_{*p}: T_pM \to T_{F(p)}N\ ,\ [F_{*p}(X)](f) = X(F^*f) = X(f \circ F)[/tex]

I understand the first function, [itex]F^*[/itex]; it maps [itex]f[/itex], a function on [itex]C^\infty(F(p))[/itex], to [itex]f \circ F[/itex], a function on [itex]C^\infty(p)[/itex].

However, I don't understand the second one, [itex]F_{*p}[/itex]. Since [itex]X(f) \in T_pM[/itex], it follows that [itex]f \in C^\infty (p)[/itex]. But then how is
[tex][F_{*p}(X)](f) = X(F^*f)[/tex]
defined? After all, in the definition of [itex]F_{*p}(X)[/itex], [itex]f[/itex] is a function on [itex] C^\infty (p)[/itex], not [itex]C^\infty(F(p))[/itex], so how can we evaluate [itex]F^*f[/itex]?
Analytically, a tangent vector at a point,p, on a manifold is a linear operator that acts on differentiable functions defined in an open neighborhood of p. A function,g, on N composed with F is a function on M. So a tangent vector at p now acts on the composition of g with F. But this may also be viewed at an action on g at F(p).
Identity
#7
Feb5-12, 03:41 PM
P: 150
Thanks everyone :)
quasar987
#8
Feb5-12, 05:22 PM
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Quote Quote by Identity View Post
Sorry, I've never heard of the term 'germ' before, can you explain please?
I assumed that by [itex]C^{\infty}(p)[/itex] you mean the set of real-valued functions f that are defined and smooth on some neighborhood U of p, modulo the equivalence relations according to which f~g iff f and g coincide on some small nbhd of p.

If so, then the elements of [itex]C^{\infty}(p)[/itex] are called germs of smooth functions.
Identity
#9
Feb6-12, 12:58 PM
P: 150
Just to make sure I've got it, in

[tex][F_{*p}(X)](f) = X(f\circ F)[/tex]
[itex]f[/itex] is kind of placeholder, in the sense that the [itex]f[/itex] on the LHS is an arbitrary function in [itex]C^\infty(p)[/itex] and the [itex]f[/itex] on the RHS is an arbitrary function in [itex]C^\infty (F(p))[/itex]


So on the left and right sides of the equation, [itex]f[/itex] does not represent functions with the same germs. I think this is where I got confused.
quasar987
#10
Feb6-12, 02:57 PM
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Quote Quote by Identity View Post
Just to make sure I've got it, in

[tex][F_{*p}(X)](f) = X(f\circ F)[/tex]
[itex]f[/itex] is kind of placeholder, in the sense that the [itex]f[/itex] on the LHS is an arbitrary function in [itex]C^\infty(p)[/itex] and the [itex]f[/itex] on the RHS is an arbitrary function in [itex]C^\infty (F(p))[/itex]
The f in the LHS is the same as the f in the RHS, and in both case, it is a function in [itex]C^\infty (F(p))[/itex]. Indeed, it better be so that f o F is in [itex]C^\infty(p)[/itex] so that X(f o F) makes sense!


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