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Need help understanding the pushforwardby Identity
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#1
Feb512, 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]? 


#2
Feb512, 12:26 PM

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P: 2,020

X(f) isn't in T_{p}M  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 T_{F(p)}N, i.e., it should be a point derivation at F(p) on N, i.e., you need to know how to evaluate F_{p}(X) at smooth germ f at F(p). 


#3
Feb512, 12:30 PM

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PF Gold
P: 4,771

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_{*p}X at F(p) (derivation on the germs of smooth function at F(p)). So what is this vector F_{*p}X then? How does it act on a germ f at F(p)? This is what the formula is telling you: it says (F_{*p}X)(f) is just X(F*f). And this makes sense, since F*f =f o F is indeed a germ of functions at p.



#4
Feb512, 12:44 PM

P: 150

Need help understanding the pushforward
Sorry, I've never heard of the term 'germ' before, can you explain please?



#5
Feb512, 01:04 PM

P: 57




#6
Feb512, 02:03 PM

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#7
Feb512, 03:41 PM

P: 150

Thanks everyone :)



#8
Feb512, 05:22 PM

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PF Gold
P: 4,771

If so, then the elements of [itex]C^{\infty}(p)[/itex] are called germs of smooth functions. 


#9
Feb612, 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. 


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