Need help understanding the pushforward

[Identity]

In my notes, the following two functions are defined:

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

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

However, I don't understand the second one, F_{*p}. Since X(f) \in T_pM, it follows that f \in C^\infty (p). But then how is
[F_{*p}(X)](f) = X(F^*f)
defined? After all, in the definition of F_{*p}(X), f is a function on C^\infty (p), not C^\infty(F(p)), so how can we evaluate F^*f?

 

[morphism]

X(f) isn't in T_pM -- X is.

F_\ast takes X to F_\ast(X). The question now is, what is F_\ast(X)? 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).

 

[quasar987]

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]

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

 

[joebohr]

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

A germ is essentially just a local topological structure.

 

[lavinia]

In my notes, the following two functions are defined:

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

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

However, I don't understand the second one, F_{*p}. Since X(f) \in T_pM, it follows that f \in C^\infty (p). But then how is
[F_{*p}(X)](f) = X(F^*f)
defined? After all, in the definition of F_{*p}(X), f is a function on C^\infty (p), not C^\infty(F(p)), so how can we evaluate F^*f?

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]

Thanks everyone :)

 

[quasar987]

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

I assumed that by C^{\infty}(p) 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 C^{\infty}(p) are called germs of smooth functions.

 

[Identity]

Just to make sure I've got it, in

[F_{*p}(X)](f) = X(f\circ F)
f is kind of placeholder, in the sense that the f on the LHS is an arbitrary function in C^\infty(p) and the f on the RHS is an arbitrary function in C^\infty (F(p))So on the left and right sides of the equation, f does not represent functions with the same germs. I think this is where I got confused.

 

[quasar987]

Just to make sure I've got it, in

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

The f in the LHS is the same as the f in the RHS, and in both case, it is a function in C^\infty (F(p)). Indeed, it better be so that f o F is in C^\infty(p) so that X(f o F) makes sense!