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!