# Need help understanding the pushforward

by Identity
Tags: pushforward
 P: 150 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 $F^*f$?
 Sci Advisor HW Helper P: 2,020 X(f) isn't in TpM -- 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 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).
 Sci Advisor HW Helper PF Gold P: 4,768 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.
P: 150

## Need help understanding the pushforward

Sorry, I've never heard of the term 'germ' before, can you explain please?
P: 57
 Quote by Identity Sorry, I've never heard of the term 'germ' before, can you explain please?
A germ is essentially just a local topological structure.
Sci Advisor
P: 1,716
 Quote by 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 $F^*f$?
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).
 P: 150 Thanks everyone :)
Sci Advisor
HW Helper
PF Gold
P: 4,768
 Quote by Identity 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.
 P: 150 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.
Sci Advisor
HW Helper
PF Gold
P: 4,768
 Quote by 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 $C^\infty (F(p))$
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!

 Related Discussions Introductory Physics Homework 3 Differential Geometry 3 Differential Geometry 3 General Math 4 Introductory Physics Homework 2