How to calculate the composition of pullbacks and pushforwards?

[Eclair_de_XII]

Homework Statement

"If $f:\mathbb{R}^n \rightarrow \mathbb{R}^m$ and $g:\mathbb{R}^m\rightarrow \mathbb{R}^p$, show that $(g\circ f)_*=g_*\circ f_*$ and $(g\circ f)^* = f^* \circ g^*$."

Homework Equations

Pushforward: $f_*(v_a)=(Df(a)(v))_{f(a)}$
Pullback: $f^*(\omega)(a)(v_a)=\omega(f(a))(f_*(v_a))=\omega(f(a))(Df(a)(v))_{f(a)}$

The Attempt at a Solution

This is my attempt for the first part:

$(g\circ f)_*(v_a)=(D(g\circ f)(a)(v))_{(g\circ f)(a)}=(Dg(f(a))Df(a)(v))_{g(f(a))}$

I don't know how to proceed from here, and I'm pretty sure that the subscript on the last term in my expression is wrong.

 

[fresh_42]

Homework Statement

"If $f:\mathbb{R}^n \rightarrow \mathbb{R}^m$ and $g:\mathbb{R}^m\rightarrow \mathbb{R}^p$, show that $(g\circ f)_*=g_*\circ f_*$ and $(g\circ f)^* = f^* \circ g^*$."

Homework Equations

Pushforward: $f_*(v_a)=(Df(a)(v))_{f(a)}$
Pullback: $f^*(\omega)(a)(v_a)=\omega(f(a))(f_*(v_a))=\omega(f(a))(Df(a)(v))_{f(a)}$

The Attempt at a Solution

This is my attempt for the first part:

$(g\circ f)_*(v_a)=(D(g\circ f)(a)(v))_{(g\circ f)(a)}=(Dg(f(a))Df(a)(v))_{g(f(a))}$

I don't know how to proceed from here, and I'm pretty sure that the subscript on the last term in my expression is wrong.

You're done with the pushforwards.
$(g\circ f)_*(v_a)=(D(g\circ f)(a)(v_a))_{(g\circ f)(a)}=\underbrace{(Dg(f(a))_{g(f(a))}}_{g_*}) ( \underbrace{Df(a))_{f(a)}}_{f_*} )(v_a)$

 

[Eclair_de_XII]

Okay, so I have for pullback:

$(f\circ g)^*(\omega)(a)(v_a)=\omega((f\circ g)(a))((f\circ g)_*(v_a))=\omega(f(g(a)))(Dg(f(a))_{g(f(a))})(Df(a))_{f(a)}$

Sorry, I don't know how to proceed for this one, as well.

 

[fresh_42]

Let me first add a remark on your first solution. You have a bit too many, i.e. unnecessary notations of the point of evaluation. Usually students don't realize, that they actually talk about $f\,'(x)_{x=c}$ when they write $f\,'(x)$. Your notation not only avoids this mistake, you doubled the avoidance.

$D$ is the differential operator. It applies to functions $f$. The result is a tangent (vector space) bundle, which when evaluated at a certain point, turns into a specific tangent (vector) space, which we denote by $Df(a)=D_af$. To write $D_a(f)(a)$ is a double which might confuse readers. Now since our functions are multidimensional, we have tangents in various directions $v$. A tangent in direction of $v$ at a point $a$ is therefore $D_af(v)=D_af \cdot v$ since our approximation by tangents is a linear operator, a (matrix) multiplication by $v$. In sum we have:

• $Df =D(f)$ = tangent (vector space) bundle: all tangent spaces at all possible locations
• $D_af=D_a(f)$ = tangent space at a specific point of evaluation $a$, a certain tangent space from the bundle; vector space because or tangents have more than one possible direction, as our function is multidimensional, and so are its tangents
• $D_af(v)=D_a(f)(v)=D_af \cdot v$ = tangent from our tangent (vector) space at $a$ in direction $v$

Thus your notation $f_*(v_a)=(Df(a)(v))_{f(a)}$ is a bit overloaded. Better is $f_*(a)(v)=(D_af)(v)=D_af \cdot v$ or even shorter $f_*(a)=D_af\,.$ To write $f(a)$ in the index is already confusing. E.g. for $f\, : \,x \longmapsto x^2$ we write $f\,'(a)=\left. \dfrac{d}{dx}\right|_{x=a}f$ and not $f\,'(a)=\left. \dfrac{d}{dx}\right|_{a^2}f\,.$

If you find time, you might want to read this little series: https://www.physicsforums.com/insights/the-pantheon-of-derivatives-i/
(Vector fields start in part 2, pullbacks in part 3)

 

[fresh_42]

Okay, so I have for pullback:

$(f\circ g)^*(\omega)(a)(v_a)=\omega((f\circ g)(a))((f\circ g)_*(v_a))=\omega(f(g(a)))(Dg(f(a))_{g(f(a))})(Df(a))_{f(a)}$

Sorry, I don't know how to proceed for this one, as well.

First let's drop all variables which are not needed. We have to show that $(f\circ g)^*(\omega) = g^*(f^*(\omega))$. I suggest to start with the expression on the right, and if you like, you can appoint evaluation points and directions afterwards. You should first draw a diagram to see which function maps from where to where, in order to avoid mistake.

 

[Eclair_de_XII]

Okay, I'm a bit confused on what $\omega$ is supposed to be in this context. From what I've learned, it's a covector field that assigns to each $a\in T_a\mathbb{R}^n$ a real number. But it can't be right, since $f^*$ is a function from $\mathbb{R}^m\rightarrow \mathbb{R}^n$.

 

[fresh_42]

Okay, I'm a bit confused on what $\omega$ is supposed to be in this context. From what I've learned, it's a covector field that assigns to each $a\in T_a\mathbb{R}^n$ a real number. But it can't be right, since $f^*$ is a function from $\mathbb{R}^m\rightarrow \mathbb{R}^n$.

If you have your manifolds to be $\mathbb{R}^n$, then it's difficult to see what is manifold and what is a tangent vector or covector.

For more clarity, let's say we have a smooth function $f\, : \,M \longrightarrow N$. Now $\omega_q$ assigns a real number to every tangent at $q\in N$, i.e. $\omega_q\, : \,T_qN \longrightarrow \mathbb{R}$ and $\omega_q$ is an element of the dual vector space of $T_qN$. The differential form $\omega$ now is the function $\omega\, : \,N \longrightarrow (T_{(.)}N)^*$, i.e. point to cotangent.

This means $\omega_b(v_b)\in \mathbb{R}$. We have two functions here. Firstly we have $\omega$ which assigns a cotangent space to a point of $b \in N\, : \,\omega(b)=\omega_b \in (T_bN)^*$. Secondly, this image $\omega_b$ is itself a function, since cotangent means the dual space of linear maps $T_bN \longrightarrow \mathbb{R}$, so $\omega_b(v_b) \in \mathbb{R}$. And to make confusion complete, we now consider points $a=f(b)\in M$ since we want to pull back $\omega$ along $f$ to have the analog situation on $M$.

That was what I meant by: Draw a diagram to make clear where you are at a certain point in the formula!
$$((f^*(\omega))_{a})(v_a)=\omega_{f(a)}(f_*(v_a))=\omega_{f(a)}(D_af(v_a))\in \mathbb{R}$$
Have a look in https://www.physicsforums.com/insights/pantheon-derivatives-part-iii/ section Cotangents and 1-Forms
What we have here is basically the transpose (${.}^\tau$) of the Jacobian matrix at a certain point and you are requested to show that
$$(J_x \cdot J_y)^\tau=J_y^\tau \cdot J_x^\tau$$
since for $A\, : \,V\longrightarrow W$ we get $A^\tau\, : \,W^*\longrightarrow V^*$ where $V^*=\operatorname{Hom}(V,\mathbb{R})$.
It's linear algebra.