Computing $ι^∗(dy^j)_p( \frac{\partial}{\partial x^l}_p)$ and $ι^∗(dz^k)_p( \frac{\partial}{\partial x^l}_p)$on $Graph(F) = \{(a, F(a)); a \in V \}$

Question

Let $V \subseteq \Bbb R^m$ open, and $F : V \to \Bbb R^n$ smooth. Consider the graph of $F$

$\operatorname{graph}(F) = \{(a, F(a)); a \in V \}$

which is an embedded submanifold of $\Bbb R^m \times \Bbb R^n$. This one has an atlas with 'one' map $(\operatorname{graph}(F); \varphi)$, where $\varphi(a, F(a)) = a$ for all $a \in V $. We denote the components of $\varphi$ with $x^1,... , x^m$. Thus we get for $l = 1, ..., m$ the 1-form $dx^l \in \Omega^1(\operatorname{graph}(F))$ and for each $a \in V$ the basis vector $\frac{\partial}{\partial x^l}|_ {(a,F (a))} \in T_{(a,F (a))} \operatorname{graph}(F)$.

We write $y^1,... , y^m$. for the standard coordinate functions on $\Bbb R^m$, and $z^1,... , z^n$. for the standard coordinate functions on $\Bbb R^n$. We take this as functions on $\Bbb R^m\times \Bbb R^n$. Via the inclusion map $ι: \operatorname{graph}(F) \to \Bbb R^m\times \Bbb R^n$ define these one forms $ι^∗(dy^j); ι^∗(dz^k) \in \Omega^1(\operatorname{graph}(F))$

a)Prove that for all $j,l \in \{1, ..., m\}$ and all $a \in V$,

$ι^∗(dy^j)_{(a,F (a))}( \frac{\partial}{\partial x^l}_{(a,F (a))})= \delta^j_l$ where $\delta^j_l$ is the Kronecker $\delta$.

b) Prove that for all $k \in \{1, ..., m\},l \in \{1, ..., m\}$ and all $a \in V$,

$ι^∗(dz^k)_{(a,F (a))}( \frac{\partial}{\partial x^l}_{(a,F (a))})= \frac{\partial F^k}{\partial y^l}(a)$ where $F^k$ is the k-th component of $F$

My try

Denote $p:=(a,F(a))$

a)

$ι^∗(dy^j)_p( \frac{\partial}{\partial x^l}|_p)$

$ =(dy^j_p )(di_p( \frac{\partial}{\partial x^l}|_p))$, where I used thee definition of pullback

$ =(di_p( \frac{\partial}{\partial x^l}|_p))(y^j)$, where I used that $dg(v)=vg$, for a function $g$ and a tangent vector $v$

$ =( \frac{\partial}{\partial x^l}|_p))(y^j\circ i)$, where I used the definition $(d\tilde F_pv)g=v(g \circ \tilde F)$

$= \frac{\partial x^j}{\partial x^l}|_p=\delta^j_l ...(1)$

I have a doubt in the last step,what is the correct way to do it? I think I am confusing coordinate functions with coordinates, but I don't know how to resolve it. I found that

$y^j\circ i: Graph(F)\xrightarrow{i}\Bbb R^m \times\Bbb R^n\xrightarrow{y^j}\Bbb R: (a^1,...,a^m,F^1(a),..,F^n(a))\mapsto (a^1,...,a^m,F^1(a),..,F^n(a)) \mapsto a^j$,

so this composition is indeed a coordinate function because it spits out the jìth component of the imput, and since the domain is in the Graph(F) how coordinate functions are $x^j$, does it mean I have to express the function using the $x^j$ as coordinates? That is what I did in the last step an got the correct answer, but I doubt this is the correct reason, after old the independent variable should be a dummy, shouldn't it? On the other hand seeing the $x^l$ in $\frac{\partial }{\partial x^l}$ kind of tells me that they are not dummy.

b) In this second part I did pretty much the same, but I could not get to the correct answer, surely because of the same problem. How do I fix this? These is what I did:

$ι^∗(dz^k)_p( \frac{\partial}{\partial x^l}|_p)$

$ =(dz^k_p )(di_p( \frac{\partial}{\partial x^l}|_p))$, where I used the definition of pullback

$ =(di_p( \frac{\partial}{\partial x^l}|_p))(z^k)$, where I used that $dg(v)=vg$, for a function $g$ and a tangent vector $v$

$ =( \frac{\partial}{\partial x^l}|_p))(z^k\circ i)$, where I used the definition $(d\tilde F_pv)g=v(g \circ \tilde F)$

$= \frac{\partial F^k(x^1,..x^m)}{\partial x^l}|_p...(2)$

Also here there is a problem in the last step. I think $(z^k\circ i)(a^1,..a^m)=z^k(a^1,..a^m)=F^k(a^1,..a^m)$ but I thought I had to plug in $F^k(x^1,..x^m)$, since the domain of the composition is Graph(F) which has coordinates $x^i$ The answer should had been $\frac{\partial F^k}{\partial y^l}(a)$ though

Answer 1

A few suggestions:

1. Do this first with $m=1$ and $n=2$.
2. You don't really want to try to work with the graph of $F$ directly. View it as a abstract manifold and the map \begin{align*} i: V &\rightarrow \operatorname{graph}(F)\\ a &\mapsto (a,F(a)) \end{align*} as a coordinate chart. Pull everything back to $V$ and do your calculations there.
3. I find it easier to do computations involving pullback of $1$-forms using the original definition of the pullback of the differential of a function: $$ i^*df(v) = \left.\frac{d}{dt}\right|_{t=0}f(i(c(t))), $$ where $c: I \rightarrow V$ is a curve such that $c'(0) = v$ (here, $c(t) = x + tv$ works). Here, you have explicit formulas for $v$, $f$, $c$, and $i$ (in terms of $F$).