1
$\begingroup$

Let $f : U \rightarrow V$ be a smooth map between open sets $U \subset \mathbb{R}^{m}$ and $V \subset \mathbb{R}^{n}$. Let us write $f=\left(f_{1}, \ldots, f_{n}\right)$ and let's use different variables for domain and codomain $\left(x^{1}, \ldots, x^{n}\right) \in \mathbb{R}^{n}$ e $\left(y^{1}, \ldots, y^{m}\right) \in \mathbb{R}^{m}$. I am asked to prove that $$f^{*}\left(d x^{i}\right)=\frac{\partial f_{i}}{\partial y^{j}} d y^{j}=d f_{i}$$ Let $v \in T_p \mathbb{R}^{n} = \mathbb{R}^{n}$. My proof goes like this $$\left(f^{*}\left(d x^{i}\right)\right)(p) (v) = d x^{i}_{f(p)} (df_p(v)) = d (x^{i}\circ f)_p(v) = d (f_i)_p (v)$$ and hence $$f^{*}\left(d x^{i}\right) = d f_{i} = \frac{\partial f_{i}}{\partial y^{j}} d y^{j}$$

However, I am supposed to prove $f^{*}\left(d x^{i}\right)=\frac{\partial f_{i}}{\partial y^{j}} d y^{j}$ directly. I managed to get $\forall p \in U \ \forall v \in T_p U \ \forall g \in C^\infty (\mathbb{R})$

\begin{align} \left(f^{*}\left(d x^{i}\right)\right)(p) (v) (g) & = d x^{i}_{f(p)} \left(df_p \left(\sum_j v^j \frac{\partial}{\partial y^j} \right) \right) (g) \\ & = \sum_j v^j \left( d (f_i)_p \left( \frac{\partial}{\partial y^j} \right) \right) (g) \\ & = \sum_j v^j \frac{\partial g \circ f_i}{\partial y^j} (p) \\ & = \sum_j dy^j_p(v) g'(f_i(p)) \frac{\partial f_i}{\partial y^j} (p) \end{align}

but I'm not really sure whether this is right. Does anyone know how to go on from here?

| cite | improve this question | |
$\endgroup$
  • $\begingroup$ What you're writing doesn't make sense. Evaluating a $1$-form on a tangent vector gives a number. It only makes sense to apply a tangent vector to a function ($g$) to compute a directional derivative. What does "directly" mean? If you know that $d$ commutes with pullback, that gives it to you directly: $f^*(dx^i) = d(f^*x_i) = df_i$. $\endgroup$ – Ted Shifrin May 11 '19 at 17:46
  • $\begingroup$ Thanks for your comment! I was making use of a smooth function from $\mathbb{R}$ to $\mathbb{R}$ in order to apply the definition of differential $df_p(v)(g) = v(g\circ f)$. I am aware that a 1-form on a tangent vector gives a number, but I was hoping I could regard that number as an element of the tangent space of $\mathbb{R}$ through the canonic identification $a \mapsto a \frac{\partial}{\partial x}$ and use the definition of differential. Do you have some tips on how to compute $d (f_i)_p \left( \frac{\partial}{\partial y^j} \right)$ otherwise? $\endgroup$ – Nicolò Cavalleri May 11 '19 at 18:20
  • $\begingroup$ Nope, it's a number, not a tangent vector to $\Bbb R$. “Directly” should mean working just with differential forms (as I much prefer), rather than evaluating on tangent vectors. I gave you the argument, or are we trying to prove $d$ commutes with pullback? $\endgroup$ – Ted Shifrin May 11 '19 at 18:24
  • $\begingroup$ Actually, I am not quite sure of what "directly" means as it is in the text of the problem which I am trying to solve but I interpreted it as "in coordinates" being the rhs in coordinates. $\endgroup$ – Nicolò Cavalleri May 11 '19 at 19:56
  • $\begingroup$ But let me see if I get this right: the differential of a map goes from a tangent space to a tangent space, so it maps vectors to vectors. A differential form is at each point is an antisymmetric tensor so yes I agree it gives me a number. However, I always thought I could interpret $df$ for a map $f: M \to \mathbb{R}$ as both a differential and a differential form because of the canonic identification from $\mathbb{R}$ to its tangent space which I wrote in the last message. Is this not correct? $\endgroup$ – Nicolò Cavalleri May 11 '19 at 19:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.