$U \subset \mathbb{R}^k$ and $V \subset \mathbb{R}^l$ be open subsets. Let $f: V \to U$ to smooth. I am wondering how can I show $$ (f^*dx_i)(\frac{\partial}{\partial y^j}) = dx_i(f_*(\frac{\partial}{\partial y^j}))?$$

I was informed that by definition, $$f^*\omega = \omega \circ f_*.$$ That's neat, then I think I got my problem solved.

However just out of curiosity, how could this definition equivalent to the definition I am given from Guillemin and Pollack?

Definition of $\mathbf{A^*T}:$ Suppose $A: V \to W$ is a linear map. Then the transpose map $A^*: W^* \to V^*$ extends to the exterior algebras, $A^*: \Lambda^p(W^*) \to \Lambda^p(V^*)$ for all $p>0$. If $T \in \Lambda^p(W^*)$, just define $A^* T \in \Lambda^p(V^*)$ by $$A^*T(v_1, \dots, v_p) = T(Av_1, \dots, Av_p)$$ for all vectors $v_1, \dots, v_p \in V$.

Definition of $\mathbf{f^*\omega(x)}:$ If $f: X \to Y$ is a smooth map and $\omega$ is a $p$-form on $Y$, define a $p$-form $f^*\omega$ on $X$ as follows: $$f^*\omega(x) = (df_x)^*\omega[f(x)].$$

  • $\begingroup$ Which definitions are you using? Often, the definition of $f*\omega$ is exactly $\omega\circ f_*$. $\endgroup$ – Neal Aug 6 '13 at 4:09
  • $\begingroup$ This one, @Neal. Definition of $\mathbf{f^*\omega(x)}:$ If $f: X \to Y$ is a smooth map and $\omega$ is a $p$-form on $Y$, define a $p$-form $f^*\omega$ on $X$ as follows: $$f^*\omega(x) = (df_x)^*\omega[f(x)].$$ $\endgroup$ – WishingFish Aug 6 '13 at 4:10
  • $\begingroup$ Push forward (not even given in GP, I found it from John Lee): $$(F_*X)(f) = X(f \circ F).$$ $\endgroup$ – WishingFish Aug 6 '13 at 4:13
  • $\begingroup$ @WishingFish I show below that the push-forward definition you give above is equivalent to the definition I used in response to your other question sometime today ( I forget which one). I think, this question has already been answered in view of this. $\endgroup$ – James S. Cook Aug 6 '13 at 6:07

This will be an issue soon if it is not already. You will soon realize that two seemingly different definitions of the push-forward are in use. I seek to show the following are actually the same:

  1. the coordinate free definition $(F_{*}X)(f) = X(f \circ F)$

  2. the coordinate-based definition: if $F: M \rightarrow N$ is a differentiable mapping of manifolds and $(x^i)$ are coordinates($i=1,\dots , m$) on $M$ and $(y^j)$ are coordinates ($j=1,\dots , n$) on $N$ which contain $p \in M$ and $F(p) \in N$ respective then: $$ (dF)_p \left( \sum_{i=1}^{m}X^i\frac{\partial}{\partial x^i}\bigg|_p \right) = \sum_{i=1}^{m}\sum_{j=1}^{n} X^i\frac{ \partial(y^j \circ F)}{\partial x^i}\frac{\partial}{\partial y^j}\bigg|_{F(p)}$$

Notice the absence of the $f$ in this definition (2.) as compared to (1.) $(F_{*}X)(f) = X(f \circ F)$. To show equivalence, we can expand on Lee's definition, we assume here that $X=\sum_{i=1}^{m}X^i\frac{\partial}{\partial x^i}\bigg|_p$ thus: $$ X(f \circ F) = \sum_{i=1}^{m}X^i\frac{\partial}{\partial x^i}\bigg|_p(f \circ F) $$ note $f$ is a function on $N$ hence the chain rule gives: $$ X(f \circ F) = \sum_{i=1}^{m}X^i\sum_{j=1}^n \frac{\partial f }{\partial y^j}(F(p))\frac{\partial (y^j \circ F)}{\partial x^i} $$ But, we can write this as: $$ X(f \circ F) = \left( \ \sum_{i=1}^{m}X^i\sum_{j=1}^n \frac{\partial (y^j \circ F)}{\partial x^i} \frac{\partial }{\partial y^j}\bigg|_{F(p)} \right)(f)$$ This is precisely what I claimed was the "definition" of the push-forward in coordinates. Observe that this is merely an application of the chain-rule separated from Lee's definition. Of course, the coordinate-free definition is the natural choice, but calculation often warrants use of coordinates.


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