Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$U \subset \mathbb{R}^k$ 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: $$f^*\omega(x) = (df_x)^*\omega[f(x)].$$

share|improve this question
    
Which definitions are you using? Often, the definition of $f*\omega$ is exactly $\omega\circ f_*$. –  Neal Aug 6 '13 at 4:09
    
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: $$f^*\omega(x) = (df_x)^*\omega[f(x)].$$ –  WishingFish Aug 6 '13 at 4:10
    
Push forward (not even given in GP, I found it from John Lee): $$(F_*X)(f) = X(f \circ F).$$ –  WishingFish Aug 6 '13 at 4:13
    
@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. –  James S. Cook Aug 6 '13 at 6:07

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.