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Definition. 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: $$f^*\omega(x) = (df_x)^*\omega[f(x)].$$

According to Daniel Robert-Nicoud's nice answer to $f: X \to Y$ is a smooth map and $\omega$ is a $p$-form on $Y$, what is $\omega[f(x)]$?, locally, differential form can be written as $$\omega_\alpha(y) = \alpha(y)dx^{i_1}\wedge\ldots\wedge dx^{i_p}$$ with $\alpha$ a smooth function. Then $$f^*\omega_\alpha(x) = (df_x)^*[(\alpha\circ f)(x)dx^{i_1}\wedge\ldots\wedge dx^{i_p}].$$

Hence, if $$\omega_\alpha(y) = \alpha(y)dx^{i_1}\wedge\ldots\wedge dx^{i_p},$$ $$\theta_\beta(y) = \beta(y)dx^{j_1}\wedge\ldots\wedge dx^{j_q},$$

Do I get $\omega \wedge \theta$ such that $$\omega_\alpha \wedge \theta_\beta(y)= \gamma(y)dx^{k_1}\wedge\ldots\wedge dx^{k_{p+q}}?$$ If so, how can I prove it?

And can I write $\gamma$ in terms of $\alpha$ and $\beta$?

Thank you~~~

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WishingFish
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1 Answers1

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I suppose generally a $p$-form is a sum of such terms, but if we can understand how one such element pulls-back then linearity extends to $\sum_{i_1, \dots , i_p}\alpha_{i_1,\dots , i_p} dy^{i_1} \wedge \cdots \wedge dy^{i_p}$. That said, to calculate $\gamma$ you just have to sort out the sign needed to arrange the indices on the wedge of $\omega_{\alpha} \wedge \theta_{\beta}$. I prefer the notation, assuming $I \cap J = \emptyset$, $$ (\alpha dy^I) \wedge (\beta dy^J) = \alpha \beta dy^I\wedge dy^J = \alpha \beta (-1)^{\sigma(I,J)}dy^K $$ here $\alpha\beta$ merely indicated the product of the scalar-valued functions $\alpha$ and $\beta$ and I have suppressed the $y$-dependence as it has little to do with the question. Here $\sigma(I,J)$ is the number of transpositions needed to rearrange $(I|J)$ into $K$.

Of course, you could just leave the $I$ and $J$ unchanged in which case the $\gamma = \alpha \beta$. I rearranged them because in some of what you are interested in reading there will be a supposition that the indices are arranged in increasing order so if $I = (1,2,5)$ and $J = (3,6)$ then you'll want $(1,2,5)(3,6) \rightarrow K = (1,2,3,5,6)$ which requires flipping $3$ and $5$ hence $\sigma(I,J) = 1$. Ok, usually we use "sgn" of a permutation to get this sign so my notation is a bit nonstandard.

James S. Cook
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  • Thank you so much James, you are really leading me through!!!!! – WishingFish Jul 31 '13 at 03:58
  • I know $T \wedge S = (-1)^{|T||S|} S \wedge T$. But we are not commuting $dy^I, dy^J$ here, why still? – WishingFish Jul 31 '13 at 04:01
  • hmm, you really know what I am thinking! I see you answered before I post the comment above asking about it.... :) – WishingFish Jul 31 '13 at 04:02
  • So $\sigma(I,J)$ is the sign function bookkeeping how many commute actually happened to combine the wedge sorted, rather than commute the whole $dy^I$ and $dy^J$ - now it makes sense. thanks :) – WishingFish Jul 31 '13 at 04:04
  • @WishingFish I probably would do better to write $\sigma(I,J)=K$ and $dy^I \wedge dy^J= sgn(\sigma)dy^K$, I think that notation is a standard notation. – James S. Cook Jul 31 '13 at 04:22
  • Got it. Thanks~~~~ James~~~ – WishingFish Jul 31 '13 at 04:28
  • Hi James... Sorry I still get stuck at the end :#) Do you mind taking a look? Thank you in either ways~http://math.stackexchange.com/questions/456135/pullback-identity-based-on-definition – WishingFish Jul 31 '13 at 04:45