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I am sorry to ask this but I was reading this question, because I want to solve that too, the thing is that it has been answered a long time ago and I don't understand the answer given there.

Question

The question ask about expressing

$$f^*\left(\sum_{j_1,\dots,j_k} a_{j_1\dots j_k}dy^{j_1}\otimes\cdots\otimes dy^{j_k}\right)$$

in terms of $dx^i$ given that $f\colon M^n\to N^m$ is a map between manifolds, with $(x,U)$ and $(y,V)$ coordinates systems around $p$ and $f(p)$. But I can't figure out how the answer gives the require formula or the actual computations?.

In fact I have another question, How can one manage the formula

$$ (f^* dy^j)(p)=\sum_{i=1}^n\frac{\partial (y^j\circ f)}{\partial x^i}(p)\cdot dx^i(p) $$

I mean How can one prove that result? Because that seems useful, and I have the same problem I don't know how $f^{\ast}$ acts.So Can someone help me with with those issues?

Thanks in advance

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user162343
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  • The answer you're looking for depends some on the order in which one has defined things, but if one accepts the naturality of the exterior derivative, then $f^* dy_p = d(f^* y)p = d(y \circ f)_p$. If we apply this to $\partial{x^i}\vert_p$, then $d(y \circ f)p(\partial{x^i}\vert_p) = \partial_{x^i}\vert_p(y \circ f)$, and expanding this using the chain rule gives exactly the $i$th coefficient of the sum. – Travis Willse Apr 19 '16 at 16:51
  • This is for both formulas or just for one ? – user162343 Apr 19 '16 at 16:52
  • Just the second, but knowing how to pull back a $1$-form, as in the second equation, tells you how to pull back a (covariant) $k$-tensor. – Travis Willse Apr 19 '16 at 16:54
  • So can you provide a detailed answer for both results ? – user162343 Apr 19 '16 at 18:53
  • Then what can e done? – user162343 Apr 20 '16 at 01:12

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