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My differential geometry course introduced the following formula for computing pullbacks by a $k$-form:

$f^{*}\alpha = \sum (f \circ a_J) df^{j_{1}} \wedge \cdots \wedge df^{j_{l}}$.

My instructor computed the pullback $f^{*} \alpha$, where $f(x_1, x_2, x_3) = (x_1^2 + x_2^2, \sin(x_3))$, and $\alpha = y_1^2 dy_1 + y_2^3 dy_2$. He then got as the answer before simplification:

$f^{*}\alpha = (x_1^2 + x_2^2)^2 d(x_1^2 + x_2^2) + (\sin(x_3))^3 d(\sin(x_3))$.

Could someone please explain how he obtained this result from the general formula written above? I am slightly confused on the indices.

Thanks.

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  • $\begingroup$ That should be $f^{*}\alpha = (x_1^2 + x_2^2)^2 d(x_1^2 + x_2^2) + (\sin(x_3))^3 d(\sin(x_3))$. You basically just substitute $x_1^2 + x_2^2$ for $y_1$ and $\sin(x_3)$ for $y_2$: that's what the composition $f \circ \alpha_J$ is. $\endgroup$ – André 3000 Oct 9 '17 at 22:58
  • $\begingroup$ Thanks. I fixed the missing exponent in the question. So, when you say composition of $f \circ a_J$, do we take this composition component-wise? since $f = (f_1, f_2)$ from above? $\endgroup$ – Thomas Moore Oct 9 '17 at 23:00
  • $\begingroup$ Actually, I guess it should be $\alpha_J \circ f$, not $f \circ \alpha_J$. In your example, $f$ is a map $\mathbb{R}^3 \to \mathbb{R}^2$, and each $\alpha_J$ is a map $\mathbb{R}^2 \to \mathbb{R}$. But yes, componentwise is right. Take a look at p. 137 of Lee's Introduction To Smooth Manifolds. $\endgroup$ – André 3000 Oct 9 '17 at 23:07
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    $\begingroup$ Thomas, if you're just learning the differential forms game, it might help you to watch some of my YouTube videos. Start with Day 24 of MATH 3510 at the very bottom and then proceed to Days 25, 26, ...., as needed above. $\endgroup$ – Ted Shifrin Oct 9 '17 at 23:23

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