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Let $M$ be a smooth manifold and let $\omega = \sum_{(i_1, \dots, i_n)}f_{(i_1, \dots, i_p)} dx^{i_1} \wedge \dots \wedge dx^{i_p}$ be a differential $p$-form. Let $d$ denote the exterior derivative.

I am trying to show that $d(dx^k) = 0$. To this end, I tried to substitute into the definition:

$$ d(dx^k) = \sum_{l=1}^n {\partial x^k \over \partial x^l} dx^l$$

But ${\partial x^k \over \partial x^l}= \delta_{kl}$ so that $\sum_{l=1}^n {\partial x^k \over \partial x^l} dx^l = dx^k$ and I am going in circles.

How can I show that $d(dx^k) = 0$?

The definition of the exterior derivative is

$$ d \omega = \sum_{(i_1, \dots , i_p)} df_{(i_1, \dots, i_p)}\wedge dx^{i_1}\wedge \dots \wedge dx^{i_p}$$

where $df = \sum_{k=1}^n {\partial f \over \partial x^k}dx^k$ and the $dx^k$ are a local basis for the differential $1$-forms, as far as I understood the explanation given in an answer to my other question here.

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1 Answers1

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When you write $dx^k$ in your standard form $\sum f_i dx^i$, you have $f_i = \delta_i^k$ (i.e., $1$ for $i=k$ and $0$ otherwise). Then $df_i=0$ and $d(dx^k)=\sum df_i\wedge dx^i =0$, as desired.

Ted Shifrin
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  • Those of you downvoting me would do everyone a service if you would kindly explain what dissatisfies you about the answer. And @self-learner, if this doesn't help you, please ask. – Ted Shifrin Oct 22 '14 at 13:27
  • Thank you, it does help me. I did not downvote your answer (please verify this by visiting the votes tab on my profile). After reading your answer I think I can write an even shorter proof: $d(dx^k) = d (1 \cdot dx^k)) = {\partial 1 \over \partial x^k} dx^k = 0$. – self-learner Oct 23 '14 at 00:30
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    I wasn't accusing you at all ... I just wanted to make sure you were happier. What you wrote isn't right. Literally, it needs to be $$d(1 dx^k) = \sum_j \frac{\partial(1)}{\partial x^j} dx^j\wedge dx^k,$$ since, after all, you need the answer to be a $2$-form. Zero, 'tis true. – Ted Shifrin Oct 23 '14 at 02:46
  • Yes you are right. Thank you for pointing it out! – self-learner Oct 23 '14 at 06:20
  • @TedShifrin I was wondering if this result holds for any basis 1 forms? meaning if $f$ was a coordinate change (Jacobian), or something more exotic (like a verbein/tetrad in General relativity). I'd expect coordinate basis as exact 1-forms and non-coordinate basis as closed 1-forms to both adhere to this? – R. Rankin Jul 02 '22 at 21:49
  • @R.Rankin Why must (local) basis $1$-forms be closed? This is far from the case. – Ted Shifrin Jul 02 '22 at 22:19
  • I was thinking for coordinate basis the components are always (exterior) derivatives of functions so they're exact. I wasn't sure about the non-coordinate/holonomic case. But I thought they might be closed but not exact, since they can't be written in terms of a jacobian. That would highlight topological information? Oh! And apologies, I am strictly thinking of orthonormal basis! – R. Rankin Jul 02 '22 at 23:05
  • Sure, for coordinates, you're fine (although it can only be an orthonormal basis when the space has zero curvature). If you know any differential geometry at all, you can take an orthonormal coframe $\omega_1,\omega_2$ for a Riemannian surface and their exterior derivatives will be $0$ if and only if the metric is flat. – Ted Shifrin Jul 02 '22 at 23:19
  • Gotcha, my brain was stuck on topology for some reason, I didn't even consider Cartans structure equations! – R. Rankin Jul 03 '22 at 07:25