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I am studying differential forms from Rudin's PMA. Here's the definition of $k$-form enter image description here

Also he proves the anticommutative relation: $dx_1 \land dx_2=-dx_2\land dx_1$

Does the following expressions make sense: $dx_1\land cdx_2$ where $c\in \mathbb{R}^1$? I guess NO since the definition 10.11 different from what I wrote.

Does the following is true $dx_1\land cdx_2=cdx_1\land dx_2$? If yes how to prove it?

EDIT: First of all, I am going to prove that $d(cx_1)\land dx_2=cdx_1\land dx_2$. Let $\omega_1=d(cx_1)\land dx_2$ and $\omega_2=cdx_1\land dx_2$. Since $$\omega_1(\Phi)=\int \limits_{\Phi}\omega_1=\int \limits_{D}\dfrac{\partial(c\phi_1,\phi_2)}{\partial(u_1,u_2)}d\mathbb {u}=\int \limits_{D}c\dfrac{\partial(\phi_1,\phi_2)}{\partial(u_1,u_2)}d\mathbb {u}=\int \limits_{\Phi}\omega_2=\omega_2(\Phi).$$ Hence $\omega_1=\omega_2$ then $$d(cx_1)\land dx_2=cdx_1\land dx_2$$ $$dx_1\land d(cx_2)=cdx_1\land dx_2.$$ How to conclude that $dx_1\land cdx_2=cdx_1\land dx_2$ from above relations?

RFZ
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  • You mean $dx_1\wedge cdx_2=cdx_1\wedge dx_2$, right? – Ben Sheller Apr 21 '16 at 20:10
  • @BenS., I am just asking is this true or not? If yes how to prove it? – RFZ Apr 21 '16 at 20:24
  • @BenS., I fixed typo. – RFZ Apr 21 '16 at 20:45
  • You may want to check out https://en.wikipedia.org/wiki/Differential_form#Intrinsic_definitions basically, a differential form can also be defined as a totally antisymmetric tensor field, in which case you can move the constants around the same way as you would for any other tensor product. – Ben Sheller Apr 22 '16 at 03:07
  • @BenS., but what does mean $dx_1 \land cdx_2$? The problem for me is that constant lies after sign $\land$. How to move it left? In definition 10.11 is another situation. – RFZ Apr 22 '16 at 05:06
  • @BenS., I edit my post. – RFZ Apr 22 '16 at 06:25

1 Answers1

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The expression $dx_1 \wedge cdx_2$ does make sense. It's not in the form given in definition 10.11, but if you skip ahead and take a look at section 10.17 where the (exterior/wedge) product of differential forms is defined, you'll notice that for the $1$-forms $\omega = dx_1$ and $\lambda = cdx_2$, both of which are in the canonical form of definition 10.11, we have

$$dx_1 \wedge cdx_2 = \omega \wedge \lambda,$$

and that is defined as

$$c\,dx_1 \wedge dx_2.$$

Daniel Fischer
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  • Dear Daniel! Thanks a lot for your permanent help on my questions! – RFZ Apr 24 '16 at 06:43
  • Dear Daniel! Sorry but can you help please with this question http://math.stackexchange.com/questions/1762429/confusing-moment-in-theorem-10-27-from-pma-rudin? – RFZ Apr 28 '16 at 09:14