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I just read section 4 of the manifold book by Loring Tu.

I can understand the $1$-form without questions as covector fields. A $k$-form on $U$ is a function that assigns every $p\in U$ an alternating $k$-linear function on $T_p(\mathbb{R}^n)$. I understand that we need to assign a $k$-linear function for integration to make sense. However, why do we need the alternating instead of just $k$-linear function?

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    $\begingroup$ When you integrate a 1 variable function $f(x)$ from $a$ to $a$, what would you like the answer to be? $\endgroup$
    – Arkady
    Jul 9, 2022 at 2:50
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    $\begingroup$ @Arkady Thanks so much for the hint! Is it because we want that $\int_S w=-\int_{-S}w$ where $-S$ denoted the same surface $S$ with the opposite orientation, and $w$ a $2$-form? This is exactly exterior algebra. $\endgroup$
    – quuuuuin
    Jul 9, 2022 at 3:11
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    $\begingroup$ ding ding ding. $\endgroup$
    – Arkady
    Jul 9, 2022 at 4:19

1 Answer 1

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Suppose $O, U \subset \mathbb{R}^n$ and $F : O \to U$ is a diffeomorphism, to be interpreted as a change of coordinates from $x \in U$ to $y \in O$. We integrate alternating forms because the transformation law for coordinate transformation $x = F(y)$ of an $n$-form $\alpha = a(x)dx_1 \wedge \dots \wedge dx_n$ on an open set $U \subset \mathbb{R}^n$ reads $$F^*(a(x)dx_1 \wedge \dots \wedge dx_n) = a(F(y))\det DF(y) dy_1\wedge \dots \wedge dy_n,$$ while the change of variables formula for integration reads $a(x)dx = a(F(y)) |\det DF(y)|\,dy$. These formulas say that $\int_{U}\alpha := \int_{U}a(x)\,dx$ is independent of the coordinate system, provided that $\det DF > 0$, which is to say that $F$ preserves orientation.

This construction wouldn't work for a general $n$-tensor $A = a(x)dx_1 \otimes \dots \otimes dx_n$ that is not alternating because the transformation formula no longer matches the change of variables theorem.

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  • $\begingroup$ Thanks for your comments and that makes sense! $\endgroup$
    – quuuuuin
    Jul 10, 2022 at 13:44

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