# Differential forms on $\mathbb{R}^n$

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?

• When you integrate a 1 variable function $f(x)$ from $a$ to $a$, what would you like the answer to be? Jul 9, 2022 at 2:50
• @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. Jul 9, 2022 at 3:11
• ding ding ding. Jul 9, 2022 at 4:19

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.