# Wedge product of $\beta \wedge dx$

I'm still struggling to understand the wedge product (and also differential forms in general) and am therefore trying to find / come up with actual examples. Say I have differential forms $$\alpha = dx + dy + dz$$ and $$\beta = 2dx - dy + dz$$, how would I wedge $$\alpha \wedge \beta \wedge dz$$. Sorry if this is not a good question, I'm just really lacking good intuition, yet.

Use the fact that the wedge product is associative, bilinear and anti-symmetric, and treat $$dx, dy, dz$$ as formal basis elements in a vector space.

In particular, anti-symmetry implies that $$dz\wedge dz=0$$, and bilinearity implies then that any part that contains $$dz$$ in the wedge $$\alpha\wedge\beta$$ will be killed when you compute $$\alpha\wedge\beta\wedge dz$$. Therefore you can disregard it from the beginning.

Note that at the end of the day, the wedge product is only defined formally on the basis elements - that is, $$dx\wedge dy$$ is not something that gets evaluated into something simpler like a binary operation. It stays $$dx\wedge dy$$. In your particular case, in the end you will get an integer times the form $$dx\wedge dy\wedge dz$$.

• So in my case I would just have $2dy \wedge dx + 2dz \wedge dx -dx \wedge dy + dz \wedge dy + dx \wedge dz + dy \wedge dz$?
– MJP
Dec 10 '19 at 7:59
• @MJP No what you have in the end is just one term: an integer times $dx\wedge dy\wedge dz$. Dec 10 '19 at 8:04
• @MJP I'd forgotten to write that the wedge product is associative, which means that you can compute $\alpha\wedge\beta$ first, and then wedge that against $dz$. Dec 10 '19 at 8:12

Just treat it as normal multiplication, but with the added twist that it’s antisymmetric ($$dx\wedge dy=-dy\wedge dx$$) and that $$dx\wedge dx=0$$, so that any term that contains two or more of the same variable vanishes. In particular, it distributes over addition, so you compute the wedge product of two differential forms pretty much the same way that you’d multiply two polynomials. For instance, $$(dx+dy)\wedge(2\,dx-dy) = dx\wedge(2\,dx-dy)+dy\wedge(2\,dx-dy) = -dx\wedge dy+2\,dy\wedge dx = -3\,dx\wedge dy.$$ Notice that when wedging with, say, $$dx$$, you can ignore any term that also contains $$dx$$ since their product will vanish.

Let $$M$$ be a smooth manifold and $$A^k(V)$$ be the space of alternating $$k$$-form, i.e. an alternating $$k$$-tensor $$V^k \to \mathbb{R}$$.

A $$k$$-form $$\omega$$ on $$M$$ is an assignment of an element $$\omega(p) \in A^k(T_pM)$$ to each $$p$$. If $$\varphi$$ is a real function on $$M$$ we define the product $$\varphi \omega$$ pointwise by $$\varphi \omega(p) = \varphi(p) \omega(p)$$, and if $$\theta$$ is an $$l$$-form on $$M$$, we define the wedge product $$k+l$$-form $$\theta \wedge \omega$$ on $$M$$, pointwise by $$(\theta \wedge \omega)(p)=\theta(p) \wedge \omega(p)$$ for each $$p$$.

In particular, given a chart $$\sigma: U \to M$$, the elements $$dx_{i_1} \wedge \cdots \wedge dx_{i_k}$$, where $$1\leq i_1 < \cdots < i_k \leq m$$, are differential forms on the open subset $$\sigma(U)$$ of $$M$$. For each $$p \in \sigma(U)$$, a basis for $$A^k(T_p M)$$ is obtained from these elements. Therefore, every $$k$$-form $$\omega$$ on $$M$$ has a unique expression on $$\sigma(U)$$:

$$\omega = \sum_{I={i_1,\dots, i_k}} a_I dx_{i_1} \wedge \cdots \wedge dx_{i_k}$$

where $$a_I: \sigma(U) \to \mathbb{R}$$.

We call $$\omega$$ smooth if all the functions $$a_I$$ are smooth, for each chart $$\sigma$$ in an atlas of $$M$$. Thus a smooth $$k$$-form is called a differential $$k$$-form. The space of differential $$k$$-forms on $$M$$ is denoted by $$A^k(M)$$.

Let $$f:M \to N$$ be a smooth map of manifolds, and let $$\omega \in A^k(N)$$. We define $$f^\ast \omega \in A^k(M)$$, called the pullback of $$\omega$$ by

$$f^\ast \omega(p) (v_1, \dots, v_k) = \omega(p)(df_p(v_1), \dots, df_p(v_k))$$

for all $$v_1$$, $$\dots$$, $$v_k \in T_p M$$.

Finally, if $$\omega = f_1 dg_1 + f_2 dg_2 + \cdots + f_m dg_m$$, then the exterior derivative of $$f$$ is $$d\omega = df_1 \wedge dg_1 + df_2 \wedge dg_2 + \cdots + df_m \wedge dg_m$$. This exterior derivative is a map $$d: A^k(M)\to A^{k+1}(M)$$ with the following properties:

• If $$k=0$$, it agrees with the differential $$d$$ on functions,
• it is linear,
• $$d(\varphi \omega) = d\varphi \wedge \omega + \varphi d\omega$$ for $$\varphi \in C^\infty(M)$$, $$w \in A^k(M)$$,
• $$d(\omega_1 \wedge \omega_2) = d\omega_1 \wedge \omega_2 + (-1)^k w_1 \wedge d\omega_2$$ for $$\omega_1 \in A^k(M)$$, $$\omega^2 \in A^l(M)$$,
• $$d(df_1\wedge \cdots \wedge df_k) =0$$ for all $$f_1$$, $$\dots$$, $$f_k \in C^\infty(M)$$,
• $$d(d\omega) = 0$$ for all $$\omega \in A^k(M)$$.