# $\wedge$-product, bilinear

I want to show, that the $\wedge$-product is bilinear.

It is $\displaystyle (\omega^k\wedge\eta^l)(v_1,\dotso,v_{k+l}):=\frac{1}{k!l!} \sum_{\sigma\in S_{k+l}} \operatorname{sgn} (\sigma) \omega^k (v_1, \dotso, v_k) \eta(v_{k+1},\dotso,v_{k+l})$

I have to show, that for forms of appropriate degree it is:

1) $(\omega_1^k + \omega_2^k) \wedge \eta^l = \omega^k_1 \wedge \eta^l + \omega_2^k \wedge\eta^l$

2) $\omega^k \wedge (\eta_1^l+\eta_2^l) = \omega^k \wedge \eta_1^l + \omega^k \wedge \eta_2^l$

3)$(\alpha\omega^k)\wedge\eta^l=\omega^k\wedge(\alpha\eta^l)=\alpha(\omega^k\wedge\eta^l)$ with $\alpha\in\mathbb{K}$

3) is completly trivial and 2) should be the same as 1). So I have a question to 1). My proof goes as follows:

$$(\omega_1^k+\omega_2^k)\wedge\eta^l=\frac{1}{k!l!}\sum_{\sigma\in S_{k+l}} \operatorname{sgn}(\sigma)(\omega_1^k+\omega_2^k)(v_1,\dotso, v_k)\eta^l(v_{k+1},\dotso,v_{k+l})$$

The forms are already multilinear. Therefore we have

$$(\omega_1^k+\omega_2^k)(v_1,\dotso, v_k) = \omega_1^k (v_1, \dotso, v_k) + \omega_2^k (v_1,\dotso,v_k)$$

and it what we have to show becomes trivial. But I am unsure if the step above is true.

• Shouldn't this follow from the tensor product being bilinear? Aug 20 '16 at 23:56
• Yea, and you are not even using multilinearity of the forms. Aug 20 '16 at 23:58
• We have not used the tensor product yet. In my opinion the statement follows from the fact, that the forms are already bilinear. Aug 20 '16 at 23:58

The wedge product is defined as follows. Take two forms $\omega,\eta$. Their tensor product is the form $\omega\otimes \eta$ such that (for tuples $v,w$ of vectors of the correct length)

$$(\omega\otimes \eta)(v,w) = \omega(v)\eta(w)$$

If $\omega$ is a form of degree $k$ and $\sigma$ is a permutation of degree $k$ there is a another form $\sigma\cdot \omega$ so that $\sigma\cdot \omega(w) = \omega(\sigma^{-1}w)$ where $\sigma^{-1}=\tau$ acts on $w$ by shuffling its coordinates according to the order dictated by $\tau$.

Now $\omega\wedge \eta$ is obtained from $\omega\otimes\eta$ by applying the element $$\varepsilon = \frac{1}{p!q!}\sum_{\sigma\in S_{p+q}}(-1)^\sigma\sigma$$ which behaves linearly with respect to sums of tensors. Thus the fact that the wedge product is bilinear follows from the fact the tensor product of forms is bilinear.

For example, it is evident from the definition of the tensor product that $$(\omega_1+\omega_2)\otimes \eta = \omega_1\otimes \eta+\omega_2 \otimes \eta,$$

and applying the linear transformation $\varepsilon$ gives that $$(\omega_1+\omega_2)\wedge \eta = \omega_1\wedge \eta+\omega_2 \wedge \eta.$$

• Thank you. I am working currently with a book and a pdf on the internet on this topic. Both do not mention tensor products and I have never worked with them in the first place, so it is a little bit odd to me. Both sources just state, that this fact follows from the definition of $\wedge$. Aug 21 '16 at 0:58
• Yes, and I am saying that too. My point is to clarify why it follows from the definition. Aug 21 '16 at 1:08
• If I get your answer right, the tensorproduct $\otimes$ of forms is defined by $(\omega\otimes\eta)(v,w):=\omega(v)\eta(w)$. Now, if $\omega$ is a form of degree $k$ and $\sigma$ is a permutation of degree $k$ there is another form $\sigma\cdot\omega$, so that $\sigma\circ\omega(w_1,\dotso, w_k)=\omega(w_{\sigma^{-1}1},\dotso,w_{\sigma^{-1}k})$. What do you mean by $(-1)^\sigma$? Just $sgn(\sigma)$? Aug 22 '16 at 0:00
• Would this be correct? $((\omega_1+\omega_2)\otimes\eta)(v,w)=(\omega_1+\omega_2)(v)\eta(w)$ by definition of the tensorproduct of forms. Now $\omega_1(v)\eta(w)+\omega_2(v)\eta(w)=\omega_1\otimes\eta+\omega_2\otimes\eta$. Aug 22 '16 at 0:03
• Yes to all of that, except for some missing $v$s and $w$s. Aug 22 '16 at 0:10