# Associativity property of wedge products

Regarding notation, I have a bit that says: The linear function $$e_i^* \in V^*$$ determined by $$e_i^*(e_j) = \delta_{ij}$$ form the basis of $$V^*$$, which is called the dual basis for $$\epsilon = (e_1, ..., e_n)$$. I thought that would explain it better than me trying to explain it myself. A quick side question, if I was to explain that, I was just going to say that $$e_i^*$$ is a vector in the dual basis and so $$e_i$$ is a vector in the real vector space. Is that correct?

In my lecture notes, regarding the associativity property, it says

$$( \alpha \wedge \beta) \wedge \gamma = \alpha \wedge (\beta \wedge \gamma) = \mathrm{Alt}(\alpha \otimes \beta \otimes \gamma)$$

But in an example, it goes onto say

If $$\alpha = e_1^*, \beta = e_2^*, \gamma = e_3^*$$ act on $$(e_1, e_2, e_3)$$, then compute $$(e_1^* \wedge e_2^*) \wedge e_3^* (e_1, e_2, e_3) = \mathrm{Alt}[(e_1^* \wedge e_2^*) \otimes e_3^*)(e_1, e_2, e_3)]$$

$$= \frac{1}{3!} \Sigma_{\sigma \in S_3} (-1)^{\sigma} (e_1^*, e_2^*)(e_{\sigma(1)}, e_{\sigma(2)} \cdot \underline{e_3^*(e_{\sigma(3)})}$$

My questions are this:

• Why, in the 'Alt' brackets in the question, does it say $$(e_1^* \wedge e_2^*) \otimes e_3^*)(e_1, e_2, e_3)$$. From the property, shouldn't it be $$(e_1^* \otimes e_2^* \otimes e_3^*)(e_1, e_2, e_3)$$? EDIT: It turns out my lecturer later goes on to work out the same question but calculating $$\mathrm{Alt}((e_1^* \otimes e_2^* \otimes e_3^*)(e_1, e_2, e_3))$$ to show that they are the same.
• In the underlined bit, in my notes it says that if $$\sigma(3) = 3$$, then this factor is non-zero and therfore equal to $$1$$. If $$\sigma(3) = 1, 2$$, then this factor plays no part in the final answer. Here, I want to know, why is $$0$$ if $$\sigma(3) = 1, 2$$ and $$1$$ if $$\sigma(3) = 3)$$? Is this '$$1$$' thing the same for all factors when computinf the wedge product?

I hope I've explained all this correctly. It's a newish topic and so I'm still not very comfortable with some of the words and notation so please let me know if something confuses you and I will try and correct it. Thank you.

• Your notes are using the "wrong" definition of the wedge product. The "correct" definition is on exterior powers, which are a quotient and not a subspace of tensor powers. See mathoverflow.net/questions/54343/… for a discussion. – Qiaochu Yuan Mar 5 '13 at 18:16
• And since it's a quotient, it inherits associativity from the algebra it is a quotient of (the tensor algebra). – rschwieb Mar 5 '13 at 18:42

## 1 Answer

If I understand the question correctly, the only thing you're having issues with is why your underlined term $e_3^*(e_{\sigma(3)})$ is zero if $\sigma(3) = 1,2$ and is $1$ if $\sigma(3) = 3$. But if you go back to the top of your post, this is precisely the definition of the dual basis vector $e_3^*$. Things don't always work out so nicely, of course. You are taking a wedge product of some very simple functions, and applying them to particularly easy vectors.

As for your first question (which I guess your instructor helped you with), you should see that this is just a consequence of the associativity of your wedge product (what you wrote in your first gray box). What you were confused about is why it says $\operatorname{Alt}[(e_1^* \wedge e_2^*) \otimes e_3^*]$ rather than $\operatorname{Alt}(e_1^* \otimes e_2^* \otimes e_3^*)$. But the first expression here is precisely $(e_1^* \wedge e_2^*) \wedge e_3^*$, so these things are equal (I assume your definition of the wedge product is $\alpha \wedge \beta = \operatorname{Alt}(\alpha \otimes \beta)$).

I hope this was helpful for you!