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.

  • 1
    $\begingroup$ 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. $\endgroup$ Commented Mar 5, 2013 at 18:16
  • $\begingroup$ And since it's a quotient, it inherits associativity from the algebra it is a quotient of (the tensor algebra). $\endgroup$
    – rschwieb
    Commented Mar 5, 2013 at 18:42

1 Answer 1


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!


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .