What is $v_1\wedge v_2 \wedge \cdots \wedge v_k$?

If $T \in \wedge^k(V)$ and $S \in \wedge^l(V)$, then the definition of their wedge product is $$T \wedge S := \frac{(k+l)!}{ k ! l!}\text{Alt}(T \otimes S) \in \wedge^{k+l}(V).$$

Meanwhile, we have that $$\text{Alt}(T) = \frac{1}{k!}\sum_{\pi \in S_p} {\operatorname{sgn} \, (\pi)} T^\pi.$$ Now my professor defines for $v_i \in V$

$$v_1\wedge v_2 \wedge \cdots \wedge v_k := \mathrm{Alt} (v_1 \otimes v_2 ...\otimes v_k) .$$

My question: do we need to define $v_1\wedge v_2 \wedge \cdots \wedge v_k$? I think we can derive it by two preceding definitions, but unfortunately I'm getting $k! \, \operatorname{Alt} (v_1 \otimes v_2 ...\otimes v_k)$. Is that $k!$ I'm getting a miscalculation?

• I would say that you have a miscalculation somewhere. A sensible definition of wedge product will be associative, and if you're getting a constant out in front like that, then it isn't associative. I imagine that the $(k+\ell)!/k!\ell!$ is designed specifically to overcome this problem. Jul 30 '16 at 17:51
• Your two definitions are clearly inconsistent; just take $k = 2$ in the second expression and $k = \ell = 1$ in the first. There are a few different conventions you could take here, depending on whether you're thinking of exterior powers as a quotient or a subspace of tensor powers. Thinking in terms of quotients is much more convenient (and gets rid of all of these pesky factorial factors) but for some purposes, sometimes people think in terms of subspaces, and then things get messy. Jul 31 '16 at 20:52

You have not miscalculated. It's just that the exterior product of alternating tensors is defined in two different ways and you have been mixing them without noticing. Let me explain.

One definition of $$\wedge$$ is:

given $$\omega\in\bigwedge^r{(\mathbb{R}^n)^*}, \eta\in\bigwedge^s{(\mathbb{R}^n)^*}$$ two alternating tensors and given $$e^1,...,e^n$$ a basis of $$\mathbb{R}^n$$,

$$\omega\wedge\eta = \text{Alt}(\omega\otimes\eta) = \frac{1}{(r+s)!}\sum_{\sigma\in\mathcal{S}_{r+s}}{\text{sgn}(\sigma)\space\sigma(\omega_{i_1,...,i_r}\eta_{i_{r+1}...,i_{r+s}}e^1\otimes\ldots\otimes e^{r+s})} = \frac{\omega_{i_1,...,i_r}\eta_{i_{r+1},...,i_{r+s}}}{(r+s)!}e^1\wedge\ldots\wedge e^{r+s}$$

With this definition, the exterior product (of alt. tensors) of two vectors coincides with the "exterior product of vectors", i.e. the tensor you get by doing $$v_1$$ wedged with $$v_2$$ is precisely $$v_1\wedge v_2$$.

The determinant as an $$n$$-form turns out to be $$n!\space e^1\wedge\ldots\wedge e^n$$. This is not very convenient when dealing with differential forms. So, there's this other definition (which I've called $$\wedge'$$ to distinguish it from the other):

$$\omega\wedge'\eta = \frac{\omega_{i_1,...,i_r}}{r!}\frac{\eta_{i_{r+1},...,i_{r+s}}}{s!}e^1\wedge\ldots\wedge e^{r+s}$$

You can think of this definition as the one you'd get had you applied the $$\text{Alt}$$ operator before wedging and not after wedging. We notice that

$$\omega\wedge'\eta = \frac{(r+s)!}{r!s!}\omega\wedge\eta = \frac{(r+s)!}{r!s!}\text{Alt}(\omega\otimes\eta)$$

The apparent miscalculation you were getting is just the result of using this second definition, which is not compatible with the exterior product of vectors.

In fact, what you've found is that

$$e^1\wedge'\ldots\wedge'e^n = n! \space e^1 \wedge\ldots\wedge e^n = \text{det}$$

and, if I didn't misunderstand this when I learnt it, $$e^i = \text{d}x^i$$, which makes the element of volume $$\text{d}x^1 \ldots \text{d}x^n$$ the determinant (as an $$n$$-form).

EDIT: as Qiaochu Yuan pointed out, these two definitions come from considering the set of alternating tensors either a quotient or a subspace of the set of tensors you're considering. Seems like I came a bit late to answer this, almost two years, but this could be useful for somebody else, so it hasn't been a waste of time to do this, anyway.