This question is related to my previous question. The $n$-th exterior power of a vector space (over some field of characteristic zero) $U$ is a pair $(\wedge^{n}, \bigwedge^{n}U)$ where $\bigwedge^{n}U$ is a vector space over the same field and $\wedge^{n}: \overbrace{U\times \cdots \times U}^{\text{$n$ times}} \to \bigwedge^{n}U$ is an alternating $n$-linear map satisfying the universal property.
It can be proved that the $n$-th exterior power of $U$, if it exists, is unique up to isomorphisms. To prove that it in fact exists, one can explicitly construct a pair with the desired properties. In the literature, this is most commonly done by taking $\bigwedge^{n}U = \bigotimes^{n}U/K_{n}(U)$, that is, the quotient space of the tensor algebra with the subspace $K_{n}(U)$ generated by products of the form $u_{1}\otimes \cdots \otimes u_{n}$ with $u_{i} = u_{j}$ for at least one pair of indices $i \neq j$.
One important result then is that such $\bigwedge^{n}U$ is isomorphic to $\text{Alt}^{n}(U)$, the space of alternating $n$-tensors. In other words, one can identify:
$$\wedge^{n}(u_{1},…,u_{n}) = u_{1}\wedge\cdots \wedge u_{n} = \frac{1}{n!}\sum_{\sigma \in S_{n}}\text{sign}(\sigma)u_{\sigma(1)}\otimes \cdots \otimes u_{\sigma(n)}.$$
There is one thing that bothers me and, after a quick search on MSE, it seems that it bothers a lot of other students as well. I still did not find a clear explanation of it, this is why I am asking this question.
For simplicity, assume that $U$ is a finite-dimensional vector space over $\mathbb{C}$. Let $f: U\times \cdots \times U \to \text{Alt}^{n}(U)$ be given by:
$$f(u_{1},…,u_{n}) := \frac{1}{n!}\sum_{\sigma \in S_{n}}\text{sign}(\sigma)u_{\sigma(1)}\otimes \cdots \otimes u_{\sigma(n)}$$
be the alternating map. As the name suggests, this is an alternating $n$-linear map and, from my previous question, it satisfies the universal property. In other words, $f$ can be used to define $\bigwedge^{n}U$; simply take $\bigwedge^{n}U = \text{Alt}^{n}(U)$ and $\wedge^{n} = f$.
However, this is not the standard construction and many posts on MSE claim that this is not a good way of defining $\bigwedge^{n}U$. But I don't understand why.
To be fair, some of the questions I saw about this topic have a more abstract setting; one takes $U$ to be a $\mathbb{K}$-module. I don't have much background in algebra yet, but maybe in these more abstract settings these constructions are not equivalent for some reason. Concerning this topic and in the case of vector spaces, I have the following questions:
- Are these constructions equivalent or not? Can I use the alternating map $f$ to define my exterior power $\bigwedge^{n}U$? If so, does it work only in finite dimension or not? (It does not seem to me that it depends on the dimension but I might be missing something).
- Is there any real advantage of defining $\bigwedge^{n}U$ as $\bigotimes^{n}U/K_{n}(U)$ instead of $\bigwedge^{n}U = \text{Alt}^{n}(U)$?
Best Answer
I can think of a couple of reasons:
The definition of the alternating tensors only works over characteristic $0$ (or characteristic $> n$, if $n$ is fixed) since you are dividing by $n!$.
This is less formal, but perhaps more important: you don't want to mix the directions of universal properties. Here's what I mean. By definition, tensor products are "easier" to map out of: the universal property is an exact characterization of maps out of $\bigotimes^n U \to V$, they are precisely multilinear maps $U^{\times n} \to V$. On the other hand, subspaces are "easier" to map into: given $V \subset W$, a linear map $Z \to V$ is exactly a linear map $Z \to W$ that takes every element of $Z$ to $V$. For either of these, maps in the other direction, ie into a tensor product or out of a subspace, are not so easy to characterize. Since your alternating algebra is defined as a subspace of a tensor product, this will make mapping both into and out of it "harder". On the other hand, a quotient is also "easy" to map out of, so maps out of the exterior power are easy to characterize, indeed given by its universal property: a map from $\bigwedge^n \to V$ is precisely a multilinear map $U^{\times n} \to V$ that vanishes on tensors with repeated factors.
Sort of related to 2, given an element of the tensor power $\bigotimes^n U$, it is not so obvious to determine if it is in the subspace generated by alternating tensors. But on the other hand: given two elements of $\bigotimes^n U$, it is not so obvious if they have the same image in $\bigwedge^n U$. So this is not necessarily an advantage or disadvantage.