Two definitions of antisymmetrization of a tensor?

Question

I am currently learning about tensors and the exterior product, and I have found some contradictory information. I have seen some sources define the antisymmetrization of a tensor as the following:

Let $T \in V^{\otimes k}$ and $\sigma \in S_k$ be a permutation. The antisymmetrization of $T$ is defined as $$\mathscr{A}(T) := \frac{1}{k!} \sum_{\sigma \in S_k} \text{sgn}(\sigma) \sigma(T)$$

This is defined such that $\mathscr{A}(\mathscr{A}(T)) = \mathscr{A}(T)$.

However, I have seen many texts define this operation in the same way but omitting the factor of $1/k!$. Let's call this version of the operation $\mathscr{A}'$. This definition leads to the identity $\mathscr{A}'(\mathscr{A}'(T)) = k! \mathscr{A}'(T)$

This has led to a lot of confusion regarding my understanding of the exterior product as the exterior product is defined as the following operation:

Let $\omega^1 \in \textstyle \bigwedge^k (V)$ and $\omega^2 \in \textstyle \bigwedge^\ell (V)$. The exterior product of $\omega^1$ and $\omega^2$ is defined as $$\omega^1 \wedge \omega^2 := \frac{1}{k!\ell!}\mathscr{A}(\omega^1 \otimes \omega^2)$$

Can someone please clear up this confusion?

Answer 1

Your convention for the factor in front of the antisymmetrization determines a corresponding convention for the factor in front of the exterior product. The factor is uniquely determined by the requirement that it makes the exterior product associative, so in some sense it doesn't matter, but different conventions will make different equations appear more or less nice.

I believe but have not checked carefully that if you use $A$ then the factor should be ${k+\ell \choose k}$ and if you use $A'$ then the factor should be $\frac{1}{k! \ell!}$ (so what you wrote is not quite right, and you can see that it doesn't do the right thing when $k = 0$).

This can all be avoided by choosing an alternate definition of the exterior power based on quotienting the tensor product rather than considering the subspace of antisymmetric tensors. With this definition the wedge product is just directly inherited from the tensor product and there's no need to antisymmetrize or consider any factors anywhere, and it inherits associativity from associativity of the tensor product.

If I had to pick between $A$ and $A'$ then I would go with $A$. An operation called "antisymmetrization" should not do anything to a tensor that is already antisymmetric, so it makes sense to require $A(A(T)) = A(T)$ (idempotence).

Answer 2

See my answer here. In short (and also to expand on it a bit), using you definition of $\mathscr A$ with the $1/k!$ out front lets us define a wedge product such that $$ v_1\wedge\dotsb\wedge v_k := \mathscr A(v_1\otimes\dotsb\otimes v_k),\quad X\wedge Y := \mathscr A(X\otimes Y) $$ where $v_1,\dotsc,v_k\in V$ are vectors and $X,Y$ alternating tensors. This is very natural because $\mathscr A^2 = \mathscr A$ makes it a (vector space) projection; and also because the natural projection $\pi : {\bigotimes}V\to{\bigwedge}V$ derived from the universal property of the tensor algebra ${\bigotimes}V$ satisfies $\pi\circ\mathscr A = \pi$, meaning that in a sense the action of $\pi$ on a tensor "already" involves applying $\mathscr A$.

Now, there are two exterior algebras we're interested in: ${\bigwedge}V$ and ${\bigwedge}V^*$. They also interact: there is a single (up to some signs) canonical way to define a bilinear pairing ${\bigwedge}V^*\times{\bigwedge}V\to\mathbb R$, or (what is the same thing) to define a linear isomorphism ${\bigwedge}V^*\cong({\bigwedge}V)^*$.

If you commit to using $\mathscr A$ to embed ${\bigwedge V}$ in ${\bigotimes}V$, then you are forced by this bilinear pairing to embed ${\bigwedge}V^*$ in ${\bigotimes}V^*$ as follows: $$ \omega_1\wedge\dotsb\wedge\omega_k := \mathscr A_0(\omega_1\otimes\dotsb\otimes\omega_k),\quad \Omega\wedge\Phi := \frac1{k!l!}\mathscr A_0(\Omega\otimes\Phi) = \frac{(k+l)!}{k!l!}\mathscr A(\Omega\otimes\Phi) $$ where $\omega_1,\dotsc,\omega_k\in V^*$, $\Omega$ is an alternating $k$-tensor, $\Phi$ is an alternating $l$-tensor, and $$ \mathscr A_0(\omega_1\otimes\dotsb\otimes\omega_k) = \sum_{\sigma\in S_k}\mathrm{sgn}(\sigma)\omega_{\sigma(1)}\otimes\dotsb\otimes\omega_{\sigma(k)} $$ is the unnormalized antisymmetrization operation.

A way to summarize this is to say that

• Elements of ${\bigwedge}V$, thought of as multivectors representing subspaces of $V$, embed into the tensor algebra via $\mathscr A$.
• Elements of ${\bigwedge}V^*$ thought of as differential forms acting on mulvivectors/subspaces of $V$ embed into the tensor algebra via $\mathscr A_0$.

Answer 3

Rewritten shorter but more rigorous answer:

You have to distinguish between a vector space $V$ and its dual $V^*$, which is a derived object.

$V^*\otimes V^*$ is easier to define than $V\otimes V$. It is the space of bilinear functions on $V$ and the tensor product of $\theta^1,\theta^2 \in V^*$ is $$ (\theta^1\otimes\theta^2)(v_1,v_2) = \theta^1(v_1)\theta^2(v_2). $$ Then $\Lambda^2V^*$ is naturally defined as the subspace of $V^*\otimes V^*$ that contains all antisymmetric bilinear functions on $V$.

The tensor product of $V$ with itself can now be defined as $$V\otimes V = (V^*\otimes V^*)^* $$ and the tensor product of $v, w \in V$ by $$(v\otimes w)(f) = f(v,w). $$

Recall that if $S$ is a subspace of a vector space $X$, then $$S^* = X^*/S^\perp.$$ We can therefore define $$ \Lambda^2V = (\Lambda^2V^*)^* = (V^*\otimes V^*)^*/(\Lambda^2V^*)^\perp = (V\otimes V)/S^2V,$$ where $S^2V = (\Lambda^2V^*)^\perp$.

The wedge product on $V$ now has the natural definition $$ v_1\wedge v_2 = [v_1\otimes v_2] = \frac{1}{2}[v_1\otimes v_2 - v_2\otimes v_1], $$ where $[\cdot]$ denotes the equivalence class. We then define the wedge product on $V^*$ by requiring that, if $(b_1, \dots, b_n)$ is a basis of $V$ and $(\beta^1, \dots, \beta^n)$ is a basis of $V^*$, then $$ (\beta^1\wedge\beta^2)(b_1\wedge b_2) = 1. $$ This implies that $$ \theta^1\wedge\theta^2 = \theta^1\otimes\theta^2-\theta^2\otimes\theta^1. $$ and$$ (\beta^j\wedge\beta^k)(b_j,b_k) = (\beta^j\wedge\beta^k)(b_j\wedge b_k) = 1, $$ This is the right normalization for the geometric interpretation in terms of areas of parallelograms.

All of this can be extended in a straightforward way to $k$-tensors and the wedge product of an exterior $k$-tensor with an exterior $l$-tensor.

Everything above can be shown to be functorial and satisfy appropriate universal properties.