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Deane
Deane
02:51
The asymmetry is the crucial point. In any application of exterior algebras, where you do not want to assume there is an inner product (or nondegenerate quadratic form), it is not possible to define in a basis-independent way $\bigwedge^2V$ as a subspace of $V\otimes V$ and $\bigwedge^2V^*$ as a subspace of $V^*\otimes V^*$ and have $\bigwedge^2V$ and $\bigwedge^2V^*$ be dual to each other (i.e., have a basis-independent pairing).
This stems from the fact that if $S$ is a subspace of a vector space $X$, $S^*$ is not a subspace of $X^*$.
Deane
Deane
03:07
So, despite the natural isomorphism $(V^*)^*=V$, in many settings (such as differential geometry), the vector spaces $V$ and $V^*$ do not play symmetric roles. You start with a vector space $V$ that is fundamental (e.g., tangent vectors on a manifold) and then view $V^*$ as being defined as linear functions on $V$ (e.g., cotangent vectors). So it is OK to have $\bigwedge^kV^*$ be a subspace of $(V^*)^{\otimes k}$ but $\bigwedge^kV$ be a quotient space of $V^{\otimes k}$.
 
13 hours later…
Nicholas Todoroff
Nicholas Todoroff
15:44
@Deane We agree here! This is what I was saying. This failure happens iff we are in (sufficiently small) positive characteristic! In sufficiently high positive characteristic we can uniformly define ${\bigwedge^2}V$ and ${\bigwedge^2}V^*$ as spaces of tensors, and we do get a pairing between them from the canonical tensor pairing in a completely basis-independent way.
But what I'm also saying is that this won't be exactly the pairing you expect! It will be the pairing you describe with some extra $k!$ factors. The question then is "why do we not expect this?" Or maybe "why is this arguably equally natural pairing undesirable, and actually not as natural?"
The answer I give to that question is that, by virtue of both ${\bigwedge^2}V$ and ${\bigwedge^2}V^*$ being "exterior" structures, we can derive the pairing I called $\Psi$ and this disagrees with the tensor pairing.
Your construction instead is saying ${\bigwedge^2}V$ should be defined exactly as the structure that pairs with your definition of ${\bigwedge^2}V^*$ to give us $\Psi$, and that's why I find it to be an unsatisfying definition: it doesn't answer the question of "why $\Psi$?"
What I was pointing out with the asymmetry is that this question seems to still be lurking in the background in the form of the question "recognizing that $(V^*)^*\cong V$, how do we cohere $({\bigwedge^2}V^*)^*$ and ${\bigwedge^2}(V^*)^*$?"
Deane
Deane
16:01
Could we stick to characteristic 0?
And could you remind me what $\Psi$ is? The definition of $\bigwedge^2 V$ is forced on you once you decide that $\bigwedge^2V^*$ is defined as a subspace of $V^*\otimes V^*$ and want $\bigwedge^2V^* = (\bigwedge^2V)^*$. The ad hoc assumption is that you want this duality. In most areas of differential geometry, we get around this by never using $\Lambda^2V$ at all. We simply assume that differential forms are multilinear functions of the tangent space and work with them that way.
There's almost never a need to take wedge products of tangent vectors, especially if you want basis-independent (i.e., coordinate independent) definitions and theorems.
Without using any inner product.
 
3 hours later…
Nicholas Todoroff
Nicholas Todoroff
19:19
$\Psi$, for me, is defined using the whole exterior algebra ${\bigwedge}V$. There are three derivations I know of, all of which agree. (2) and (3) are the ones I mentioned previously.

1. Every $\omega \in V^*$ extends uniquely to an antiderivation $i_\omega$ on ${\bigwedge}V$. So we have a map $V^* \to End({\bigwedge}V)$ which extends to ${\bigwedge}V^* \to End({\bigwedge}V)$, giving the interior product $\Omega\mathbin\lrcorner X$ for $\Omega \in {\bigwedge}V^*$ and $X \in {\bigwedge}V$. The pairing is now the scalar part $\Psi(\Omega, X) = \langle\Omega\mathbin\lrcorner X\rangle_0$. This…
That last display equation should be $$W^* \cong (V^*)^*\oplus V^* \cong V\oplus V^* \cong W.$$
"In most areas of differential geometry, we get around this by..."
I agree, and practically speaking all we need is some pairing and then geometric considerations (i.e. integration over cubes, something we "know" how to do)
show us how to adjust this pairing if necessary. I wouldn't say the above view of $\Psi$ is any sort of practical consideration, if anything it is aesthetic.
"There's almost never a need to take wedge products of tangent vectors, especially if you want basis-independent (i.e., coordinate independent) definitions and theorems."
Can you give a specific example?

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