@LSpice, I added the missing text. I think you have a point with the $\frac{1}{2}$.This is supposed to all work for a (free?) nodule over $\mathbb{Z}$, right? I gotta work out the details on that.

@LSpice, apologies! I I hope it's fixed now.

Regarding the $1/2$ issue, more generally in positive characteristics what happens is that we always have a pairing $\Psi:{\bigwedge}V^*\times{\bigwedge}V\to\mathbb K$ but ${\bigwedge}V$ cannot be embedded in the tensor algebra because we can't construct an associative wedge product there. However, we can still embed exterior powers into tensor powers by using the unnormalized alternating tensors. But the pairing we get from this embedding is $k!\Psi$ on alternating $k$-tensors, and this is zero for some $k$ in positive characteristic, so we cannot recover $\Psi$ this way.

I don't think that $\Psi$ can be derived naturally descending from the tensor algebra in this case. This embedding of exterior powers into tensor powers is also not compatible with the projection ${\bigotimes}V\to{\bigwedge}$ coming from the universal property of the tensor algebra ${\bigotimes}V$.

@NicholasTodoroff, where or what is $\Psi$? Here's a short summary of how I now see things: 1) Define $V^*\otimes V^*$ as space of bilinear functions on $V$ 2) Define $\Lambda^2V^*\subset V^*\otimes V^*$ as subspace of antisymmetric bilinear functions. 3) There is a canonical isomorphism $V^*\otimes V^*=(V\otimes V)^*$. 4) The natural dual of $\Lambda^2V^*$ is $$(\Lambda^2V^*)^* = (V^*\otimes V^*)^*/(\Lambda^2V^*)^\perp=(V\otimes V)/(\Lambda^2V^*)^\perp.$$ Define $S^2V= (\Lambda^2V^*)^\perp$ and $$\Lambda^2V=(\Lambda^2V^*)^* = (V\otimes V)/S^2V.$$

@Deane $\Psi$ can be derived a number of ways. One way that works even with $V$ a module is to use the canonical coalgebra structure on ${\bigwedge}V$ to construct a product on $({\bigwedge}V)^*$; then $V^*$ naturally embeds in $({\bigwedge}V)^*$ by declaring its action on non-vector multivectors to be zero. Then the universal property gives the isomorphism ${\bigwedge}V^*\cong({\bigwedge}V)^*$, which is $\Psi$. See Quadratic Mappings and Clifford Algebras by Micali and Helmstetter.

When the characteristic is not two, my favorite way is to construct the Clifford algebra $Cl(V^*\oplus V)$ with bilinear form given by $$(V^*\oplus V)^* \cong (V^*)^*\oplus V^* \cong V\oplus V^*\cong V^*\oplus V.$$ Both ${\bigwedge}V$ and ${\bigwedge}V^*$ embed naturally in $Cl(V^*\oplus V)$. Now let $\langle X\rangle = tr(X)/tr(1)$ where $tr(X)$ is the trace of $Y\mapsto XY$ in $Cl(V^*\oplus W)$. Now $\Psi$ is the restriction of the form $(X,Y)\mapsto\langle XY\rangle$.

@NicholasTodoroff, since I define $\Lambda^2V = (\Lambda^2V^*)^*$, the pairing is there by definition.

@Deane Ok. I think I was confused because you've written some isomorphisms as equalities. Well, if we declare something to be true, then it is. I suppose I think of these objects as defined by their universal properties, and I'm most interested in the structures that arise from that. When you work with a construction like this, what I can't help thinking is "is this inherent to the 'exterior' structure, or is this 'merely' a result of the particular construction?" But if that doesn't bother you, then I don't think there's anything wrong with this approach, at least for exterior powers.

As for how you define the wedge product $v\wedge w$ of ${\bigwedge^2}V$ defined like this, you have to say something like: it is the element $X \in ({\bigwedge^2}V^*)^*$ defined by $X(\Phi) = \Phi(v\otimes w)$ for all $\Phi \in {\bigwedge^2}V$.

@NicholasTodoroff, as I say in the last sentence, these can all be defined as functors between the appropriate categories and therefore satisfy the relevant universal properties. So they are not ad hoc. I chose not to define everything in terms of universal properties, because it's longer and harder to follow. Instead, I provided explicit definitions. The universal properties can be verified easily, for example, using bases for all of the vector spaces.

I don't understand your last comment. Maybe there's a typo? If $v, w \in V$, then $v\wedge w \in (\Lambda^2V^*)^*$ is defined as follows: For any $\Theta \in \Lambda^2V^*$, $$(v\wedge w)(\Theta) = \Theta(v,w). $$ This is equivalent to the definition I give in my answer.

Ok. But latex isn’t rendered here?

@Deane