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In Gravitation (Misner, Thorne & Wheeler), the wedge product over a vector space $V$ is introduced as just the antisymmetrised tensor product: for $u, v \in V$, they define $$u \wedge v = u \otimes v - v \otimes u$$ I'm trying to work out how to fit this together with the characterisation of the exterior algebra $\Lambda V$ as the quotient of the tensor algebra $T(V)$ by the two-sided ideal $I$ generated by tensor products of the form $v \otimes v$. In that characterisation (at least in Wikipedia), the wedge product is defined by $$[u] \wedge [v] = [u \otimes v]$$ where square brackets indicate equivalence classes, for the equivalence relation $\sim$ induced by the ideal (i.e., $u \sim v$ iff $v = u + i$, for some $i \in I$). At first, I took it that the way they fit together was that we would be able to show that

$$u \otimes v \sim u \otimes v - v \otimes u$$

But so far as I'm able to tell, this isn't the case: what we can show instead is that $u \otimes v \sim -v \otimes u$, and hence that $$u \otimes v \sim \frac{1}{2} (u \otimes v - v \otimes u)$$ Having browsed a bit, this answer suggests to me that what's going on is just that there are two ways of relating the exterior algebra to the tensor algebra: the MTW characterisation amounts to treating $\Lambda V$ as a subalgebra of $T(V)$ (specifically, the subalgebra generated by antisymmetric tensor products), rather than as a quotient algebra of $T(V)$. Is that correct? If so, what are the advantages of a subalgebra vs a quotient algebra construction, and vice versa?

(Apologies if this is overly duplicative of the linked discussion - it was a bit too abstract for me to be sure I'd followed it correctly.)

anygivenpoint
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2 Answers2

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Over a real vector space $V$ one can regard the second exterior power $\Lambda^2 V$ as either a subspace of $V\otimes V$ (the space of antisymmetric tensors) or as a quotient space of $V\otimes V$ (by factoring out the subspace generated by the $v\otimes v$). Over $\Bbb R$ these definitions are equivalent; the equivalence class $[v\otimes w]$ corresponding to the tensor $\frac12(v\otimes w-w\otimes v)$.

But over a vector space of characteristic 2, or over a ring where $2$ is not invertible, these definitions can vary. I would say the right definition is as a quotient of $V\otimes V$. But let $V$ be a vector space of dimension $n$ over the field $\Bbb F_2$. Here antisymmetry is the same as symmetry, so the antisymmetric tensors in $V\otimes V$ have dimension $\frac12n(n+1)$. However the "right" definition as a quotient gives the dimension of $\Lambda^2(V)$ as $\frac12n(n-1)$.

Of course all these considerations extend to $\Lambda^k(V)$. Over a field of characteristic zero, it matters not which approach we take.

Angina Seng
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From your computation, the two definition are equivalent apart from a factor of $2$.

Rust Z
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