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Let $V$ be a vector space. We know we have the associated exterior algebra which is $$ \bigwedge V\equiv\bigoplus_{n=0}^{\infty}\bigwedge^nV $$ where $\bigwedge^nV$ is the $n$th exterior power of $V$.

If $(e_j)_{j=1}^{\dim{V}}$ is a basis for $V$, then we know $(e_{j_1}\wedge\dots\wedge e_{j_n})_{ 1\leq j_1<\dots<j_n\leq \dim{V}}$ is a basis for $\bigwedge^nV$.

If we anti-symmetrize such elements: $$ \sum_{\sigma\in S_n} \mathrm{sign}(\sigma)e_{j_1}\wedge\dots\wedge e_{j_n}$$ we get the so-called Slater determinant associated with $j_1,\dots,j_n$, which does not care about the ordering of the indices (via anti-symmetry) and so the various Slater determinants form the so-called "occupation basis" which is labelled not by a sequence $j_1,\dots,j_n$ but rather by stating whether a state $j\in\{1,\dots,\dim{V}\}$ appears in the determinant or not, i.e., specifying a sequence $l_1,\dots,l_d\subseteq\{0,1\}$ such that $l_1+\dots+l_d=n$.

Once we have the occupation basis though, we can forget about the constraint $l_1+\dots+l_d=n$ that places us in a particular direct summand $\bigwedge^nV$ and instead we may just think of any sequence $l_1,\dots,l_d\subseteq\{0,1\}$ as specifying a basis vector of $\bigwedge V$, and hence we have set up an isomorphism $$ \bigwedge V\cong\bigotimes_{k=1}^{\dim V}\mathbb{C}^2\,. $$

My question is: does any of this make sense when $V$ is infinite dimensional? How to make sense of $\bigotimes_{k=1}^{\infty}\mathbb{C}^2$? I suppose one has to ask for some topological properties on $V$, e.g., that it be a Hilbert space? Given a Hilbert space $\mathcal{H}$, is there a canonical name for $\bigotimes_{v\in B}\mathbb{C}^2$ where $B$ is a basis of $\mathcal{H}$? Is that object indeed isomorphic to $\bigwedge\mathcal{H}$?

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I've seen this construction twice. The first in Alice Rogers book, Supermanifolds and also Bryce DeWitts book, also titled the same.

DeWitt just introduces the infinite exterior algebra through a countably infinite basis. If I recall correctly, the same goes for Rogers.

I don't see why the usual algebraic construction of the exterior algebra shouldn't go through even if the dimension of the underlying space $V$ is infinite. This goes by taking the free vector space on all elements of $V$ and then quotienting by the ideal generated by the exterior relation.

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