Exterior algebra of normed spaces

Question

This question is related to my prior question, but this one is aimed, even though it's more general. If $V$ is a vector space, we define the exterior algebra of $V$ do be: $$\bigwedge V := \bigoplus_{n=0}^{\infty}\bigwedge^{n}V $$ where $\bigwedge^{n}V$ is the $n$-fold exterior power of $V$ and I used the identification $\bigwedge^{0}V = \mathbb{C}$ and $\bigwedge V = V$. Thus, an element $v \in \bigwedge V$ is a sequence $v=(v_{0},v_{1},...)$, with $v_{n}\in \bigwedge^{n}V$, with all but finitely many nonzero entries.. $\bigwedge^{n}V$ can be realized as the subspace of all skew-symmetric tensors of $\overbrace{V\otimes \cdots \otimes V}^{\text{n times}}$.

My question is: If $V$ is replaced by a normed vector space $U$, is $\bigwedge^{n}U$ defined in the same algebraic way as before? I know that I can induce a norm on $\bigwedge^{n}U$ from $U$ but I realy don't know anything about the construction of exterior powers of normed vector spaces. Is that any different from the algebraic one? Also, if $U$ is Banach, is $\bigwedge U$ also Banach or we need to complete it?

Answer 1

If $V$ is a Hilbert space there is a standard notion of alternating tensor. First, we have a definition of full tensor products of Hilbert spaces such that if $\{e_i\}$ is an orthonormal basis of $V$ then $\{e_i \otimes e_j\}$ is an orthonormal basis of $V \otimes V$ (and similarly for more than two factors). Then we have a notion of symmetric and antisymmetric parts of $V^{\otimes n}$ coming from the natural action of $S_n$ on this space. The antisymmetric part is considered as a space of alternating tensor over $V$, and the exterior algebra is taken to be the direct sum of the alternating tensor powers of $V$. This is the antisymmetric or "bosonic" Fock space.

I don't know about other Banach spaces. Anthony Quas gives a reference for this in the comments. But in general there are many reasonable ways to norm tensor products of Banach spaces, so I don't think it's fair to expect there to be a really canonical answer.