# Exterior algebra of normed spaces

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?

• If $H$ is a Hilbert space, then probably the right definition for $\bigwedge H$ is the antisymmetric Fock space over $H$. Jun 26, 2020 at 0:40
• In my paper with Philippe Thieullen and Mohammed Zarrabi, we construct in a somewhat Bourbaki style exterior powers of a Banach space. See Appendix A4 of math.uvic.ca/faculty/aquas/papers/paper53.pdf Jun 26, 2020 at 0:59
• Both of these seem like great answers (possibly with a little expansion)! Jun 26, 2020 at 16:13
• @AnthonyQuas I've not had time to read this appendix in detail, but can't one just take the usual n-fold projective tensor power of a Banach space E (much studied) and then mod out by the (closure of) the usual subspace of degenerate tensors? This is how one gets e.g. the symmetric (projective) tensor powers in BSp world Jun 27, 2020 at 2:10
• @YemonChoi: I am really not too expert in this area. Our goal was to have a framework which makes it easy to calculate growth rates of finite-dimensional volumes for random matrix products. I suspect the quotient formulation you’re proposing might be hard to calculate with. Jun 27, 2020 at 3:01

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.