# 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 $H$. – Nik Weaver Jun 26 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 – Anthony Quas Jun 26 at 0:59
• Both of these seem like great answers (possibly with a little expansion)! – LSpice Jun 26 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 – Yemon Choi Jun 27 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. – Anthony Quas Jun 27 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.