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?

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    $\begingroup$ If $H$ is a Hilbert space, then probably the right definition for $\bigwedge H$ is the antisymmetric Fock space over $H$. $\endgroup$
    – Nik Weaver
    Jun 26, 2020 at 0:40
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    $\begingroup$ 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 $\endgroup$ Jun 26, 2020 at 0:59
  • $\begingroup$ Both of these seem like great answers (possibly with a little expansion)! $\endgroup$
    – LSpice
    Jun 26, 2020 at 16:13
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    $\begingroup$ @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 $\endgroup$
    – Yemon Choi
    Jun 27, 2020 at 2:10
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    $\begingroup$ @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. $\endgroup$ Jun 27, 2020 at 3:01

1 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.


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