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I was reading Chapter 6.2 of Martingales in Banach Spaces by Gilles Pisier. The result is used in the context: $L_2(G) = \bigotimes\limits_{k\geq0}L_2(\mathbb{T})$, where $G=\prod_{k\geq0}\mathbb{T}$ and all groups are equipped with the usual topology and normalized Haar measure. I have the following questions:

  1. Is the following explanation correct?

    The infinite tensor product is difined as an inductive limit. More precisely, it is inductive limit of the functor $ \mathbb Z_{\geq0}\to(\mathsf{Hil}),n\mapsto L_2(\mathbb T)^{\otimes n}$, where morphisms are trivial embeddings.

  2. Can we explain the universal property of $\bigotimes\limits_{k\geq0}L_2(\mathbb{T})$ as an analogue of the Theorem stated in this post, and achieve $H_k=I\otimes\cdots\otimes I\otimes H\otimes I\otimes\cdots$ by universal property? I tried as follows, is this reasonable? Using the unviersal property.

Some mappings are omitted.

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    $\begingroup$ @AnneBauval Sorry for ambiguity. I have editted the statement and hope it is clear now $\endgroup$
    – Confusion
    Commented Jul 3 at 14:41
  • $\begingroup$ Sorry as well: I am incompetent and can only provide these two hopefully useful links: math.stackexchange.com/a/4714584 and en.wikipedia.org/wiki/… $\endgroup$
    – Anne Bauval
    Commented Jul 3 at 15:00
  • $\begingroup$ The inductive limit part would be correct if you specify what the “trivial embeddings” you’re referring to are. After all, there is no canonical embedding of $H^{\otimes n}$ into $H^{\otimes n+1}$, for an arbitrary Hilbert space $H$. This is why infinite tensor products are defined with a pre-specified unit vector $h\in H$, because then you do have an embedding $H^{\otimes n}\ni v\mapsto v\otimes h\in H^{\otimes n+1}$. If you do that, then yes, the infinite tensor product is the inductive limit (categorically speaking, colimit), if your category is Hilbert spaces with contractive linear maps. $\endgroup$
    – David Gao
    Commented Jul 3 at 19:08
  • $\begingroup$ @AnneBauval Thanks. I feel the definition more reasonable. $\endgroup$
    – Confusion
    Commented Jul 4 at 4:44
  • $\begingroup$ @DavidGao Thanks. For embeddings I mean choosing $1\in L_2(\mathbb T)$ . More precisely, $L_2(\mathbb T)^{\otimes n}\to L_2(\mathbb T)^{\otimes (n+1)}$ , $f\mapsto f\otimes 1$ , where $1\in L_2(\mathbb{T})$ is the constant map. $\endgroup$
    – Confusion
    Commented Jul 4 at 4:56

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