I am reading Blackadar's book on operator algebras, and came upon the notion of spatial tensor product of $C^*$-algebras. It turns out to be the same as the minimal tensor product. To me, the term ''minimal'' makes sense. But why is it called ''spatial'' in the first place? What is the ''space'' being alluded to?


1 Answer 1


When you represent your C*-algebras faithfully on Hilbert spaces, say $A \subseteq B(H), B \subseteq B(K)$, the spatial/minimal tensor product is just the norm closure of algebraic tensor product $A \odot B \subseteq B(H \otimes K)$. In my mind, this is why its called spatial - it is defined as some algebra of operators on the underlying Hilbert space $H \otimes K$. It just so happens that the resulting C*-algebra is independent of the choices of faithful representations and so you have "the" spatial tensor product or "the" minimal tensor product.

  • $\begingroup$ Blackadar defines the minimal tensor as $||\sum_ja_j\otimes b_j||_{min}=\sup ||\sum_j \pi_A(a_j)\otimes \pi_B(b_j)||$ over all representations $\pi_A$ and $\pi_B$ of $A$ and $B$ respectively. Is this equivalent to the formulation you presented? It's not immediately obvious to me, unfortunately, and I can't seem to find a reference for it. $\endgroup$
    – chhro
    Jan 27, 2021 at 14:37

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