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I'm currently enrolled in an operator spaces course and I'm finding it difficult to understand why we study them in the first place. Functional analysis is motivated well enough for me and even though I don't have a firm grasp on them, I guess I can see why $C^{\ast}$-algebras are studied as well (purely from a quantum mechanics point of view). However I fail to see what the motivation is for studying operator spaces or why they're useful/important. What is their motivation? What led people to be interested in them and what is so special about completely bounded/positive maps that we study them?

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What do you mean by an operator space? Can you please provide an example? – timur Sep 26 '13 at 17:59
This book by Pisier can provide some motivation (see the 3 problems in the introduction). It turns out that OS provide a natural framework to investigate these natural FA problems. And 0.3 was solved by Pisier that way. The other 2 remain unsolved. – 1015 Sep 26 '13 at 18:27
+1 for Julien. If somebody can explain why operator spaces are worth studying, Pisier is the guy. – Etienne Sep 26 '13 at 18:46
@julien, nice to see you here again! – Norbert Sep 26 '13 at 22:27
On the technical side, this other book of Pisier can be helpful too, once you have found enough motivation for studying OS. Paulsen's book is useful too. Both have exercises and solutions (by Mrinal Raghupathi here for Paulsen's book). – 1015 Sep 27 '13 at 15:51
up vote 18 down vote accepted

In short the list of reasons is:

1) The modern trend in math today is to develop non-commutative analogues of well known theories. Vaguely speaking operator spaces are normed spaces over "non-commutative" scalars, in fact over matricies.

2) There were several long standing problems, that were solved via methods of operator space theory. As soon as a problem is embedded in its natural environment the solution comes in a natural way. For example, thanks to D. P. Blecher, we have a criterion for a Banach algebra $A$ to be isomorphic as algebra to a closed subalgebra of $\mathcal{B}(H)$: multiplication in Banach algebra must be completely bounded for some embedding (as normed space) of $A$ into some space of bounded operators.

3) Mathematicians not only prove theorems but also guess good definitions. Whether a definition is good or not will become evident only after its usage in theorems. It turns out that some notions is better to define withing the scope of operator space theory. For example criterion of amenability of Fourier algebra $A(G)$ for locally compact group $G$ was very unnatural. But if we consider not a simple amenability but an operator amenability then we get a nice characterization in the sense of B. Johnson: Fourier algebra is operator ameanable iff its group is ameanable.

4) Operator spaces posses some unexpected interesting properties. For example in this theory we have a tensor product which both injective and projective and what is more it is not commutative! Also, thanks to W. F. Steinspring, we have quite explicit description of maps between operator spaces. For classical normed spaces this problem is hopeless.

The most I've written here is a a copy-paste from the book Quantum Functional Analysis: Non-coordinate approach by A. Ya. Helemskii.

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Hi Norbert, +1! – 1015 Sep 27 '13 at 12:04
Thanks, @Norbert. I guess it's become pretty clear to me that my background is a bit too weak to appreciate or even understand operator spaces. – Cameron Williams Sep 27 '13 at 15:11

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