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I am interested in the $\ell^1$ analogue of direct sums for Operator spaces, e.g. Operator Space Dictionary. Briefly, and operator space is either a concrete subspace of $B(H)$, the operators on a Hilbert space, or equivalently a normed space $E$ with norms on the matrix spaces $M_n(E)$, satisfying Ruan's axioms. The analogue of the $\ell^\infty$ direct sum can be stated as $M_n(E\oplus F) = M_n(E)\oplus M_n(F)$, which is compatible with the diagonal embedding $B(H)\oplus B(K)\rightarrow B(H\oplus K)$.

As a Banach space, the dual of $E\oplus F$ is $E^*\oplus_1 F^*$ the $\ell^1$ sum. We use this idea to define an operator space structure on $E\oplus_1 F$ be embedding it into $(E^*\oplus F^*)^*$. One can then show, in increasing order of difficulty (IMHO):

  • $E\oplus_1 F$ has the universal property that if $u:E\rightarrow X, v:F\rightarrow X$ are complete contractions, then $u\oplus v: E\oplus_1 F\rightarrow X$ is a complete contraction.
  • $E\rightarrow E\oplus_1 F$ (and for $F$) is a complete isometry, and $E\oplus_1 F\rightarrow E$ (and for $F$) is a complete quotient map.
  • $(E\oplus_1 F)^* = E^* \oplus F^*$
  • $(E\oplus F)^* = E^*\oplus_1 F^*$. This in particular had me stumped for a bit; I needed to use the fact that $M_n(E)^{**} = M_n(E^{**})$.

I am struggling to find references. Points (2) and (3) above are covered in these notes of Blecher. The books by Paulsen and Effros & Ruan seem not to consider $\oplus_1$. The book of Blecher & Le Merdy leaves all the proofs to the reader. The original paper of Blecher also does not give details for points (3) and (4). Pisier's book instead defines $\oplus_1$ using point (1) (the universal property) but leaves (3) and (4) as exercises.

I would like a reference to a clear proof of (3) and (4).

Alternatively, am I missing some genuinely "easy" argument?

In particular, just using the universal properties, I can show that $(E\oplus_1 F)^* = E^*\oplus F^*$. How can one give an analogous proof that $(E\oplus F)^* = E^*\oplus_1 F^*$?

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  • $\begingroup$ belated comment about the last part: there is a sort of meta-argument why one can expect a universal-property proof of the first isomorphism but might have to work harder for the second one. Namely, duals of coproducts will be products of the duals, but duals of products need not be coproducts of the duals $\endgroup$
    – Yemon Choi
    Commented May 7, 2019 at 3:25
  • $\begingroup$ This is illustrated by considering the category of Banach spaces and linear contractions, when the coproduct of countably many copies of the ground field is $\ell_1$ while the product of countably many copies of the ground field is $\ell_\infty$, and we have $(\ell_1)^*\cong\ell_\infty$ but $(\ell_\infty)^* \not\cong \ell_1$ $\endgroup$
    – Yemon Choi
    Commented May 7, 2019 at 3:27
  • $\begingroup$ All this is related to the POV where the dual space functor (let's say for $\newcommand{\BSp}{\sf BSp}\BSp$ but probably this works for OpSp as well) is viewed as $D: \BSp \to \BSp^{op}$ with corresponding $D^{op}:\BSp^{op}\to\BSp$, and then $D$ is left adjoint to $D^{op}$ but $D^{op}$ is not left adjoint to $D$. Now left adjoints preserve colimits, and since binary coproduct in $\BSp$ is $\oplus_1$ while binary product is $\oplus$, you get $D(E\oplus_1 F) = (DE)\oplus (DF)$. But $D^{op}$ need not send colimits in $\BSp^{op}$ (limits in $\BSp$) to colimits in $\BSp$. $\endgroup$
    – Yemon Choi
    Commented May 7, 2019 at 3:42
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    $\begingroup$ @Matthew Daws Is there any follow-up on this? I'm interested in the same thing, so maybe you discovered some useful references? $\endgroup$
    – J. De Ro
    Commented Sep 12, 2023 at 19:08

1 Answer 1

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Maybe I can answer my own question, given it's been 4 years.

I didn't find any references, so I wrote up some brief notes, just to have a self-contained proof. You can find this at this link, to GitHub.

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  • $\begingroup$ Thanks for writing this down. Looks very good, and exactly what I was looking for! $\endgroup$
    – J. De Ro
    Commented Sep 19, 2023 at 14:05

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