Haagerup tensor product $\otimes_{\rm h}$ is both injective and projective.

Pisier, Gilles, Introduction to operator space theory, London Mathematical Society Lecture Note Series 294. Cambridge: Cambridge University Press (ISBN 0-521-81165-1/pbk). vii, 478 p. (2003). ZBL1093.46001.

Can the following be true?

Let $q_i : E_i \rightarrow F_i$ be complete quotient maps of operator spaces. Then \begin{equation} {\rm Ker} \, q_1 \otimes_{\rm h} q_2 = {\rm Ker} \, q_1 \otimes_{\rm h} E_2 + E_1 \otimes_{\rm h} {\rm Ker} \, q_2. \end{equation}

I have some hints which tilt me towards believing it is true. Denote by U the operator space on the right. Then $U \subset {\rm Ker} \, q_1 \otimes_{\rm h} q_2$. Then the product map drops to a map on $(E_1 \otimes_{\rm h} E_2)/U$ and we need to prove that this map is injective. As for the algebraic tensor product, one finds an inverse map from $F_1 \otimes_{\rm h} F_2 \simeq E_1/{\rm Ker}\, q_1 \otimes_{\rm h} E_2/{\rm Ker} \, q_2$ to $(E_1 \otimes_{\rm h} E_2)/U$. For this one starts from the bilinear map: $(\hat e_1,\hat e_2) \mapsto \widehat{e_1 \otimes e_2}$, where the hats are the obvious classes. It is well defined and completely bounded. Hence it defines a linear map on the Haagerup product. Then one checks that it is the inverse map we looked for.

I am hesitant because this implies $(E \otimes_{\rm h} F)/(G \otimes_{\rm h} F) \simeq (E/G) \otimes_{\rm h} F$ for any sub-space $G$ of $E$.

  • 1
    $\begingroup$ Welcome to the site! As a fellow appreciator of operator spaces it would hurt if you put some references to the books or notes you're readings and the definitions you're using. This stuff is not well known. Regarding your last question. The Haagerup tensor product is projective. So there's no contradiction. I haven't figured out if the rest of your argument is correct. $\endgroup$ May 21, 2020 at 12:31
  • $\begingroup$ From being projective, one can only extract that the right side of the last equation is metric isomorphic to (E \otimes F)/Ker (q_1 \otimes 1). As such, that statement is far more stronger than one can derive straight from projectivity. $\endgroup$ May 21, 2020 at 14:48

1 Answer 1


I found the answer in the literature (see Complete boundedness of multiple operator integrals by Clément Coine, https://arxiv.org/abs/1908.07879). Specifically, in Proposition 3(iv), it is shown that the Ker I asked about is just the norm-closure of the linear space written in the statement. Of course, sums of closed linear spaces are not generally closed, so the closure should have been there in the first place.


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