Failure of the trilinear operator space Grothendieck theorem

Abstract

Failure of the trilinear operator space Grothendieck theorem, Discrete Analysis 2019:8, 16 pp. Let $\beta:\ell_\infty^n\times \ell_\infty^n\to\mathbb C$ be a bilinear form. We define its operator norm by the formula $$\|\beta\|=\sup\{|\beta(x,y)|:\|x\|_\infty=\|y\|_\infty=1\}.$$ Let $(A_{ij})$ be the matrix of $\beta$, so that $$\beta(x,y)=\sum_{ij}A_{ij}x_iy_j.$$ This second formula can be generalized in interesting ways. For instance, we can replace the coefficients $x_i$ and $y_j$ of $x$ and $y$ by vectors that live in a Hilbert space $H$, and replace the formula by $$\tilde\beta(x,y)=\sum_{ij}A_{ij}\langle x_i,y_j\rangle.$$ It is natural to define $\|x\|_\infty$ to be $\max_i\|x_i\|_2$ when $x$ is a vector-valued sequence like this, which allows us to define a norm for $\tilde\beta$ using the same formula (appropriately interpreted) as for $\beta$. An important inequality of Grothendieck states that there is an absolute constant $K$ such that $\|\beta\|\leq\|\tilde\beta\|\leq K\|\beta\|$. Grothendieck also showed (using the Hahn-Banach theorem) that a consequence of this inequality is that there always exists a matrix $(\tilde A_{ij})$ of operator norm at most $K\|\beta\|$ and unit vectors $\lambda,\mu\in\mathbb C^n$ such that $A_{ij}=\lambda_i\tilde A_{ij}\mu_j$ for every $i,j$. Writing $\lambda,\mu$ also for the multipliers that multiply the $i$th coordinate by $\lambda_i$ and $\mu_i$, respectively, and $\tilde\alpha$ for the linear map with matrix $(\tilde A_{ij})$, we obtain from this the formula $$\beta(x,y)=\langle\tilde\alpha\lambda x,\mu y\rangle$$ Note that $\lambda$ and $\mu$ are linear maps from $\ell_\infty^n$ to $\ell_2^n$ and that they have norm 1. Setting $\Psi_1=\tilde\alpha\lambda$ and $\Psi_2=\mu$, we therefore have a factorization $$\beta(x,y)=\langle\Psi_1x,\Psi_2y\rangle$$ where $\Psi_1$ and $\Psi_2$ are linear maps from $\ell_\infty^n$ to $\ell_2^n$ with $\|\Psi_1\|\|\Psi_2\|\leq K\|\beta\|$. The space $\ell_\infty^n$ with pointwise multiplication is a prototypical example of a commutative $C^*$-algebra, and the result above generalizes straightforwardly to all commutative $C^*$-algebras. Grothendieck conjectured that it could also be generalized to noncommutative $C^*$-algebras, a conjecture that was eventually proved many years later by Gilles Pisier under certain extra assumptions and Uffe Haagerup in full generality. A further direction of generalization concerns a concept known as an _operator space_, which, very roughly, is what you get if you take account not just of standard operator norms but also of a sequence of related norms on matrices. Observe that if $\mathcal A$ and $\mathcal B$ are two algebras and $\beta:\mathcal A\times\mathcal B\to\mathbb C$ is a bilinear form, then for each positive integer $d$ we can build out of $\beta$ a bilinear map $\beta_d:M_d(\mathcal A)\times M_d(\mathcal B)\to M_d(\mathbb C)$ in a natural way, where $M_d(\mathcal A)$ and $M_d(\mathcal B)$ are the spaces of $d\times d$ matrices with coefficients in $\mathcal A$ and $\mathcal B$, respectively. Indeed, if $A\in M_d(\mathcal A)$ and $B\in M_d(\mathcal B)$, then we define $$\beta_d(A,B)_{ij}=\sum_k\beta(A_{ik},B_{kj}).$$ This can be thought of as standard matrix multiplication, but using the bilinear form $\beta$ in order to "multiply" individual coefficients together. One then says that the bilinear form $\beta$ is _completely bounded_ if the norms of the bilinear maps $\beta_d$ form a bounded sequence, and the supremum of this sequence is called the _completely bounded_ norm $\|\beta\|_{\mathrm{cb}}$. The completely bounded norm has the drawback that it is not invariant under taking the transpose (in a sense that is not hard to make precise). However, there is a simple symmetrization procedure that gives rise to a symmetric version, denoted $\|.\|_{\mathrm{sym}}$. While matrix multiplication is an obvious bilinear map on $M_d(\mathbb C)\times M_d(\mathbb C)$, it is not the only natural one. Another is the tensor product, which takes $M_d(\mathbb C)\times M_d(\mathbb C)$ to $M_{d^2}(\mathbb C)$. We can describe it explicitly by identifying $\{1,\dots,d\}^2$ with $\{1,\dots,d^2\}$ in a natural way, and then using the formula $$(A\otimes B)_{(i_1,i_2),(j_1,j_2)}=A_{i_1j_1}B_{i_2j_2}.$$ As with matrix multiplication, this then has an easy algebra-valued generalization. With $\mathcal A,\mathcal B,\beta,A,B$ as before, we can define $$\tilde\beta_d(A,B)_{(i_1,i_2),(j_1,j_2)}=\beta(A_{i_1j_1},B_{i_2j_2}).$$ This allows us to associate with the bilinear form $\beta$ another sequence of norms, namely the operator norms of the $\tilde\beta_d$. If this sequence is bounded, then $\beta$ is said to be _jointly completely bounded_, and the supremum is denoted by $\|\beta\|_{\mathrm{jcb}}$. It turns out that an appropriate operator-space generalization of Grothendieck's theorem is the statement that $\|\beta\|_{\mathrm{jcb}}\leq\|\beta\|_{\mathrm{sym}}\leq K\|\beta\|_{\mathrm{jcb}}$ for some absolute constant $K$, which, it turns out, can be taken to be 2. Another natural direction of generalization is from bilinear forms to trilinear forms. It is not obvious what the right statement should be -- for example, with the first formulations above, it is not obvious what trilinear form should play the role of the inner product in a Hilbert space. However, interest in this possibility grew with the operator space version, since the definitions just given can be generalized fairly straightforwardly to trilinear forms, which suggested that perhaps a satisfactory trilinear Grothendieck statement had been found. However, this paper dashes that hope, in the process answering a question of Gilles Pisier, by giving a counterexample. Interestingly, and very appropriately for a paper in Discrete Analysis, the proof makes use of ideas from additive combinatorics. In particular, it makes use of a noncommutative version of the generalized von Neumann inequality. This allows the authors to bound the jointly completely bounded norm of a trilinear form built out of a function $f$, defined on a finite Abelian group, in terms of the $U^3$-norm of $f$. This is coupled with an argument of Varopoulos that provides a significantly larger lower bound for the symmetrized completely bounded norm.