Grothendieck constant is norm of Strassen matrix multiplication tensor

We show that two important quantities from two disparate areas of complexity theory --- Strassen's exponent of matrix multiplication $\omega$ and Grothendieck's constant $K_G$ --- are intimately related. They are different measures of size for the same underlying object --- the matrix multiplication tensor, i.e., the $3$-tensor or bilinear operator $\mu_{l,m,n} : \mathbb{F}^{l \times m} \times \mathbb{F}^{m \times n} \to \mathbb{F}^{l \times n}$, $(A,B) \mapsto AB$ defined by matrix-matrix product over $\mathbb{F} = \mathbb{R}$ or $\mathbb{C}$. It is well-known that Strassen's exponent of matrix multiplication is the greatest lower bound on (the log of) a tensor rank of $\mu_{l,m,n}$. We will show that Grothendieck's constant is the least upper bound on a tensor norm of $\mu_{l,m,n}$, taken over all $l, m, n \in \mathbb{N}$. Aside from relating the two celebrated quantities, this insight allows us to rewrite Grothendieck's inequality as a norm inequality $$\lVert\mu_{l,m,n}\rVert_{1,2,\infty} =\max_{X,Y,M\neq0}\frac{|\operatorname{tr}(XMY)|}{\lVert X\rVert_{1,2}\lVert Y\rVert_{2,\infty}\lVert M\rVert_{\infty,1}}\le K_G.$$ We prove that Grothendieck's inequality is unique: If we generalize the $(1,2,\infty)$-norm to arbitrary $p,q, r \in [1, \infty]$, $$\lVert\mu_{l,m,n}\rVert_{p,q,r}=\max_{X,Y,M\neq0}\frac{|\operatorname{tr}(XMY)|}{\|X\|_{p,q}\|Y\|_{q,r}\|M\|_{r,p}},$$ then $(p,q,r )=(1,2,\infty)$ is, up to cyclic permutations, the only choice for which $\lVert\mu_{l,m,n}\rVert_{p,q,r}$ is uniformly bounded by a constant independent of $l,m,n$.