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$definedbymatrix-matrixproductover$$\mathbb{F}$ = $\mathbb{R}$$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}$$. Aside from relating the two celebrated quantities, this insight allows us to rewrite Grothendieck’s inequality as a norm inequality $$\parallel {\mu }_{l,m,n}{\parallel }_{1,2,\mathrm{\infty }}=\underset{X,Y,M\mathrm{\ne }0}{\max }\frac{\mathrm{\mid }\mathrm{tr}(XMY)\mathrm{\mid }}{\parallel X{\parallel }_{1,2}\parallel Y{\parallel }_{2,\mathrm{\infty }}\parallel M{\parallel }_{\mathrm{\infty },1}}\le K_G\mathrm{.}$$ We prove that Grothendieck’s inequality is unique: If we generalize the $(1,2,\mathrm{\infty })$-norm to arbitrary $p,q,r\in [1,\mathrm{\infty }]$, $$\parallel {\mu }_{l,m,n}{\parallel }_{p,q,r}=\underset{X,Y,M\mathrm{\ne }0}{\max }\frac{\mathrm{\mid }\mathrm{tr}(XMY)\mathrm{\mid }}{\mathrm{\parallel }X{\mathrm{\parallel }}_{p,q}\mathrm{\parallel }Y{\mathrm{\parallel }}_{q,r}\mathrm{\parallel }M{\mathrm{\parallel }}_{r,p}},$$ then $(p,q,r)=(1,2,\mathrm{\infty })$ is, up to cyclic permutations, the only choice for which $\parallel {\mu }_{l,m,n}{\parallel }_{p,q,r}$ is uniformly bounded by a constant independent of $l,m,n$.