TITLE: Categorical duals in Banach spaces
QUESTION [6 upvotes]: Near the bottom of the nlab page for Banach space I see "To be described: duals (p+q=pq)".
Are $(\mathbb{R}^n)_p$ and $(\mathbb{R}^n)_q$ dual objects in the closed symmetric monoidal category of Banach spaces and linear contractions (with the tensor product described on that page)?
Edit: take n=2, p=1, q=∞.  Then the question becomes whether $V \times V$ (which is $V^2$ with the $l_\infty$ norm) is isomorphic to $(\mathbb{R}^2)_\infty \otimes V$.  But it seems to me that the functor $V \mapsto V \times V$ does not even commute with coproducts... is that right?

REPLY [2 votes]: My suspicion is "no", because if I recall correctly the map $I \to V \otimes V^*$ naturally lands in the injective tensor product, not the projective tensor product, and it is the latter which appears as the ``correct'' tensor product for the SMC category of Banach spaces and linear contractions.
In the toy example given, $V\oplus V$ with the sup norm is the same as continuous maps from a 2-point set to $V$, equipped with sup-norm, and I'm pretty sure that this is indeed isometrically linearly isomorphic to ${\mathbb R}^2 \check{\otimes} V$ i.e. the injective tensor product.
EDIT: as Reid points out my remarks above assume without justification that the inj. t.p. does differ from the proj t.p. in the specific case being considered. I think this is indeed the case. Take $V$ to be ${\mathbb R}^2$ with usual Euclidean norm. The projective tensor product of $V$ with $V^\*$ can be identified with $M_2({\mathbb R})$ equipped with the trace class norm; the injective tensor product would lead to the `same' underlying vector space, equipped with the operator norm. The 2 x 2 identity matrix has trace class norm 2 and operator norm 1, so the two norms are genuinely different.
My answer is still not as clear as it should be, because due to a sluggish and temperamental internet connection I'm having trouble looking up just what the axioms for categorical duals in a SMC are. But if I recall correctly the natural map from $I \to V \otimes V^\*$ should be given by multiplying a scalar by the vector $e_1\otimes e_1 + e_2\otimes e_2$ where $e_1,e_2$ is an o.n. basis of ${\mathbb R}^2$ -- and that vector does not have norm 1 in the proj t.p. althought it does have norm 1 in the inj t.p.