MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
[I deleted a somewhat rambling and confused comment.] I was wondering if that note was only referring to the fact that (ℝ^n)_p and (ℝ^n)_q are isometrically isomorphically each other's dual spaces. There isn't a strict adherence on that page to the category with linear contractions; for instance, it is stated without specifying the category that (ℝ^n)_p and (ℝ^n)_q are isomorphic. – Jonas Meyer Jan 5 '10 at 9:32
+1 for making me think about duals in category theory and analysis. – Loop Space Jan 5 '10 at 11:31
@Jonas: yeah, that had me confused for a while too. – Reid Barton Jan 5 '10 at 15:03
While the question is good, there are several senses of "dual" used in category theory. We (authors of the nLab article) meant a pair of Banach spaces V, W equipped with an isometric isomorphism W --> hom(V, k) [where k = R or C as appropriate], where V is reflexive, so that the transpose V --> hom(W, k) is also an isometric isomorphism. I think this is a standard functional analysis sense of "V and W are dual to one another as Banach spaces", and it's also one of the meanings of "dual" used for smc cats, even if it's not the "compact closed" sense of the question above. – Todd Trimble Jan 5 '10 at 23:06
Thanks, Todd, for the clarification. (My main motivation in asking this question was actually to use Ban as a test case for a statement about categories with duals in the "compact closed" sense. I wasn't trying to suggest that there aren't other useful notions of "dual", or doubting the statements on the nlab page, although I can see how it could read that way.) – Reid Barton Jan 5 '10 at 23:37
up vote 2 down vote accepted

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.

share|cite|improve this answer
And the injective and projective tensor products are non-isometric in this case for general V, right? – Reid Barton Jan 5 '10 at 17:46
I think so; see the updated entry. – Yemon Choi Jan 5 '10 at 18:02
Here is a concrete counterexample for someone to sanity check. It seems that "the set of extreme points of the unit ball" is an isomorphism invariant of a Banach space, and that it preserves coproducts and products, at least the ones used in forming $(R^{\amalg n})^{\times 2}$ and $(R^{\times 2})^{\amalg n}$. It then sends these spaces to sets of cardinality $(2n)^2$ and $4n$, respectively, which are distinct for $n \ge 2$. This shows that $V \mapsto V \times V$ doesn't commute with coproducts. – Reid Barton Jan 5 '10 at 18:15
(Just to spell out exactly how this is related to my original question: Asking for a dual for an object $X$ of a closed symmetric monoidal category is the same as asking for an object $X^*$ and an identification $\mathbf{Hom}(X, {-}) = X^* \otimes {-}$. In particular (using closedness again) $\mathbf{Hom}(X, {-})$ must commute with colimits. In my case $X = (\mathbb{R}^2)_1 = \mathbb{R} \amalg \mathbb{R}$, and so $\mathbf{Hom}(X, V) = V \times V$. – Reid Barton Jan 5 '10 at 18:30

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.