Categorical duals in Banach spaces - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T21:49:02Z http://mathoverflow.net/feeds/question/10789 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/10789/categorical-duals-in-banach-spaces Categorical duals in Banach spaces Reid Barton 2010-01-05T06:07:56Z 2010-01-05T18:02:24Z <p>Near the bottom of <a href="http://ncatlab.org/nlab/show/Banach+space" rel="nofollow">the nlab page for Banach space</a> I see "To be described: duals (p+q=pq)".</p> <p>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)?</p> <p><strong>Edit</strong>: take n=2, p=1, q=&infin;. 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?</p> http://mathoverflow.net/questions/10789/categorical-duals-in-banach-spaces/10822#10822 Answer by Yemon Choi for Categorical duals in Banach spaces Yemon Choi 2010-01-05T17:25:11Z 2010-01-05T18:02:24Z <p>My suspicion is "no", because if I recall correctly the map $I \to V \otimes V^*$ naturally lands in the <em>injective tensor product</em>, not the <em>projective tensor product</em>, and it is the latter which appears as the correct'' tensor product for the SMC category of Banach spaces and linear contractions.</p> <p>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.</p> <p>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 <em>think</em> 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.</p> <p>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.</p>