What functor is adjoint to the tensor product of 2-vector spaces? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T07:53:42Z http://mathoverflow.net/feeds/question/21373 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/21373/what-functor-is-adjoint-to-the-tensor-product-of-2-vector-spaces What functor is adjoint to the tensor product of 2-vector spaces? Theo Johnson-Freyd 2010-04-14T18:52:30Z 2010-04-14T21:20:30Z <p>This is a refinement of my (naive, poorly asked) question <a href="http://mathoverflow.net/questions/17951/what-tensor-product-of-chain-complexes-satisfies-the-usual-universal-property" rel="nofollow">here</a>. The reference for my question is <a href="http://math.ucr.edu/home/baez/hda6.pdf" rel="nofollow">Baez and Crans, HDA6</a>.</p> <h2>Background: category objects, etc.</h2> <p>Let <code>$\mathcal V$</code> be a category. A <strong>category object</strong> internal to <code>$\mathcal V$</code> consists of the following data and properties:</p> <ol> <li>Objects <code>$C_0,C_1 \in \mathcal V$</code> and morphisms <code>$s,t: C_1 \to C_0$</code> and <code>$i: C_0 \to C_1$</code>.</li> <li>Such that <code>$s\circ i = t\circ i = \text{id}_{C_0}$</code> and the pull-back <code>$C_1 \underset{C_0}{\times} C_1 = C_1 \underset{\displaystyle ^{\searrow\!\!^{\scriptstyle s}} {C_0} ^{^{\scriptstyle t} \!\!\swarrow} }{\times} C_1$</code> exists.</li> <li>A morphism <code>$m: C_1 \underset{C_0}\times C_1 \to C_1$</code> such that <code>$s\circ m = s_R$</code> and <code>$t\circ m = t_L$</code>, where <code>$s_R: C_1 \underset{C_0}\times C_1 \to C_0$</code> is the "<code>$s$</code>" projection from the right factor, and similarly for <code>$t_L$</code>.</li> <li>And such that the obvious "associativity" square (two ways to get from <code>$C_1 \underset{C_0}\times C_1 \underset{C_0}\times C_1$</code> to <code>$C_1$</code>) and "identity" triangles (three ways to get from <code>$C_1 = C_1 \underset{C_0}\times C_0 = C_0 \underset{C_0}\times C_1$</code> to <code>$C_1$</code>) commute.</li> </ol> <p>For example, a category object in <code>$\mathcal V = {\rm SET}$</code> is a small category.</p> <p>For this post, I will be interested in <code>$\mathcal V = {\rm VECT}$</code>, the category of vector spaces over your favorite field. I will call your favorite field "<code>$\mathbb R$</code>". A category object in <code>${\rm VECT}$</code> is a <strong>2-vector space</strong>.</p> <p>2-vector spaces are relatively mild things. Indeed, it turns out that in <code>$\rm VECT$</code> the data and properties of 1-2 above uniquely determine a map <code>$m$</code> satisfying 3-4.</p> <p>By the general yoga known as "commutativity of internalization", a 2-vector space is the same as a "vector space object in <code>$\rm CAT$</code>". More precisely, let <code>$\rm CAT$</code> be the category of small categories. Then it makes sense to talk about "field objects" &mdash; like a category object, a field object consists of some objects, some maps, some pull-backs, and some more maps, and some commuting diagrams. In particular, by thinking of <code>$\mathbb R$</code> as a discrete category (<code>$\mathbb R_0 = \mathbb R = \mathbb R_1$</code> and <code>$s = t = i = {\rm id}$</code>), it is in fact a field object. Then with some more diagrams, we can talk about "vector space objects over <code>$\mathbb R$</code>" internal to <code>$\rm CAT$</code>, and it is straightforward to check that these are the same as 2-vector spaces.</p> <h2>Background: tensor products</h2> <p>I know of two natural approaches to define "tensor products":</p> <ol> <li>Define a notion of "bilinear map", such that the assignment <code>$X,Y,Z \mapsto \{\text{bilinear maps }X\times Y \to Z\}$</code> is contravariant in the first two spots and covariant in the last. Then set <code>$X\otimes Y$</code> to be the object (if it exists) that represents the function <code>$Z \mapsto \{\text{bilinear maps }X\times Y \to Z\}$</code>.</li> <li>Define a notion of "internal hom", i.e. a (nice) functor <code>$\underline{\rm Hom}: \mathcal V^{\rm op} \times \mathcal V \to \mathcal V$</code>. For each <code>$X\in \mathcal V$</code>, define the functor <code>$-\otimes X$</code> by declaring that it is left adjoint to <code>$\underline{\rm Hom}(X,-)$</code>.</li> </ol> <p>Approach 1 is the way that tensor products are introduced in grade school. Approach 2 is I think more standard in the real world. We can implement each in the case of 2-vector spaces:</p> <ol> <li>The trick for approach 1 is that the notion of "bilinear" depends on more than just the category. So realize the category of 2-vector spaces as the category of vector spaces objects in <code>$\rm CAT$</code>. Recall that a morphism of 2-vector spaces is simply a morphism of underlying categories so that some diagrams commute. Then we can say the following. Let <code>$X,Y,Z$</code> be 2-vector spaces. Then a morphism of underlying categories <code>$X \times Y \to Z$</code> is <strong>bilinear</strong> if a bunch of diagrams commute (these diagrams refer to the vector-space-object structures of <code>$X,Y,Z$</code>, and are precisely the diagrams that you learned in grade school). By reproducing the proofs from vector spaces internal to <code>$\rm SET$</code>, this in fact defines a functor <code>$\otimes$</code>.</li> <li>Given 2-vector spaces <code>$X,Y$</code>, there is a category whose objects are linear functors <code>$X \to Y$</code> and whose morphisms are linear natural transformations of functors, and this category has a natural structure as a 2-vector space. Moreover, the corresponding notion of <code>$\underline{\rm Hom}$</code> is correctly functorial, and has an adjoint. So this approach defines a functor <code>$\otimes$</code>.</li> </ol> <p>However:</p> <blockquote> <p>The two "tensor products" defined in 1-2 above do not agree.</p> </blockquote> <p>Writing 2-vector spaces <code>$X = (X_1 \rightrightarrows X_0)$</code> and <code>$Y = (Y_1 \rightrightarrows Y_0)$</code> as category objects in <code>$\rm VECT$</code>, approach 1 gives <code>$(X\otimes Y)_a = X_a \otimes Y_a$</code> for $a=0,1$, with the tensor products of the structure maps. Approach 2 also has <code>$(X\otimes Y)_0 = X_0 \otimes Y_0$</code>, but <code>$(X\otimes Y)_1 \cong$</code> <code>$$X_0 \otimes Y_0 \oplus \text{coker}\Bigl( \bigl(\ker(X_1 \overset s \to X_0) \otimes \ker(Y_1 \overset s \to Y_0)\bigr) \overset{t\otimes{\rm id} - {\rm id}\otimes t}\longrightarrow \bigl( X_0 \otimes \ker(Y_1 \overset s \to Y_0) \oplus \ker(X_1 \overset s \to X_0) \otimes Y_0\bigr) \Bigr)$$</code> The structure maps are: <code>$s$</code> is the projection onto the first factor, and <code>$t$</code> is the sum of the same projection and the map <code>${\rm id}\otimes t + t\otimes {\rm id}$</code> from the second factor (it is well-defined out of the cokernel).</p> <p>(There is probably a way to simplify the above description. The trick is that, as explained in HDA6, the category of 2-vector spaces is equivalent as a category to the category of 2-term chain complexes, and this is the natural "internal Hom" over there.)</p> <p>Anyway, a dimension count shows that these two "tensor" constructions are inequivalent in general.</p> <p>Hence:</p> <h2>Question</h2> <p>Does the tensor product of 2-vector spaces given in "approach 1" above &mdash; the tensor product defined as representing "bilinear functors" &mdash; under this tensor product, does the functor <code>$\otimes X$</code> have a right adjoint?</p> http://mathoverflow.net/questions/21373/what-functor-is-adjoint-to-the-tensor-product-of-2-vector-spaces/21390#21390 Answer by Reid Barton for What functor is adjoint to the tensor product of 2-vector spaces? Reid Barton 2010-04-14T21:20:30Z 2010-04-14T21:20:30Z <p>I'll denote your category of 2-vector spaces by 2Vect. By your preliminary remarks, 2Vect is actually the category of Vect-valued presheaves on &Delta;<sub>&le;1</sub> where &Delta;<sub>&le;1</sub> denotes the full subcategory of &Delta; on the objects [0] and [1]. Therefore, colimits in 2Vect are computed objectwise under this identification. So the functor &ndash; &otimes; X certainly has a right adjoint <strong>Hom</strong>(X, &ndash;) (by the adjoint functor theorem for locally presentable categories). This adjunction also respects the Vect enrichment.</p> <p>To compute this adjoint, we can use the "Vect-enriched Yoneda lemma": writing &Delta;<sup>i</sup> for the 2-vector space (&Delta;<sup>i</sup>)<sub>j</sub> = Hom<sub>&Delta;<sub>&le;1</sub></sub>([j], [i]) &bull; &#8477;, we have hom<sub>2Vect</sub>(&Delta;<sup>i</sup>, X) = X<sub>i</sub> as vector spaces, where hom denotes the Vect-enriched Hom. So</p> <p><strong>Hom</strong>(X, Y)<sub>0</sub> = hom(&Delta;<sup>0</sup>, <strong>Hom</strong>(X, Y)) = hom(&Delta;<sup>0</sup> &otimes; X, Y) = hom(X, Y)</p> <p>since &Delta;<sup>0</sup> happens to be the unit for this &otimes;, but</p> <p><strong>Hom</strong>(X, Y)<sub>1</sub> = hom(&Delta;<sup>1</sup> &otimes; X, Y)</p> <p>will have a more complicated formula which you'll have to work out (my guess is it will look similarly complicated to your expression for the other tensor product).</p>