Do rational numbers admit a categorification which respects the following "duality"? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T17:28:05Z http://mathoverflow.net/feeds/question/22860 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/22860/do-rational-numbers-admit-a-categorification-which-respects-the-following-dualit Do rational numbers admit a categorification which respects the following "duality"? Steven Gubkin 2010-04-28T14:57:00Z 2010-04-30T14:35:45Z <p>I need to give a lot of quite basic background to this question because I think (at least from conversing with fellow graduate students) that most mathematicians have not <em>really</em> thought about fractions for a long time. I think that there is an interesting germ of an idea in here somewhere, but I cannot exactly pinpoint it. Essentially there seems to be two canonical ways to solve division problems and there does not seem to be a "natural isomorphism" relating the two ways. I am interested in framing this duality formally: is there a "categorification" of the rational numbers where this duality can be precisely framed?</p> <p>I TA a class for future elementary school teachers. The idea is to go back and really understand elementary school mathematics at a deep level. Hopefully this understanding gets passed on to the next generation.</p> <p>We were discussing division of fractions. Rather than say "Yours is not to wonder why, just invert and multiply", we try to make sense of this question physically, and then use reasoning to solve the problem. Take (3/4) / (2/3). When doing this there seems to be two reasonable interpretations:</p> <p>1) 3/4 of a cup of milk fills 2/3 of a container. How much milk (in cups) does it take to fill the entire container?</p> <p>This is a "How many in each group" division problem, analogous to converting 6/3 into the question "If I have six objects divided into three equal groups, how many objects will be in each group?"</p> <p>The solution that stares you in the face if you draw a picture of this situation is the following: 3/4 of a cup fills 2 thirds of a container. That means there must be 3/8 cups of milk in each third of the container. One container must have 9/8 cups of milk then, because this is 3 of these thirds. Note that the solution involved first dividing and then multiplying.</p> <p>2) I have 3/4 cups of milk, and I have bottles which each hold 2/3 of a cup. How many bottles can I fill?</p> <p>This is a "How many groups" division problem, analogous to converting 6/3 into the question "If I have six objects divided into groups of two, how many groups do I have?"</p> <p>The solution suggested by this situation is the following: 3/4 of a cup of milk is actually 9 twelfths of a cup . Each twelfth is an eighth of a bottle. So I have 9/8 of a bottle. This solution involved first multiplying and then dividing.</p> <p>Now I come to my question. This pattern persists! Every real world example of a "how many in each group" division problem suggests a solution by first dividing and then multiplying, whereas each "How many groups" division problem involves first multiplying and then dividing. It seems that solving the problem in the other order does not admit a conceptual realization in terms of the original problem. This is interesting to me! It suggests that the two solution methods are fundamentally different somehow. The standard approach to rational numbers (natural numbers get grothendieck grouped into integers, which get ring of fractioned into rational numbers) ignores this kind of distinction. Is there a "categorification" of the rational numbers which preserves the duality between these two types of question?</p> <p>UPDATE 1: In the category of sets, if you wanted to express $(\frac{6}{2})(3) = \frac{(6)(3)}{2}$ you would have to do something like this:</p> <p>Let $A$ be a set with 6 elements, $B$ a set with 3 elements, $\sim$ an equivalence relation on A where each equivalence class has 2 elements, and $\cong$ an equivalence relation on $A \times B$ where each equivalence class has 2 elements. Then there is no canonical morphism from $(A/ \sim) \times B \to (A \times B)/ \cong$. This seems to explain things somewhat on the level of integers, but we are talking fractions here.</p> <p>UPDATE 2: Qiaochu points out in a comment to his answer that the order of operations is not the most essential thing here. You can solve the first problem by observing that 9/4 cups of milk fill 2 containers, so 9/8 must fill one. Torsors give a formal distinction between the two situations, but it still feels like UPDATE 1 should go through in a suitable category of "fractional sets".</p> <p>UPDATE 3: For a very nice discussion of the ideas related to Theo's answer see <a href="http://golem.ph.utexas.edu/category/2008/12/groupoidification_made_easy.html" rel="nofollow">http://golem.ph.utexas.edu/category/2008/12/groupoidification_made_easy.html</a></p> http://mathoverflow.net/questions/22860/do-rational-numbers-admit-a-categorification-which-respects-the-following-dualit/22866#22866 Answer by Qiaochu Yuan for Do rational numbers admit a categorification which respects the following "duality"? Qiaochu Yuan 2010-04-28T15:40:33Z 2010-04-28T15:51:04Z <p>Let me expand on the answer I gave in meta. In my mind the appropriate "categorification" begins with the observation that "cups" is a unit, and in the first approach you endow only the numerator with units while in the second approach you endow both the numerator and denominator with units. This is formalized as follows. Quantities with units which take values in $\mathbb{Q}$ are <a href="http://math.ucr.edu/home/baez/torsors.html" rel="nofollow">torsors</a> for the nonzero rationals $G$ under multiplication, e.g. $G$-sets $X$ which are free and transitive. The choice of unit corresponds to a choice of element of $X$ against which all other elements are measured.</p> <p>In the first approach, you give a group element $g = \frac{2}{3}$ and a set element $x = \frac{3}{4} \text{ cup}$ (where $\text{cup}$ is the distinguished element of $X$) and are asked to find the unique $y \in X$ such that $gy = x$. In the second approach, you give two set elements $x = \frac{3}{4} \text{ cup}, y = \frac{2}{3} \text{ cup}$ and are asked to find the unique $g \in G$ such that $gy = x$. So I think the key here is that the group action, as a function $G \times X \to X$, treats $G$ and $X$ nearly the same, but not canonically so. (I haven't thought this through, but it's also significant that in the first problem one can just write $y = g^{-1} x$, whereas in the second problem it is actually necessary, or at least natural, to express $x$ and $y$ in terms of $\text{cup}$ to determine $g$.)</p> http://mathoverflow.net/questions/22860/do-rational-numbers-admit-a-categorification-which-respects-the-following-dualit/22879#22879 Answer by Theo Johnson-Freyd for Do rational numbers admit a categorification which respects the following "duality"? Theo Johnson-Freyd 2010-04-28T16:57:35Z 2010-04-28T16:57:35Z <p>I don't claim to have a full answer, but I wanted to put it here so that others might help elaborate it.</p> <p>I think that the "classical" (aside: at a recent seminar on Heegaard Floer homology, the word "classical" was defined as "posted on the arXiv") categorification of the positive rationals is to the world of finite groupoids. This is, in any case, the approach pushed by John Baez and collaborators in their HDA and Groupoidification series. You may know this already, but I'll review the basic structure for other readers. So recall that a finite groupoid is precisely: a finite set of objects, and for each pair of objects a finite set of ways that they are isomorphic, and some composition rules so that isomorphism is transitive. The simplest examples of groupoids are: discrete groupoids, aka "sets", in which each object is isomorphic to itself in exactly one way, and not isomorphic to any other object; and "point mod G", in which there is a single object, and a finite group G worth of ways that it is isomorphic to itself (G should be a group so that isomorphism is correctly transitive). There is a good notion of "equivalence" of groupoids, in which, for example, the groupoid with only one object, which is isomorphic to itself in only one way, is equivalent to the groupoid with two objects which are each isomorphic to themselves in precisely one way and which are isomorphic to each other in precisely one way. Groupoids have a good notion of "disjoint union", and in fact every finite groupoid is equivalent to a groupoid that is a disjoint union of various "point mod G"s for various groups G, just as every finite set is a disjoint union of points.</p> <p>What Baez and Dolan described was the correct notion of "cardinality" of a groupoid. Namely, cardinality is additive under disjoint union, preserved under equivalence of groupoids, and the cardinality of "point mod G" is 1/|G|. (They prove that this uniquely defines a cardinality.) There is also a notion of cartesian product of groupoids, and this notion of cardinality is multiplicative under products. Moreover, if a group G acts on another groupoid X, then there is a "quotient groupoid" X//G, which has the same objects as X and extra isomorphisms for the G action. If the action is free, then X//G is equivalent to the usual "coarse" quotient X/G. Moreover, the cardinality of X//G is the cardinality of X divided by the number of group elements of G.</p> <p>Since there are groups of arbitrary positive-integer cardinality, there are groupoids of arbitrary nonnegative-rational cardinality. In this sense, {Groupoids} categorifies {Rationals}.</p> <p>But there are at least two reasons to be suspicious of this proposed categorification. First of all, although addition and multiplication categorify as we expect them to --- to disjoint union and cartesian product --- division is a bit strange. It is not inverse to multiplication except on one side: if you take a groupoid A, multiply it by a set B, and then divide by a group that acts freely and transitively on B, you get a groupoid equivalent to A; but if you divide first, you cannot multiply back. Similarly, it is not generally true that X//G is equivalent to X times pt//G. Maybe there is some sort of "semidirect product" or "crossed product" of groupoids that really does unify it all --- I don't know. In any case, when Baez explains his division, he treats numerator and denominator very differently: the numerator is something like a set, whereas the denominator is something like a group.</p> <p>Second, it is not true that {Rationals} are the "grothendieck group" of {Groupoids}, at least not without some heavy duty groupifying. At the very least, there are many inequivalent groupoids with the same cardinality. Maybe this is a plus from the point of view of your question: {Groupoids} may be able to tell apart the two kinds of division that you sketch.</p> http://mathoverflow.net/questions/22860/do-rational-numbers-admit-a-categorification-which-respects-the-following-dualit/23112#23112 Answer by Thomas Kragh for Do rational numbers admit a categorification which respects the following "duality"? Thomas Kragh 2010-04-30T14:35:45Z 2010-04-30T14:35:45Z <p>Another way of categorifying $\mathbb{Q}_+$ is as follows:</p> <p>Let $\mathcal{F}$ be the category of finite non-empty sets and bijections. This is a symmetric monoidal category with respect to $\times$. Let $\mathbb{N}$ be viewed as a symmetric monoidal category, such that the objects are $\mathbb{N}$ and we only have identity morphism. The product is given by the usual product $\cdot$ in $\mathbb{N}$.</p> <p>Then there is an obvious symmetric monoidal functor $G \colon \mathcal{F} \to \mathbb{N}$, which simply counts the elements in the set. Taking Grayson Quillen construction on both sides, we get a symmetric monoidal functor:</p> <p>$G' \colon \mathcal{F}^{-1}\mathcal{F} \to \mathbb{Q}_+$,</p> <p>whic may be viewed as a categorification (note that this is also the $\pi_0$ functor). The Grayson Quillen construction may be thought of as a categorified version of the Grothendeick construction. Objects in $\mathcal{F}^{-1}\mathcal{F}$ are pairs of objects in $\mathcal{F}$ which I suggestively write as $A/B$</p> <p>If we only use the skeletal category of $\mathcal{F}$ given by objects $s_n=\{1,\dots,n\}$ for each $n\in \mathbb{N}$ we get that $G$ is an isomorphism on the set of objects. This is not true for $G'$, because we have implicitly divided out by isomorphisms in the image (i.e. $p/q=np/(nq)$ in $\mathbb{Q}_+$ not just isomorphic objects as they are in the Grayson Quillen construction on $\mathbb{N}$). We cannot do this on the domain of $G'$ because $s_p/s_q$ is NOT isomorphic to $s_{np}/s_{nq}$. However, there are many maps from $s_p/s_q$ to $s_{np}/s_{nq}$, which identifies them when taking $\pi_0$.</p> <p>I should note that I believe (but am not sure) that this can be made <strong>bi</strong>-monoidal by taking the category of finite sets and bijections as a bimonoidal category and localizing with respect to $\times$ over the full subcategory of finite and non-empty sets.</p>