two essentially different concretizaions - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T04:33:53Z http://mathoverflow.net/feeds/question/79253 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/79253/two-essentially-different-concretizaions two essentially different concretizaions KotelKanim 2011-10-27T10:27:12Z 2011-10-27T17:48:33Z <p>It is sometimes emphasized that a "concrete category" is not a property of a category $C$, but rather a structure, i.e. a faithful functor from $C$ to $Set$. Thus, When people talk about a concrete category $C$ they <em>really</em> mean $C$ together with some implicitly defined and naturally understood functor from $C$ to $Set$ (most commonly, when the objects of $C$ are sets with some extra structure and the morphisms are some functions between those sets). This emphasize suggests that for a given category $C$, there might be several inequivalent concretizaions (of course there might also be none, but this is less important for this discussion) where I take "equivalent" to mean the two concretizing functors are naturally isomorphic (this seems natural, but is it the "correct" definition?). </p> <p>In light of this, I would like to see an interesting as possible example of a category $C$ with two inequivalent concretizations. I guess there are tailor made examples with possibly even a finite category, though I most confess I didn't try to find such myself, so it might be also interesting to see such an example, but the most satisfying example would be of a category of some sort of "real life" mathematical structure with two meaningful inequivalent concretizations each giving some different intuition about the category (perhaps this is to much to ask, but it is only to clarify what I mean by "interesting").</p> http://mathoverflow.net/questions/79253/two-essentially-different-concretizaions/79259#79259 Answer by Theo Johnson-Freyd for two essentially different concretizaions Theo Johnson-Freyd 2011-10-27T11:39:46Z 2011-10-27T11:39:46Z <p>One of course has many examples. To define the terms, a <em>concretization</em> of a category $C$ should be a faithful functor $C \to \mathrm{Set}$. In examples, concretizing functors are never full, so I will not ask this, but I do not mind asking, say, that the concretizing functor reflect isomorphisms. Maybe you want moreover for "concretization" to have some adjointness properties with some "free" or "discrete" functor $\mathrm{Set} \to C$, but maybe not. But even if you do ask for this, then there are still many such functors. For example, take any concretizing functor $F : C \to \mathrm{Set}$; then probably the functor $2\times F$, which assigns to $x\in C$ the set $F(x) \sqcup F(x) = \lbrace 0,1\rbrace \times F(x)$, is also concretizing.</p> <p>An example that comes up in nature is the following. I can concretize the category of finite-dimensional Lie algebras by assigning to each Lie algebra the underlying set of its underlying vector space. Instead, I can embed the category of finite-dimensional Lie algebras into the category of Lie groups (by assigning to each Lie algebra its connected simply-connected group), and then I can concretize the category of Lie groups by assigning to each its underlying set. To see that these functors are not isomorphic, consider their actions on any non-trivial morphism $\mathbb R \to \mathfrak{so}(3)$. Under the first concretization, this morphism turns into an injection from the set $\mathfrak c$ with cardinality continuum into itself. Under the second concretization, the morphism also becomes a map $\mathfrak c \to \mathfrak c$, but it is not an injection.</p> http://mathoverflow.net/questions/79253/two-essentially-different-concretizaions/79260#79260 Answer by Todd Trimble for two essentially different concretizaions Todd Trimble 2011-10-27T12:40:25Z 2011-10-27T12:40:25Z <p>Here's an example with "natural" content. </p> <p>Consider the category $Mat$ whose objects are natural numbers $n$ and whose morphisms $n \to m$ are $m \times n$ matrices with real entries. The "usual" concrete representation takes $n$ to (the underlying set of) $\mathbb{R}^n$. </p> <p>But, for any finite $k$, the field $\mathbb{R}$ is Morita equivalent to the algebra $M_k$ of $k \times k$ matrices over $\mathbb{R}$, i.e., to $\hom(V, V)$ where $V$ is a $k$-dimensional vector space. What this means is that the category of finitely generated left projective modules over $\mathbb{R}$ is equivalent to the category of finitely generated left projective modules over $M_k$; the equivalence takes $\mathbb{R}^n$ to $V^n$, i.e., a vector space $W$ to the $M_k$-module $V \otimes_{\mathbb{R}} W$. </p> <p>We have $Mat(m, n) \cong M_k-Mod(V^m, V^n)$. Thus, we may consider the functor $n \mapsto U(V^n)$ (the underlying set of the module $V^n$) as a completely natural concrete representation of $Mat$, no more and no less natural than the first, and yet it is completely different. </p> <hr> <p>Many more such examples can be obtained by exploiting general Morita theory. The other example I was going to write about is based on an abstract equivalence between the category of Boolean algebras and the category of $k$-<a href="http://www.ams.org/journals/tran/1960-095-02/S0002-9947-1960-0112855-8/S0002-9947-1960-0112855-8.pdf" rel="nofollow">Post algebras</a> for some fixed $k$, where the "usual" underlying set of a finite Boolean algebra has cardinality $2^n$ whereas the usual underlying set of the corresponding Post algebra under this equivalence has cardinality $k^n$. Thus, two functors </p> <p>$$U: C \to Set$$ </p> <p>which lay equal claim to being considered "the" underlying set functor, but different sets according to whether we think of an object of $C$ as a Boolean algebra or a Post algebra. (I have written about this in other MO answers, explaining the connection with Morita theory, <a href="http://mathoverflow.net/questions/6538/generalizations-of-boolean-posets-lattices/64541#64541" rel="nofollow">here</a> and <a href="http://mathoverflow.net/questions/69086/lawvere-theories-versus-classical-universal-algebra/69097#69097" rel="nofollow">here</a>.) </p> http://mathoverflow.net/questions/79253/two-essentially-different-concretizaions/79287#79287 Answer by Qiaochu Yuan for two essentially different concretizaions Qiaochu Yuan 2011-10-27T17:22:55Z 2011-10-27T17:48:33Z <p>Two representable functors $\text{Hom}(A, -), \text{Hom}(B, -)$ are naturally isomorphic if and only if $A, B$ are isomorphic, and it is not hard to come up with many examples of categories with two non-isomorphic objects $A, B$ such that $\text{Hom}(A, -), \text{Hom}(B, -)$ are both faithful. In some sense the obvious examples of this form are not "interesting," though. In all of the examples of $A, B$ I can think of there is an object $X$ such that $\text{Hom}(X, -)$ is faithful and $A = X \sqcup A', B = X \sqcup B'$ for some $A', B'$. Can anyone think of examples not of this form? </p> <hr> <p>The following example might be "interesting." On the one hand $\text{Set}$ has the trivial concretization given by the identity functor $\text{Set} \to \text{Set}$ (which is representable and given by $\text{Hom}(1, -)$). On the other hand there is also a concretization sending a set $A$ to the power set $2^A$ and sending a function $f : A \to B$ to the function $$f : 2^A \ni S \mapsto \{ f(x) : x \in S \} \in 2^B.$$</p> <p>This functor is not representable in $\text{Set}$, but $\text{Set}$ embeds into $\text{Rel}$ (the category of sets and relations) and there this functor is $\text{Hom}(1, -)$ again. </p>