Are semi-direct products categorical limits? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T19:02:06Z http://mathoverflow.net/feeds/question/96078 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/96078/are-semi-direct-products-categorical-limits Are semi-direct products categorical limits? Makhalan Duff 2012-05-05T17:39:45Z 2012-05-07T19:35:33Z <p>Products, are very elementary forms of categorical limits. My question is whether in the category of groups, semi-direct products are categorical limits.</p> <p>As was pointed in: <a href="http://unapologetic.wordpress.com/2007/03/08/split-exact-sequences-and-semidirect-products/" rel="nofollow">http://unapologetic.wordpress.com/2007/03/08/split-exact-sequences-and-semidirect-products/</a></p> <p>Bourbaki (General Topology, Prop. 27) gives a universal property:</p> <p>Let $f \colon N \to G$, $g \colon H \to G$ be two homomorphisms into a group $G$, such that $f(\phi_h(n)) = g(h)f(n)g(h^{-1})$ for all $n \in N$, $h \in H$. Then there is a unique homomorphism $k \colon N \rtimes H \to G$ extending $f$ and $g$ in the usual sense.</p> <p>However, I remain unsatisfied. The condition $f(\phi_h(n)) = g(h)f(n)g(h^{-1})$ is a condition on elements of groups, rather than a condition that says that some diagram is commutative.</p> <p>So the question remains: are semi-direct products in the category of groups categorical limits?</p> http://mathoverflow.net/questions/96078/are-semi-direct-products-categorical-limits/96081#96081 Answer by Martin Brandenburg for Are semi-direct products categorical limits? Martin Brandenburg 2012-05-05T19:45:00Z 2012-05-07T07:27:48Z <p>This is a partial answer, summing up some of my comments.</p> <p>The semi-direct product is not a limit, but rather it is a colimit. The reason is that the universal property cited above describes maps <em>on</em> the semi-direct product. In the special case that $\phi$ is the trivial action, the semi-direct product becomes the direct product $N \times H$ and the universal property is <em>not</em> just the usual universal property as a product, but rather as a representing object of the pairs of morphisms on $N,H$ which commute pointwise. In a general semi-direct product, this commutation is twisted by an action of $H$ on $N$.</p> <p>So basically the idea is that we have the coproduct $N * H$ of the two groups (which is usually called the free product, which is quite unfortunate), and we impose the relation $h n h^{-1} = \phi_h(n)$. The universal property of $N \rtimes H$ is equivalent to the isomorphism</p> <p>$$N \rtimes H = (N * H) / \{h n h^{-1}= \phi_h(n)\}_{h \in H, n \in N},$$</p> <p>which exhibits $N \rtimes H$ as a special colimit of some diagram associated to $N,H,\phi$. However, this still uses elements in the relations. I think we cannot get rid of these elements, unless we use $2$-colimits. See below. Actually this isomorphism is used very often in group theory in order to recoqnize groups given by some presentation as a semi-direct product. For example, the dihedral group $D_n = \langle r,s : r^n = s^2 = 1, srs=r^{-1} \rangle$ is $\mathbb{Z}/n \rtimes \mathbb{Z}/2$.</p> <p>On the other hand, there is a purely category-theoretic construction which is due to Grothendieck: Let $I$ be a small category and $F : I \to \mathsf{Cat}$ be a diagram of small categories. The Grothendieck construction $\int^I F$ is the category of pairs $(i,x)$, where $i$ is an object of $I$ and $x$ is an object of $F(i)$. A morphism $(i,x) \to (j,y)$ is a pair $(a,f)$, consisting of a morphism $f : i \to j$ and a morphism $a : F(f)(x) \to y$ in $F(j)$. The composition is defined by the rule</p> <p>$(a_2,f_2) \circ (a_1,f_1) = (a_2 \circ F(f_2)(a_1),f_2 \circ f_1)$.</p> <p>Now if $H$ is a monoid, considered as a category with just one object $*$, and $F : H \to \mathsf{Cat}$ is a diagram such that $F(*)=N$ is just a monoid, then $F$ corresponds to a homomorphism of monoids $H \to \mathrm{End}(N)$ and the Grothendieck construction $\int^H N$ has just one object, thus corresponds to a monoid, namely what is usually called the semi-direct product $N \rtimes H$. This is shown by the multiplication rule above.</p> <p>Back to the general case of a diagram $F : I \to \mathsf{Cat}$, the Grothendieck construction $\int^I F$ is the <em>lax 2-colimit</em> of $F$. I don't know the original reference right now, but a very comprehensive account on that is the Appendix A in "The stack of microlocal sheaves" by I. Waschkies. The choice of the morphism $a : F(f)(x) \to y$ in the definition above is precisely the reason for the "2" here. If it was the identity, we would get the usual colimit.</p> <p>Thus, the semi-direct product $N \rtimes H$ is the lax $2$-colimit of the diagram $N : H \to \mathrm{Cat}$.</p> http://mathoverflow.net/questions/96078/are-semi-direct-products-categorical-limits/96256#96256 Answer by Andreas Thom for Are semi-direct products categorical limits? Andreas Thom 2012-05-07T19:35:33Z 2012-05-07T19:35:33Z <p>There is (another ?) description of the crossed product in categorical terms. </p> <p>Let ${\rm Mor}(Gp)$ be the category whose objects are homomorphisms of groups and morphisms are commutative diagrams. Let $C$ be the category of "groups acting on groups" whose objects are pairs of groups $(H,G)$ together with a homomorphism $H \to {\rm Aut}(G)$. Morphisms in this category are equivariant homomorphisms.</p> <p>Now, there is a natural forgetful functor $T \colon {\rm Mor}(Gp) \to C$ which sends $H \to G$ to the pair $(H,G)$ with the homomorphism $H \to {\rm Aut}(G)$ given by conjugation. Now, almost by definition, the crossed product is the left-adjoint of this forgetful functor. Indeed, the left adjoint is easily seen to map $(H,G)$ with $H \to {\rm Aut}(G)$ to the inclusion $H \to G \rtimes H$.</p> <p>Being a left-adjoint, the "crossed product" maps colimits to colimits.</p>