Products, are very elementary forms of categorical limits. My question is whether in the category of groups, semi-direct products are categorical limits.

As was pointed in: http://unapologetic.wordpress.com/2007/03/08/split-exact-sequences-and-semidirect-products/

Bourbaki (General Topology, Prop. 27) gives a universal property:

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.

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.

So the question remains: are semi-direct products in the category of groups categorical limits?

  • 4
    $\begingroup$ It's a certain colimit. Do you know the Grothendieck construction for fibrations? $\endgroup$ – Martin Brandenburg May 5 '12 at 17:43
  • 1
    $\begingroup$ Colimit? Are you sure about what you're saying? After all products are a particular case of semi-direct products, and they are limits not colimits. I don't know anything about Grothendieck's construction for fibrations... $\endgroup$ – Makhalan Duff May 5 '12 at 17:46
  • 11
    $\begingroup$ Sure, colimit, because you describe maps on the semi-direct product. Actually this universal property of the semi-direct products in the special case of products does not give you the usual universal property of a categorical product: It gives you that group morphisms $N \times H \to G$ are given by pairs of group morphisms $N \to G$, $H \to G$ which commute pointwise. In other words, $N \times H = N * H / \langle\langle nhn^{-1} h^{-1} \rangle\rangle$, and here you already see the colimit. In a general semi-direct product, this commutator is twisted. $\endgroup$ – Martin Brandenburg May 5 '12 at 17:52
  • 3
    $\begingroup$ Semi-direct products involve an action $\phi: H \to Aut(N)$ in addition to the two factor groups $H,N$. So the first step would be to figure out how to describe this action in purely category-theoretic terms, without referencing individual elements. I don't know how to do this, but suspect that if one can achieve this, then describing the semi-direct product category-theoretically should be straightforward. $\endgroup$ – Terry Tao May 5 '12 at 18:09
  • 3
    $\begingroup$ Semidirect products of profinite groups are special because a compactness argument lets you prove one can obtain the action as an inverse limit of actions of finite quotients. $\endgroup$ – Benjamin Steinberg May 5 '12 at 21:47

This is a partial answer, summing up some of my comments.

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 on 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 not 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 $N$.

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

$$N \rtimes H = (N * H) / \{h n h^{-1}= \phi_h(n)\}_{h \in H, n \in N},$$

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 $\mathbb{Z}/n \rtimes \mathbb{Z}/2$.

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

$(a_2,f_2) \circ (a_1,f_1) = (a_2 \circ F(f_2)(a_1),f_2 \circ f_1)$.

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.

Back to the general case of a diagram $F : I \to \mathsf{Cat}$, the Grothendieck construction $\int^I F$ is the lax 2-colimit 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.

Thus, the semi-direct product $N \rtimes H$ is the lax $2$-colimit of the diagram $N : H \to \mathrm{Cat}$.  

| cite | improve this answer | |
  • $\begingroup$ In your answere $H$ have to be a group (instead change $Aut(N)$ by $End(N)$). If $H$ and $N$ are internal groups, and we can internalizing $Aut(N)$ with a natural map $Aut(N)\times N\to N$ (this is possible in enought good enriched $V$-categories), then the semidirected composition has a internal formulation, furthermore I think that a action $\phi: L\to Aut(L)$ could be a pseudo-functor (consider groups as 2-cetegories, with one object on one (identity) morphism), then I guess that the semidirect-products is some kind of a (co)lax-colimit ...(sorry for my bad English) $\endgroup$ – Buschi Sergio May 5 '12 at 20:20
  • $\begingroup$ If you want, a traditional source on limits (and variations of these) on 2-categories is J. W. Gray "adjointness for 2-categories" (LNM 391). $\endgroup$ – Buschi Sergio May 6 '12 at 10:35

There is (another ?) description of the crossed product in categorical terms.

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.

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 $H \to G \rtimes H$.

Being a left-adjoint, the "crossed product" maps colimits to colimits.

| cite | improve this answer | |
  • 7
    $\begingroup$ This is a very concise categorical description! $\endgroup$ – Martin Brandenburg May 3 '13 at 15:12
  • 2
    $\begingroup$ Just checking: are the morphisms in $C$ just homomorphisms $(H, G) \to (H, G')$, or are we permitted to vary the group which is acting as well? That is, can we have a pair of group homomorphism $\phi: H \to H'$ and $\psi: G \to G'$ such that $\psi(h.g) = \phi(h).\psi(g)$ as a morphism in $C$? $\endgroup$ – ಠ_ಠ Mar 10 '16 at 11:27
  • $\begingroup$ Yes, exactly, $H$ need not be fixed. $\endgroup$ – Andreas Thom Mar 10 '16 at 14:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.