2
$\begingroup$

For a vector space $V$, I have constructed $AGL(V) = V \rtimes GL(V)$ as the elements $(v, A) \in V \times GL(V)$ (Cartesian product of sets, not a direct product) with multiplication $(v, A) (w, B) = (v + Aw, AB)$. This then has a faithful action on $V$ given by $$x^{(v,A)} = A^{-1}(x + v)$$ which is fairly ugly.

It was suggested to me that letting $GL(V)$ act on $V$ from the right (that is, write $V$ as row vectors rather than column vectors) would be nicer, but I'm struggling to make a construction that is actually nicer.

I'm theoretically trying to get to an action like $$x^{(v,A)} = xA + v, \tag{$\ast$}$$ (this is the action that e.g. [DM96, section 2.8] and [Neu87, section 2] use, but neither spell out the full group) but every multiplication I define is either invalid with the group actions, seems to not be a valid multiplication for a semidirect product, or is ugly.

E.g.

  1. Taking the obvious $(v,A)(w,B) = (v+wA, AB)$ does not work with $(\ast)$ (nor does it work with $(x + v)A^{\pm 1}$).
  2. The "multiplication" $(v,A)(w, B) = (Bv + w, AB)$ does work with $(\ast)$, $$x^{(v,A)(w,B)} = (xA + v)^{(w,B)} = (xA + v)B + w = x(AB) + (vB + w) = x^{(vB + w, AB)}.$$ However, AFAIK, a semidirect product should be adjusting the first element of second term by (some image in $\mathrm{Aut}(V)$ of) the second element of the first time, that is $(a,b)(c,d) = (ac^b, bd)$, and this is the reverse: $(a,b)(c,d) = (a^d c, bd)$. (Maybe I am being too inflexible with the definition of a semidirect product?)
  3. Defining $(v,A)(w,B) = (v + wA^{-1}, AB)$ works with $x^{(v, A)} = (x + v)A$, but this does not seem any clearer than my original one, and also doesn't match the natural action above.

I feel like I am missing something simple; could someone suggest a more natural representation & action of $AGL(V)$ (from either the left or right)? (I imagine there may be a canonical one that is so 'obvious' it is not worth mentioning in anything but the most basic text.)

[DM96] J. Dixon and B. Mortimer. Permutation Groups. Graduate Texts in Mathematics. Springer New York, 1996.
[Neu87] P. Neumann. “Some algorithms for computing with finite permutation groups”. In: Proceedings of Groups St Andrews 1985. Ed. by E. F. Robertson and C. M. Campbell. Cambridge Books Online. Cambridge University Press, 1987, pp. 59–92.

$\endgroup$
3
$\begingroup$

Using your first multiplication $\mathrm{AGL}(V)$ acts on $V$ from the left via:

$(v,A)*z:=Az+v$

This is an action since $$(w,B)*((v,A)*z)=(w,B)*(Az+v)=BAz+Bv+w=(Bv+w,BA)*z=((w,B)(v,A))*z$$ and feels fairly natural to me.

$\endgroup$
  • $\begingroup$ Ah, of course, thanks. This is the opposite to my 2., so I have a mismatch between the right action $x^g$ and the natural way in which $\mathrm{AGL}(V)$ acts... this is annoying: I have a non-trivial essay using $x^g$, and that notation is the precedent in this field, so I don't think it feasible to switch to a left-action. Hm... $\endgroup$ – huon Oct 24 '14 at 10:48
  • $\begingroup$ Well, the difference between a right and a left action is merely technical so it won't matter that much. On the other hand you can always think of the space $V$ as "rows" let $\mathrm{GL}(V)$ act on it from the right (in the "natural" way) and define your multiplication in $\mathrm{AGL}$ accordingly to get a nicer action from the right... $\endgroup$ – Sebastian Schoennenbeck Oct 24 '14 at 11:39
  • $\begingroup$ That's a good point. (I haven't forgotten about accepting, I just want to think for a bit more.) $\endgroup$ – huon Oct 24 '14 at 12:19

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.