Does L_g pass to the quotient G/H?

[PsychonautQQ]

Say H is a subgroup of topological group G. Let L_g: G--->G be denote a map of G acting on itself by a left translation of g. Show that L_g passes(descends) to the quotient G/H.

I am a bit confused here, for L_g to pass to the quotient G/H, it would have to be constant on the fibers of G/H. This means that if q: G--->G/H is the quotient map, then q(a)=q(b) implies that L_g(a) = L_g(b); I don't believe this implication is true. If q(a) = q(b) then ar = bs for some a,s in H. but if L_g(a) = L_g(b) this means that ag = bg.

Input anyone?

 

[fresh_42]

Say H is a subgroup of topological group G. Let L_g: G--->G be denote a map of G acting on itself by a left translation of g. Show that L_g passes(descends) to the quotient G/H.

I am a bit confused here, for L_g to pass to the quotient G/H, it would have to be constant on the fibers of G/H. This means that if q: G--->G/H is the quotient map, then q(a)=q(b) implies that L_g(a) = L_g(b); I don't believe this implication is true. If q(a) = q(b) then ar = bs for some a,s in H. but if L_g(a) = L_g(b) this means that ag = bg.

Input anyone?

We don't have to show $ag = bg$ and neither $ga=gb$ which I use here as you spoke of left multiplication. We must show that $q(a)=aH=bH=q(b)$ implies $g.q(a) := q(ga)=gaH\stackrel{!}{=}gbH=q(gb) =: g.q(b)$. The cosets have to be equal, not the elements, so $gah \in gbH$ for any $h \in H$ has to be shown.

Of course we'll get in trouble if we want to identify the induced operation $\overline{L_g}$ on $G/H$ with the left multiplication $L_{q(g)}$ in $G/H$ because $H$ isn't required to be normal.

 

[PsychonautQQ]

Ooooooh so the left translation of G acting on elements of G whilst in their cosets, right?

 

[fresh_42]

Ooooooh so the left translation of G acting on elements of G whilst in their cosets, right?

Not sure I know what you mean here. The left translation $L_g\, : \,x \longmapsto g\cdot x$ induces a map $\overline{L_g}\, : \,xH \longmapsto g\cdot (xH)=(g\cdot x)H$ which is well-defined as two representatives of the same coset $xh=y$ lead to the same (co-)set $gxH=gyH$ using $H\cdot H = H$. However, since $G/H$ is no group, it is not the left multiplication as $L_g$ is in $G$. So $\overline{L_g}$ is purely set theoretically defined.

 

[PsychonautQQ]

Okay, what you are saying is making a lot of sense. I guess when it said "passing to the quotient" I thought that the translation L_g: G-->G needed to equal the composition of some map from L_g to the quotient space and then the quotient space back to L_g... like usually when we say 'passes' or 'descends' to the quotient it means something along these lines i feel like.