4
$\begingroup$

Let $\Gamma$ be a group and $\Gamma'\leq \Gamma$ be the subgroup. Let $Y$ be $\Gamma'$ module. Define induction $Ind^\Gamma_{\Gamma'}Y=<f:\Gamma\to Y\vert\forall\gamma\in\Gamma,\gamma'\in\Gamma',f(\gamma'\gamma)=\gamma'f(\gamma)>$ as abelian group generated by elements prescribed.

Then $\Gamma$ action on $Ind^\Gamma_{\Gamma'}$ is defined by $\gamma\in\Gamma, (\gamma f)(x)=f(x\gamma)$.

$\textbf{Q:}$ When I try to verify the group action for $\gamma_i\in\Gamma$ $[(\gamma_1\gamma_2)f](x)=f(x\gamma_1\gamma_2)$ and $(\gamma_1)[(\gamma_2)f]=\gamma_1(f(x\gamma_2))=f(x\gamma_2\gamma_1)$, I cannot have $(\gamma_1\gamma_2)f=\gamma_1(\gamma_2(f))$. Have I done something wrong? Neukirch defined $\Gamma$ action by $(\gamma f)(x)=f(\gamma x)$. Then I have no trouble to see $\Gamma$ action here by $(\gamma_1\gamma_2f)(x)=f(\gamma_1\gamma_2x)=\gamma_1(f(\gamma_2x))=\gamma_1((\gamma_2)f)(x)$


Per Arnaud Mortier's comment, I think I figured out what is wrong in my mind setting. $\gamma f(x)=f(x\gamma)$ is setting left action of $f$ to right action on $x$. $(\gamma\gamma' )f(x)=f(x\cdot(\gamma\gamma'))=f((x\gamma')\gamma)=\gamma f(x\gamma')=\gamma(\gamma' f)(x)$

For $\gamma f(x)=f(\gamma x)$(Neukirch's definition), I need $(\gamma\gamma')f(x)=\gamma(\gamma' f)(x)=\gamma' f(\gamma x)=f(\gamma'\gamma x)$. I need to change left action to right action again here.

Ref. Lectures on Algebraic Geometry by Gunter Harder 2.2.4 Exercise 7.

$\endgroup$
  • $\begingroup$ There is a notion of right action and left action, and the syntax should be adapted accordingly. $\endgroup$ – Arnaud Mortier Jul 26 '18 at 23:24
  • $\begingroup$ @ArnaudMortier Yes. That is what I thought. My guess is $Hom(-,Y)$ is contravariant. That is why you want right action turned into left action. $\endgroup$ – user45765 Jul 26 '18 at 23:28
  • $\begingroup$ @ArnaudMortier Thanks for the hint. I think I figured it out. $\endgroup$ – user45765 Jul 26 '18 at 23:48
  • $\begingroup$ Dear @user45765, In the derivation you have added there are two errors that cancel one another to give you the desired result. Namely the second and fourth equalities are incorrect. $\endgroup$ – Keenan Kidwell Jul 27 '18 at 3:26
1
$\begingroup$

The correct computation to verify that $(\gamma_1\gamma_2)f=\gamma_1(\gamma_2f)$ goes as follows. Given $x\in\Gamma$, we have

$$((\gamma_1\gamma_2)f)(x)=f(x(\gamma_1\gamma_2))=f((x\gamma_1)\gamma_2) =(\gamma_2 f)(x\gamma_1)=(\gamma_1(\gamma_2f))(x)\text{.}$$

Note also that in your definition of the induced $\Gamma$-module, you do not need to include "abelian group generated by [the] elements prescribed," because the indicated functions already form a group under pointwise addition (using the operation on $Y$) of functions.

$\endgroup$

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.