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.

  • $\begingroup$ 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

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.


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