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Calculate the following actions:

$(i)$ Find an action of the group $GL(n,\mathbb{R})$ on the Euclidean space $\mathbb{R}^n$.

$(ii)$ Find an action of the group $GL(n,\mathbb{C})$ on the Euclidean space $\mathbb{C}^n$.

$(iii)$ Find an action of the group $\operatorname{Aut}(\mathbb{Z})$ on the Euclidean space $\mathbb{C}^n$.

Answer:

$(i)$

Let $A \in GL(n, \mathbb{R})$ and $ v \in \mathbb{R}^n$, then the action is defined by $A \cdot v=Av$

i.e, the group act as a linear transformation.

$(ii)$

The similar action can be described in this case also.

$(iii)$

We know $ \operatorname{Aut}(\mathbb{Z})=\{1,-1\}$.

Also the automorphism group $Aut(G)$ acts on $G$ by $ \psi \cdot g=\psi(g), \ \ \psi \in Aut(G)$ and $g \in G$.

But I do not know what is the action of $\operatorname{Aut}(\mathbb{Z})$ on $\mathbb{C}^n$ .

Is it Frobenius action or something else?

Kindly explain and answer this part $(iii)$.

Bernard
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MAS
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    If you're simply asked to find an action, with no further specifications, I would just define the trivial action for all 3 groups and call it a day. – Viktor Vaughn Jun 09 '19 at 20:11

1 Answers1

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Since $\operatorname{Aut}(\mathbb Z)\simeq\{1,-1\}$, you can make it act on $\mathbb C^n$ by making $1$ act as $\operatorname{Id}$ and $-1$ act as $-\operatorname{Id}$, for instance.

José Carlos Santos
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    Or as the identity and coordinatewise complex conjugation. – Arturo Magidin Jun 09 '19 at 18:02
  • @Jose, If I change $Aut(\mathbb{Z})$ by $Aut(\mathbb{C})$, then what will happen? – MAS Jun 09 '19 at 18:05
  • Then you will get a new question. I suggest that you post it as such. – José Carlos Santos Jun 09 '19 at 18:11
  • @JoséCarlosSantos, yes but I just want to know, how much change will happen. is it going to be difficult? why not just hints me at least? – MAS Jun 09 '19 at 18:15
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    I believe that it is difficult, due to the existenc of wild automorphisms of $\mathbb C$. – José Carlos Santos Jun 09 '19 at 18:55
  • @ArturoMagidin, would you please explain "coordinate wise complex conjugation" ? – MAS Jun 10 '19 at 01:47
  • @M.A.SARKAR: Excuse me, but what is there to explain? If you do not follow that, then the gaps in your knowledge are too great for a comment to fix. – Arturo Magidin Jun 10 '19 at 02:39
  • @ArturoMagidin, Yes I have little knowledge in Group Action, I am studying it now. Can you explain it shortly or refer me some notes or book ? I know any group acts on itself in two way, one is by left or right multiplication g→gh and another through conjugation g→ghg−1. Is this you are talking about ? Or you are talking about the following: Let $V$ be a vector space over $\mathbb{C}$, then any antilinear map $\phi:V \to V$ satisfying $\phi^2=Id,$ $\phi(zv)=\bar z \phi(v), \ v \in V, \ z \in \mathbb{C}$ and $\phi(v_1+v_2)=\phi(v_1)+\phi(v_2)$ is the complex conjugation. Is it this? – MAS Jun 10 '19 at 10:46
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    @M.A.SARKAR Again, I can't teach you group actions on a comment. Go talk to your professor; they get paid to teach you. – Arturo Magidin Jun 10 '19 at 15:53