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I am reading a document it says

A group realization is a map from elements of G to transformations of a space M that is a group homomorphism, i.e. it preserves the group multiplication law. Thus if $T:G \rightarrow T(M): g \mapsto T(g)$ where $T(g)$ is some transformation on M, then T is a homomorphism if $T(g_1 g_2) = T(g_1)T(g_2)$.

I have got a bad feeling about this definition.

If T(M) is transformations on M then how can $T$ also be a map from $G$ to $T(M)$?

This document page 4.

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It's not a bad definition, it's a bad notation, because $T$ is playing two roles: it is the name of the transformation group of $M$, and it is also the name of the function from $G$ to the transformation group of $M$. Instead of calling the transformation group $T(M)$, call it something else, for example, $\mathrm{Tr}(M)$. Then $T\colon G\to\mathrm{Tr}(M)$ is a function from $G$ to $\mathrm{Tr}(M)$. – Arturo Magidin Mar 18 '11 at 19:36
I suspected it. Would you post modified definition as an answer? Thanks! :) – Pratik Deoghare Mar 18 '11 at 19:41
Done. – Arturo Magidin Mar 18 '11 at 19:51
up vote 1 down vote accepted

The problem here is really just the notation: $T$ is playing two roles, both the name of the group homomorphism from $G$ to the transformation group of $M$, and as part of the name of the group of transformations of $M$. Since the notes seem to often give special names to the group of transformations, depending on who $M$ is, a simple fix is just to change the generic name for this group.

Instead of using $T(M)$ to denote "a group of transformations $M$", call it something else, say $\mathrm{Tr}(M)$. Then $T\colon G\to \mathrm{Tr}(M)$ is a realization of $G$ if $T$ is a homomorphism from $G$ to a group of transformations of the a space $M$.

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Why community wiki? – Pratik Deoghare Mar 18 '11 at 19:54
@MachineCharmer: I posted it as a comment because I thought it was not "worthy enough" of full answer status. It's really just a simple observation. By making it a "community wiki" answer, you can mark it as accepted (so the site knows the answer has been answered), but I don't get any reputation from the (in my view rather minor) contribution. – Arturo Magidin Mar 18 '11 at 19:56

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