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I'm reading through my notes on representation theory of $S_n$ and $GL(V)$, and have come unstuck on a definition which I can't understand - furthermore I can't seem to find any information on it online, so I can't read up on it to figure out what's going on. I'd be very grateful for an explanation or some direction to somewhere online I can find out more about it. Our notes contain the following definition:

Suppose $V$ is an $m$-dimensional vector space over $\mathbb{C}$, and $n \in \mathbb{Z}$: then we denote the 1-dimensional $\mathbb{C}GL(V)$ module corresponding to the representation $GL(V) \to \mathbb{C}^*$, $g \to (\det g)^n$ is denoted $\det ^n$. So, for example, $\det ^ 1 \cong \Lambda^mV$, where $\Lambda^m V$ denotes the $m$-th exterior power of V, $\det^n \cong (\det^1 )^{\otimes n}$ if $n \geq 0$ and $\det^n \cong (\det^{-n})^*$ if $n \leq 0$.

So, I don't really get what this module actually is: I know what the exterior powers are, but why is it that the module described is equal to the exterior power for $n = 1$, and equal to the other objects stated for other values of $n$? I can't find anything useful on these "det modules", so if anyone could explain step-by-step what's being said here, that would be extremely helpful. Obviously I have the basic representation theory background (what tensor products, $\mathbb{C}GL(V)$ are etc), but if you could keep explanations relatively simple here if possible I'd be very grateful. Many thanks in advance.

(Edit: just in case it isn't possible to tell what these modules are from my given portion of the notes, they later go on to prove "the 1-dimensional rational $\mathbb{C}GL(V)$-modules are precisely the $\det^n,\,n \in \mathbb{Z}$" if that helps provide some context.)

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There is a canonical isomorphism from $\mathbb C^\times$ to the group of automorphisms $GL(\mathbb C)$ of the $1$-dimensional vector space. The map $\det:GL(V)\to\mathbb C^\times$ can then be viewed as a homomorphism $\det:GL(V)\to GL(\mathbb C)$, so it endows $\mathbb C$ with the structure of a $GL(V)$-module. This the the module which we usually write simply $\det$.

Now if $n\geq1$, we can consider the $n$th tensor power $\mathbb C^{\otimes n}$ of this module. This is what we write $\det^n$. If $n<0$, we denote $\det^n$ the dual of $\det^{-n}$.

Using the isomorphism $\mathbb C^\times\cong GL(\mathbb C)$ mentioned above, the module $\det^n$ corresponds to the homomorphism $g\in GL(V)\mapsto (\det g)^n\in\mathbb C^\times$.

One justification for this notation is that $\det^{n+m}\cong\det^n\otimes\det^m$ for all $n,m\in\mathbb Z$.

Why is $\det$ isomorphic to $\Lambda^nV$? Simply because the two are isomorphic!

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Thanks for your response. So, you're saying for $n = 1$ we can treat both $\mathbb{C}$ and $\Lambda^m(V)$ as $GL(V)$-modules, with the "action" $\rho$ of $GL(V)$ in $\mathbb{C}$ given by $\rho(g) z = \det (g) z$, and the action in the latter given by $\rho(g) e_1 \wedge e_2 \wedge \ldots e_m = (ge_1) \wedge \ldots (ge_m)$, $e_i \in V$? I'm probably being stupid but I can't see what the isomorphism is exactly? More generally, could you possibly elaborate on why $\det^n = \mathbb{C}^{\otimes n}$ is the module corresponding to $\rho:g \to (\det g)^n$ when $n>0$, and the dual of this for $n<0$? –  Spyam Mar 28 '12 at 5:56
    
I'll add later the idea behind the isomorphism with the exterior power, buy you should be able to check my claim about the tensor powers of the $\det$ module if you know the definitions! –  Mariano Suárez-Alvarez Mar 28 '12 at 5:58
    
Ok, yes I think I can check those sorry, I was thinking of $\rho(g)(z_1 \times \ldots z_n) := (gz_1)\times \ldots (gz_n)$ rather than $\rho(g)(z_1 \times \ldots z_n) := (\rho(g)z_1)\times \ldots (\rho(g)z_n)$, and got myself confused. I'll look forward to your comment on the isomorphism, still. –  Spyam Mar 28 '12 at 6:23
    
did you ever have a chance to think about the isomorphism here? Many thanks for your help :) –  Spyam Mar 30 '12 at 20:30
    
Hi, sorry to bother you again, but I was just wondering if you might be able to explain the isomorphism here like you said at some point, if it's not too much trouble? –  Spyam Apr 27 '12 at 0:00
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