Exterior Power of a Vector Space

Let $G$ be a finite group, $V$ an $n$-dimensional vector space over $\mathbb{C}$, and $\tau: G \rightarrow GL(V)$ a representation such that $\tau(g)$ has determinant 1 for all $g \in G$. Why is it that $\wedge^{n}(V) = \mathbb{C}$?

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Tangential remark: \wedge is not supposed to be used like that (it is a binary operator). Some people write exterior powers using \bigwedge, which gives $\bigwedge$ and which at least gets the size correct. – Mariano Suárez-Alvarez Feb 10 '12 at 21:01

An element $g\in G$ acts on $\Lambda^bV$ by multiplication by $\det\tau(g)$, which is $1$.

Indeed, let $B=\{v_1,\dots,v_n\}$ is a basis of $V$ and suppose $g\cdot v_i=\sum_{j=1}^ng_{i,j}v_j$ for each $i\in\{1,\dots,n\}$, so that the matrix $(g_{i,j})_{1\leq i,j\leq n}$ is the matrix of $\tau(g)$ with respect to the basis $B$. The element $\omega=v_1\wedge\cdots\wedge v_n\in\Lambda^nV$ is a non-zero element in that $1$-dimensional space, so that $\{\omega\}$ is a basis of $\Lambda^nV$, and \begin{align} g\cdot\omega&=g\cdot v_1\wedge\cdots\wedge v_n\\ &=(g\cdot v_1)\wedge\cdots\wedge(g\cdot v_n)\\ &=\sum_{1\leq i_1,i_2,\dots,i_n\leq n}(g_{1,i_1}v_{i_1})\wedge\cdots\wedge (g_{n,i_n}v_{i_n})\\ &= \det(\tau(g)) \omega \end{align}

Of course, one needs to justify the last equality. We start by rewriting the sum as $$\sum_{1\leq i_1,i_2,\dots,i_n\leq n}g_{1,i_1}\cdots g_{n,i_n}\;v_{i_1}\wedge\cdots\wedge v_{i_n}.$$ A term corresponding to a choice of $i_1$, $\dots$, $i_n$ in which we have repetitions is zero, so we can sum only over the choices of $i_1$, $\dots$, $i_n$ for which $(i_1,\dots,i_n)$ is a permutation of $(1,2,\dots,n)$, which we call $\pi$. If that is the case, then one should know that $$v_{i_1}\wedge\cdots\wedge v_{i_n}=\operatorname{sgn}(\pi)\;v_1\wedge\cdots\wedge v_n.$$ Doing this in all terms, we see that the sum is in fact equal to $\det(\tau(g))\;\omega$, in view of the definition of the determinant as a sum over permutations.

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Dear Mariano, I wasn't advised that you had posted an answer while I was finishing mine. However sometimes I am advised (and on MO I am always advised, so I suppose it is not my browser). Is it because in this case only three minutes had elapsed between your answer and mine? – Georges Elencwajg Feb 10 '12 at 20:55
I am answering the question «Why is $\Lambda^nV$ a trivial representation?» If you intended to ask «why is $\Lambda^nV$ of dimension $1$?» too, you should probably make it explicit in the question :) – Mariano Suárez-Alvarez Feb 10 '12 at 20:57
@GeorgesElencwajg, it seems to depend on the position of the planets... It would surely be useful for that information to be given as fast as possible, to avoid wasted work. Maybe you can post a question/feature request in meta about this? – Mariano Suárez-Alvarez Feb 10 '12 at 20:59
Dear Mariano, I've asked at meta.Thanks for the suggestion. – Georges Elencwajg Feb 10 '12 at 21:44

It is not strictly true that $\Lambda^n V=\mathbb C$, but only that $\Lambda^n V$ is a $1$-dimensional complex vector space.

However the automorphism $\tau_g:V\to V$ has an $n$-th exterior power which is an endomorphism
$\Lambda^n (\tau_g):\Lambda^n V\to \Lambda^n V$ of a one-dimensional vector space and thus is multiplication by a complex number . That number is $det (\tau_g)\in \mathbb C$.

Note
At the level of elementary representation theory, it is permissible to be more down-to-earth and identify $V$ with $\mathbb C^n$, $GL(V)$ with $GL_n(\mathbb C)$, $\Lambda^n V$ with $\mathbb C$, etc.
However once you study more advanced material, that sort of confusion will lead to disaster: for example if $V$ is a rank-$n$ vector bundle, it is a terrible mistake to think that its determinant bundle is the trivial bundle with fibre $\mathbb C$.
So I encourage you to try and progressively learn the more canonical view as soon as possible.

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sorry guys i did not understand one thing,why it shows me math processing error in red form? – dato datuashvili Feb 10 '12 at 20:57