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In the Fulton/Harris book "Representation theory: a first course", section 1.3:

We are looking at any representation W of $S_3$, by just looking at the abelian group $U_3 \sim Z/3Z \subset S_3$. Let $\tau$ be any generator of $U_3$.

Why is that the book says "the space W is spanned by the eigenvectors of the action $\tau$", and "whose eigenvalues are of cause all the powers of $\omega = e^{2\pi i/3}$? Can someone elaborate on this?

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2 Answers 2

$\tau$ has order 3, so its image in $GL(W)$ has order 3. (or the degenerate case, 1) This image has minimal polynomial being a factor of $x^3 - 1$, which means it's diagonalizable, with possible eigenvalues being the three third roots of unity.

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+1: This works well enough, if we know that $W$ is finite-dimensional. That is probably the case of interest to the OP. Otherwise you can always reduce it to at most 3-dimensional subspaces (given a vectors $x\in W$ consider the $U_3$-invariant span of $\{w=\tau^3w,\tau w, \tau^2 w\}$, or go with the trick in my answer (more or less the same thing). –  Jyrki Lahtonen Sep 12 '11 at 12:25

You have $U_3$ acting on $W$ (presumably a vector space ove the complex numbers). If $w\in W$ is arbitrary, we can write $$ w=\frac13\left([w+\tau w+\tau^2w]+[w+\omega^2\tau w+\omega\tau^2w]+[w+\omega \tau w+\omega^2\tau^2w]\right), $$ because $1+\omega+\omega^2=0$. Here the vectors surrounded by square brackets are eigenvectors of $\tau$. The first belongs to eigenvalue $\lambda=1$, the second to $\lambda=\omega$ and the last to $\lambda=\omega^2$.

In other words $W$ decomposes into a direct sum of representations of $U_3$. These are all known to be 1-dimensional, and thus $\tau$ acts on any irreducible $U_3$-module by scalar multiplication. After you have learned a bit more representation theory, you will observe that you can replace $U_3$ with any finite abelian group. Some people would view this decomposition as discrete Fourier analysis. The representation theory of finite cyclic (actually all finite abelian) groups is surprisingly ubiquitous.

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