Invariant subspace and the irreducibility of a representation.

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Below is one definition for the irreducible representation:

For a group $G$, a $G-$representation $V$ is said to be irreducible if there is no nontrivial proper subspace of it that is $G-$invariant.

In the $Q2\ (c)$ of this note: https://www2.math.ethz.ch/education/bachelor/lectures/hs2013/math/algebra1/Solution10.pdf, it talked about the irreducibility of the hyperplane. That is:

Consider $G=S_{4}$, and $V=\mathbb{C}^{4}$ a permuted representation associated to the action of $S_{4}$ on the canonical coordinates.

Then, in the part $(b)$, it showed that

$V$ is the direct sum $V=V_{1}\oplus V_{2}$ of two irreducible representations $V_{1}$ and $V_{2}$ that are not isomorphic.

In part $(c)$, it asks the reader to show that $W$ is $S_{4}-$invariant and then conclude from here that $W$ is irreducible.

I have no problem with the invariance. But then, the note seems directly using $(b)$ to conclude that $V=W+W^{\perp}$, and $W$ is irreducible without talking about why.

It is indeed true that $W$ is $S_{4}$-invariant, but then how do I know if it has a subspace that is also $S_{4}-$invariant?

Thank you!

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This representation of $ S_4 $ clearly has a single one-dimensional irrep in its decomposition, since any element spanning such an irrep must be fixed by any even permutation in $ S_4 $ (since the only one-dimensional irreps of $ S_4 $ are those of its abelianization $ C_2 $, i.e. the trivial and the sign representations) and then it's trivial to show this element must be $ (1, 1, 1, 1) $ up to multiplication by a scalar.

This forces there to be only one other irrep in the direct sum decomposition of dimension $ 3 $, since dimension $ > 3 $ doesn't fit into the four dimensional space $ \mathbb C^4 $ and dimension $ 2 $ would require the decomposition to be like $ 1 \oplus 1 \oplus 2 $, implying two irreps of dimension $ 1 $, which we know to be impossible by the above argument. This means the representation looks like $ 1 \oplus 3 $, and the two irreducible components are clearly not isomorphic since they have different dimensions.