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Let $\rho : \mathbb Z \to \mathrm{GL}_2(\mathbb C)$ be the representation defined by $\rho(1) = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$. I'd like to show that $\rho$ is not completely reducible.

  1. I have one preliminary question (which is probably a silly one) - for what vector space $V$ is $\mathrm{GL}(V) \cong \mathrm{GL}_2(\mathbb C)$?

    Firstly, I noted that $\rho(1)$ has an eigenvector, so the representation is not irreducible. So if it were completely reducible, it would have to break up as a direct sum of two $1$-dimensional sub representations. But a 1-dimensional subrep is given by an eigenvector - but $\rho$ only has one eigenvalue, which has a $1$-dimensional eigenspace. So this can't happen.

  2. Is this reasoning OK?

    Once I've shown that the representation isn't irreducible, the problem is equivalent to showing that $\rho(1)$ cannot be diagonalised (which I've done by showing that the sum of the dimensions of the eigenspaces is $1$, not 2).

    Depending on the answer to question (1), I could have reduced (excuse the pun) the amount of work by considering Jordan Normal Form ($\rho(1)$ is in JNF but isn't diagonal, so isn't diagonalisable).

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    $\begingroup$ I think I am being silly with question i). $V = \mathbb C^2$ as a $\mathbb C$-vector space, right? $\endgroup$
    – Matt
    Mar 17, 2012 at 16:01
  • $\begingroup$ So the Jordan Normal Form argument does apply $\endgroup$
    – Matt
    Mar 17, 2012 at 16:02
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    $\begingroup$ Indeed, the vector space is $\mathbb C^2$., and yes, the JNF argument does work, too. $\endgroup$
    – Mariano Suárez-Álvarez
    Mar 17, 2012 at 16:04
  • $\begingroup$ @Matt : The only such $V$'s you will find are those who are isomorphic to $\mathbb C^2$. This is the classical example where we have a counter-example to justify representation theory working only with finite groups. When in infinite group, the whole idea of computing characters to work with the representation is pointless. $\endgroup$
    – Patrick Da Silva
    Mar 17, 2012 at 16:04
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    $\begingroup$ @Patrick: Well... pointless is a bit too strong :) For example, in the correct context, the finite dimensional representation theory of many infinite groups (like semisimple Lie groups, say) is completely controled by characters. $\endgroup$
    – Mariano Suárez-Álvarez
    Mar 17, 2012 at 16:15

1 Answer 1

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This CW answer intends to remove the question from the unanswered queue.

As you already noted yourself in the comments you can take $V=\mathbb{C}^2$ and then a Jordan normal form argument is perfectly fine.

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