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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.

i) 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.

ii) 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).

Dependant on the answer to question i), 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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I think I am being silly with question i). $V = \mathbb C^2$ as a $\mathbb C$-vector space, right? – Matt Mar 17 '12 at 16:01
So the Jordan Normal Form argument does apply – Matt Mar 17 '12 at 16:02
Indeed, the vector space is $\mathbb C^2$., and yes, the JNF argument does work, too. – Mariano Suárez-Alvarez Mar 17 '12 at 16:04
@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. – Patrick Da Silva Mar 17 '12 at 16:04
@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. – Mariano Suárez-Alvarez Mar 17 '12 at 16:15

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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