# Irreducible representations of $\mathbb{Z}$

I'm wondering what are the irreducible representations of the group ($\mathbb{Z}$,+).

Knowing that for $\mathbb{Z}_n$ the 1-dimensional representations are the nth roots of unity, I considered taking the limit when $n \to \infty$ but I think I'm ending up with the U(1) group instead of $\mathbb{Z}$ and I have no other idea...

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You are probably looking for the complex representations (is that right?). Firstly I claim any finite-dimensional simple rep is one-dimensional. Let $G=\langle g\rangle$ be infinite cyclic, this will stand in place of $\mathbb{Z}$. If $g$ acts on some nonzero finite-dimensional complex vector space $V$ by a linear map $\rho(g)$ then this linear map has an eigenvector, which spans a one-dimensional subrep. If the rep was irreducible, this subrep must be all of $V$, so $\dim V=1$.

Now note that if $\lambda \in \mathbb{C}^*$ (I mean the non-zero complex numbers) then $g \mapsto \lambda$ induces a homomorphism $\rho_\lambda: G \to \mathsf{GL}_1(\mathbb{C})$, i.e. a representation of $G$. Clearly any one-dimensional representation arises this way. Furthermore if $\lambda \neq \mu$ then the representations afforded by $\rho_\lambda$ and $\rho_\mu$ are not isomorphic: not much conjugacy happens in $\mathsf{GL}_1$ !

In conclusion, the simple representations are in natural bijection with $\mathbb{C}^*$: this is called the character group.

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A terminological remark: some people (especially in older literature) use the term quasi-characters to denote characters with values in $\mathbb C^*$ but not necessarily in the circle group $S^1$, restricting the term character to mean what others might call a unitary character, i.e. a character with values in the unit circle $S^1$. – Matt E Nov 25 '11 at 21:45
Thanks a lot... With this result, I'm guessing : is that where the z-transform comes from ? – AlexPof Nov 26 '11 at 7:18
I don't think this is related to the Z-transform – m_t_ Nov 26 '11 at 11:09

If $\rho\colon \mathbb Z\to\mathrm{GL}(n,\mathbb C)$ is a complex representation, then because $\rho$ is a homomorphism, it is entirely determined by $\rho(1)$ (since $\rho(n)=\rho(1)^n$). So if $\rho$ and $\rho'$ are two non-isomorphic representations of $\mathbb Z$, then $\rho(1)$ and $\rho'(1)$ determine two non-conjugate matrices in $\mathrm{GL}(n,\mathbb C)$. Since two matrices are conjugate over $\mathbb C$ iff they have the same Jordan canonical form, we get that the $n$-dimensional complex representations of $\mathbb Z$ (up to isomorphism) are in one-to-one correspondence with the $n\times n$ Jordan canonical forms.

EDIT: I didn't notice you asked for irreducible representations of $\mathbb Z$. In this case, I guess see mt_'s answer.

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Of these representations, which ones are irreducible? – Greg Martin Nov 25 '11 at 21:38
As mt_ said, all the irreducible representations are 1 dimensional, so if $n>1$, then $\rho$ will not be irreducible. However, if $\rho(1)$ is diagonalizable, then $\rho$ decomposes into the direct sum of the one dimensional representations corresponding to its eigenvalues. – SL2 Nov 25 '11 at 22:07
Right, but what if $\rho(1)$ is not diagonalizable? Does it also decompose into the direct sum of one-dimensionals? Does every representation? I'm asking because it's clear I'm missing part of the picture, not because I think there's a problem with the answer. – Greg Martin Nov 26 '11 at 6:38
Representations don't have to decompose. Let $G = \langle g \rangle$ act on a 2d vector space by a 2x2 Jordan block with eigenvalue one. It has a composition series consisting of 1d reps, but it does not split as a direct sum of 1d reps (to see this, show that there is a unique 1d submodule that is not complemented). – m_t_ Nov 26 '11 at 11:03