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I am asked to determine all irreducible $FG$-modules of dimension $n$ for $G=C_k$, the cyclic group of order $k\in \mathbb N$ in the following cases:

  • $n=1, F= \mathbb R$
  • $n=2, F= \mathbb R$
  • $n=1, F= \mathbb C$
  • $n=2, F= \mathbb C$

I honestly do not know where to start. This seems like a classification question, can someone help me to get started?

  • $C_k-$cyclic group of order $k$? If so, what do you know about the dimension of irreducible representations of abelian groups? – Want to learn Apr 25 '20 at 17:06
  • https://math.stackexchange.com/questions/1661134/deduce-that-the-number-of-inequivalent-degree-1-complex-representations-of-g – Want to learn Apr 25 '20 at 17:13
  • $\rho$ is an irreducible representation $\iff $ dim$(V)=1$? where $V$ denotes the FG-module. –  Apr 25 '20 at 17:14
  • Yes. It shows the second and fourth option there will be no irreps. Remaining cases you can count how many of them are there, by looking at the possible images of $1$. – Want to learn Apr 25 '20 at 17:15
  • Well I know that $\rho(e_G)=1 \in \text{GL}(n, \mathbb R)$, but how do I use this? –  Apr 25 '20 at 17:21
  • When $n=1$, what is $GL_n(\Bbb R)$? – Want to learn Apr 25 '20 at 17:22
  • That is simply $\mathbb R$ –  Apr 25 '20 at 17:23
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    No. invertible matrices. So it is $\Bbb R^* = \Bbb R \backslash {0}$. – Want to learn Apr 25 '20 at 17:24
  • https://math.stackexchange.com/questions/354926/complex-finite-dimensional-irreducible-representation-of-abelian-group – Want to learn Apr 25 '20 at 17:25
  • Oh yes, my bad. Excluding zero. So the multiplicative group of real numbers. –  Apr 25 '20 at 17:26
  • So we know that $\rho(e) = 1 \in \mathbb R^$ for $e$ the identity permutation. Or for the other option (third case) $\rho(e) = 1 \in \mathbb C^$ –  Apr 25 '20 at 17:30
  • Everything I was told you about was over $\Bbb C$. When $n = 1$, $GL_n(\Bbb C) = \Bbb C^*$. – Want to learn Apr 25 '20 at 17:30
  • https://math.stackexchange.com/questions/3119335/prove-that-every-irreducible-real-representation-of-an-abelian-group-is-one-or-t – Want to learn Apr 25 '20 at 17:36
  • What I understand is that the theorem only holds for algebraically closed fields, such as $\mathbb C$. So in this case we have that a $\mathbb C G$ module is irreducible if and only if its dimension is 1. And in the fourth example there are none. Then since a representation over $\mathbb C$ can be turned into one over $\mathbb R^2$, we know that in $\mathbb R^2$ irreducible representations can be of dimension $2$. –  Apr 26 '20 at 10:44
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    Indeed, it is not true that an irreducible representation of an abelian group over $\mathbb{R}$ has to be $1$-dimensional. It can be $2$-dimensional. – Captain Lama Apr 26 '20 at 12:45
  • Easy example: $C_3$ acts irreducibly on $\mathbb{R}^2$ by rotations of angle $2\pi/3$. – Captain Lama Apr 26 '20 at 12:46
  • But then $C_k$ acts on $\mathbb R^2$ by rotations of an angle of $2 \pi/k$. How do you know this is irreducible? Irreducibility is defined for modules as being the property that its only submodules are the trivial ones. –  Apr 26 '20 at 13:19

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This follows on from the discussion in the comments.

Let $V_k$ be the $2$-dimensional $\mathbb{R}C_k$-representation that you described. We shall show that $V_k$ is irreducible when $k > 2$, and reducible when $k = 1,2$

Suppose that $k>2$ and suppose, for the sake of contradiction, that $V$ is reducible, that is it contains some non-zero proper subrepresentation $U \subseteq V$. Since $V$ is $2$-dimensional, that forces $U$ to be one-dimensional. In other words any non-zero $u \in U$ is simultaneously an eigenvector for the action of $C_k$ (or alternatively an eigenvector for its generator since it is cyclic). But this generator has no eigenvectors when $k > 2$, and so we have reached our contradiction.

When $k = 1,2$ the vector $\begin{pmatrix} 1 \\ 0\end{pmatrix}$ is an eigenvector of the generator of $C_k$, and so generates a non-trivial one-dimensional submodule of $V_k$.

For an extension, why does this argument fail if we replace $\mathbb{R}$ with $\mathbb{C}$?

Captain Lama
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Adam Higgins
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