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The Cyclic Groups $(\mathbb Z_n,+)$ have various representations. This answer asserts that the only irreducible representations are

  • 1-dimensional, with matrix a real $n$th root of unity;
  • 2-dimensional, with matrix $\left(\begin{smallmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{smallmatrix}\right)$ where $n\theta \equiv 0 \pmod{2\pi}$ but $\sin\theta \neq 0$.

But what about other minimal unitary matrix representations?

Take as an example $\mathbb Z_6$, which has other representations. Such as the 6 cyclic permutations of

$$\pmatrix{1&0&0&0&0&0\\0&1&0&0&0&0\\0&0&1&0&0&0\\0&0&0&1&0&0\\0&0&0&0&1&0\\0&0&0&0&0&1} $$

constructed by repeatedly moving the leftmost column to the righthand side of the matrix and shifting all of the other columns left by 1 space.

Or the representation of 6 5x5 matrices constructed from the direct sum of the $\mathbb Z_2$ and $\mathbb Z_3$ representations

$$\left\{\pmatrix{1&0\\0&1},\pmatrix{0&1\\1&0}\right\} $$ and $$\left\{\pmatrix{1&0&0\\0&1&0\\0&0&1},\pmatrix{0&0&1\\1&0&0\\0&1&0},\pmatrix{0&1&0\\0&0&1\\1&0&0}\right\} $$ respectively.

Neither of these representations can be written as a direct sum of other representations of $\mathbb Z_6$. In other words, as far as I can tell, these representations have only trivial subrepresentations; therefore, why should they not be considered irreducible (as per the linked question)?

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  • $\begingroup$ They are reducible, meaning they can be written as direct sums of other representations. It takes a bit of cleverness. $\endgroup$ – Qiaochu Yuan Oct 14 '17 at 1:27
  • $\begingroup$ Are you working over the real numbers or the complex numbers? $\endgroup$ – Trevor Gunn Oct 14 '17 at 1:27
  • $\begingroup$ @TrevorGunn I suppose I was thinking about the reals when I wrote the question, but I wouldn't mind an answer that explains how they are reducible using complex values. I still don't see how that would work. $\endgroup$ – Geoffrey Oct 14 '17 at 1:34
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    $\begingroup$ @QiaochuYuan If you'd care to elaborate in an answer, it would be appreciated! $\endgroup$ – Geoffrey Oct 14 '17 at 1:35
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With $\mathbb{Z}_6$ acting on $\mathbb{R}^6$ by cyclic coordinate shifts, the irreducible subrepresentations are

  • The 1D span of $(1,1,1,1,1,1)$
  • The 1D span of $(1,-1,1,-1,1,-1)$
  • The 2D space $\{(a,b,c,a,b,c):a+b+c=0\}$
  • The 2D space $\{(a,-b,c,-a,b,-c):a+b+c=0\}$

With $\mathbb{Z}_6\cong\mathbb{Z}_2\times\mathbb{Z}_3$ acting on $\mathbb{R}^2\oplus\mathbb{R}^3$ by cyclic coordinate shifts in each component, they are

  • The 1D span of $(1,1,0,0,0)$
  • The 1D span of $(1,-1,0,0,0)$
  • The 1D span of $(0,0,1,1,1)$
  • The 2D space $\{(0,0,a,b,c):a+b+c=0\}$
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  • $\begingroup$ Maybe I'm missing something, doesn't $\mathbb R^2 \times 0$ have the subrepresentation $\mathbb R\cdot(1, 1) \times 0$? $\endgroup$ – Jim Oct 14 '17 at 2:49
  • $\begingroup$ @Jim Oops yes. ${}$ $\endgroup$ – anon Oct 14 '17 at 2:51
  • $\begingroup$ Oh, I think I understand this now. If you can find a set of eigenvectors which simultaneously block-diagonalize all of the matrices in the representation, then you will have decomposed the representation into a direct sum space. Right? Is there any advice you can give me about identifying what eigenvectors to use for these irreducible subspaces? $\endgroup$ – Geoffrey Oct 14 '17 at 3:41
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    $\begingroup$ @Geoffrey I wouldn't use the term "eigenvector" - none of the vectors in any of the 2D spaces I gave are eigenvectors for instance. But yes, every group element acts block diagonally with respect to a decomposition into subreps. The point is that the properties characterizing a subrep must be invariant with respect to cyclic shifts. For instance, the property of having the same coordinate, or the property of coordinates summing to zero, or the property of coordinates alternating in sign, etc. $\endgroup$ – anon Oct 14 '17 at 4:28
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    $\begingroup$ @Geoffrey That's presumably because you aren't using an orthonormal basis for said 2D rep. If you use an orthonormal basis, like $$ \begin{array}{r} \alpha= & \frac{1}{2}(+1,-1,0,+1,-1,0) \\ \beta= &\frac{1}{2\sqrt{3}}(-1,-1,2,-1,-1,2) \end{array} $$ then the generators should act as rotations by an angle of $2\pi/3$. $\endgroup$ – anon Oct 14 '17 at 6:04

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