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I must be missing something stupid here. A representation $D^{(\mu)}$ of $G$ is irreducible iff the following holds $$\frac{1}{[G]}\sum_{g\in G} |\mathcal{X}^{(\mu)}(g)|^2=1$$ where $\mathcal{X}^{(\mu)}(g)= \text{tr} D^{(\mu)}(g)$ is the character, i.e. the trace of the matrix representation.

I tried working an example. I have the irreducible (over $\mathbb{R}$) representation of $C_3$ given by $$ D(c) = \begin{bmatrix}0 & 1\\-1 &-1\end{bmatrix} \quad D(c^2) = \begin{bmatrix}-1 & -1\\1 & 0\end{bmatrix} \quad D(c^3) = \begin{bmatrix}1 & 0\\0 &1\end{bmatrix} $$ I have $$ \mathcal{X}(c) = -1 \quad \mathcal{X}(c^2) = -1 \quad \mathcal{X}(c^3) = 2 $$ But then $$\frac{1}{3}\left( |-1|^2+|-1|^2+|2|^2 \right)=2\neq 1 $$

What am I doing wrong?

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  • $\begingroup$ What is $\frac{1}{[g]}$ here? $\endgroup$ – Rudi_Birnbaum Mar 25 '18 at 13:13
  • $\begingroup$ @R_Berger The reciprocal of the group order of $G$. The theorem is call the dimensionality theorem. Here is a source: mathworld.wolfram.com/IrreducibleRepresentation.html $\endgroup$ – berrygreen Mar 25 '18 at 13:16
  • $\begingroup$ Conusing to use $g$ here since you use it also for the elements. $\endgroup$ – Rudi_Birnbaum Mar 25 '18 at 13:17
  • $\begingroup$ @R_Berger, sorry but I know that, and I still do not understand, so maybe I have some misconceptions of the definitions, I have one such irreducible representation; so you are saying that my representation is actually reducible in $\mathbb{C}$ and therefore the theorem does not work here; but it is irreducible in $\mathbb{R}$, so is there some way of making the theorem work in $\mathbb{R}$? $\endgroup$ – berrygreen Mar 25 '18 at 13:27
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    $\begingroup$ @R_Berger, so basically the irreducible representations are the cube roots $z$ of 1 such that $z^3=1$. But you need a 2 dimensional matrix to represent these in the $\mathbb{R}^n$. Makes sense. So it is irreducible in $\mathbb{R}^{2\times 2}$ but not in $\mathbb{C}^{2\times 2}$. thanks all is clear now $\endgroup$ – berrygreen Mar 25 '18 at 13:49
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The value you calculate when you see the representation as a real representation or as a complex representation is the same, simply because the traces are the same. Here $D(c)$ doesn't have an eigenvector in $\mathbb{R}^2$ so the representation is irreducible as a real representation, but it has one in $\mathbb{C}^2$, so it's a reducible representation : the value we get in $\mathbb{C}$ should be different from $1$. But the value is the same as that in $\mathbb{R}$ !

So we get a result that's different from $1$ in $\mathbb{R}$ as well, but there the representation is irreducible. So where did we go wrong ?

Well obviously the theorem you mention isn't wrong, but if it could be applied as you do, it would be contradictory. The problem here is that $\mathbb{R}$ is not algebraically closed, and so the theorem doesn't apply to it.

In fact for any nonalgebraically closed field $k$ you can find a similar issue where a representation is irreducible in $k$, but not in an algebraically closed field $K$ containing $k$. The value $\frac{1}{|G|} \displaystyle\sum_{g\in G}|\chi(g)|^2$ will therefore be different from $1$ in $K$, but then it will also be different from $1$ in $k$. This shows that the theorem cannot hold for all fields;and in its proof you actually see that you use the fact that the underlying field is algebraically closed (like $\mathbb{C}$)

What this implies is that checking whether a real representation is irreducible is more complicated than checking whether a complex representation is irreducible.

To perhaps see this more concretely, take the example of abelian groups (as $C_3$): over $\mathbb{C}$ (or any algebraically closed field), irreducible representations of such groups are all $1$-dimensional: obviously (as your example shows) this isn't true for $\mathbb{R}$.

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The criteria (formula) you are using is a consequence of "Schur's lemma". Stating that the irreducible characters are orthonormal. To be precise it states:

Is $G$ a finite group, $K$ a field with characterisics $ch(K)$ which does not divide $[G]$ and $\varphi$ and $\psi$ two different irreducible characters of $G$ over the field $K,$ then $$<\varphi,\psi>_{G}=\delta_{\varphi,\psi}$$ if $K$ is an algebraically closed field.

So you need to work over $\mathbb{C}$ not $\mathbb{R}$.

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