There are two questions which are interrelated hence I want to mention them over here.

Given the irrep $\Gamma^{(3)}$ of group C3V, which is of 2 dim. It can be diagonalized further, into a simple diagonal matrix. We still say this set is irrep because a common matrix S does not diagonalize them simultaneously.

I want to ask if I am right on this one?

Now, if I am right on this one.

If, I create a reducible form of C3v group by

$\Gamma^{(2)}\bigoplus\Gamma^{(3)}$ then this set also cannot be diagonalized simultaneously, then why do not we call this representation as irreducible.

I would like to clear the confusion. I may be wrong somewhere, may in the calculation or understanding the concept of block diagonalization.

I have been reading from Physics textbooks and unfortunately they do not properly emphasize on this. Please consider my limited knowledge of advanced abstract mathematics while answering.

  • $\begingroup$ Why do you say " It can be diagonalized further, into a simple diagonal matrix. " $\endgroup$ – ancientmathematician Oct 27 '18 at 10:43
  • $\begingroup$ @ancientmathematician I mean to say that each of the matrix in the representation set can be diagonalized further for example $ \Gamma^{(3)}(\sigma_v)= \begin{bmatrix} \frac{1}{2} & -\frac{\sqrt(3)}{2} \\ -\frac{\sqrt(3)}{2} & -\frac{1}{2} \end{bmatrix} $ can be diagonalized to M= \begin{bmatrix} 1& 0 \\ 0 & -1 \end{bmatrix} $\endgroup$ – Chetan Waghela Oct 27 '18 at 11:05
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    $\begingroup$ OK, you can say "each element can be diagonalised" (no surprise) but you shouldn't say "it [the representation] can be diagonalised": it can't. Your second representation can't be "called ... irreducible" because it is clearly reducible. $\endgroup$ – ancientmathematician Oct 27 '18 at 11:10
  • $\begingroup$ @ancientmathematician Thanks, I now understand that we intend to diagonalize the whole representation set by same matrix. If it happens then it is reducible. But then this arises confusion in the second part of my questions. Can you please elaborate a bit more. $\endgroup$ – Chetan Waghela Oct 27 '18 at 11:16
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    $\begingroup$ Almost. One dimensional irreducibles are simultaneously reduced. Character theory gives you the tools to decompose representations into irreducibles. Bye. $\endgroup$ – ancientmathematician Oct 27 '18 at 11:32

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