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From just the character table and the basis functions of the irreducible representations, how do I know whether a representation is faithful or unfaithful?

For the 1-D representation it is trivial to know the answer, of course, so I am only talking about the 2-dimensional ones.

For example, the hexagon group, $D_6$:

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How do I know whether $\Gamma_5$ and $\Gamma_6$ are faithful or unfaithful?

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The numbers in the right hand section of the table are called the character values. A representation is faithful if and only if the number in the $E$ column of that row only appears once in that row. So $\Gamma_1$ is not faithful, since all the columns have the same value as $E$ (namely, $1$). For nearly the same reason, $\Gamma_2$ is not faithful (1 appears four times, rather than once).

The only faithful representations listed are $\Gamma_6$ and $\chi^{\text{atom sites}} = \Gamma_6 + \Gamma_5 + \Gamma_3 + \Gamma_1$.

If a representation $\Gamma_i$ is faithful, so is $\Gamma_i + \Gamma_j$ for any representation $\Gamma_j$.

By the way, $\Gamma_1$ and $\Gamma_2$ are both 1-dimensional (at least to mathematicians). The dimension is that number in the $E$ column, also called the degree.

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  • $\begingroup$ Thanks. Is there a physical meaning as to why the representation is unfaithful if the character value of the identity is repeated? $\endgroup$ – SuperCiocia May 28 '14 at 15:09
  • $\begingroup$ Also, could I argue that Γ5 is unfaithful from the fact that it has quadratic basis functions? $\endgroup$ – SuperCiocia May 28 '14 at 15:10
  • $\begingroup$ And yes sorry I meant Γ5 and Γ6, the 2-D ones $\endgroup$ – SuperCiocia May 28 '14 at 15:13
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    $\begingroup$ I'm not familiar with physics. The character value is the trace of a matrix (the sum of the diagonals). If $z$ is the trace of an $n \times n$ matrix that represents a finite group element, then $|z| \leq n$ and $|z|=n$ iff $z$ is a scalar multiple of the identity (so if your matrices are real, then it is $\pm$ the identity matrix). Even more $z=n$ (the value of $E$) if and only if the matrix is the identity (the matrix for $E$). Smaller numbers $z$ might be the trace of many matrices, but the large ones with $|z|=n$ are very special. $\endgroup$ – Jack Schmidt May 28 '14 at 15:28
  • $\begingroup$ If you like eigenvalues: the trace is the sum of the eigenvalues. Since the matrices all satisfy $M^{12}=1$, their eigenvalues all satisfy $\lambda^{12} =1$, so $|\lambda|=1$. There are $n$ of them, so $|z| \leq n |\lambda| = n$. The only way to actually get $n$ exactly is if all the eigenvalues are the same, so the matrix $M$ is just the number $\lambda$ (times the identity matrix). The only way to get $z=n$ is if all the $\lambda=1$. $\endgroup$ – Jack Schmidt May 28 '14 at 15:31

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