1
$\begingroup$

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, $D_6$:

enter image description here

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

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.