I am working through the paper Non-Abelian Discrete Symmetries in Particle Physics, which can be found here. I am confused about the following highlighted passage found on pg. 8:

enter image description here

I understand this follows from the fact that there are two one dimensional irreps and one two dimensional irreps. My main question is asking if my following interpretation of this is correct:

The above statement is saying that the representation of each group $g\in G$ element can be written as the direct sum

\begin{align} D(g)=\mathbf{1}_{g}\oplus\mathbf{1'}_g\oplus \mathbf{2}_g. \end{align}

I don't think this is correct, but I'm a bit stuck at the moment. Could somebody please clarify how I go about finding $\mathbf{1}, \mathbf{1'}$, and $\mathbf{2}$, and how exactly that corresponds to the representation of the group elements?


Some context might be helpful:

In the paper you refer to, $m_n$ denotes the number of $n$-dimensional irreps of $S_3$. The passage you are reading to is trying to count how many irreps there are of each dimension.

The paper quotes two well-known facts:

  • The order of the group equals $m_1 + 4m_2 + 9m_3 + 16m_4 + \dots $

  • The number of conjugacy classes equals $m_1 + m_2 + m_3 + m_4 + \dots$

Since the order of $S_3$ is $6$, and the number of conjugacy classes is $3$, the only possibility is that $$ m_1 = 2, \ \ \ m_2 = 1, \ \ \ m_3 = m_4 = \dots = 0.$$

The irreps of $S_3$ are:

  • The one-dimensional "identity" irrep $\mathbf{1}$, where every element is represented by $1$.
  • The one-dimensioanl "alternating" irrep $\mathbf{1}'$, where every even element is represented by $1$ and every odd element is represented by $-1$.
  • The two-dimensional irrep $\mathbf{2}$, obtained by viewing elements of $S_3$ as $2\times 2$ matrices describing symmetries of a triangle.

Your statement that "the representation of each group element $g \in G$ can be written as the direct sum $\mathbf{1}_g \oplus \mathbf{1}'_g \oplus \mathbf{2}_g$" doesn't make a great deal of sense to me, I'm afraid! There is no such thing as "the representation". There are many possible representations of $S_3$: any direct sum of $\mathbf{1}$, $\mathbf{1}'$ and $\mathbf{2}$ will do.

  • $\begingroup$ Thank you, that was quite helpful. I just have a few follow up questions. (1) How does one arrive at the fact that $\mathbf{1'}$ was the "alternating" irrep?(2) I have the exact same question for the the ire $\mathbf{2}$. (3) Lastly, when you say that "any direct sum of $\mathbf{1}, \mathbf{1'}$ and $\mathbf{2}$ will do, what exactly do you mean, and how is does that differ from what I wrote down in the post? $\endgroup$ – InertialObserver Mar 18 '17 at 0:04
  • $\begingroup$ I don't quite follow... $\mathbb{1}'$ is just the standard notation for the alternating irrep. And the "alternating irrep" is just the standard name for this irrep. $\endgroup$ – Kenny Wong Mar 18 '17 at 0:05
  • $\begingroup$ I mean that $\mathbb{1}$ is a representation, $\mathbb{1}'$ is a representation, $\mathbb{2}$ is a representation, $\mathbb{1} \oplus \mathbb{1}$ is a representation, $\mathbb{1} \oplus \mathbb{2}$ is a representation, $\mathbb{1}' \oplus \mathbb{2} \oplus \mathbb{2} \oplus \mathbb{2}$ is a representation, $\mathbb{1} \oplus \mathbb{1} \oplus \mathbb{1}' \oplus \mathbb{2} \oplus \mathbb{2} \oplus \mathbb{2}$ is a representation... You get the idea. $\endgroup$ – Kenny Wong Mar 18 '17 at 0:09
  • $\begingroup$ I see. I suppose if I were to rephrase it it would be as follows: When use the characters to solve for certain information, I arrive at the fact that there are two one dimensional solutions, and one two dimensional irreps and that's it. I don't know anything about what maps to what (and hence I don't know that it is the alternating irrep). It seems like, should I not ask this question, I would be told to "just remember it is the alternating rep", if that helps clarify. $\endgroup$ – InertialObserver Mar 18 '17 at 0:10
  • $\begingroup$ I do get the idea of your previous comment, yes. Thank you that was helpful. $\endgroup$ – InertialObserver Mar 18 '17 at 0:11

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.