Concept question about the dimensionality of a representation in group theory here: Look at 3.1(c) of problem set, from group theory application to the physics of condensed matter of M.S.Dresselhaus:

3.1(c) Given the point group T,verify that the equality \begin{align} \sum_j{l_j^2} =h \end{align}

What is the meaning of the two sets of characters given for the two-dimensional irreducible representation E?

It seems that E with 2 sets of characters is called two-dimensional irreducible representation. But when verifying the equality above, where lj should be the dimensionality of the IR Γj, and E is viewed as TWO representations and each has a dimension lj=1(referring to the character of group T and p40 of Dresselhaus's book),so that we have : \begin{align} \sum_j{l_j^2} =h \\ 1^2+1^2+1^2+3^2 = 12 \end{align} to finish the proof. Instead of: \begin{align} 1^2+2^2+3^2=14 \neq 12 \end{align}

So, my question is what is the meaning of a 2 dimension representation? Becoz it seems that there are two definitions of dimension here.

The character table of group T attached here, from Dresselhaus's book The character table of group T attached here, from Dresselhaus's book

  • $\begingroup$ The dimension of a representation is the dimension of the vector space on which the group is acting. $\endgroup$
    – WillO
    Jan 30, 2016 at 5:19
  • $\begingroup$ Thank you, can we tell the dimension from the character table? Does it mean the 'E' is a 2 dimensional representation with 2 sets of characters and 'T' is a three dimensional representation with just one set of characters? But in this case, the equality mentioned above doesn't hold? $\endgroup$
    – HRY
    Jan 30, 2016 at 7:39
  • $\begingroup$ The character of a representation is its trace (each group element corresponds to a linear transformation). As such, the character evaluated in the neutral element is the dimension of the representation. $\endgroup$
    – doetoe
    Jan 30, 2016 at 11:10
  • 1
    $\begingroup$ @doetoe yes I agree with that, but the author says 'E' is a two dimensional representation while on the character table its neutral element is TWO '1'. And if E is two-dimensional, it doesn't coincide with the equality \begin{align} \sum_j{l_j^2} =h \end{align} mentioned above.That is exactly what I am confused about. $\endgroup$
    – HRY
    Jan 30, 2016 at 15:21
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    $\begingroup$ But they are orthogonal. The inner product is $1\cdot 1 + 3 (1\cdot 1) + 4(\omega\cdot \omega)+4(\omega^2\cdot\omega^2)=0$. (Don't forget that the conjugate of $\omega$ is $\omega^2$.) $\endgroup$
    – WillO
    Jan 30, 2016 at 16:45

1 Answer 1


Comments to the question (v3):

  1. The point group in question is the chiral tetrahedral symmetry group $T$ of order 12, i.e. the symmetry group of the tetrahedron.

  2. Problem 3.1(c) confusingly talks about a 2-dimensional irreducible representation $E$, which is in fact the reducible sum of a 1-dimensional representation and its complex conjugate representation.

  3. The above 1-dimensional representation assigns a $e^{i\phi}$-phase factor to $\phi$-rotations around one of the four corners of the tetrahedron.


  1. M.S. Dresselhaus, G. Dresselhaus, & A. Jorio, Group Theory: Application to the Physics of Condensed Matter, 2008.
  • $\begingroup$ Thank you, is there any meaning to put two irreducible representations together as 'E'? $\endgroup$
    – HRY
    Jan 31, 2016 at 3:31

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