On page 96 of M. S. Dresselhaus's Applications of Group Theory to the Physics of Solids, it is said that

Any arbitrary function $F$ can be written as a linear combination of a complete set of basis functions $f^{\Gamma_{n'}}_{j'}$
$F= \sum_{n'}\sum_{j'} f^{\Gamma_{n'}}_{j'} |\Gamma_{n'}, j'\rangle$

in which $|\Gamma_{n'}, j'\rangle$ represents the $j'\text{-th}$ basis function of the $n'\text{-th}$ irreducible representation of some finite group.

I don't understand why an arbitrary function can be expanded.

Here is my understanding:

Given a finite group $G$, by studying its character table, we know the invariant subspaces of the group. Finding the set of basis functions of each of the invariant subspaces of $G$ gives us the set of basis functions {$|\Gamma_{n'}, j'\rangle$}.

What is so special about this set of functions? Clearly, the size of the group determines the size of the set {$|\Gamma_{n'}, j'\rangle$}, does it mean that a smaller set of {$|\Gamma_{n'}, j'\rangle$} has the same 'expressive power' as a larger set?


Inspired by ZeroTheHero's answer, here is what I understood:

A real-valued function defined on space $X$ is a map $$\text{A point in } X \rightarrow \Bbb R$$ Given a finite group $G$, we construct a linear representation of which in space $X$. Since the representation can be decomposed into the irreducible representations of $G$, space $X$ can be decomposed into the invariant subspaces of $G$. So we say any arbitrary function F can be written as a linear combination of a complete set of basis functions.

Given a space $X$, there are many sets of basis functions that we can find. The invariant subspaces of a group (as long as a linear representation of the group can be constructed in $X$) just provides a way to partition the space and simultaneously leads us to those basis functions.

  • $\begingroup$ Minor comment to the post (v1): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$
    – Qmechanic
    Commented Oct 2, 2018 at 14:17
  • 1
    $\begingroup$ I voted to migrate this to Math.SE. $\endgroup$ Commented Oct 2, 2018 at 14:40

1 Answer 1


In a large number of situations, the actual functions of interest not transform by a single irreducible representation, and it's useful to write them in terms of irreducible pieces since these transform only amongst functions within the same irreducible irrep.

The simplest example would be the function $f=x_1+y_2$, which is clearly not invariant under permutation since $P_{12}f= x_2+y_1$. However, the functions $$ f_+= x_1+y_2+ x_2+y_1\, ,\qquad f_-= x_1+y_2 -(x_2+y_1) $$ transform irreducibly under permutation (here up to a sign) of the indices, and indeed $$ x_1+y_2 = \frac{1}{2}(f_++f_-) $$ Another obvious example would be the two-particle spin state $$ \vert +\rangle_1\vert -\rangle_2 = \frac{1}{2} \left(\vert +\rangle_1\vert -\rangle_2+ \vert +\rangle_1\vert -\rangle_2\right)+ \frac{1}{2} \left(\vert +\rangle_1\vert -\rangle_2- \vert +\rangle_1\vert -\rangle_2\right) $$ where $\frac{1}{2} \left(\vert +\rangle_1\vert -\rangle_2 \pm \vert +\rangle_1\vert -\rangle_2\right)$ are symmetric/antisymmetric w/r to permutations.

This holds quite generally. Maybe the better known example would be how an generic function $f(\theta,\varphi)$ can be expanded as a sum of spherical harmonics $Y_{\ell,m}(\theta,\varphi)$, keeping in mind the spherical harmonics are basis functions for $SO(3)$ irreps.

  • $\begingroup$ This is very helpful! I have added my understanding into the body of my post, could you check whether I understand it correctly? $\endgroup$ Commented Oct 3, 2018 at 2:25

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