How can I compute the representation matrices of a point group under given basis functions?

Question

Take the $C_{3v}$ point group for example:

Rot[θ_] := RotationMatrix[θ]
E0 = IdentityMatrix[2];
C1 = Rot[2π/3];
C2 = C1.C1;
σ1 = ReflectionMatrix[Rot[2π/3].{1, 0}];
σ2 = C1.σ1.Inverse[C1];
σ3 = C2.σ1.Inverse[C2];
C3v = {E0, C1, C2, σ1, σ2, σ3};

So $C_{3v}$ has the following 6 elements: $ \left\{ \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}, \begin{pmatrix} -\frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \end{pmatrix}, \begin{pmatrix} -\frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \end{pmatrix}, \begin{pmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \end{pmatrix}, \begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \end{pmatrix}, \begin{pmatrix} -1 & 0 \\ 0 & 1 \\ \end{pmatrix} \right\} $

For given basis functions $\{f_n(\mathbf{r})\mid n=1\cdots N\}$, there is a representation $\{D(g)\mid g\in C_{3v}\}$, in which the $N\times N$ matrix $D(g)$ is defined as $$ f_i(g^{-1}\mathbf{r})=\sum_{j=1}^N f_j(\mathbf{r})D_{ji}(g)$$ where $\mathbf{r}=${x,y} .

If the basis functions are just monomials, e.g. $$f_1(\mathbf{r})=x^2,~~f_2(\mathbf{r})=y^2,~~f_3(\mathbf{r})=xy$$ I can compute the representation matrices using the Coefficient function

f1[{x_, y_}] := x^2
f2[{x_, y_}] := y^2
f3[{x_, y_}] := x y
basis = {f1, f2, f3};
Rep[g_] := Block[{i, j, ig, x, y, r}, 
  ig = Inverse[g]; r = {x, y};
  Table[Coefficient[basis[[j]][ig.r], basis[[i]][r]], {i, 3}, {j, 3}]
  ]
repC3v = Rep /@ C3v;
MatrixForm /@ repC3v

This gives the following results: $$ \left\{\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix},\begin{pmatrix} \frac{1}{4} & \frac{3}{4} & \frac{\sqrt{3}}{4} \\ \frac{3}{4} & \frac{1}{4} & -\frac{\sqrt{3}}{4} \\ -\frac{\sqrt{3}}{2} & \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \end{pmatrix},\begin{pmatrix} \frac{1}{4} & \frac{3}{4} & -\frac{\sqrt{3}}{4} \\ \frac{3}{4} & \frac{1}{4} & \frac{\sqrt{3}}{4} \\ \frac{\sqrt{3}}{2} & -\frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \end{pmatrix},\begin{pmatrix} \frac{1}{4} & \frac{3}{4} & \frac{\sqrt{3}}{4} \\ \frac{3}{4} & \frac{1}{4} & -\frac{\sqrt{3}}{4} \\ \frac{\sqrt{3}}{2} & -\frac{\sqrt{3}}{2} & \frac{1}{2} \\ \end{pmatrix},\begin{pmatrix} \frac{1}{4} & \frac{3}{4} & -\frac{\sqrt{3}}{4} \\ \frac{3}{4} & \frac{1}{4} & \frac{\sqrt{3}}{4} \\ -\frac{\sqrt{3}}{2} & \frac{\sqrt{3}}{2} & \frac{1}{2} \\ \end{pmatrix},\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \\ \end{pmatrix}\right\} $$

However, if the basis functions are general functions, not monomials, how can I get the coefficient $a_i$ of function $f_i(\mathbf{r})$ in the transformed function $f_k(g^{-1}\mathbf{r})=\sum_i a_i f_i(\mathbf{r})$?

For example, if

F1[{x_, y_}] := (x + I y)^2
F2[{x_, y_}] := (x - I y)^2
F3[{x_, y_}] := (x + y)^2

How can I get the representation matrices under this basis?

NOTE: I mean a general method, not just a transformation from basis f1~f3 to F1~F3.

Answer 1

You were almost there. Just add the following to your code:

iC3v = Inverse /@ C3v; 
sa = SolveAlways[Flatten@
       Table[basis[[i]][iC3v[[k]].{x, y}] == Sum[basis[[j]][{x, y}] d[k, j, i], {j, 3}], 
            {i, 3}, {k, 6}],
     {x, y}];
MatrixForm /@ Table[d[k, i, j], {k, 6}, {i, 3}, {j, 3}] /. sa

And you get your expected result:

$\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right),\left( \begin{array}{ccc} \frac{1}{4} & \frac{3}{4} & \frac{\sqrt{3}}{4} \\ \frac{3}{4} & \frac{1}{4} & -\frac{\sqrt{3}}{4} \\ -\frac{\sqrt{3}}{2} & \frac{\sqrt{3}}{2} & -\frac{1}{2} \end{array} \right),\left( \begin{array}{ccc} \frac{1}{4} & \frac{3}{4} & -\frac{\sqrt{3}}{4} \\ \frac{3}{4} & \frac{1}{4} & \frac{\sqrt{3}}{4} \\ \frac{\sqrt{3}}{2} & -\frac{\sqrt{3}}{2} & -\frac{1}{2} \end{array} \right),\left( \begin{array}{ccc} \frac{1}{4} & \frac{3}{4} & \frac{\sqrt{3}}{4} \\ \frac{3}{4} & \frac{1}{4} & -\frac{\sqrt{3}}{4} \\ \frac{\sqrt{3}}{2} & -\frac{\sqrt{3}}{2} & \frac{1}{2} \end{array} \right),\left( \begin{array}{ccc} \frac{1}{4} & \frac{3}{4} & -\frac{\sqrt{3}}{4} \\ \frac{3}{4} & \frac{1}{4} & \frac{\sqrt{3}}{4} \\ -\frac{\sqrt{3}}{2} & \frac{\sqrt{3}}{2} & \frac{1}{2} \end{array} \right),\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array} \right)$

Edit

For your Complex base:

F1[{x_, y_}] := (x + I y)^2
F2[{x_, y_}] := (x - I y)^2
F3[{x_, y_}] := (x + y)^2
basis = {F1, F2, F3};

gives:

$\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right),\left( \begin{array}{ccc} -\frac{1}{2}+\frac{i \sqrt{3}}{2} & 0 & \frac{\sqrt{3}}{4}+\frac{3 i}{4} \\ 0 & -\frac{1}{2}-\frac{i \sqrt{3}}{2} & \frac{\sqrt{3}}{4}-\frac{3 i}{4} \\ 0 & 0 & 1 \end{array} \right),\left( \begin{array}{ccc} -\frac{1}{2}-\frac{i \sqrt{3}}{2} & 0 & -\frac{\sqrt{3}}{4}+\frac{3 i}{4} \\ 0 & -\frac{1}{2}+\frac{i \sqrt{3}}{2} & -\frac{\sqrt{3}}{4}-\frac{3 i}{4} \\ 0 & 0 & 1 \end{array} \right),\left( \begin{array}{ccc} 0 & -\frac{1}{2}-\frac{i \sqrt{3}}{2} & \frac{\sqrt{3}}{4}+\frac{i}{4} \\ -\frac{1}{2}+\frac{i \sqrt{3}}{2} & 0 & \frac{\sqrt{3}}{4}-\frac{i}{4} \\ 0 & 0 & 1 \end{array} \right),\left( \begin{array}{ccc} 0 & -\frac{1}{2}+\frac{i \sqrt{3}}{2} & -\frac{\sqrt{3}}{4}+\frac{i}{4} \\ -\frac{1}{2}-\frac{i \sqrt{3}}{2} & 0 & -\frac{\sqrt{3}}{4}-\frac{i}{4} \\ 0 & 0 & 1 \end{array} \right),\left( \begin{array}{ccc} 0 & 1 & i \\ 1 & 0 & -i \\ 0 & 0 & 1 \end{array} \right)$

Edit

Although it could probably be improved with some Representation Theory knowledge, here you have a generalization:

calcRep[group_List, bas_List] :=
  Module[{r, inv, x, kk, ii, jj},
   {kk, ii, jj} = Dimensions@group ;
   blen = Length@bas;
   inv = Inverse /@ group;
   r = Array[x@# &, ii];
   Return[
    Block[{d}, 
     Table[d[k, i, j], {k, 6}, {i, 3}, {j, 3}] /. 
      SolveAlways[Flatten@Table[bas[[i]][inv[[k]].r] == Sum[bas[[j]]@r d[k,j,i], {j, blen}], 
                          {i, blen}, {k, kk}], 
       r]]];
   ];
calcRep[C3v, basis]

Edit

Answer 2

As per whuber's comment, given a function $h$ and a set of functions $f_i$, where you know that

$$h(\mathbf{r})=\sum_i a_i f_i(\mathbf{r})$$

you wish to know what the $a_i$ are?

A dumb, but possibly effective way is just to throw some points at it and solve. For example, with

F1[{x_, y_}] := (x + I y)^2
F2[{x_, y_}] := (x - I y)^2
F3[{x_, y_}] := (x + y)^2

let

h = Expand[4 F1[{x, y}] - 3 F2[{x, y}] + (7 - 3 I) F3[{x, y}]]

(8 - 3 I) x^2 + (14 + 8 I) x y + (6 - 3 I) y^2

then

eq = a1 F1[{x, y}] + a2 F2[{x, y}] + a3 F3[{x, y}] == h;
Solve[{eq /. {x -> 1, y -> 1}, eq /. {x -> 0, y -> 1}, eq /. {x -> 1, y -> 0}}]

gives

{a1 -> 4, a2 -> -3, a3 -> 7 - 3 I}}