Why does (2z²-x²-y², x²-y²) have -1 trace in the character table of Td for C3 rotation? I'm trying to understand the irreducible rep for $(2z^2-x^2-y^2, x^2-y^2)$ in the $T_\mathrm{d}$ point group. Specifically, for $C_3$ rotations, a trace of $-1$ is listed in the character table for this group. For any matrix, one can rotate the coordinates (basis for the matrix) - i.e. an orthogonal transformation, yielding a matrix with different entries, but the trace will remain the same. So in the case of $120^\circ$ rotations around an axis through a vertex and opposite face of the tetrahedron, one can rotate the coordinates so that the $z$-axis aligns with rotation axis making the calculation of the trace easier.

Any rotation in this basis will leave $2z^2-x^2-y^2$ unchanged, ie. a $1$ on the diagonal. However, the projection of a rotated $x^2-y^2$ back to its own basis will yield $(\cos 2\theta)(x^2-y^2)$. The trace will then be $1 + \cos 2\theta$.
Can anyone help me where I went wrong?
I think I know now.

Choose rotation axis that makes equal angles with the $x$-, $y$-, and $z$-axes so that the $C_3$ rotation essentially permutes the axes. Without loss of generality, I could choose a $C_3$ rotation that permutes $x$ to $z$, $z$ to $y$, and $y$ to $x$. Rotation in either direction will give the same trace with this $C_3$ rotation. With this rotation (permutation), $(2z^2-x^2-y^2,x^2-y^2)$ becomes $(2y^2-z^2-x^2,z^2-x^2)$. Multiplication of basis function vector a by $2\times 2$ matrix in this representation yields

$$\begin{pmatrix}
  -1/2 &amp; -3/2\\
  1/2 &amp; -1/2 
\end{pmatrix}
\begin{pmatrix}
  2z^2-x^2-y^2 \\
  x^2-y^2
\end{pmatrix}
= 
\begin{pmatrix}
  2y^2-z^2-x^2\\
  z^2-x^2
\end{pmatrix}
$$

The trace of this matrix is $-1$ as in the character table. Choosing any axis via orthogonal transformation of this matrix yields same trace.

The tricky thing about this example was that I needed to use a rotation in a three-dimensional vector space and project it into a rotation in a two-dimensional vector space.