Episode #125 of the Stack Overflow podcast is here. We talk Tilde Club and mechanical keyboards. Listen now
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I think I know now. Choose

Choose rotation axis that makes equal angles with xthe $x$-, y$y$-, and z axes$z$-axes so that C3the $C_3$ rotation essentially permutes the axes. Without loss of generality, I could choose C3a $C_3$ rotation that permutes x$x$ to z$z$, z$z$ to y$y$, and y$y$ to x$x$. Rotation in either direction will give the same trace with this C3$C_3$ 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 2x2$2\times 2$ matrix in this representation yields $\bigl( \begin{smallmatrix} -1/2 & -3/2\\ 1/2 & -1/2 \end{smallmatrix} \bigr)\bigl( \begin{smallmatrix} 2z^2-x^2-y^2 \\ x^2-y^2 \end{smallmatrix} \bigr)$=$\bigl( \begin{smallmatrix} 2y^2-z^2-x^2\\ z^2-x^2 \end{smallmatrix} \bigr)$.

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

The trace of this matrix is -1$-1$ 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-dimensional vector space and project it into a rotation in a two dimensional-dimensional vector space.

I think I know now. Choose rotation axis that makes equal angles with x, y, and z axes so that C3 rotation essentially permutes the axes. Without loss of generality, I could choose C3 rotation that permutes x to z, z to y, and y to x. Rotation in either direction will give same trace with this C3 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 by 2x2 matrix in this representation yields $\bigl( \begin{smallmatrix} -1/2 & -3/2\\ 1/2 & -1/2 \end{smallmatrix} \bigr)\bigl( \begin{smallmatrix} 2z^2-x^2-y^2 \\ x^2-y^2 \end{smallmatrix} \bigr)$=$\bigl( \begin{smallmatrix} 2y^2-z^2-x^2\\ z^2-x^2 \end{smallmatrix} \bigr)$. 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.

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

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

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.

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I think I know now. Choose rotation axis that makes equal angles with x, y, and z axes so that C3 rotation essentially permutes the axes. Without loss of generality, I could choose C3 rotation that permutes x to z, z to y, and y to x. Rotation in either direction will give same trace with this C3 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 by 2x2 matrix in this representation yields $\bigl( \begin{smallmatrix} -1/2 & -3/2\\ 1/2 & -1/2 \end{smallmatrix} \bigr)\bigl( \begin{smallmatrix} 2z^2-x^2-y^2 \\ x^2-y^2 \end{smallmatrix} \bigr)$=$\bigl( \begin{smallmatrix} 2y^2-z^2-x^2\\ z^2-x^2 \end{smallmatrix} \bigr)$. 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.

I think I know now. Choose rotation axis that makes equal angles with x, y, and z axes so that C3 rotation essentially permutes the axes. Without loss of generality, I could choose C3 rotation that permutes x to z, z to y, and y to x. Rotation in either direction will give same trace with this C3 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 by 2x2 matrix in this representation yields $\bigl( \begin{smallmatrix} -1/2 & -3/2\\ 1/2 & -1/2 \end{smallmatrix} \bigr)\bigl( \begin{smallmatrix} 2z^2-x^2-y^2 \\ x^2-y^2 \end{smallmatrix} \bigr)$=$\bigl( \begin{smallmatrix} 2y^2-z^2-x^2\\ z^2-x^2 \end{smallmatrix} \bigr)$. The trace of this matrix is -1 as in the character table. Choosing any axis via orthogonal transformation of this matrix yields same trace.

I think I know now. Choose rotation axis that makes equal angles with x, y, and z axes so that C3 rotation essentially permutes the axes. Without loss of generality, I could choose C3 rotation that permutes x to z, z to y, and y to x. Rotation in either direction will give same trace with this C3 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 by 2x2 matrix in this representation yields $\bigl( \begin{smallmatrix} -1/2 & -3/2\\ 1/2 & -1/2 \end{smallmatrix} \bigr)\bigl( \begin{smallmatrix} 2z^2-x^2-y^2 \\ x^2-y^2 \end{smallmatrix} \bigr)$=$\bigl( \begin{smallmatrix} 2y^2-z^2-x^2\\ z^2-x^2 \end{smallmatrix} \bigr)$. 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.

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I think I know now. Choose rotation axis that makes equal angles with x, y, and z axes so that C3 rotation essentially permutes the axes. Without loss of generality, I could choose C3 rotation that permutes x to z, z to y, and y to x. Rotation in either direction will give same trace with this C3 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 by 2x2 matrix in this basisrepresentation yields $\bigl( \begin{smallmatrix} -1/2 & -3/2\\ 1/2 & -1/2 \end{smallmatrix} \bigr)\bigl( \begin{smallmatrix} 2z^2-x^2-y^2 \\ x^2-y^2 \end{smallmatrix} \bigr)$=$\bigl( \begin{smallmatrix} 2y^2-z^2-x^2\\ z^2-x^2 \end{smallmatrix} \bigr)$. The trace of this matrix is -1 as in the character table. Choosing any axis via orthogonal transformation of this matrix yields same trace.

I think I know now. Choose rotation axis that makes equal angles with x, y, and z axes so that C3 rotation essentially permutes the axes. Without loss of generality, I could choose C3 rotation that permutes x to z, z to y, and y to x. Rotation in either direction will give same trace with this C3 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 vector in this basis yields $\bigl( \begin{smallmatrix} -1/2 & -3/2\\ 1/2 & -1/2 \end{smallmatrix} \bigr)\bigl( \begin{smallmatrix} 2z^2-x^2-y^2 \\ x^2-y^2 \end{smallmatrix} \bigr)$=$\bigl( \begin{smallmatrix} 2y^2-z^2-x^2\\ z^2-x^2 \end{smallmatrix} \bigr)$. The trace of this matrix is -1 as in the character table. Choosing any axis via orthogonal transformation of this matrix yields same trace.

I think I know now. Choose rotation axis that makes equal angles with x, y, and z axes so that C3 rotation essentially permutes the axes. Without loss of generality, I could choose C3 rotation that permutes x to z, z to y, and y to x. Rotation in either direction will give same trace with this C3 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 by 2x2 matrix in this representation yields $\bigl( \begin{smallmatrix} -1/2 & -3/2\\ 1/2 & -1/2 \end{smallmatrix} \bigr)\bigl( \begin{smallmatrix} 2z^2-x^2-y^2 \\ x^2-y^2 \end{smallmatrix} \bigr)$=$\bigl( \begin{smallmatrix} 2y^2-z^2-x^2\\ z^2-x^2 \end{smallmatrix} \bigr)$. The trace of this matrix is -1 as in the character table. Choosing any axis via orthogonal transformation of this matrix yields same trace.

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