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 5 added 2 characters in body edited Dec 25 '15 at 7:24 orthocresol♦ 43.8k77 gold badges139139 silver badges267267 bronze badges 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. 4 added 177 characters in body edited Nov 25 '15 at 6:50 dualredlaugh 17322 bronze badges 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. 3 deleted 7 characters in body edited Nov 25 '15 at 6:35 dualredlaugh 17322 bronze badges 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. 2 deleted 7 characters in body edited Nov 25 '15 at 6:30 dualredlaugh 17322 bronze badges 1 answered Nov 25 '15 at 6:01 dualredlaugh 17322 bronze badges