Coordinate transformation of a tensor in 2 dimensions

[physicus]

Homework Statement

Given a symmetric tensor T_{\mu\nu} on the flat Euclidean plane (g_{\mu\nu}=\delta_{\mu\nu}), we want to change to complex coordinates z=x+iy, \,\overline{z}=x-iy.
Show, that the components of the tensor in this basis are given by:
T_{zz}=\frac{1}{4}(T_{00}-2iT_{10}-T_{11}),\, T_{\overline{z}\overline{z}}=\frac{1}{4}(T_{00}+2iT_{10}-T_{11}),\, T_{\overline{z}z}=T_{z\overline{z}}=\frac{1}{4}(T_{00}+T_{11})

Homework Equations

T_{\mu\nu}=T_{\nu\mu}
T'_{\alpha\beta} = U_{\alpha}{}^{\mu}U_{\beta}{}^{\nu}T_{\mu\nu}

The Attempt at a Solution

In priciple, this should be an easy linear algebra problem. I just don't get the right result.
I use T'for the tensor in the new coordinates. Then, there should be a transformation matrix U, s.t.
T'_{\alpha\beta} = U_{\alpha}{}^{\mu}U_{\beta}{}^{\nu}T_{\mu\nu}
But what is U? Neiter with the transformation matrix from the new to the old coordinates \begin{pmatrix}1 & 1 \\ i & -i\end{pmatrix} nor with the transformation matrix from the old to the new coordinates \begin{pmatrix}\frac{1}{2} & \frac{-i}{2} \\ \frac{1}{2} & \frac{i}{2} \end{pmatrix} it is working. What am I doing wrong?

physicus

 

[clamtrox]

It's difficult to say without seeing the calculation. You do remember that there are two transformation matrices, right? (one for each index)

 

[physicus]

Thanks, but I still get a wrong result.

The matrix U = \begin{pmatrix}\frac{1}{2} & \frac{-i}{2} \\ \frac{1}{2} & \frac{i}{2} \end{pmatrix} transforms from the old basis x,y to the new basis z=x+iy, \overline{z}=x-iy.
The original metrix is g_{\mu\nu}=\delta_{\mu\nu}. The transformed metric is g'_{\mu\nu}=\begin{pmatrix} 0 & \frac{1}{2} \\ \frac{1}{2} & 0 \end{pmatrix}
Therefore, the tensor in the new basis is given by
T'^{\alpha}{}_\beta = U^{\alpha}{}_\mu T^{\mu}{}_\nu (U^{-1})^\nu{}_\beta = \begin{pmatrix}\frac{1}{2}T^0{}_0+\frac{1}{2}T^1{}_1 & \frac{1}{2}T^0{}_0-\frac{i}{2}T^0{}_1- {\frac{i}{2}}T^1{}_0-T^1{}_1 \\ \frac{1}{2}T^0{}_0+ \frac{i}{2}T^0{}_1+ \frac{i}{2}T^1{}_0-T^1{}_1 & \frac{1}{2}T^0{}_0+\frac{1}{2}T^1{}_1 \end{pmatrix}

Now, I use the metric to lower the indices. Since the original metric ist Euclidean one can simply lower the indices of Twithout further factors, but not for T'.
T'_{00}=g'_{00}T'^0{}_0+g'_{01}T'^1{}_0 = \frac{1}{2}T'^1{}_0 = \frac{1}{2}(\frac{1}{2}T_{00}+ \frac{i}{2}T_{01}+ \frac{i}{2}T_{10}-T_{11}) = \frac{1}{4}(T_{00}+2iT_{10}-T_{11})
I have used the symmetry of T_{\mu\nu}.
The plus sign in the middle of the expression is noch correct. Does someone see my mistake?

physicus