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 $T$without 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