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In a homework assignment, I am asked to show that the components of three dimensional spinorial tensor obey the following relationship $$\Psi_{12}=-\Psi_{1}^{1}=-\Psi^{21}, \Psi_{11}=\Psi_{1}^{2}=\Psi^{22}$$

I am given the transformation rule $\Psi_{\alpha \beta}=\zeta_{\alpha \gamma} \zeta_{\beta \delta} \Psi^{\gamma \delta} \quad \Psi_{\beta}^{\alpha}=\zeta_{\beta \gamma} \Psi^{\alpha \gamma}$, where $$\zeta_{\alpha \beta}=\left(\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right)=\mathrm{i} \sigma_{2}$$ is the spinor's metric tensor. And that $\zeta_{\alpha \beta}=-\zeta^{\alpha \beta}$. I'm not sure how to show this based on the information given.

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Given those relations, you also have $\Psi_{\alpha\beta}=\zeta_{\alpha\gamma}\Psi^{\gamma}_\beta$. Then

$$\begin{align}\Psi_{12}=&\zeta_{1\gamma}\Psi^{\gamma}_2=\overbrace{\zeta_{11}}^{=0}\Psi^{1}_2+\overbrace{\zeta_{12}}^{=1}\Psi^{2}_2=\\=&\Psi^{2}_2~~~~~~~~\text{(check if this is equal to $-\Psi^1_1$)}\\=&\zeta_{2\gamma}\Psi^{2\gamma}=\overbrace{\zeta_{21}}^{=-1}\Psi^{21}+\overbrace{\zeta_{22}}^{=0}\Psi^{22}=\\=&-\Psi^{21}\end{align}$$

Try the second one, here is a spoiler:

$$\begin{align}\Psi_{11}=&\zeta_{1\gamma}\Psi^{\gamma}_1=\overbrace{\zeta_{11}}^{=0}\Psi^{1}_1+\overbrace{\zeta_{12}}^{=1}\Psi^{2}_1=\\=&\Psi^{2}_1\\=&\zeta_{1\gamma}\Psi^{2\gamma}=\zeta_{11}\Psi^{21}+\zeta_{12}\Psi^{22}=\\=&\Psi^{22}\end{align}$$

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