Product between Pauli 4-vectors ${\sigma^\mu}\bar{\sigma}^\nu-{\sigma}^\nu\bar{\sigma}^\mu$

Question I'm not getting properly how to perform the product $${\sigma^\mu}\bar{\sigma}^\nu-{\sigma}^\nu\bar{\sigma}^\mu={\sigma^\mu}\eta^{\mu\nu}\bar{\sigma}_\nu-{\sigma}^\nu\eta^{\mu\nu}\bar{\sigma}_\mu=(I,\sigma^i)\begin{pmatrix}1&0\\0&-I_3\end{pmatrix}\begin{pmatrix}1\\-\sigma^j\end{pmatrix}-(I,\sigma^j)\begin{pmatrix}1&0\\0&-I_3\end{pmatrix}\begin{pmatrix}1\\-\sigma^i\end{pmatrix}=[\sigma^i,\sigma^j]=2i\epsilon^{ijk}\sigma^k$$ I'm missing the term containing $$\eta$$.

Context: I want to show that $$e^{\frac{i}{2}\omega_{\mu\nu}J^{\mu\nu}}=e^{\frac{i}{2}\omega_{\mu\nu}\frac{\sigma^{\mu\nu}}{2}}$$ i.e. that it's possible to write the Lorentz transformation for a Dirac 4-spinor in this fashion $$\psi(x)\longrightarrow\psi'(x')=\left(e^{\frac{i}{2}\omega_{\mu\nu}\frac{\sigma^{\mu\nu}}{2}}\right)\psi(x)$$

starting from the known transformation relation in the chiral representation $$\begin{pmatrix}\psi'_L(x') \\ \psi'_R(x')\end{pmatrix}=\begin{pmatrix}\Lambda_L & 0 \\ 0 &\Lambda_R\end{pmatrix}\begin{pmatrix}\psi_L(x) \\ \psi_R(x)\end{pmatrix}=\begin{pmatrix}e^{(-i\vec{\theta}-\vec{\eta})\cdot\frac{\vec{\sigma}}{2}} & 0 \\ 0 &e^{(-i\vec{\theta}+\vec{\eta})\cdot\frac{\vec{\sigma}}{2}} \end{pmatrix}\begin{pmatrix}\psi_L(x) \\ \psi_R(x)\end{pmatrix}$$

where $$x'=\Lambda \cdot x$$ is the Lorentz-transformed point in Minkowsi space,$$\sigma^{\mu\nu}=\frac{i}{2}\left[\gamma^\mu,\gamma^\nu\right]$$ and the Dirac matrices in the chiral representation $$\gamma^\mu=\begin{pmatrix}0&\sigma^\mu \\\bar{\sigma}^\mu &0\end{pmatrix}$$ $$\sigma^\mu=\left(I,\sigma^i\right)$$ $$\bar{\sigma}^\mu=\left(I,-\sigma^i\right)$$ I is the 2x2 identity and $$\sigma^i$$ the 3 spatial Pauli matrices. $$\theta^i=\frac{1}{2}\epsilon^{ijk}\omega^{jk}$$ $$\eta^i=\omega^{i0}$$

Precisely i tried this way: $$\sigma^{\mu\nu}=\frac{i}{2}\left[\gamma^\mu,\gamma^\nu\right]=\frac{i}{2}\begin{pmatrix} {\sigma^\mu}\bar{\sigma}^\nu-{\sigma}^\nu\bar{\sigma}^\mu & 0 \\ 0 &\bar{\sigma}^\mu{\sigma}^\nu-\bar{\sigma}^\nu{\sigma}^\mu\end{pmatrix}$$

Now, I'm not getting properly how to perform the product $${\sigma^\mu}\bar{\sigma}^\nu-{\sigma}^\nu\bar{\sigma}^\mu={\sigma^\mu}\eta^{\mu\nu}\bar{\sigma}_\nu-{\sigma}^\nu\eta^{\mu\nu}\bar{\sigma}_\mu=(I,\sigma^i)\begin{pmatrix}1&0\\0&-I_3\end{pmatrix}\begin{pmatrix}1\\-\sigma^j\end{pmatrix}-(I,\sigma^j)\begin{pmatrix}1&0\\0&-I_3\end{pmatrix}\begin{pmatrix}1\\-\sigma^i\end{pmatrix}=[\sigma^i,\sigma^j]=2i\epsilon^{ijk}\sigma^k$$

So I'm missing the term containing $$\eta$$. How to untangle?

Your leading formula is nonsensical and wrong. Instead, inspect the ten (six) 2$$\times$$2 matrices $$\Omega^{\mu\nu}\equiv {\sigma^\mu}\bar{\sigma}^\nu-{\sigma}^\nu\bar{\sigma}^\mu$$ and plug in the definitions to obtain $$\Omega^{00}=0,\\ \Omega^{ 0j}=-2\sigma^j ,\\ \Omega^{i0 }=2\sigma^i , \\ \Omega^{ij}= -[\sigma^i,\sigma^j]=2i\epsilon ^{jik} \sigma^k.$$
In your notation, then, $$\omega_{\mu\nu} \sigma^{\mu\nu}= \begin{bmatrix} 2i\omega_{\!j0~}\sigma^j + \omega_{ij}\epsilon^{ijk}\sigma^k &0\\ 0& -2i\omega_{\!j0~}\sigma^j+ \omega_{ij}\epsilon^{ijk}\sigma^k \end{bmatrix}.$$