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Let $\omega^{ab}$ be antisymmetric in the indices $a$ and $b$. Why we have

$$\omega^{ab}(\theta_{ab}-\theta_{ba})=2\omega^{ab}\theta_{ab}$$

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    $\begingroup$ Hint: you can rename dummy indices $\endgroup$ – Nihar Karve Apr 5 at 17:28
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$$\omega^{ab}(\theta_{ab}-\theta_{ba})=\omega^{ab}(\theta_{ab})-\omega^{ab}(\theta_{ba})=\omega^{ab}(\theta_{ab})+\omega^{ba}(\theta_{ba})=2\omega^{ab}(\theta_{ab})$$ where the last but one equality holds since $w$ is antisymmetric and the last equality is due only to rename the dummy indices (so the indices summed over) as suggested in a comment; in particular you have to rename $a$ as $b$ and viceversa, in the second term given by $\omega^{ba}(\theta_{ba})$.

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