Multiple skew brackets in tensor summation, e.g. $U_{[ab]}V^{[ab]}$

Question

I'm wondering what does $U_{[ab]}V^{[ab]}$ or $U_{[ab]}V^{(ab)}$ usually mean? I thought the double brackets meant to apply the $\text{sgn}(\pi)$ twice so $$U_{[ab]}V^{[ab]}=\frac{1}{2}\sum_{a,b}\left(U_{ab}\text{sgn}(\pi(ab))\cdot V^{ab}\text{sgn}(\pi(ab))\right)$$

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For contexts, I've been asked to prove $\phi_{i} \psi_{j} v_{1}^{[i} v_{2}^{j]}=\phi_{[i} \psi_{j]} v_{1}^{[i} v_{2}^{j]}$ in differential forms.

I tried $\phi_{i} \psi_{j} v_{1}^{[i} v_{2}^{j]}=\frac{1}{2}(\phi_1\psi_2v_1^1v_2^2-\phi_2\psi_1v_1^2v_2^1)$

$\phi_{[i} \psi_{j]} v_{1}^{[i} v_{2}^{j]}=\frac{1}{2}(\phi_1\psi_2v_1^1v_2^2+\phi_2\psi_1v_1^2v_2^1)$

And they equal because the second term $=0$?

Answer 1

You can use a simpler definition $$U_{[ab]}=\tfrac 12(U_{ab}-U_{ba}).$$ To proof that $$U_{ab}V^{[ab]}=U_{[ab]}V^{[ab]}$$ for any tensor $U$ you can first split $U$ into its symmetric and antisymmetric part $$U_{ab}=U_{(ab)}+U_{[ab]}$$ and then show that $$U_{(ab)}V^{[ab]}=0.$$ Similarly there's an identity for a symmetric contraction: $$U_{ab}V^{(ab)}=U_{(ab)}V^{(ab)}$$

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Hint: to show $U_{(ab)}V^{[ab]}=0$ you can use $A=-A\implies A=0$