Proving Lorentz Metric on Real Type (1,0;1,0) Tensors in Wald Ch. 13

[ergospherical]

In ch. 13, pg.349 of Wald it's asked to prove that $g_{AA'BB'} = \epsilon_{AB} \bar{\epsilon}_{A'B'}$ is a Lorentz metric on $V$ (containing the real elements of the vector space $Y$ of type $(1,0;1,0)$ tensors). Given the basis $t^{AA'} = \dfrac{1}{\sqrt{2}}(o^A \bar{o}^{A'} + \iota^A \bar{\iota}^{A'})$, $x^{AA'} = \dfrac{1}{\sqrt{2}}(o^A \bar{\iota}^{A'} + \iota^A \bar{o}^{A'})$, $y^{AA'} = \dfrac{i}{\sqrt{2}}(o^A \bar{\iota}^{A'} - \iota^A \bar{o}^{A'})$, $z^{AA'} = \dfrac{1}{\sqrt{2}}(o^A \bar{o}^{A'} - \iota^A \bar{\iota}^{A'})$, where $o_A \iota^A = 1$ by definition. I need to show that these are Minkowski orthonormal. So for example\begin{align*}
\epsilon_{AB} \bar{\epsilon}_{A'B'} t^{AA'} t^{BB'} &= \dfrac{1}{2} o_B o_{B'} o^B \bar{o}^{B'} + \dfrac{1}{2} o_B \bar{o}_{B'} \iota^B \bar{\iota}^{B'} + \dfrac{1}{2} \iota_B \bar{\iota}_{B'} o^B \bar{o}^{B'} + \dfrac{1}{2} \iota_B \bar{\iota}_{B'} \iota^B \bar{\iota}^{B'}
\end{align*}should be equal to $1$. Since $o_B \iota^B = 1$ it also follows that
\begin{align*}
1 = o_B \iota^B = \epsilon_{AB} o^A \iota^B = - \epsilon_{BA} o^A \iota^B = -o^A \iota_A
\end{align*}therefore the middle two terms are both equal to $\dfrac{1}{2}$. What I can't see is what to write for $\iota_A \iota^A$ and $o_A o^A$; they should be zero (right...?), but why?

 

[martinbn]

##o

What I can't see is what to write for $\iota_A \iota^A$ and $o_A o^A$; they should be zero (right...?), but why?

$o_A o^A=\epsilon_{BA}o^B o^A=0$ The two form $\epsilon$ is antysymmetric. Everything is orthogonal to itself.

 

[ergospherical]

$o_A o^A=\epsilon_{BA}o^B o^A=0$ The two form $\epsilon$ is antysymmetric. Everything is orthogonal to itself.

ahhh course, thanks.