Lorentz metric on real type (1,0;1,0) tensors

  • #1
ergospherical
Gold Member
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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?
 
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Answers and Replies

  • #2
martinbn
Science Advisor
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##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.
 
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  • #3
ergospherical
Gold Member
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##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.
 

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