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 Exercise [11.13] 
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Joined: 12 Mar 2008, 10:57
Posts: 69
Location: India
Post Exercise [11.13]
P=P_1\wedge{P_2\wedge{...\wedge{P_p}}}
Q=Q_1\wedge{Q_2\wedge{...\wedge{Q_q}}}
P\wedge{Q}
=(P_1\wedge{P_2\wedge{...\wedge{P_p}}})\wedge{(Q_1\wedge{Q_2\wedge{...\wedge{Q_q}}})}
=P_1\wedge{P_2\wedge{...\wedge{P_p}}}\wedge{Q_1\wedge{Q_2\wedge{...\wedge{Q_q}}}}
=(-1)^{p}Q_{1}\wedge{P_1\wedge{P_2\wedge{...\wedge{P_p}}}}\wedge{Q_2\wedge{Q_3\wedge{...\wedge{Q_q}}}}
=(-1)^{p}(-1)^{p}Q_{1}\wedge{Q_2}\wedge{P_1\wedge{P_2\wedge{...\wedge{P_p}}}}\wedge{Q_3\wedge{Q_4\wedge{...\wedge{Q_q}}}}
=(-1)^{p}(-1)^{p}...(q times)Q_1\wedge{Q_2\wedge{...\wedge{Q_q}}}\wedge{P_1\wedge{P_2\wedge{...\wedge{P_p}}}}
=(-1)^{pq}Q\wedge{P}
Hence,
If p and q are both odd then pq is odd and,
P\wedge{Q}=-Q\wedge{P}
If one of them is even then pq is even and,
P\wedge{Q}=Q\wedge{P}


31 May 2008, 10:46
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Joined: 07 May 2009, 16:45
Posts: 54
Post Re: Exercise [11.13]
A question. Just a point of clarification.

The grade of P, or Q depends on how many basis \boldsymbol\eta_is are used to wedge product the thing together. So in you aswer to the exercise, when you say P_1, for example, you mean P_1 = p_1\boldsymbol\eta_1, where p_1 is a scalar. Yes?


19 Jul 2009, 22:33
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