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Anticommutativity of the Wedge Product ... ... Tu, Proposition 3.21 ... ...

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,887
Hobart, Tasmania
I am reading Loring W.Tu's book: "An Introduction to Manifolds" (Second Edition) ...

I need help in order to fully understand Tu's Proposition 3.21 ... ...

Proposition 3.21 reads as follows:



Tu - 1 - Proposition 3.2 ... ... PART 1 ... ... .png
Tu - 2 - Proposition 3.21 ... ... PART 2 ... ... .png



In the above proof by Tu we read the following:

" ... ...

... \(\displaystyle = \sum_{ \sigma_{ k + l } } ( \text{ sgn } \sigma ) f( v_{ \sigma \tau (l+1) }, \cdot \cdot \cdot , v_{ \sigma \tau (l+k) }) g ( v_{ \sigma \tau (1) }, \cdot \cdot \cdot , v_{ \sigma \tau (l) })\)


\(\displaystyle = ( \text{ sgn } \tau ) \sum_{ \sigma_{ k + l } } ( \text{ sgn } \sigma \tau ) g ( v_{ \sigma \tau (1) }, \cdot \cdot \cdot , v_{ \sigma \tau (l) }) f( v_{ \sigma \tau (l+1) }, \cdot \cdot \cdot , v_{ \sigma \tau (l+k) }) \)


... ... ... "



Can someone please explain/demonstrate how/why we have that


\(\displaystyle \sum_{ \sigma_{ k + l } } ( \text{ sgn } \sigma ) f( v_{ \sigma \tau (l+1) }, \cdot \cdot \cdot , v_{ \sigma \tau (l+k) }) g ( v_{ \sigma \tau (1) }, \cdot \cdot \cdot , v_{ \sigma \tau (l) })\)


\(\displaystyle = ( \text{ sgn } \tau ) \sum_{ \sigma_{ k + l } } ( \text{ sgn } \sigma \tau ) g ( v_{ \sigma \tau (1) }, \cdot \cdot \cdot , v_{ \sigma \tau (l) }) f( v_{ \sigma \tau (l+1) }, \cdot \cdot \cdot , v_{ \sigma \tau (l+k) })\)





Help will be much appreciated ... ...

Peter
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,887
Hobart, Tasmania
I am reading Loring W.Tu's book: "An Introduction to Manifolds" (Second Edition) ...

I need help in order to fully understand Tu's Proposition 3.21 ... ...

Proposition 3.21 reads as follows:








In the above proof by Tu we read the following:

" ... ...

... \(\displaystyle = \sum_{ \sigma_{ k + l } } ( \text{ sgn } \sigma ) f( v_{ \sigma \tau (l+1) }, \cdot \cdot \cdot , v_{ \sigma \tau (l+k) }) g ( v_{ \sigma \tau (1) }, \cdot \cdot \cdot , v_{ \sigma \tau (l) })\)


\(\displaystyle = ( \text{ sgn } \tau ) \sum_{ \sigma_{ k + l } } ( \text{ sgn } \sigma \tau ) g ( v_{ \sigma \tau (1) }, \cdot \cdot \cdot , v_{ \sigma \tau (l) }) f( v_{ \sigma \tau (l+1) }, \cdot \cdot \cdot , v_{ \sigma \tau (l+k) }) \)


... ... ... "



Can someone please explain/demonstrate how/why we have that


\(\displaystyle \sum_{ \sigma_{ k + l } } ( \text{ sgn } \sigma ) f( v_{ \sigma \tau (l+1) }, \cdot \cdot \cdot , v_{ \sigma \tau (l+k) }) g ( v_{ \sigma \tau (1) }, \cdot \cdot \cdot , v_{ \sigma \tau (l) })\)


\(\displaystyle = ( \text{ sgn } \tau ) \sum_{ \sigma_{ k + l } } ( \text{ sgn } \sigma \tau ) g ( v_{ \sigma \tau (1) }, \cdot \cdot \cdot , v_{ \sigma \tau (l) }) f( v_{ \sigma \tau (l+1) }, \cdot \cdot \cdot , v_{ \sigma \tau (l+k) })\)





Help will be much appreciated ... ...

Peter



I have been reflecting on my question/problem in the above post ... ... and I think I have resolved the problem ...


Since \(\displaystyle \text{ sgn } \tau\) is \(\displaystyle +1\) or \(\displaystyle -1\) then \(\displaystyle \text{ sgn } \tau \tau = 1\) ...


Therefore we have


\(\displaystyle \text{ sgn } \sigma = \text{ sgn } \sigma \tau \tau = \text{ sgn } \tau \text{ sgn } \sigma \tau\)


which answers the question ... ...


Is that correct ... ... ?

Peter