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The Alternation Operator ... Browder, Proposition 12.20 ... ...

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,887
Hobart, Tasmania
I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ...

I am currently reading Chapter 12: Multilinear Algebra ... ...

I need some further help in order to fully understand the proof of Theorem 12.20 on page 275 ... ...


The relevant text reads as follows:


Browder ... Proposition 12.20 ... .png



In the above proof by Browder we read the following:

" ... ... \(\displaystyle ^{\tau }{ ( A \alpha ) } = ^{\tau }{ ( } \sum_{ \sigma \text{ in } S_r } \varepsilon ( \sigma ) ^{ \sigma }{ \alpha } ) = \sum_{ \sigma \text{ in } S_r } \varepsilon ( \tau ) \varepsilon ( \tau \sigma ) ^{ \tau \sigma }{ \alpha } \) ... ... ... "



My question is as follows:


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


\(\displaystyle ^{\tau }{ ( } \sum_{ \sigma \text{ in } S_r } \varepsilon ( \sigma ) ^{ \sigma }{ \alpha } ) = \sum_{ \sigma \text{ in } S_r } \varepsilon ( \tau ) \varepsilon ( \tau \sigma ) ^{ \tau \sigma }{ \alpha }\)



Help will be much appreciated ... ...

Peter


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So that MHB readers can see and understand Browder's definitions and notation in the text given above I am providing the text of Browder's section on alternating tensors ... as follows:



Browder - 1 - Start of Section 12.2 ... ... PART 1 .... .png
Browder - 2 - Start of Section 12.2 ... ... PART 2 .... .png
Browder - 3 - Start of Section 12.2 ... ... PART 3 .... .png
Browder - 4 - Start of Section 12.2 ... ... PART 4  ....  .png



Hope that helps ...

Peter
 
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