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

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,886
Hobart, Tasmania
I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ...

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

I have another question regarding the proof of Theorem 12.20 on page 275 ... ...


The relevant text reads as follows:



Browder ... Proposition 12.20 ... .png





At the end of the above proof by Browder we read the following:

" ... ... Finally, (d) is proved in exactly the same way as (a). ... ... "


Despite the above statement by Browder I am unable to formulate a valid proof of (d) ...

Can someone please help by demonstrating a formal and rigorous proof of (d) ... ... ?




Help will be much appreciated ...

Peter


=========================================================================================


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
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,886
Hobart, Tasmania
I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ...

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

I have another question regarding the proof of Theorem 12.20 on page 275 ... ...


The relevant text reads as follows:









At the end of the above proof by Browder we read the following:

" ... ... Finally, (d) is proved in exactly the same way as (a). ... ... "


Despite the above statement by Browder I am unable to formulate a valid proof of (d) ...

Can someone please help by demonstrating a formal and rigorous proof of (d) ... ... ?




Help will be much appreciated ...

Peter


=========================================================================================


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:










Hope that helps ...

Peter


I have reflected on my post above ... and now believe I have a proof of (d) ...

... that is a proof that


(d) \(\displaystyle A \ ^{ \tau }{ \alpha } = \varepsilon ( \tau ) A \alpha\)



I think that the following is a valid proof ...


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


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


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


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


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



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


\(\displaystyle = \varepsilon ( \tau ) A \alpha\)



Can someone please indicate whether the above proof is correct and/or critique the above proof pointing out errors or shortcomings ...

Peter


*** EDIT ***

I have just realised that I used Proposition 12.18 (see scanned pages above) in the above proof ... but Proposition 12.18 assumes the tensor concerned is alternating ... so my proof is invalid ...

Hmm ... still need some help ...

Peter
 
Last edited:

GJA

Well-known member
MHB Math Scholar
Jan 16, 2013
255
Hi Peter ,

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


\(\displaystyle = \sum_{ \sigma \text{ in } S_r } \varepsilon ( \sigma ) \ ^{ \sigma \tau }{ \alpha }\)
Try using $\varepsilon(\sigma)=\varepsilon(\sigma)\varepsilon(\tau)^{2}=\varepsilon(\tau)\varepsilon(\sigma\tau)$ at this point and see where that takes you.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,886
Hobart, Tasmania
Hi Peter ,



Try using $\varepsilon(\sigma)=\varepsilon(\sigma)\varepsilon(\tau)^{2}=\varepsilon(\tau)\varepsilon(\sigma\tau)$ at this point and see where that takes you.


Well ... that certainly helped ...

Now have worked through proof ... obvious now with your help!!!

Thanks ...

Peter