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

#### Peter

##### Well-known member
MHB Site Helper
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) ...

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:    Hope that helps ...

Peter

#### Peter

##### Well-known member
MHB Site Helper
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) ...

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
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
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