Tensors & the Alternation Operator .... Browder, Propn 12.25

[Math Amateur]

I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ...

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

I need some help in order to fully understand the proof of Proposition 12.2 on pages 277 - 278 ... ...Proposition 12.2 and its proof read as follows:

In the above proof by Browder (near the end of the proof) we read the following:

" ... ... To see also that ##A( \beta \otimes \alpha ) = 0##, we observe that ## \beta \otimes \alpha = \ ^{ \sigma }( \alpha \otimes \beta )## where ##\sigma## is the permutation which sends ##(1, \cdot \cdot \cdot , r+s )## to ##(r+1, \cdot \cdot \cdot , r+s, 1 \cdot \cdot \cdot , r )## ... ... "My question ... or more accurately problem ... is that given ##\sigma## as defined by Browder I cannot verify that ##\beta \otimes \alpha = \ ^{ \sigma }( \alpha \otimes \beta )## is true ...My working is as follows:

Let ##\alpha \in \bigwedge^r## and let ##\beta \in \bigwedge^s## ... ...

Then we have ...

##\beta \otimes \alpha (v_1, \cdot \cdot \cdot , v_{ r+s } ) = \beta ( v_1, \cdot \cdot \cdot , v_s ) \alpha ( v_{ s+1 }, \cdot \cdot \cdot , v_{ r+s } )##

and

##\alpha \otimes \beta (v_1, \cdot \cdot \cdot , v_{ r+s } ) = \alpha ( v_1 , \cdot \cdot \cdot , v_r ) \beta ( v_{ r+1 }, \cdot \cdot \cdot , v_{ r+s } )##Now consider ##\sigma## where ...

... ##\sigma## sends ##(1, \cdot \cdot \cdot , r+s )## to ##(r+1, \cdot \cdot \cdot , r+s, 1 \cdot \cdot \cdot , r )##We have

##^{ \sigma } ( \alpha \otimes \beta ) (v_1, \cdot \cdot \cdot , v_{ r+s } )##

##= \alpha \otimes \beta ( v_{ r+1 }, \cdot \cdot \cdot , v_{ r+s }, v_1, \cdot \cdot \cdot , v_r )## ... ... hmm ... this does not appear to be correct ...BUT ... ... if we consider ##\sigma_1## where ...

... ##\sigma_1## sends ##(1, \cdot \cdot \cdot , r+s )## to ##(s+1, \cdot \cdot \cdot , r+s, 1 \cdot \cdot \cdot , s )##

then we have

##^{ \sigma_1 } ( \alpha \otimes \beta ) (v_1, \cdot \cdot \cdot , v_{ r+s } )##

##= ( \alpha \otimes \beta ) ( v_{ s+1 }, \cdot \cdot \cdot , v_{ r+s }, v_1, \cdot \cdot \cdot , v_s )##

##= \alpha ( v_{ s+1 }, \cdot \cdot \cdot , v_{ r+s } ) \beta ( v_1, \cdot \cdot \cdot , v_s )##

##= \beta \otimes \alpha## ...
Given that my working differs from Browder ... I suspect I have made an error ...

Can someone please point out the error(s) in my working ...
Help will be much appreciated ...

Peter

 

[andrewkirk]

Let ##\alpha \in \bigwedge^r## and let ##\beta \in \bigwedge^s## ... ...

Then we have ...

##\beta \otimes \alpha (v_1, \cdot \cdot \cdot , v_{ r+s } ) = \beta ( v_1, \cdot \cdot \cdot , v_s ) \alpha ( v_{ s+1 }, \cdot \cdot \cdot , v_{ r+s } )##

and

##\alpha \otimes \beta (v_1, \cdot \cdot \cdot , v_{ r+s } ) = \alpha ( v_1 , \cdot \cdot \cdot , v_r ) \beta ( v_{ r+1 }, \cdot \cdot \cdot , v_{ r+s } )##Now consider ##\sigma## where ...

... ##\sigma## sends ##(1, \cdot \cdot \cdot , r+s )## to ##(r+1, \cdot \cdot \cdot , r+s, 1 \cdot \cdot \cdot , r )##We have

##^{ \sigma } ( \alpha \otimes \beta ) (v_1, \cdot \cdot \cdot , v_{ r+s } )##

##= \alpha \otimes \beta ( v_{ r+1 }, \cdot \cdot \cdot , v_{ r+s }, v_1, \cdot \cdot \cdot , v_r )## ... ... hmm ... this does not appear to be correct ...

It's fine. Just take it a few lines further:

\begin{align*}
{}^\sigma(\alpha\otimes\beta) (v_1, \cdot \cdot \cdot v_{ r+s })
&=
\alpha \otimes \beta ( v_{ r+1 }, \cdot \cdot \cdot , v_{ r+s }, v_1, \cdot \cdot \cdot , v_r )\\
&=
\alpha( v_{ r+1 }, \cdot \cdot \cdot , v_{ r+s }) \beta (v_1, \cdot \cdot \cdot , v_r )\\
&=
\beta (v_1, \cdot \cdot \cdot , v_r )\alpha( v_{ r+1 }, \cdot \cdot \cdot , v_{ r+s })\\
&=
\beta\otimes\alpha (v_1, \cdot \cdot \cdot , v_r, v_{ r+1 }, \cdot \cdot \cdot , v_{ r+s })\\
&=
\beta\otimes\alpha (v_1, \cdot \cdot \cdot v_{ r+s })
\end{align*}

 

[Math Amateur]

It's fine. Just take it a few lines further:

\begin{align*}
{}^\sigma(\alpha\otimes\beta) (v_1, \cdot \cdot \cdot v_{ r+s })
&=
\alpha \otimes \beta ( v_{ r+1 }, \cdot \cdot \cdot , v_{ r+s }, v_1, \cdot \cdot \cdot , v_r )\\
&=
\alpha( v_{ r+1 }, \cdot \cdot \cdot , v_{ r+s }) \beta (v_1, \cdot \cdot \cdot , v_r )\\
&=
\beta (v_1, \cdot \cdot \cdot , v_r )\alpha( v_{ r+1 }, \cdot \cdot \cdot , v_{ r+s })\\
&=
\beta\otimes\alpha (v_1, \cdot \cdot \cdot , v_r, v_{ r+1 }, \cdot \cdot \cdot , v_{ r+s })\\
&=
\beta\otimes\alpha (v_1, \cdot \cdot \cdot v_{ r+s })
\end{align*}

Thanks Andrew ...

Appreciate your help ...

BUT ... just a clarification ...

You write:##^\sigma(\alpha\otimes\beta) (v_1, \cdot \cdot \cdot v_{ r+s })##

## = \alpha \otimes \beta ( v_{ r+1 }, \cdot \cdot \cdot , v_{ r+s }, v_1, \cdot \cdot \cdot , v_r )##

## = \alpha( v_{ r+1 }, \cdot \cdot \cdot , v_{ r+s }) \beta (v_1, \cdot \cdot \cdot , v_r )##

Here you have the multilinear function (tensor) ##\alpha## with ##s## variables ...

... but ##\alpha## is of rank ##r## ... ?

Can you clarify ... ?

 

[andrewkirk]

It's possible that what the author means by ##{}^\sigma(\alpha\otimes\beta)## is the converse of what I had assumed.

What is your understanding of what the author wants the ##\sigma## operator to mean? Under the author's definitions, do we have, for a tensor ##\eta## or rank ##m##:

$$({}^\sigma(\eta))(v_1,...,v(m)) = \eta(\sigma(v_1),...,\sigma(v_m))$$
or
$$({}^\sigma(\eta))(v_1,...,v(m)) = \eta(\sigma^{-1}(v_1),...,\sigma^{-1}(v_m))$$

If it's the latter then the proof works - but what you and I have written will need to be rewritten. If not, it doesn't.

 

[Math Amateur]

It's possible that what the author means by ##{}^\sigma(\alpha\otimes\beta)## is the converse of what I had assumed.

What is your understanding of what the author wants the ##\sigma## operator to mean? Under the author's definitions, do we have, for a tensor ##\eta## or rank ##m##:

$$({}^\sigma(\eta))(v_1,...,v(m)) = \eta(\sigma(v_1),...,\sigma(v_m))$$
or
$$({}^\sigma(\eta))(v_1,...,v(m)) = \eta(\sigma^{-1}(v_1),...,\sigma^{-1}(v_m))$$

If it's the latter then the proof works - but what you and I have written will need to be rewritten. If not, it doesn't.

Hi Andrew ...

Thanks again for your help ...The definition of ##^{\sigma} \alpha## is definition 12.17 on page 274 of Browder and reads as follows:12.17 Definition. Let ##\alpha## be a tensor of rank ##r## and ##\sigma \in S_r##. We define a new tensor ##^{\sigma} \alpha## of rank ##r## by the formula

## ^{\sigma} \alpha ( v_1, \cdot \cdot \cdot , v_r ) = \alpha ( v_{ \sigma(1) }, \cdot \cdot \cdot , v_{ \sigma(r) } ) ##

for all ##v_1, \cdot \cdot \cdot , v_r \in V##
... so it is your first possibility ...

But .. I think that that leaves us with the same problem of a rank ##r## tensor expressed with ##s## input vectors ...

Can you resolve this difficulty ... or is Browder in error with his description of ##\sigma## ... ?Can you please help further ...

Peter

 

[andrewkirk]

Thanks for that.

Given that definition, I think the author has got a key line the wrong way around in his proof. It doesn't wreck the proof. It just needs correction.

I think, at the bottom, where he wrote

##\beta\otimes\alpha = {}^\sigma(\alpha\otimes\beta)##, where ##\sigma## is the permutation which sends ##(1, ...,r+s)## to ##(r+1, ...,r+s,1,...,r))##

he should have written

"where ##\sigma## is the permutation which sends ##(1, ...,r+s)## to ##(s+1, ...,r+s,1,...,s))##"

I haven't checked it, but I feel pretty confident that that amendment will not derail his proof.

 

[Math Amateur]

Thanks for that.

Given that definition, I think the author has got a key line the wrong way around in his proof. It doesn't wreck the proof. It just needs correction.

I think, at the bottom, where he wrote
he should have written

"where ##\sigma## is the permutation which sends ##(1, ...,r+s)## to ##(s+1, ...,r+s,1,...,s))##"

I haven't checked it, but I feel pretty confident that that amendment will not derail his proof.

Thanks Andrew ...

That resolves the issue ...

Thanks again ...

Peter