Why Use Odd Notation for Tensors in Linear Algebra?

[Zorba]

I've been looking through my notes for the last few weeks and i still do not see the reason for this use of notation that my lecturer uses, for example

We denote by M^{*} \otimes M \otimes M^{*} the vector space of all tensors of type M \times M^{*} \times M \rightarrow \mathbb{R}, where M is a finite dimensional real vector space, and M* is the dual space of M.

So why not just say instead:

We denote by M \otimes M^{*} \otimes M the vector space of all tensors of type M \times M^{*} \times M \rightarrow \mathbb{R}

which seems far more natural to me than the first one (and less confusing too...), is there some reason for using the former rather than latter?

 

[dextercioby]

Because when discussing tensors, the tensor product of vector spaces is used thus the notation with the \otimes instead of the ordinary \times.

If the answer I've given you is not useful, could you, please, rephrase your question ? I may have not understood it properly.

 

[tiny-tim]

Hi Zorba!

Using just one M, isn't M* defined as all thingys from M to R ?

 

[Zorba]

Because when discussing tensors, the tensor product of vector spaces is used thus the notation with the \otimes instead of the ordinary \times.

If the answer I've given you is not useful, could you, please, rephrase your question ? I may have not understood it properly.

What I mean is, if we are considering M^{*} \otimes M \otimes M^{*} which means M \times M^{*} \times M \rightarrow \mathbb{R}, so the argument is of the from (x,f,y) where f are linear forms, so since f are elements of M* and x,y is in M, then why don't we write M \otimes M^{*} \otimes M instead?

 

[klackity]

Maybe this will clear things up:

Suppose instead of dealing with a three-place tensor, we are dealing with just a one-place tensor. Let T be such a one-place tensor.

Suppose T has an argument of the form (x) where x belongs to M. Then T must be a linear functional on M. So then T itself belongs to M* (the dual of M).

We see from this that M* is the vector space of all tensors of the same type as T, namely of type M --> R.

So if M* is the vector space of all tensors of type M --> R, then shouldn't we expect that M*\otimesM* is the vector space of all tensors of type M x M --> R?

 

[Zorba]

Ah, yes I see it now, thanks for that.