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*$$\otimes$$M* is the vector space of all tensors of type M x M --> R?

 

[Zorba]

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