Tensors and Isomorphisms

  • #1
1,444
0
I cannot at all understand the theorem on p16 of the notes attached in this thread:


Surely seeing as we want an isomoprhism between V and V**, [itex]\Phi[/itex] should act on an element of V i.e. a vector X and take it to an element of V** (i don't know what such an element would look like though!). But anyway I am immediately lost because the map seems to act not only on a vector but also on a covector [itex]\omega[/itex] which won't be in V.

There is some blurb at the beginning of Wald on this as well but to be honest I think it was, if possible, even less helpful!

An how does it define an isomorphism?

Aaaargh!
 

Answers and Replies

  • #2
1,444
4
I cannot at all understand the theorem on p16 of the notes attached in this thread:
In Harvey
Which notes? Can you be more specific? Link?
 
  • #4
1,444
4
I do not see any theorem on p. 16.
 
  • #6
1,444
4
You want to construct the isomorphism [itex]\Phi[/itex] from V to V**. So you want to associate to each vector X in V an element [itex]\Phi(X)[/itex] in V**. But V** is the dual of V*. That is a linear functional on V*. To define a linear functional on V* you say how it acts on elements of V*. So, you chose [itex]\omega[/itex] in V* and you want to define a number that [itex]\Phi(X)[/itex] will associate to [itex]\omega[/itex]. You can write it as follows:

[tex]\Phi(X):\omega \mapsto \omega(X)[/tex]

This is the definition of [itex]\Phi(X)[/itex]. [itex]\Phi(X)[/tex] by definition associates with each [itex]\omega[/itex] it's value on [tex]X[/tex]. Makes sense? And you are supposed to check, writing it down on paper and understanding (so simple that may be difficult!) why this is really a number that linearly depends on [itex]\omega[/itex]. Another way of writing the same is

[tex]\Phi(X)(\omega)=\omega(X)[/tex]

or

[tex](\Phi(X))(\omega)=\omega(X)[/tex]
 
Last edited:
  • #7
1,444
0
You want to construct the isomorphism [itex]\Phi[/itex] from V to V**. So you want to associate to each vector X in V an element [itex]\Phi(X)[/itex] in V**. But V** is the dual of V*. That is a linear functional on V*. To define a linear functional on V* you say how it acts on elements of V*. So, you chose [itex]\omega[/itex] in V* and you want to define a number that [itex]\Phi(X)[/itex] will associate to [itex]\omega[/itex]. You can write it as follows:

[tex]\Phi(X):\omega \mapsto \omega(X)[/tex]

This is the definition of [itex]\Phi(X)[/itex]. [itex]\Phi(X)[/tex] by definition associates with each [itex]\omega[/itex] it's value on [tex]X[/tex]. Makes sense? And you are supposed to check, writing it down on paper and understanding (so simple that may be difficult!) why this is really a number that linearly depends on [itex]\omega[/itex]. Another way of writing the same is

[tex]\Phi(X)(\omega)=\omega(X)[/tex]

or

[tex](\Phi(X))(\omega)=\omega(X)[/tex]
But since [itex]\Phi[/itex] is an isomorphism between V and V**, shouldn't it act on elements in V? I can kind of see that it does since you've written [itex]\Phi(X)[/itex] but it seems like you've got it acting on [itex]\omega[/itex] when you write [itex]\Phi(X) : \omega \mapsto \Phi(X)(\omega)[/itex], no?

Also, when we think of V** as the space of linear functions from V* to R, is our isomorphism basically saying that to each X in V, we associate [itex]\Phi(X)(\omega)=\omega(X)[/itex] which is a linear function from V* to R i.e. an element of V**?

Thanks!
 
  • #8
1,444
4
But since [itex]\Phi[/itex] is an isomorphism between V and V**, shouldn't it act on elements in V? I can kind of see that it does since you've written [itex]\Phi(X)[/itex] but it seems like you've got it acting on [itex]\omega[/itex] when you write [itex]\Phi(X) : \omega \mapsto \Phi(X)(\omega)[/itex], no?
Map from V to V**. [itex]\Phi[/itex] is the map. [itex]X[/itex] is in V. So [itex]\Phi(X)[/itex] must be in V**. How is ane element of V** defined? By saying how it acts on V*. So to define [itex]\Phi(X)[/itex] we must tell what it does with elements of V*. This is what we do.

The story has some similarity with what you can do with functions. Say, you play with functions on set A. You know what are functions, yes? But do you know that each point a of A defines a function on the set of all functions? Think about this:

[tex]a(f)=f(a)[/tex]

The above is the definition. Very precise one.
In the above a is constant, f is the variable.

Contemplate it for a while and you will understand how tricky and clever the math notation can be.
 
  • #9
1,444
4
There is another way of doing the same. For [itex]X\in V[/tex] let us denote by [itex]X^{**}[/itex] the element of V** that we want to associate with X. We define X** as follows

[tex]X^{**}(\omega)=\omega(X)[/tex]

Is X** well defined? Is it indeed an element of V**? Yes? If so, we will denote the map [itex]X\mapsto X^{**}[/itex] by [itex]\Phi[/itex]. Better?
 

Related Threads on Tensors and Isomorphisms

Replies
0
Views
4K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
22
Views
2K
  • Last Post
Replies
1
Views
784
  • Last Post
Replies
10
Views
1K
  • Last Post
Replies
20
Views
3K
  • Last Post
Replies
0
Views
585
Replies
4
Views
833
  • Last Post
Replies
1
Views
2K
Top