Understanding (0,1)-Tensor in Relativity

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In summary: I have also read things about the invariance of a one-form and now I have a doubt: if I have v1 and v2 two different vectors in V, and u a one-form V->R, then it is not necessarily the case that u(v1)=u(v2), right ? Again, here I actually consider "one-form" as synonym from "function from V to R".Yes, that's correct. A one-form is a function from vectors to scalars, but not all functions from vectors to scalars are one-forms. A one-form must be linear (satisfy certain properties under addition and scalar multiplication) and must map each vector to a unique scalar. So in general, two different vectors
  • #1
emma83
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I have just starting learning relativity and I have still a problem with the definition and notation of tensors.
So a (r,s)-tensor takes r vectors, s one-forms and gives a scalar.

Then I understand that a (1,0)-tensor takes 1 vector (e.g. from V) and gives a scalar which is exactly the definition of a one-form (in V*), which corresponds to the mapping (V->R).

But I am still uncomfortable with the symmetrical situation, i.e. that a (0,1)-tensor is a vector. A (0,1)-tensor takes a one-form (e.g. u \in V*) and gives a scalar. But the one-form u given as argument is itself a mapping from V to R, so in a sense my (0,1)-tensor is a mapping ((V -> R) -> R) and instinctively I would "reduce" it to (V->R) or (V->RxR) but I cannot figure out how at the end it gives something which is again in V.

I am probably wrong in the vector spaces I consider, am I ?

Thanks a lot for your help.
 
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  • #2
Welcome to Physics Forums and good question!

This is a somewhat subtle point.

The operation of taking duals applies to all vector spaces. In particular, applying it to the vector space [itex]V*[/itex] results in [itex]V**[/itex], the dual of [itex]V*[/itex] and the double dual of [itex]V*[/itex]. Since [itex]V**[/itex] is the dual of [itex]V*[/itex], [itex]V**[/itex] is the space of (0,1) tensors.

Your questions really is "Why can [itex]V**[/itex] be identified with [itex]V[/itex]?"

Consider a mapping (defined later)

[tex]i : V \rightarrow V**,[/tex]

so that for every [itex]v \in V[/itex], [itex]i \left( v \right)[/itex] is in [itex]V**[/itex], i.e.,

[tex]i \left( v \right) : V* \rightarrow \mathb{R}.[/tex]

Define the mapping [itex]i[/itex] by

[tex]i \left( v \right) \left( f \right) = f \left( v \right)[/itex]

for every [itex]v \in V[/itex] and [itex]f \in V*[/itex], and note that [itex]i[/itex], so defined is a linear mapping. It is also always a one-to-one mapping, and it is onto if and only if [itex]V[/itex] is finite-dimensional.

If [itex]V[/itex] is finite-dimensional, [itex]i[/itex] is a bijection that is independent of any choice of basis for [itex]V[/itex], so, usually no distinction is made between [itex]V[/itex] and [itex]V**[/itex], and [itex]i \left( v \right)[/itex] is written just as [itex]v[/itex].

All this of is often a bit confusing on first reading.
 
  • #3
Oh weird, when I learned this notation (0,1) and (1,0) were flipped - (1,0) was a vector and (0,1) was a dual vector.

Here's another way to explain why the dual of a dual vector is a vector (everything here will be finite-dimensional of course):
Say we've got a vector [tex]v \in V[/tex], and a dual vector [tex]\omega \in V^*[/tex].
We can write the linear mapping that [tex]\omega: V \to \mathbb{R}[/tex] as [tex]\omega(v)=\omega_\mu v^\mu[/tex] ([tex]\left\omega[/tex] eats [tex]v[/tex] and spits out a scalar).
Now, we could just as well say that it's [tex]v[/tex] that's providing the mapping: [tex]v: V^* \to \mathbb{R}[/tex] where [tex]v(\omega)=v^\mu \omega_\mu[/tex] ([tex]v[/tex] eats [tex]\omega[/tex] and spits out a scalar).
Since we have an explicit mapping from [tex]V^* \to \mathbb{R}[/tex] for an arbitrary dual-vector, this means [tex]v[/tex] must be in [tex]V^{**}[/tex], the space of linear functionals on dual vectors.

edit: anyone have any idea why my single Latex characters are all misaligned?
 
  • #4
Hello,

Thank you very much for your answers, it is getting slightly clearer. I have always seen somehow a strict "hierarchy" between arguments and functions so I will need a bit time to feel comfortable with the identification w(v) and v(w).

Actually what confuses me is that 3 different expressions are used which for me do more or less the same, i.e. transform some arguments into something else:
- a tensor "takes" something to something (e.g. a (1,2)-tensor takes 1 vector and 2 one-forms to the real numbers)
- a mapping
- a function(al)

Now can we say that a tensor is a mapping ? And are the notions mapping and function (or functional ?) synonyms ?
 
  • #5
Now can we say that a tensor is a mapping ? And are the notions mapping and function (or functional ?) synonyms ?

Yes, a tensor is mapping (an assignment of elements in one set to elements in another set). The rank will specify from what to what (eg. a (1,1) tensor can be thought of as a mapping from a vector to a dual vector, a dual vector to a vector, or a pair consisting of both a vector and a dual vector to a scalar - depending on what you decide to feed it)

Mapping, function, and functional are all very similar but have slightly different technical meanings. As I mentioned above, a mapping is just a general assignment of elements in one set to elements in another set. Functions and functionals are specific kinds of mappings. A function is mapping that associates precisely one output value with each input value. A functional is a function that takes some sort of object to a scalar (eg. a dual vector is a functional on vectors and a vector is a functional on dual vectors).
 
  • #6
Thanks Jaunty.

Can I see then function(al)s as a subset of mappings, i.e. mappings with some additional properties ? What could be a mapping that cannot be called a function ?

I have also read things about the invariance of a one-form and now I have a doubt: if I have v1 and v2 two different vectors in V, and u a one-form V->R, then it is not necessarily the case that u(v1)=u(v2), right ? Again, here I actually consider "one-form" as synonym from "function from V to R".
 
  • #7
Can I see then function(al)s as a subset of mappings, i.e. mappings with some additional properties ? What could be a mapping that cannot be called a function ?

Yes, a functional is a mapping with additional requirements (those of a function to scalars). A mapping that's not a function could be anything that takes one input element to more than one output element (eg. the graph [tex]f(x) = \pm \sqrt{x}[/tex] would assign 2 to two values, [tex]\pm \sqrt{2}[/tex]). Actually, a lot of the time people use the word mapping as a synonym for function (and physicists almost never need to make the distinction).

I have also read things about the invariance of a one-form and now I have a doubt: if I have v1 and v2 two different vectors in V, and u a one-form V->R, then it is not necessarily the case that u(v1)=u(v2), right ? Again, here I actually consider "one-form" as synonym from "function from V to R".

Yes, it's certainly not necessary for two vectors to be assigned to the same scalar by a one form.
 
  • #8
Jaunty said:
Oh weird, when I learned this notation (0,1) and (1,0) were flipped - (1,0) was a vector and (0,1) was a dual vector.

It occurred to me while I was posting that I couldn't remember which is which.
Jaunty said:
edit: anyone have any idea why my single Latex characters are all misaligned?

Try using itex and /itex for tags within lines of text. Reserve the tex and /tex tags for lines of stand-alone math.
Jaunty said:
Mapping, function, and functional are all very similar but have slightly different technical meanings.

I use mapping and function interchangeably, and what you call a mapping, I call a relation.
 
  • #9
Jaunty said:
Mapping, function, and functional are all very similar but have slightly different technical meanings. As I mentioned above, a mapping is just a general assignment of elements in one set to elements in another set. Functions and functionals are specific kinds of mappings. A function is mapping that associates precisely one output value with each input value. A functional is a function that takes some sort of object to a scalar (eg. a dual vector is a functional on vectors and a vector is a functional on dual vectors).

I don't think these are universally accepted definitions.

Generally, a function is the most general kind of thing. A function assigns each input a specific output. Functions, formally speaking, are never multi-valued.

A "map" can mean a few different things. Often, it is used synonymously with "function." The term is also used in some fields to refer to a function with special properties. In linear algebra, a linear map (often just abbreviated as "map") is a function which is linear over a vector space. That is f(au + bv) = af(u) + bf(v). In topology, though, a (continuous) map means a continuous function.

A functional, like map, is a specific kind of function. In geometry, a (linear) functional is a linear function from some set to the reals.
 
  • #10
A mapping and a function are the same thing, except that "function" is usually reserved for mappings to real numbers, and "mapping" is used when talking about arbitrary sets.
 

Related to Understanding (0,1)-Tensor in Relativity

1. What is a (0,1)-tensor in relativity?

A (0,1)-tensor in relativity is a mathematical object that is used to represent the properties of space and time in the theory of relativity. It is a mathematical concept that describes the relationship between different quantities, such as energy, mass, and momentum, in the context of Einstein's theory of general relativity.

2. How is a (0,1)-tensor different from other tensors?

A (0,1)-tensor is different from other tensors because it has one contravariant index and zero covariant indices. This means that it transforms in a specific way when coordinates are changed, unlike other tensors which may have different numbers of covariant and contravariant indices.

3. What is the significance of (0,1)-tensors in relativity?

(0,1)-tensors are significant in relativity because they provide a mathematical framework for understanding the concepts of space and time in the theory of relativity. They are used to describe the curvature of space-time and the relationship between different physical quantities, allowing for the prediction and understanding of various phenomena, such as gravitational waves and black holes.

4. How are (0,1)-tensors used in solving problems in relativity?

(0,1)-tensors are used in solving problems in relativity by providing a mathematical tool for calculating and analyzing various physical quantities and their relationships in the context of Einstein's theory of general relativity. They allow for the formulation of equations that describe the behavior of matter and energy in the presence of gravitation, leading to a deeper understanding of the nature of space and time.

5. Are (0,1)-tensors relevant in other fields of science?

Yes, (0,1)-tensors are relevant in other fields of science, particularly in the study of electromagnetism, quantum mechanics, and fluid dynamics. They are used in these fields to describe the behavior of physical quantities and their relationships in a similar way as in relativity, providing a powerful mathematical tool for understanding and predicting various phenomena.

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