Understanding (0,1)-Tensor in Relativity

[emma83]

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.

 

[George Jones]

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 V* results in V**, the dual of V* and the double dual of V*. Since V** is the dual of V*, V** is the space of (0,1) tensors.

Your questions really is "Why can V** be identified with V?"

Consider a mapping (defined later)

i : V \rightarrow V**,

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

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

Define the mapping i by

i \left( v \right) \left( f \right) = f \left( v \right)[/itex]<br /> <br /> for every v \in V and f \in V*, and note that i, so defined is a linear mapping. It is also always a one-to-one mapping, and it is onto if and only if V is finite-dimensional.<br /> <br /> If V is finite-dimensional, i is a bijection that is independent of any choice of basis for V, so, usually no distinction is made between V and V**, and i \left( v \right) is written just as v.<br /> <br /> All this of is often a bit confusing on first reading.

 

[Jaunty]

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 v \in V, and a dual vector \omega \in V^*.
We can write the linear mapping that \omega: V \to \mathbb{R} as \omega(v)=\omega_\mu v^\mu (\left\omega eats v and spits out a scalar).
Now, we could just as well say that it's v that's providing the mapping: v: V^* \to \mathbb{R} where v(\omega)=v^\mu \omega_\mu (v eats \omega and spits out a scalar).
Since we have an explicit mapping from V^* \to \mathbb{R} for an arbitrary dual-vector, this means v must be in V^{**}, the space of linear functionals on dual vectors.

edit: anyone have any idea why my single Latex characters are all misaligned?

 

[emma83]

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 ?

 

[Jaunty]

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).

 

[emma83]

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".

 

[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 ?

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 f(x) = \pm \sqrt{x} would assign 2 to two values, \pm \sqrt{2}). 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.

 

[George Jones]

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.

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.

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.

 

[Tac-Tics]

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.

 

[dx]

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.