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Confusing (0,1)-tensor

  1. Jan 25, 2009 #1
    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.
  2. jcsd
  3. Jan 25, 2009 #2

    George Jones

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    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.
  4. Jan 25, 2009 #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?
  5. Jan 25, 2009 #4

    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 ?
  6. Jan 25, 2009 #5
    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).
  7. Jan 25, 2009 #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".
  8. Jan 25, 2009 #7
    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).

    Yes, it's certainly not necessary for two vectors to be assigned to the same scalar by a one form.
  9. Jan 25, 2009 #8

    George Jones

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    It occurred to me while I was posting that I couldn't remember which is which.
    Try using itex and /itex for tags within lines of text. Reserve the tex and /tex tags for lines of stand-alone math.
    I use mapping and function interchangeably, and what you call a mapping, I call a relation.
  10. Feb 6, 2009 #9
    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.
  11. Feb 7, 2009 #10


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    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.

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