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Tensor action on Vectors.

  1. Apr 12, 2008 #1
    Hello all.

    Now i have stopped trying to visualize tensors and approach them from the roots i have made some progress but still have problems.

    I understand a tensor to be a multilinear mapping from vector spaces and their duals to the reals ( the underlying field ). How does a tensor act upon a vector to produce another vector. This seems to go against the definition. The same query applies when there are not the same number of vectors and one forms in the mapping as they cannot all be contracted to get a scalar or real number.

  2. jcsd
  3. Apr 12, 2008 #2


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    Well you're right that these situations go against the definitions.

    In what context did you see them happen?
  4. Apr 12, 2008 #3
    Hello Quasar987.

    Thanks for prompt reply.

    I will reply to your question soon. It is 5.30 AM here and well past my bedtime.

  5. Apr 13, 2008 #4


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    So your tensor wants to eat m covectors and n vectors so it can become a real boy (scalar)
    you want it to eat them all at once
    suppose only k<m covectors and l<n vectors are available
    later the m-k covectos and n-l vectors show up
    your tensor eats and is transformed
    you can construct another tensor that eats only the late comers
    and what ever late comers it eats it equals your tensor that eats everybody
    Now we can imagine T pauses briefly before eating the stuff S eats it seems then that
    T would be equal to S at that point

    There is a map V->V thus defined by
    that we may write by abuse of notation as
    Now we can define an innerproduct for tensors so that
    a (m,n) tensor can eat a (k,l) tensor producing a (m-k,n-l) tensor
    and an outer product so that
    a (m-k,n-l) tensor can merge with a (k,l) tensor producing a (m,n) tensor
    Last edited: Apr 13, 2008
  6. Apr 13, 2008 #5

    George Jones

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    Let [itex]T[/itex] be a tensor that is a bilinear mapping on pairs of vectors, so

    [tex]T: V \times V \rightarrow \mathbb{R}.[/itex]

    Use [itex]T[/itex] to define

    [tex]\tilde{T}: V \rightarrow V*.[/tex]

    For each [itex]v[/itex] in [itex]V[/itex],

    [tex]\tilde{T} \left( v \right): V \rightarrow \mathbb{R}[/tex]

    is defined by

    [tex] \left( \tilde{T} \left( v \right) \right) \left( u \right) = T \left( v,u \right)[/itex]

    for every [itex]u[/itex] in [itex]V[/itex].

    If [itex]V[/itex] has a non-degenerate "metric" tensor [itex]g[/itex], then this construction gives a natural isomorphism from [itex]V[/itex] to [itex]V*[/itex]. Compose the inverse of this mapping with [tex]\tilde{T}[/tex] to turn [itex]T[/itex] into a mapping from [itex]V[/itex] to [itex]V[/itex].

    The component version (with the summation convention) looks something like

    [tex] g^{\alpha \mu}T_{\mu \nu} v^\nu,[/tex]

    which takes the vector components [itex]v^\nu[/itex] into the components of another vector.

    My post is probably somewhat cryptic, but I can't procrastinate from my real (very pressing) work for any longer.
    Last edited: Apr 13, 2008
  7. Apr 13, 2008 #6
    Thankyou lurflurf and George Jones for your time in answering me. I will do some more work and take your examples on board and get back soonish.

  8. May 7, 2008 #7
    Hello Quasar987.

    In answer to your question of context:-

    In General Relativity by Rober M Wald page 20 he says

    -----if we fix a v in V, T(.,v)is an element of V**, which we identify with an element of V. Thus given a vector in V, T produces another vector in V in a linear fashion. In other words, we can view a tensor of type T(1,1) as a linear map from V into V and vice versa. Similarly we can view T as a linear map from V* into V*.--------

    I ( think ) i understand what he is saying but again it seems to contradict the definition of a tensor as a map into the real numbers.


  9. May 8, 2008 #8


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    We can do that if we have defined an inner product.

    I would no define a tensor as a map into the real numbers (or even some collection of scalars), but doing so only requires that instead of considering the tensor as the map we consider some map.
    we have the function
    many times (after sorting out branches)
    one might write
    but y is a variable not a function

    like wrise even if it is not true
    T induces such a mapping
    thus the source of confunsion
  10. May 9, 2008 #9
    Thaks lurflurf.

    I think that makes a bit more sense. I will give it some more thought.

  11. May 9, 2008 #10


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    duality is very confusing. suppose we are restricting attention to finite dimensional real spaces, and let L(V,W) denote the space of linear maps from V to W, and R the reals.

    then most of the confusion stems from these isomorphisms:

    L(R,V) = V, and L(L(V,R),R) = V, and L(V,V) = VtensorL(V,R).

    not to mention it gets worse when one chooses an inner product on V, which renders V and L(V,R) also isomorphic!
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