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The Tensor Product

  1. Apr 9, 2006 #1
    Can someone please explain to me what is the tensor product and any good elementary tensor algerbra books?
    Last edited: Apr 10, 2006
  2. jcsd
  3. Apr 9, 2006 #2


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    If v1 and v2 are any vectors in and n-dimensional vector space, then v1 tensor v2 is the 2-tensor whose ij entry is v1iv2j. Of course, the tensor product itself is independent of a basis. That is, we can take a 2 dimensional vector space and choose e1 and e2 to be any two independent vectors. That will be a basis for the vector space and, in terms of that basis, those two vectors are necessarily represented as (1, 0) and (0,1). In terms of that basis, the 2-tensor e1 tensor e2 would be represented as the 2 by 2 identity matrix.
  4. Apr 9, 2006 #3
    Thanks alot hall, its starting to make a bit of sense now! Could you also try and explain to me what wedge is?
    Last edited: Apr 9, 2006
  5. Apr 10, 2006 #4


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    It's a tensor product of two arbitrary differential forms in which one has to completely antisymmetrize the components of the resulting covariant tensor.

  6. Apr 10, 2006 #5


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  7. Apr 10, 2006 #6
    The wedge product is so easy to understand you have probably been playing with it for years without noticing.

    The cross product of two vectors is actually a wedge product, although a very lucky one.

    [tex](\bold{u}\times\bold{v})_i = \epsilon_{ijk}u_jv_k[/tex]

    So the vector product can be written with the Levi-Civita symbol, which takes care of the anti-symmetry.

    But we can write this as

    [tex]\epsilon_{ijk}u_jv_k = \frac{1}{2}\epsilon_{ijk}(u_jv_k - u_kv_j)[/tex]

    [tex] = \epsilon_{ijk}(u\wedge v)_{jk}[/tex]

    [tex] = \ast(u\wedge v)_i[/tex]

    So the vector product is related to the wedge product by the Hodge star operator.

    Note: The last equality arises due to the fact that the Hodge star of the identity element in 3 dimensions is:

    [tex](\ast 1)_{ijk} = \epsilon_{ijk}[/tex]
  8. Apr 11, 2006 #7
    I've had the tensor product space [tex]V \otimes W[/tex] defined to me as the unique space with the universal property that given the cross product space [tex]V \times W[/tex] with a map [tex]T[/tex] to the tensor product space and a map [tex]r[/tex] to an arbitrary vector space [tex]R[/tex], then there is a unique linear map [tex]r'[/tex] from the tensor product space to [tex]R[/tex] such that

    [tex]r' \circ T = r[/tex]

    Can someone explain to me why this is an important definition. What do they mean when they say "has the universal property that..."?
  9. Apr 12, 2006 #8


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    When a universal object has a universal property, that means it's the "best" object with that property, and all other objects with that property can be written in terms of the best one.

    In this case, we're interested in bilinear maps:

    VxW ----------------> R

    Well, the tensor product (along with its canonical map)

    VxW ---> V (x) W

    is the best possible bilinear map from VxW. Because when we have any other bilinear map (such as the one written above), we can uniquely factor it as:

    VxW ---> V (x) W ---> R

    Lots of useful things can be described in terms of universal objects (or more interesting constructions). In fact, these descriptions are often very important, because they're the way in which you'd actually use the object.

    For example, we are interested in all sorts of bilinear products on VxV. For example:

    Any inner product: VxV ---> R
    The cross product: R³xR³ ---> R³
    The outer product: R^m x R^n ---> M_(m, n)
    (M_(m, n) is the vector space of all m by n matrices)

    Each of these can be described as a bilinear map on the tensor product. In fact, because of the universal property of the tensor product, it's often easier to define new products in terms of the tensor product! For example, the wedge product on V is most easily defined by:

    v/\w := v(x)w - w(x)v

    and its values live in the appropriate subspace of V(x)V.

    (There is an equivalent definition in terms of a quotient space)

    Actually, the wedge product is another example of a universal object: the wedge product is the universal anticommutative product. If I let /\²(V) denote the space in which the wedge product lives, then we have that any anticommutative bilinear map (that is, T(v, w) = -T(w, v))

    VxV --------------> R

    can be factored uniquely as

    VxV ---> /\²(V) ---> R

    In fact, we've seen this universal property at work in the other examples: Oxymoron showed how to define the cross product in terms of the wedge product!

    Anyways, since one of the whole points of linear algebra is to study linear, bilinear, trilinear, and general multilinear maps, I would argue that the tensor product is central to the entire subject!
    Last edited: Apr 12, 2006
  10. Apr 12, 2006 #9


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    wasn't there a lengthy series of posts on here by people reading a book on this topic for undergrads? tom mattson led it. it was by dave bachman, and very intuitive and descriptive.
  11. Apr 14, 2006 #10
    I can't think of a book right now but if a and b are two tensors then the tensor product is given by

    [tex]/boldfacec[/b] = /boldface a O /boldface b [/tex]

    Where O represents the tensor product symbol. It really should have an X through it but I can't find the Latex rules that use to be posted around here. Once I find it I'll finish this post.

  12. Apr 14, 2006 #11


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    I think you mean this thread:
  13. Apr 14, 2006 #12


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    a tensor product is away of linearizing bilinear functions. I.e. you probably know what a bilinear function is from XxY-->Z, mwhere X,Y,Z are modules.

    A tensor product of X and Y is a "universal" bilinear map XxY-->T, i.e. one such that any other bilinear map XxY-->Z, has a unique factorization XxY-->T-->Z where T-->Z is linear.

    a wedge product is way of linearizing alternating bilinear maps.

    i.e. given X,Y you probably know what an alternating bilibear map XxY-->Z is.

    then you tell me what a wedge product is:....

    see how useful and trivial, category theory makes all definitions?

    them construction on the other ahnd are another matter.

    i.e. do these objects just defined actually exist?

    old fashioned people refer to describing the constructions as telling what a tensor or wedge product is. new fangled persons prefer to give the properties of the objects rather tha the detailed constructions, in order to say what they "are".

    this new fangled approach began over 100 years ago but some people are still not on board.
  14. Nov 13, 2008 #13
    Hi, can anyone also explain if there is a transformation matrix in R2 space and if it is a tensor product of two vectors, ie :
    T= V \otimes W
    where V and W are two vectors in R2. Then what do the eigen values and eigen vectors of T mean?
    Thanks in advance.
  15. Nov 13, 2008 #14


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    if you know the definitions of the words you are using, then it seems you can answer this yourself.

    i.e. if T is a transformation such that Tx = cx for some number c and non zero vector x, then x is an eigenvector and c an eigenvalue.

    since your question implied you know how to make a transformation by tensoring two things, yiou should be able to answer your own question. if not you need to clarify what you mean by the setup given in your question. i.e how does your T act?
  16. Nov 13, 2008 #15
    Mathwonk, Thank you for your reply, Yes I am aware of the basic definition of eigen values and eigen vectors, but yes I shoudl have been more clear... I am expecting a physical interpretation of eigen values and eigen vectors for this case :
    T is transformation matrix which defines a coordinates sytem transformation from a global to local normalized space like so:


    T= V \otimes W

    where V =xi/[tex]\deltai[/tex]
    and W = xi
    xi defines a orthogonal basis in local space, while [tex]\deltai[/tex]
    defines a length along global coordinate system.
    For this transformation matrix T what do their eigen values and eigen vectors mean, do they mean normalized spacing and directions respectively? I hope I am clear.
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