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Tensor and wedge products

  1. Jan 6, 2007 #1

    cristo

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    Ok, one more diff geometry question! I need to show the following:

    Show that if Cij are components of an antisymmetric tensor then [itex] C_{ij}dx^i\otimes dx^j[/itex] corresponds to the 2-form [itex]C_{ij}dx^i\wedge dx^j[/itex].

    Now, I know that the wedge product is anti symmetric, and that it can be expressed in terms of the tensor product as follows: [tex] C_{ij}dx^i\wedge dx^j=C_{ij}\frac{1}{2}(dx^i\otimes dx^j-dx^j\otimes dx^i) [/tex]

    However, I can't for the life of me show what is required!

    Any hints would be much appreciated!!
     
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  3. Jan 6, 2007 #2

    Hurkyl

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    Have you invoked the antisymmetry of C?
     
  4. Jan 6, 2007 #3

    cristo

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    Well, yes, I did try that, but on invoking the antisymmetry of C, we obtain

    [tex] C_{ij}dx^i\wedge dx^j=\frac{1}{2}(C_{ij}dx^i\otimes dx^j+C_{ji}dx^j\otimes dx^i) [/tex]

    And I can't seem to progress from there.
     
  5. Jan 6, 2007 #4
    Well, but
    [tex]\frac{1}{2}(C_{ij}dx^i\otimes dx^j+C_{ji}dx^j\otimes dx^i)=C_{ij}dx^i\otimes dx^j[/tex]

    Don't get confused with the indices, just think of what one is doing.
     
  6. Jan 6, 2007 #5

    cristo

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    Haha, I can't believe I didn't see that! Thanks!
     
  7. Jan 30, 2007 #6

    mathwonk

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    how interesting, you are looking at antisymmetric tensors as if they are a subspace of usual tensors. they can slao be looekd at as a quotient space of usual temsors, then your statement is true for allk tensors if un derstood correctly. i.e. the usual tensor space maps onto the space of alternating tensors by sending dxtensdy to dxwedgedy.

    then there also a map back from alternating tensors to usual tensors sending dxwedgedy to (1/2)(dxtensdy - dytensdx).

    the composition of these maps is the identity in one direction but not the other.

    so i was puzzled by your question, since thinking in terms of the quotient space interpretation, the answer to your question is yes, even for non antisymmetric tensors, and trivially so.

    i was confused because you did not define the word "corresponds".
     
  8. Jan 30, 2007 #7

    mathwonk

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    which in your case meant "equals".
     
  9. Jan 30, 2007 #8

    mathwonk

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    this an algebra question by the way.
     
  10. Jan 30, 2007 #9

    mathwonk

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    the algebra of tensros is a little like that of polynomials, as i have tried to explain elsewhere. think about how you write out a homogeneous polynomial, say of degree 2, as a linear combination of x^2, xy, and y^2.

    all you need to know is the coefficient of each term if we number the variabkles x,y as say x1 and x2, then we can represent the monomial x^2, i.e. x1x1 just as (1,1), and similarly, we get the representations (1,1),(1,2), (2,2), for x^2, xy, and y^2,

    i.e. a symbol a(i,j) is the coeficient of the monomial xixj.

    so if we understand summation, then a symbol like a(i,j) means the polynomial which is the sum of the terms a(i,j)xixj.

    if we want to do this for non commutative polynomials, we need also xx, xy, yx, yy. so we have coefficients a(i,j) where now there is no way to combine a(i,j) and a(j,i).

    here too we can think of usual polynomials as either a quotient or a subspace of non commutative ones.


    so we map the non commutaitve polynomials to th usal ones just by sending both xy and yx to xy. well lets make a capital letter for non commutative oens.

    so we send XY and YX both to xy.

    then we can also map back from reguklar polynomials to non commutaive ones, by sending xy to (1/2)(XY+YX). but obviously xy, where xy = yx, is a simpler object that (1/2)(XY+YX).


    this is why it more natural to view anticommutative ones also as a quotient rather than a subspace. but anyway.....

    spivaks little book calculus on manifolds explains both points of view i believe.

    but some physicists learning this stuff are apparently left without any awareness of why tensors are a generalizaton of polynomials.


    this insight also makes clear the transformation laws, as all you do to get them is substitute say X1 = (aY1+bY2), and X2 = cY1+dY2, and expand.

    again just like polynomials.
     
    Last edited: Jan 30, 2007
  11. Jan 30, 2007 #10

    mathwonk

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    the tensor sign, circle cross, just means the multiplication is not commutative, thats all.
     
  12. Jan 30, 2007 #11

    cristo

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    Thanks for the insightful comments, mathwonk. I've never come across this explanation of non-commutative polynomials before, and I can see that here, the quotient space is a more natural view. However, for the alternating tensor case, especially since I've only just been introduced to the subject, it seems easier to think of alternating tensors as a subspace of normal tensors, than to picture them as a quotient space. (I realise that this may sound wrong to you; but as a beginner I'm familiar with subspaces, but quotient spaces seem a bit "strange").

    Anyway, thanks again for your comments.
     
  13. Jan 30, 2007 #12

    mathwonk

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    well you are over haflway there when you use the wedge notation, instead of trying to manipulate tensors with antisymmetric coefficients.

    the purpose of the wedge notation is to accomplish the same thing as the quotient construction, namely take away the need to deal with expressions like (1/2)(AB-BA), and just use A^B instead.
     
  14. Jan 31, 2007 #13

    mathwonk

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    i seem to be confused here. does the OP mean to say that

    dx^dy corresponds to (1/2)(dxtensdy?.

    If so, then this seems odd from one point of view, as it would seem to imply that when acting on the standard unit block e1^e2, the value of dx^dy is 1/2 instead of 1.

    ????????? obviously i am more at home with general ideas than computations.

    but one thing becoming clear here is that some of these things are merely conventions, and not intrinsic properties.
     
  15. Jan 31, 2007 #14

    mathwonk

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    so i may be wrong here, but it seems one cannot haVE BOTH THIGNS at once, i.e. the natural correspondence desired by the OP seems to conflict with the volume form dx^dy having the expected value (one) on the standard unit square.

    what is the source for your conventions, i.e. what book are you reading?
     
  16. Jan 31, 2007 #15

    George Jones

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    No, this is not what the original post says.
     
  17. Jan 31, 2007 #16

    cristo

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    No, the question in the OP is correct, asking to show that [itex] C_{ij}dx^i\otimes dx^j[/itex] corresponds to [itex]C_{ij}dx^i\wedge dx^j[/itex].

    The question didn't come from a book, but from the lecture notes for a course on differential geometry that I'm taking. This, he states, is the way to obtain an alternating tensor from two form: [tex] dx^i\wedge dx^j=\frac{1}{2}(dx^i\otimes dx^j-dx^j\otimes dx^i) [/tex]
     
  18. Jan 31, 2007 #17

    mathwonk

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    how does saying that dx^dy corresponds to (1/2)(dxtensdy differ from saying that Cij dx^dy corresponds to Cij dxtensdy?

    what does it mean to say Cij dx^dy corresponds to Cij dxtensdy? as ia sked, is a summation intended here?

    i.e. does this mean say that dx^dy - dy^dx corresponds to

    dxtensdy - dytensdx?

    if so, then ........OOPs, some of my post seems to have disappeared.

    i meant that if he was saying that

    dx^dy - dy^dx corresponds to dxtensdy - dytensdx,
    then since dx^dy = -dy^dx,

    this would imply that 2dx^dy corresponds to dxtensdy - dytensdx,

    and this would imply that dx^dy corresponds to
    (1/2)(dxtensdy-dytensdx).

    but then the value of dx^dy on the unit square (e1,e2)

    would equal that of the tensor (1/2)(dxtensdy-dytensdx), namely 1/2.

    which seems odd.
     
  19. Jan 31, 2007 #18

    mathwonk

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    all i am saying is that there sem to be many ways to obtain an alternating tensor from a symbol of form A^B, depending on what one wants to be true.

    for the purpose of measuring volumes, i.e. integration, which is hoforms are usually used, the choice above has some curious properties, and seems to differ from the convention in books like spivak, fleming, loomis sternberg, etc........
     
  20. Jan 31, 2007 #19

    mathwonk

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    there is no hard and fast law about to do it, but there are reasons to ask that

    the form dx1^....^dxn act on a sequence of n vectors as the determinant does, i.e. to have dx1^....^dxn correspond to the determinant, not to the determinant divided by (n!).

    has your prof defined wedge multiplication between two alternating tensors? if so, how?


    i.e. if dx^dy is the wedge product of the two tensors dx and dy, how is it defined?

    is dx^dy equal to dxtensdy - dytensdx? or is it (1/2)of that?
     
  21. Jan 31, 2007 #20

    cristo

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    I remember my prof discussing this. He never defind the wedge of two tensors, but rather defined the wedge of two differential forms: [tex](f(x)dx^i\wedge dx^j \wedge \cdots)\wedge(g(x)dx^l\wedge dx^m \wedge \cdots)=f(x)g(x)(dx^i\wedge dx^j \wedge \cdots)\wedge(dx^l \wedge dx^m \wedge \cdots) [/tex]. Then later, when introducing tensors, used the expression [tex]dx^i\wedge dx^j=1/2(dx^i\otimes dx^j=dx^j\otimes dx^i)[/tex]

    He mentioned that the factor 1/2 (1/n! in general) is a convention generally used in mathematical physics texts, whereas in pure maths books the factor is 1.

    With respect to the summation; my understanding was that repeated indices were assumed to be summed over unless explicitly stated otherwise.
     
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