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2-dimensional cross product

  1. Mar 28, 2005 #1
    I'm not even sure such a thing exists :P.

    Anyway, if it does, is it correct that it is:
    Code (Text):

    Xc = -Y
    Yc = X
     
    (rotation 90 deg counter clockwise)

    Couldn't find anything about it on google...
     
  2. jcsd
  3. Mar 28, 2005 #2

    SpaceTiger

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    The two-dimensional equivalent of a cross product is a scalar:

    [tex]\hat{x} \times \hat{y} = x_{1}y_{2}-x_{2}y_{1}[/tex]

    It's also the determinant of the 2x2 row matrix formed by the vectors. I don't think it's usually used, though. Unlike dot products, cross products aren't geometrically generalizable to n dimensions.
     
  4. Mar 28, 2005 #3

    dextercioby

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    Yes,it's a particular case of a more general statement.Since in a 2D euclidian space a 1form is a vector,then the cross-product of these vectors in,of course,the Hodge dual of the wedge product of the two 1forms,which,according to the general theory

    [tex] \left[\ast\left(A\wedge B\right)\right]=\epsilon^{ij}A_{i}B_{j}=A_{1}B_{2}-A_{2}B_{1} [/tex]

    Daniel.
     
  5. Mar 28, 2005 #4

    dextercioby

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    Okay,which part of "differential geometry" didn't u get...?? :uhh:

    What is a dot product...?

    Daniel.
     
  6. Mar 28, 2005 #5

    SpaceTiger

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    In four dimensions or more, the "cross product" returns a matrix. Yes, you can come up with some geometrical object to describe this matrix, but you're no longer working with vectors, so the concept isn't the same. This is what I mean by "not geometrically generalizable". My apologies for not being more precise.


    I hope you're kidding. If not, check Mathworld.
     
  7. Mar 28, 2005 #6

    dextercioby

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    There's no matrix...Consider a 6 dim manifold and compute the Hodge dual of a wedge product of two 1forms (which are actually covectors) and u'll end up with a 4 form...What matrix are u dreaming about...?

    I wanted to know whether u know what a scalar product is...

    Daniel.

    P.S.It bears a name only for 3D euclidean space (in which 1forms=covectors-------->vectors),viz. cross product...
     
  8. Mar 28, 2005 #7

    SpaceTiger

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    Yeah, I'll get right on that. You're right, though, I should have said "tensor", not matrix.
     
  9. Mar 28, 2005 #8

    mathwonk

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    there is a rather lengthy discussion of these issues in the thread tom mattson is running on geometric interpretation of diff forms. probably located in the diff geom dept.
     
  10. Mar 28, 2005 #9

    chroot

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    A 4-form is not a tensor, either, SpaceTiger. I'm afraid dexter has you on this one; he's right.

    - Warren
     
  11. Mar 28, 2005 #10

    dextercioby

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    It is a (0,4) tensor on the manifold...:wink: "p" forms are (0,p) totally antisymmetric tensors .The em 4potential is a covector/1form,the em tensor (double covariant,antisymmetric tensor) is a 2form,its Hodge dual is a 2form as well,etc

    I didn't like that "cross products aren't geometrically generalizable to n dimensions" part.Cross product needn't be generalized,other concepts in diff.geom.need to take particular values in order to find the cross product...

    Daniel.
     
  12. Mar 28, 2005 #11

    chroot

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    Ouch. Dexter, can I borrow your brain for a few weeks? I won't hurt it, I promise.

    - Warren
     
  13. Mar 28, 2005 #12

    mathwonk

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    isn't that a matter of terminology? i.e. exterior forms are often referred to as "alternating tensors".

    i.e. for any module or vector space, its exterior algebra is a quotient of the tensor algebra by the homogeneous ideal generated by all squares of elements of degree one. no?

    so in this sense an exterior form is an equivalence class of tensors.

    there are skew - symmetric and well as symmetric tensors.

    indeed if we think dually of a tensor as a field of multilinear functionals on the tangent spaces, as is usually done in differential geometry, then adding a condition like skew symmetry, merely restricts the class of tensors to a subclass.

    in this sense antisymmetric tensors are actually a subfamily of the general tensors, hence they are actual tensors, and not just equivalence classes of them.
     
  14. Mar 28, 2005 #13

    mathwonk

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    as duscussed at length elsewhere, given a dot product, each vector subspace of R^n has an orthogonal complement, hence given a k plane in R^n, and a number, we can pass to the orthogonal complement paired with the same number.


    this prthogonal dualization process is called cross products or hodge dual, or whatever you want in various contexts.

    in oriented three space a plane and a number thought of as an oriented area, passes to a line perpendicular to that plane, and that number thought of as an oriented area, i.e. a vector in that line. that is called "cross product" of any two vectors in the original plane spanning the plane and spanniong a parallelogram with oriented area equal to the given number.


    in higher dimensions, the orthogonal complement of a k plane is an (n-k) plane, so the cross product of two vectors would be an (n-2) dimensional object. it is simpler to just think of it as a 2 dimensional object, called the wedge product of the two vectors, i.e. essentially the plane they span, plus the oriented area of the parallelogram they span.

    alternatively, one can take n-1 vectors instead of only 2, and say their cross product is the vector determined by the orthocomplement of the n-1 plane they span.
     
    Last edited: Mar 28, 2005
  15. Mar 28, 2005 #14

    SpaceTiger

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    Thanks mathwonk. I never studied differential geometry, but your explanations make the issue a good bit clearer.
     
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