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Determinant Function

  1. Mar 22, 2013 #1
    I am curious on how the determinant function determines orientation? I read about in in one of Werner greubs books and I just cannot manage to understand what it is
     
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  3. Mar 22, 2013 #2

    chiro

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    Hey Tenshou.

    Are you familiar with the cross product?

    Once you look at that, you can look at what is called the wedge product and finally its relationship to determinants.

    This approach is a lot more intuitive since you can rectify the nature of the cross product geometrically and also see that the cross product is non-commutative since a x b = -(b x a)
     
  4. Mar 22, 2013 #3

    quasar987

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    An orientation of a vector space V is a choice of a basis B for that space. Once such a choice is made, all the possible basis of V fall into one of two category: either the change of basis matrix from B' to B has positive determinant or negative determinant. In the first case, we then say B' is positively oriented, and in the second case, we say it is negatively oriented.

    For instance, on R², the "usual orientation" is to chose B=(e1,e2). Then it is relatively easy to see that according to this definition, the positively oriented basis are the ones (v1,v2) such that the shortest way to rotate v1 onto v2 is anti-clockwise, and the negatively oriented basis are the ones (v1,v2) such that the shortest way to rotate v1 onto v2 is clockwise.

    Similarly, in R³, the "usual orientation" is the one specified by (e1,e2,e3). Then you can check that the positively oriented basis are the so called "right handed ones", if you know what that means.
     
  5. Mar 22, 2013 #4
    Hi!

    The answer before this post seems to be what was asked since it explains exactly how a determinant determines orientation. of course proof is needed here to show that this is well defined, i.e. that the equivalence relation specified by bases are equivalent if the determinant of the change of bases matrix is positive is in fact an equivalence relation.

    However the wedge product seems also to be important, maybe to provide some intuition it is a good idea to note here that volume forms are involved. If you try to make a n-linear alternating function on an n-dimensional vectorspace to the base field (which is the natural, if you think about it, definition of a volume form) you will find that up to a constant you will always get the determinant function. So basically every volume form Δ is defined after you know the value of Δ on a basis. Then Δ on any basis is just this constant times the determinant of the matrix with colums the vectors of this basis in terms of the first basis, i.e. the change of basis matrix! So if I calcualte the volume of some n-dimensional 'block' by applying Δ to the vectors that form it the order of the vectors matters! orientation then means that for a matching orientation the volume turns out positive and for the opposite orientation it turns out negative.

    In a way then orientation is just a way to deal with things seeming to have negative volume, they only have negative volume if they are oriented against the orientation of the space. Of course this discussion now needs generalization to the case of manifolds. Since then you want to have volume defined in the 'same way' in every tangent space. I am not sure though if anyone wants to discuss that as well.
     
  6. Mar 22, 2013 #5
    Okay, so how is the 3-form defined on a tangent space and its manifold?

    So, when you get the determinant's value it will be positive or negative with respect to the "usual basis/orientation", or zero is they are the same orientation ?

    Thank you for you reply :D it was well needed!
    Yes I am very familiar with the cross product, I am currently teaching my self Differential Geometry, I was wondering if you could explain the wedge product a little bit, I understand it has something to do with k-differential forms, or something of the sort, but I do not understand how it connects to the cross product of the determinant( ergo I have never used it in practice) . Thank you for your response it was well needed! even more so since I am working on a problem which needs the determinant function(or cross product/ wedge product)
     
  7. Mar 23, 2013 #6

    quasar987

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    The determinant of the change of basis matrix from B to B' will never be zero. Change of basis matrices are always of nonzero determinant since they are the same thing as "the matrix of the linear isomorphism defined by sending each element of B to the corresponding element of B'" This is invertible hence so is the matrix that represents it.

    The determinant of the change of basis matrix from B to B' will be >0 or <0. In the first case, we call B' positively oriented (wrt the chose orientation B) or that it shares or determine the same orientation as B, and in the second case, we say B' is negatively oriented (wrt the chosen orientation B') or that it is of opposite orientation to B.
     
  8. Mar 23, 2013 #7

    lavinia

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    If one interchanges two adjacent rows in a matrix its determinant changes sign. This means that the determinant detects the ordering of the row vectors in the matrix. But the ordering is the same as the orientation.
     
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