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Very Basic Vector Calculus Question - dx,dy,dz and i,j,k

  1. May 16, 2012 #1
    I am learning about Stokes, Green's, and Gauss Divergence Theorems but from the angle of differential forms (the progression found in Pugh's "Real Mathematical Analysis"). This is supplemented by some more computational books, and I notice that these books frequently toss around i, j, and k, e.g. "F = (f_x)i + (f_y)j" while Pugh does not. This corresponds to "F = (f_x)dx + f_y(dy)", yes? If not, how are the two related, if at all?

  2. jcsd
  3. May 16, 2012 #2
    Actually, it seems that I am wrong and i is more like dx^dy, j is dz^dx, k is dy^z?
  4. May 17, 2012 #3
    Hi Zooxanthellae,

    Your hunch is correct, these two things are extremely related. I am going to try to provide a short answer to this, but you should try to find a good reference.

    First, what does [itex]d\xi[/itex] mean? Say [itex]\xi[/itex] is a vector living in [itex]\mathbb{R}^n[/itex]. By definition [itex]\xi[/itex] is a column vector. We are going to define what [itex]d\xi[/itex] means. [itex]d\xi[/itex] is the row vector with the same entries as [itex]\xi[/itex]. We call [itex]d\xi[/itex] a covector. Why is this useful? because now we think of [itex]d\xi[/itex] as a kind of function mapping vectors to real numbers.

    Definition: vector always means column vector.
    Definition: covector always means row vector.

    [itex]d\xi[/itex] takes any vector in [itex]\mathbb{R}^n[/itex] and gives you a number. Let [itex]v[/itex] be any vector. Then [itex]d\xi(v) := d\xi v[/itex], where this is just matrix multiplication.

    You wrote [itex]dx[/itex], and implicitly assumed you were working in [itex]\mathbb{R}^3[/itex]. [itex]dx[/itex] is then shorthand for the row vector with entries [itex][1, 0, 0][/itex].

    so then we can say such things as [itex]dx(3,5,10) = 3[/itex] or [itex]dx(-4,2,5) = -4[/itex]

    Here is where [itex]i,j,k[/itex] is related to [itex]dx,dy,dz[/itex]. [itex]i[/itex] is the column vector with entries [itex][1,0,0]^t[/itex]. [itex]i[/itex] has a covector, we could write it as [itex]di[/itex] and probably should, but we don't. Instead we write the covector as [itex]dx[/itex].

    Why would anyone talk about this? Because we want to integrate on surfaces (manifolds)!

    Say I have a one dimensional curve in [itex]\mathbb{R}^4[/itex]. I'm using dimension [itex]4[/itex] because this theory is not needed in three dimensions. I also have a vector field which permeates all space. I want to integrate the vector field along this curve. How do I do this? I need a method which translates vectors into real numbers, we need covectors! But where can we find a covector? The answer is to use the tangent (co)vectors of the line.

    Pick a point on the curve. This point has two things: a vector from the vector fields, and a (co)vector from the tangent. To integrate, I simply multiply every covector and vector pair. Now I have an integral over the curve, where the integrand is a real valued function (point on the curve goes in, [itex]d\xi(v)[/itex] comes out. And we know how to integrate this!

    Now instead of a curve, think about a two dimensional surface. At everypoint we now have a tangent plane. In three dimensions we represent a plane by the normal vector. We can't do that here; in [itex]\mathbb{R}^4[/itex] every plane has two normal vectors! We get around this by using wedges. A wedge of two vectors, say [itex]\xi_1, xi_2[/itex] is denoted [itex]\xi_1\wedge\xi_2[/itex] and means the (oriented) plane spanned by these two vectors.

    So what does [itex]d\xi_1\wedge d\xi_2[/itex] mean? Again it's a function, but now we need two vectors to get a real number! Let [itex]v_1, v_2[/itex] be vectors (remember: column vectors). Then [itex]d\xi_1\wedge d\xi_2 (v_1, v_2) := d\xi_1(v_1) \cdot d\xi_2(v_2)[/itex], remembering that [itex]d\xi_1[/itex] and [itex]d\xi_2[/itex] are covectors.

    So if we have a two dimensional surface in [itex]\mathbb{R}^4[/itex] and a two-vector field (at every point in space there are two vectors), we know how to integrate. Just as before we use our covectors, this time taken to be from the tangent plane (we get two covectors!).

    I said I would try to keep it short, but indeed I have left out a lot of material. Basis are important if you actually want to compute these integrals. Also, the covector does not have to come from the tangent space, (who says we have to anyways?) These are differential forms. When we choose the differential form coming from the tangent space, this is like computing the flux.

    I hope this is coherent and helps answer your question.
  5. May 17, 2012 #4


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    Thank you theorem4.5.9, I also benefited from your explanation.
  6. May 18, 2012 #5
    I as well.thanks t459
  7. Jan 1, 2013 #6
    This is of course very late and I'm not sure how resurrecting old threads is viewed on PF. But at any rate I was looking through my old posts and was surprised to see I didn't thank theorem4.5.9 for this (even though I used the help). So, thanks for a thoughtful response.
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