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

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SUMMARY

This discussion focuses on the relationship between differential forms and vector notation in the context of Stokes, Green's, and Gauss Divergence Theorems, as explored through Pugh's "Real Mathematical Analysis." The distinction between column vectors (i, j, k) and row vectors (dx, dy, dz) is clarified, with dx representing the row vector [1, 0, 0] in ℝ³. The conversation emphasizes the importance of covectors in integrating vector fields along curves and surfaces, particularly in higher dimensions like ℝ⁴, where traditional methods require the use of wedge products to compute integrals over two-dimensional surfaces.

PREREQUISITES
  • Understanding of differential forms and their applications
  • Familiarity with vector calculus concepts, particularly in ℝ³ and ℝ⁴
  • Knowledge of Stokes, Green's, and Gauss Divergence Theorems
  • Basic understanding of matrix multiplication and linear algebra
NEXT STEPS
  • Study the properties and applications of differential forms in advanced calculus
  • Learn about the computation of integrals using wedge products in higher dimensions
  • Explore the geometric interpretation of covectors and their role in integration
  • Investigate the implications of Stokes' Theorem in various dimensions and contexts
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Mathematicians, physics students, and anyone interested in advanced calculus, particularly those studying vector fields and differential geometry.

zooxanthellae
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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?

Thanks.
 
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Actually, it seems that I am wrong and i is more like dx^dy, j is dz^dx, k is dy^z?
 
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 d\xi mean? Say \xi is a vector living in \mathbb{R}^n. By definition \xi is a column vector. We are going to define what d\xi means. d\xi is the row vector with the same entries as \xi. We call d\xi a covector. Why is this useful? because now we think of d\xi as a kind of function mapping vectors to real numbers.

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

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

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

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

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

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

Say I have a one dimensional curve in \mathbb{R}^4. I'm using dimension 4 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, d\xi(v) 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 \mathbb{R}^4 every plane has two normal vectors! We get around this by using wedges. A wedge of two vectors, say \xi_1, xi_2 is denoted \xi_1\wedge\xi_2 and means the (oriented) plane spanned by these two vectors.

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

So if we have a two dimensional surface in \mathbb{R}^4 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.
 
Thank you theorem4.5.9, I also benefited from your explanation.
 
chiro said:
Thank you theorem4.5.9, I also benefited from your explanation.

I as well.thanks t459
 
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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