By Katz N.M.
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Additional resources for A conjecture in arithmetic theory of differential equations
We defined ′ ω before we started talking about parameterizations. Hence, the value which we M calculate for this integral should not depend on our choice of parameterization. So what happened? To analyze this completely, we need to go back to the definition of ω from M the previous section. We noted at the time that a choice was made to calculate 1 2 1 2 ωpi,j (Vi,j , Vi,j ) instead of ωpi,j (−Vi,j , Vi,j ). But was this choice correct? The answer is a resounding maybe! We are actually missing enough information to tell.
The Steps. To compute the integral of a differential n-form, ω, over a region, S, the steps are as follows: (1) Choose a parameterization, Ψ : R → S, where R is a subset of Rn (see Figure 5). z v S Ψ R u y x Figure 5. (2) Find all n vectors given by the partial derivatives of Ψ. Geometrically, these are tangent vectors to S which span its tangent space (see Figure 6). z ∂Ψ ∂u y x ∂Ψ ∂v Figure 6. , un ). (4) Integrate the resulting function over R. 6. 2. Integrating 2-forms. The best way to see the above steps in action is to look at the integral of a 2-form over a surface in R3 .
Interlude: 0-forms Let’s go back to Section 1, when we introduced coordinates for vectors. At that time we noted that if C was the graph of the function y = f (x) and p was a point of C then the tangent line to C at p lies in Tp R2 and has equation dy = m dx, for some constant, m. Of course, if p = (x0 , y0 ) then m is just the derivative of f evaluated at x0 . Now, suppose we had looked at the graph of a function of 2-variables, z = f (x, y), instead. At some point, p = (x0 , y0 , z0 ), on the graph we could look at the tangent plane, which lies in Tp R3 .