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    • CommentRowNumber1.
    • CommentAuthorStevenGubkin
    • CommentTimeMay 6th 2014
    • (edited May 6th 2014)

    This thread is a continuation of the discussion occurring here

    I think I agree with you that using an error term of εVol(P)\begin{align} \omega(p)(v_1)+\omega(p+v_1)(v_2) - \omega(p+v_2)(v_1) - \omega(p)(v_2) &=f(p)dx(v_1)+g(p)dy(v_1) + f(p+v_1)dx(v_2)+g(p+v_1)dy(v_2)-f(p+v_2)dx(v_1) - g(p+v_2)dy(v_1) - f(p)dx(v_2)-g(p)dy(v_2)\\ &=[f(p+v_1)-f(p)]dx(v_2)-[f(p+v_2)-f(p)]dx(v_1)+[g(p+v_1)-g(p)]dy(v_2)-[g(p+v_2)-g(p)]dy(v_1)\\ & [df(p)(v_1) + \epsilon|v_1|]dx(v_2)-[df(p)(v_2) + \epsilon |v_2|]dx(v_1) +[ dg(p)(v_1) + \epsilon |v_1|]dy(v_2)-[dg(p)(v_2)+\epsilon|v_2|]dy(v_1)\\ &=(df \wedge dx)(v_1,v_2)+(dg \wedge dy)(v_1,v_2)+\epsilon|v_1|dx(v_2)+\epsilon|v_2|dx(v_1) +\epsilon|v_1|dy(v_2)+\epsilon |v_2|dy(v_1)\\ \end{align}$$\epsilon Vol(P) isn’t going to fly. What does work (I believe!) are bounds of the form ε|v 1||v 2|...|v n|\epsilon|v_1||v_2|...|v_n|.

    Here is an example using the new definition.

    Let ω=f(x,y)dx+g(x,y)dy\omega = f(x,y)dx+g(x,y)dy be a one form on 2\mathbb{R}^2.


    Since (dfdx)(v 1,v 2)+(dgdy)(v 1,v 2)(df \wedge dx)(v_1,v_2)+(dg \wedge dy)(v_1,v_2) is the usual formula for dωd\omega, we will be done if we can show that ε|v 1|dx(v 2)+ε|v 2|dx(v 1)+ε|v 1|dy(v 2)+ε|v 2|dy(v 1)\epsilon|v_1|dx(v_2)+\epsilon|v_2|dx(v_1) +\epsilon|v_1|dy(v_2)+\epsilon |v_2|dy(v_1) can be bounded in absolute value by something that looks like ε|v 1||v 2|\epsilon|v_1||v_2|. But this just follows from the pythagorean theorem.

    • CommentRowNumber2.
    • CommentAuthorMike Shulman
    • CommentTimeMay 6th 2014

    FWIW, I’m pretty sure that vol(P)vol(P) does work if you use integrals around the boundary rather than just evaulation.

    When you say “the pythagorean theorem” do you just mean that the components of a vector are no bigger than its length?

    I think I believe your argument now that if ff and gg are differentiable functions, then ω=f(x,y)dx+g(x,y)dy\omega= f(x,y)\,dx+g(x,y)\,dy is a differentiable 1-form by this definition, with its usual derivative. That’s excellent. But I’m now trying to show the slightly more general-looking statement that if ff is a differentiable function and η\eta a differentiable 1-form, then d(fη)=dfη+fdη\mathrm{d}(f\,\eta) = \mathrm{d}f \wedge \eta + f\,\mathrm{d}\eta, and I’m stuck. By writing

    f x+v 1η x+v 1(v 2)f xη x(v 2)=(f x+v 1f x)(η x+v 1(v 2)η x(v 2))+f x(η x+v 1(v 2)η x(v 2))+(f x+v 1f x)η x(v 2) f_{x+v_1} \, \eta_{x+v_1}(v_2) - f_x\, \eta_x(v_2) = (f_{x+v_1} - f_x)(\eta_{x+v_1}(v_2) - \eta_x(v_2)) + f_x(\eta_{x+v_1}(v_2) - \eta_x(v_2)) + (f_{x+v_1} - f_x)\eta_x(v_2)

    and similarly swapping v 1v_1 and v 2v_2, I can write

    f xη x(v 1)+f x+v 1η x+v 1(v 2)f x+v 2η x+v 2(v 1)f xη x(v 2) f_x \, \eta_x(v_1) + f_{x+v_1}\,\eta_{x+v_1}(v_2) - f_{x+v_2}\,\eta_{x+v_2}(v_1) - f_x \,\eta_x (v_2)


    (df x(v 1)+ε|v 1|)η x(v 2)(df x(v 2)+ε|v 2|)η x(v 1)+f x(dη x(v 1,v 2)+ε|v 1||v 2|)+(f x+v 1f x)(η x+v 1(v 2)η x(v 2))(f x+v 2f x)(η x+v 2(v 1)η x(v 1)) (\mathrm{d}f_x(v_1) + \epsilon |v_1|) \eta_x(v_2) - (\mathrm{d}f_x(v_2) + \epsilon|v_2|)\eta_x(v_1) + f_x(\mathrm{d}\eta_x(v_1,v_2) + \epsilon|v_1||v_2|) + (f_{x+v_1} - f_x)(\eta_{x+v_1}(v_2) - \eta_x(v_2)) - (f_{x+v_2} - f_x)(\eta_{x+v_2}(v_1) - \eta_x(v_1))

    which is

    (dfη+fdη) x(v 1,v 2)+ε|v 1||v 2|+(f x+v 1f x)(η x+v 1(v 2)η x(v 2))(f x+v 2f x)(η x+v 2(v 1)η x(v 1)), (\mathrm{d}f\wedge \eta + f\,\mathrm{d}\eta)_x(v_1,v_2) + \epsilon |v_1||v_2| + (f_{x+v_1} - f_x)(\eta_{x+v_1}(v_2) - \eta_x(v_2)) - (f_{x+v_2} - f_x)(\eta_{x+v_2}(v_1) - \eta_x(v_1)),

    but I can’t see how to write the last two terms as ε|v 1||v 2|\epsilon |v_1||v_2|. I seem to need to combine them to invoke differentiability of η\eta, but they are multiplied by df\mathrm{d}f evaluated on different vectors, so I can’t factor that out.

    If we can get this to work, then hopefully it will generalize to the product rule d(ωη)=dωη±ωdη\mathrm{d}(\omega\wedge\eta) = \mathrm{d}\omega \wedge \eta \pm \omega \wedge \mathrm{d}\eta. After that, the next question would be whether differentiability of dω\mathrm{d}\omega is sufficient to imply ddω=0\mathrm{d}\mathrm{d}\omega = 0. I’m not sure whether to be too hopeful or not.