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1. • CommentRowNumber1.
   • CommentAuthorStevenGubkin
   • CommentTimeMay 6th 2014
   • (edited May 6th 2014)
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   Author: StevenGubkin
   Format: MarkdownItexThis thread is a continuation of the discussion occurring here http://mathoverflow.net/questions/164845/non-continuous-differentiability-for-differential-forms/ I think I agree with you that using an error term of $\epsilon Vol(P)$ isn't going to fly. What does work (I believe!) are bounds of the form $\epsilon|v_1||v_2|...|v_n|$. Here is an example using the new definition. Let $\omega = f(x,y)dx+g(x,y)dy$ be a one form on $\mathbb{R}^2$. Then $$\begin{align} \omega(p)(v_1)+\omega(p+v_1)(v_2) - \omega(p+v_2)(v_1) - \omega(p)(v_2) &=f(p)dx(v_1)+g(p)dy(v_1) + f(p+v_1)dx(v_2)+g(p+v_1)dy(v_2)-f(p+v_2)dx(v_1) - g(p+v_2)dy(v_1) - f(p)dx(v_2)-g(p)dy(v_2)\\ &=[f(p+v_1)-f(p)]dx(v_2)-[f(p+v_2)-f(p)]dx(v_1)+[g(p+v_1)-g(p)]dy(v_2)-[g(p+v_2)-g(p)]dy(v_1)\\ & [df(p)(v_1) + \epsilon|v_1|]dx(v_2)-[df(p)(v_2) + \epsilon |v_2|]dx(v_1) +[ dg(p)(v_1) + \epsilon |v_1|]dy(v_2)-[dg(p)(v_2)+\epsilon|v_2|]dy(v_1)\\ &=(df \wedge dx)(v_1,v_2)+(dg \wedge dy)(v_1,v_2)+\epsilon|v_1|dx(v_2)+\epsilon|v_2|dx(v_1) +\epsilon|v_1|dy(v_2)+\epsilon |v_2|dy(v_1)\\ \end{align}$$ Since $(df \wedge dx)(v_1,v_2)+(dg \wedge dy)(v_1,v_2)$ is the usual formula for $d\omega$, we will be done if we can show that $\epsilon|v_1|dx(v_2)+\epsilon|v_2|dx(v_1) +\epsilon|v_1|dy(v_2)+\epsilon |v_2|dy(v_1)$ can be bounded in absolute value by something that looks like $\epsilon|v_1||v_2|$. But this just follows from the pythagorean theorem.

   This thread is a continuation of the discussion occurring here http://mathoverflow.net/questions/164845/non-continuous-differentiability-for-differential-forms/

   I think I agree with you that using an error term of $\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 $\epsilon|v_1||v_2|...|v_n|$.

   Here is an example using the new definition.

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

   Then

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

2. • CommentRowNumber2.
   • CommentAuthorMike Shulman
   • CommentTimeMay 6th 2014
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   Author: Mike Shulman
   Format: MarkdownItexFWIW, I'm pretty sure that $vol(P)$ 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 $f$ and $g$ are differentiable functions, then $\omega= f(x,y)\,dx+g(x,y)\,dy$ 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 $f$ is a differentiable function and $\eta$ a differentiable 1-form, then $\mathrm{d}(f\,\eta) = \mathrm{d}f \wedge \eta + f\,\mathrm{d}\eta$, and I'm stuck. By writing $$ f_{x+v_1} \, \eta_{x+v_1}(v_2) - f_x\, \eta_x(v_2) = (f_{x+v_1} - f_x)(\eta_{x+v_1}(v_2) - \eta_x(v_2)) + f_x(\eta_{x+v_1}(v_2) - \eta_x(v_2)) + (f_{x+v_1} - f_x)\eta_x(v_2) $$ and similarly swapping $v_1$ and $v_2$, I can write $$ f_x \, \eta_x(v_1) + f_{x+v_1}\,\eta_{x+v_1}(v_2) - f_{x+v_2}\,\eta_{x+v_2}(v_1) - f_x \,\eta_x (v_2) $$ as $$ (\mathrm{d}f_x(v_1) + \epsilon |v_1|) \eta_x(v_2) - (\mathrm{d}f_x(v_2) + \epsilon|v_2|)\eta_x(v_1) + f_x(\mathrm{d}\eta_x(v_1,v_2) + \epsilon|v_1||v_2|) + (f_{x+v_1} - f_x)(\eta_{x+v_1}(v_2) - \eta_x(v_2)) - (f_{x+v_2} - f_x)(\eta_{x+v_2}(v_1) - \eta_x(v_1)) $$ which is $$ (\mathrm{d}f\wedge \eta + f\,\mathrm{d}\eta)_x(v_1,v_2) + \epsilon |v_1||v_2| + (f_{x+v_1} - f_x)(\eta_{x+v_1}(v_2) - \eta_x(v_2)) - (f_{x+v_2} - f_x)(\eta_{x+v_2}(v_1) - \eta_x(v_1)), $$ but I can't see how to write the last two terms as $\epsilon |v_1||v_2|$. I seem to need to combine them to invoke differentiability of $\eta$, but they are multiplied by $\mathrm{d}f$ 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 $\mathrm{d}(\omega\wedge\eta) = \mathrm{d}\omega \wedge \eta \pm \omega \wedge \mathrm{d}\eta$. After that, the next question would be whether differentiability of $\mathrm{d}\omega$ is sufficient to imply $\mathrm{d}\mathrm{d}\omega = 0$. I'm not sure whether to be too hopeful or not.

   FWIW, I’m pretty sure that $vol(P)$ 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 $f$ and $g$ are differentiable functions, then $\omega= f(x,y)\,dx+g(x,y)\,dy$ 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 $f$ is a differentiable function and $\eta$ a differentiable 1-form, then $\mathrm{d}(f\,\eta) = \mathrm{d}f \wedge \eta + f\,\mathrm{d}\eta$, and I’m stuck. By writing

   $$f_{x+v_1} \, \eta_{x+v_1}(v_2) - f_x\, \eta_x(v_2) = (f_{x+v_1} - f_x)(\eta_{x+v_1}(v_2) - \eta_x(v_2)) + f_x(\eta_{x+v_1}(v_2) - \eta_x(v_2)) + (f_{x+v_1} - f_x)\eta_x(v_2)$$

   and similarly swapping $v_1$ and $v_2$, I can write

   $$f_x \, \eta_x(v_1) + f_{x+v_1}\,\eta_{x+v_1}(v_2) - f_{x+v_2}\,\eta_{x+v_2}(v_1) - f_x \,\eta_x (v_2)$$

   as

   $$(\mathrm{d}f_x(v_1) + \epsilon |v_1|) \eta_x(v_2) - (\mathrm{d}f_x(v_2) + \epsilon|v_2|)\eta_x(v_1) + f_x(\mathrm{d}\eta_x(v_1,v_2) + \epsilon|v_1||v_2|) + (f_{x+v_1} - f_x)(\eta_{x+v_1}(v_2) - \eta_x(v_2)) - (f_{x+v_2} - f_x)(\eta_{x+v_2}(v_1) - \eta_x(v_1))$$

   which is

   $$(\mathrm{d}f\wedge \eta + f\,\mathrm{d}\eta)_x(v_1,v_2) + \epsilon |v_1||v_2| + (f_{x+v_1} - f_x)(\eta_{x+v_1}(v_2) - \eta_x(v_2)) - (f_{x+v_2} - f_x)(\eta_{x+v_2}(v_1) - \eta_x(v_1)),$$

   but I can’t see how to write the last two terms as $\epsilon |v_1||v_2|$. I seem to need to combine them to invoke differentiability of $\eta$, but they are multiplied by $\mathrm{d}f$ 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 $\mathrm{d}(\omega\wedge\eta) = \mathrm{d}\omega \wedge \eta \pm \omega \wedge \mathrm{d}\eta$. After that, the next question would be whether differentiability of $\mathrm{d}\omega$ is sufficient to imply $\mathrm{d}\mathrm{d}\omega = 0$. I’m not sure whether to be too hopeful or not.