Hyperflächensingularität/Zweidimensional/Quasihomogen/Geschlossene Differentialformen/Fakt/Beweis

Beweis

Es ist unter Verwendung von Fakt

$\begin{array}{cc}& d(x^i{y}^j{z}^{k}D)\\ & =x^i{y}^j{z}^{k}dD+(i{x}^{i-1}{y}^j{z}^{k}dx+j{x}^i{y}^{j-1}{z}^{k}dy+k{x}^i{y}^j{z}^{k-1}dz)\wedge D\\ & =x^i{y}^j{z}^{k}dD+(i{x}^{i-1}{y}^j{z}^{k}dx+j{x}^i{y}^{j-1}{z}^{k}dy+k{x}^i{y}^j{z}^{k-1}dz)\wedge \frac{-{\delta }_{2}ydx+{\delta }_{1}xdy}{{\partial }_3(F)}\\ & =x^i{y}^j{z}^k\frac{{\delta }_1+{\delta }_2+{\delta }_3-\delta }{{\partial }_3(F)}dx\wedge dy+(i{x}^{i-1}{y}^j{z}^{k}dx+j{x}^i{y}^{j-1}{z}^{k}dy+k{x}^i{y}^j{z}^{k-1}dz)\wedge \frac{-{\delta }_{2}ydx+{\delta }_{1}xdy}{{\partial }_3(F)}\\ & =\frac{x^i{y}^j{z}^k}{{\partial }_3(F)}(({\delta }_1+{\delta }_2+{\delta }_3-\delta )dx\wedge dy+(i\frac{dx}{x}+j\frac{dy}{y}+k\frac{dz}{z})\wedge (-{\delta }_{2}ydx+{\delta }_{1}xdy))\\ & =\frac{x^i{y}^j{z}^k}{{\partial }_3(F)}(({\delta }_1+{\delta }_2+{\delta }_3-\delta )dx\wedge dy+{\delta }_{1}idx\wedge dy+{\delta }_{2}jdx\wedge dy+\frac{k}{z}(-{\delta }_{2}ydz\wedge dx+{\delta }_{1}xdz\wedge dy))\\ & =\frac{x^i{y}^j{z}^k}{{\partial }_3(F)}(({\delta }_1+{\delta }_2+{\delta }_3-\delta +{\delta }_{1}i+{\delta }_{2}j)dx\wedge dy+\frac{k}{z}(-{\delta }_{2}y\frac{{\partial }_2(F)}{{\partial }_3(F)}dx\wedge dy-{\delta }_{1}x\frac{{\partial }_1(F)}{{\partial }_3(F)}dx\wedge dy))\\ & =\frac{x^i{y}^j{z}^k}{{\partial }_3(F)}(-\delta +{\delta }_1(i+1)+{\delta }_2(j+1)+{\delta }_3(k+1))dx\wedge dy.\end{array}$

Unter der angegebenen Bedingung ist der Vorfaktor gleich $0$ und die Form ist geschlossen. Der Grad ist in diesem Fall gleich  ${\delta }_{1}i+{\delta }_{2}j+{\delta }_{3}k-(\delta -{\delta }_3)+{\delta }_1+{\delta }_2=0$. 

Zur bewiesenen Aussage