ProofOfPoincareLemma 2004-06-12 14:49:14 2004-06-12 14:49:14 Proof Poincar\'e lemma 0 % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here Let $X$ be a smooth manifold, and let $\omega$ be a closed differential form of degree $k>0$ on $X$. For any $x\in X$, there exists a contractible neighbourhood $U\subset X$ of $x$ (i.e. $U$ is homotopy equivalent to a single point), with inclusion map $$\iota\colon U\hookrightarrow X.$$ To construct such a neighbourhood, take for example an open ball in a coordinate chart around $x$. Because of the homotopy invariance of de Rham cohomology, the $k$th de Rham cohomology group ${\rm H}^k(U)$ is isomorphic to that of a point; in particular, $$ {\rm H}^k(U)=0\quad\hbox{for all $k>0$}. $$ Since $d(\iota^*\omega)=\iota^*(d\omega)=0$, this implies that there exists a $(k-1)$-form $\eta$ on $U$ such that $d\eta=\iota^*\omega$. In the case where $X$ is a contractible manifold, such an $\eta$ exists globally since we can choose $U=X$ above.