{"id":2228,"date":"2012-05-07T23:07:46","date_gmt":"2012-05-07T15:07:46","guid":{"rendered":"http:\/\/wamath.sinaapp.com\/?p=2228"},"modified":"2012-05-07T23:07:46","modified_gmt":"2012-05-07T15:07:46","slug":"duistermaat-heckman-formula-and-botts-original-idea","status":"publish","type":"post","link":"https:\/\/lttt.vanabel.cn\/?p=2228","title":{"rendered":"Duistermaat-Heckman Formula and Bott&#8217;s Original Idea"},"content":{"rendered":"<p>\\section{Duistermaat-Heckman Formula}<br \/>\nIn this section, we consider the case of that $\\left(M^{2l},\\omega\\right)$ is a symplectic manifold. Let $(M,\\omega)$ be a symplectic manifold with $\\omega$ is a symplectic structure. It means<br \/>\n\\begin{enumerate}<br \/>\n  \\item $\\omega$ is a non-singular 2-form. i.e. If for any $Y\\in \\Gamma(TM)$ there always have $\\omega(X,Y)=0$, then $X=0$.<br \/>\n  \\item $\\rd \\omega=0$.<br \/>\n\\end{enumerate}<!--more--><br \/>\nAssume $\\omega$ is $S^1$-invariant and the $S^1$-action on $(M,\\omega)$ is Hamiltonian. It means there exists a smooth function $\\mu\\in C^{\\infty}(M)$ such that $\\rd \\mu=i_K\\omega$. the $\\mu$ is called with momentum map. We still assume that the zero set of $K$ is discrete.<br \/>\n\\begin{thm}<br \/>\n\\[\\int_{M}\\exp\\left(\\sqrt{-1}\\mu\\right)\\frac{\\omega}{(2\\pi)^ll!}<br \/>\n=\\left(\\sqrt{-1}\\right)^l\\sum_{p\\in\\mathrm{zero}(K)}\\frac{\\exp\\left(\\sqrt{-1}\\mu(p)\\right)}{\\lambda(p)}.\\]<br \/>\n\\end{thm}<br \/>\n\\begin{proof}<br \/>\nSince<br \/>\n\\begin{align*}<br \/>\n\\rd_{K}(\\omega-\\mu)=&amp;(\\rd+i_K)(\\omega-\\mu)\\\\<br \/>\n=&amp;\\rd\\omega+i_k\\omega-\\rd\\mu-i_K\\mu\\\\<br \/>\n=&amp;\\rd\\omega+i_K\\omega-i_K\\omega=\\rd\\omega=0,<br \/>\n\\end{align*}<br \/>\none sees that $\\exp\\left(\\sqrt{-1}\\mu-\\sqrt{-1}\\omega\\right)$ is also $\\rd_K$-closed. Using Berline-Vergne localization formula, one has<br \/>\n\\[<br \/>\n\\int_{M}\\exp\\left(\\sqrt{-1}\\mu-\\sqrt{-1}\\omega\\right)=<br \/>\n(2\\pi)^l\\sum_{p\\in\\mathrm{zero}(K)}\\frac{\\exp\\left(\\sqrt{-1}\\mu(p)\\right)}{\\lambda(p)}<br \/>\n\\]i.e.,<br \/>\n\\begin{align*}<br \/>\n\\int_M\\exp\\left(\\sqrt{-1}\\mu\\right)\\left(-\\sqrt{-1}\\right)^l\\frac{\\omega^l}{l!}<br \/>\n=&amp;(2\\pi)^l\\sum_{p\\in\\mathrm{zero}(K)}\\frac{\\exp\\left(\\sqrt{-1}\\mu(p)\\right)}{\\lambda(p)}\\\\<br \/>\n\\int_M\\exp\\left(\\sqrt{-1}\\mu\\right)\\frac{\\omega^l}{(2\\pi)^ll!}<br \/>\n=&amp;\\left(-\\sqrt{-1}\\right)^{-l}\\sum_{p\\in\\mathrm{zero}(K)}\\frac{\\exp\\left(\\sqrt{-1}\\mu(p)\\right)}{\\lambda(p)}\\\\<br \/>\n=&amp;\\left(\\sqrt{-1}\\right)^{l}\\sum_{p\\in\\mathrm{zero}(K)}\\frac{\\exp\\left(\\sqrt{-1}\\mu(p)\\right)}{\\lambda(p)}<br \/>\n\\end{align*}<br \/>\n\\end{proof}<br \/>\nAbout the generalization of above theorem in the case where the zero set of $K$ may not be discrete you can see[DH]:  &#8220;J.J.Duistermaat and G.Heckman, Onthevaroation inthe cohomology of the symplectic from of the reduce phase space&#8221;.<br \/>\n\\section{Bott&#8217;s Original Idea}<br \/>\nFrom Bismut&#8217;s lemma, If $\\rd_K\\Omega =0$, one has<br \/>\n\\[<br \/>\n\\int_M\\omega=\\int_M\\omega\\exp\\left(-T\\rd_K\\theta\\right), \\quad for~~any~~T&gt;0.<br \/>\n\\]when $\\mathrm{zero(K)}=\\emptyset $,<br \/>\n\\[<br \/>\n\\int_M\\omega=\\lim_{T\\rightarrow+\\infty}\\int_M\\omega\\exp\\left(-T\\rd_K\\theta\\right)=0.<br \/>\n\\]Bott&#8217;s original  proof is different from above. Now we give a description of Bott&#8217;s idea.<br \/>\nLet $\\omega$ be a $\\rd_K$-closed form on $M$ with $K$ has no zeros on $M$. Since $\\rd_K(\\theta)=|K|^2+\\rd\\theta=|K|^2\\left(1+\\frac{\\rd\\theta}{|K|^2}\\right)$, by $\\frac{1}{1+x}=\\sum_{i=0}^{\\infty}(-1)^ix^i$, we can define<br \/>\n\\[<br \/>\n\\left(\\rd_K\\theta\\right)^{-1}=<br \/>\n\\frac{1}{|K|^2}\\left[\\sum_{i=0}^{\\infty}(-1)^i\\left(\\frac{\\rd\\theta}{|K|^2}\\right)^i\\right].<br \/>\n\\]From $K$ is $S^1$-invariant, the $\\theta$ is also $S^1$-invariant. It shows $\\mathcal{L}_K\\theta=0$, i.e., $\\rd_K^2\\theta =0$. one can verify that<br \/>\n\\begin{align*}<br \/>\n\\rd_K\\left[\\rd_K^{-1}\\wedge\\theta\\wedge\\omega\\right]=&amp;<br \/>\n\\rd_K\\left(\\rd_K\\theta\\right)^{-1}\\wedge\\theta\\wedge\\omega+<br \/>\n\\left(\\rd_K\\right)^{-1}\\theta \\wedge\\rd_K\\theta\\wedge\\omega-\\left(\\rd_k\\theta\\right)^{-1} \\wedge\\theta\\wedge\\rd\\omega\\\\<br \/>\n=&amp;\\left(\\rd_K\\theta\\right)^{-2}\\wedge \\left(\\rd_K\\theta\\right)^2+\\omega-0\\\\<br \/>\n=&amp;\\omega.<br \/>\n\\end{align*}<br \/>\nFrom this formula and the Stokes formula, one can directly calculate<br \/>\n\\[<br \/>\n\\int_M\\omega=0.<br \/>\n\\] <\/p>\n","protected":false},"excerpt":{"rendered":"<p>\\section{Duistermaat-Heckman Formula} In &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"more-link\" href=\"https:\/\/lttt.vanabel.cn\/?p=2228\"> <span class=\"screen-reader-text\">Duistermaat-Heckman Formula and Bott&#8217;s Original Idea<\/span> \u9605\u8bfb\u66f4\u591a &raquo;<\/a><\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[17,12],"tags":[],"class_list":["post-2228","post","type-post","status-publish","format-standard","hentry","category-index-theory","category-mathnotes"],"_links":{"self":[{"href":"https:\/\/lttt.vanabel.cn\/index.php?rest_route=\/wp\/v2\/posts\/2228","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/lttt.vanabel.cn\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/lttt.vanabel.cn\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/lttt.vanabel.cn\/index.php?rest_route=\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/lttt.vanabel.cn\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=2228"}],"version-history":[{"count":0,"href":"https:\/\/lttt.vanabel.cn\/index.php?rest_route=\/wp\/v2\/posts\/2228\/revisions"}],"wp:attachment":[{"href":"https:\/\/lttt.vanabel.cn\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=2228"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/lttt.vanabel.cn\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=2228"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/lttt.vanabel.cn\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=2228"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}