The Unapologetic Mathematician

Mathematics for the interested outsider

The Poincaré Lemma (proof)

We can now prove the Poincaré lemma by proving its core assertion: there is a chain homotopy between the two chain maps \iota_0^* and \iota_1^* induced by the inclusions of M into either end of the homotopy cylinder M\times[0,1]. That is, we must define a map I:\Omega^k(M\times[0,1])\to\Omega^{k-1}(M) satisfying the equation


Before defining the map I, we want to show that any k-form \omega on the homotopy cylinder can be uniquely written as \omega'+dt\wedge\eta, where \omega' is a k-form and \eta is a k-1-form, both of which are “constant in time”, in a certain sense. Specifically, we can pull back the canonical vector field \frac{d}{dt} on [0,1] along the projection M\times[0,1]\to[0,1] to get a “time” vector field T on the cylinder. Then we use the interior product to assert that \iota_T\omega'=0 and \iota_T\eta=0.

But this should be clear, if we just define \eta=\iota_T\omega then we definitely have \iota_T\eta=\iota_T\iota_T\omega=0, since interior products anticommute. Then we can define \omega'=\omega-dt\wedge\eta, and calculate \iota_T\omega'=\iota_T\omega-\iota_T(dt\wedge\eta)=\eta-\eta=0, since the pairing of dt with T is 1. The uniqueness should be clear.

So now let’s define


where \iota_t is the inclusion of M into the homotopy cylinder sending p to (p,t).

Now to check that this is a chain homotopy, which is purely local around each point p\in M. This means that we can pick some coordinate patch (U,x) on M, which lifts to a coordinate patch (U\times[0,1],(\bar{x},t)) on M\times[0,1], where \bar{x}=x\circ\pi_M. Since everything in sight is linear we will consider two cases: \omega=fd\bar{x}^I, where I is some multi-index of length k; and \omega=fdt\wedge d\bar{x}^I, where I is some multi-index of length k-1.

In the first case we have I\omega=0, while d\omega=df\wedge d\bar{x}^I, which we can write as a bunch of terms not involving t at all plus \frac{\partial f}{\partial t}dt\wedge d\bar{x}^I. Therefore we calculate:

\displaystyle\begin{aligned}{}[[I(d\omega)](p)]\left(\frac{\partial}{\partial x^{j_1}},\dots,\frac{\partial}{\partial x^{j_k}}\right)&=\int\limits_0^1\frac{\partial f}{\partial t}\bigg\vert_{(p,t)}[d\bar{x}^I(p,t)]\left(\iota_{t*}\frac{\partial}{\partial x^{j_1}},\dots,\iota_{t*}\frac{\partial}{\partial x^{j_k}}\right)\,dt\\&=\int\limits_0^1\frac{\partial f}{\partial t}\bigg\vert_{(p,t)}[dx^I(p)]\left(\frac{\partial}{\partial x^{j_1}},\dots,\frac{\partial}{\partial x^{j_k}}\right)\,dt\\&=\left(\int\limits_0^1\frac{\partial f}{\partial t}\bigg\vert_{(p,t)}\,dt\right)[dx^I(p)]\left(\frac{\partial}{\partial x^{j_1}},\dots,\frac{\partial}{\partial x^{j_k}}\right)\\&=(f(p,1)-f(p,0))[dx^I(p)]\left(\frac{\partial}{\partial x^{j_1}},\dots,\frac{\partial}{\partial x^{j_k}}\right)\\&=[[\iota_1^*\omega](p)-[\iota_0^*\omega](p)]\left(\frac{\partial}{\partial x^{j_1}},\dots,\frac{\partial}{\partial x^{j_k}}\right)\end{aligned}

and we conclude that \iota_1^*\omega-\iota_0^*\omega=I(d\omega)+d(I\omega), as asserted.

Now, as to the other side. This time, since \iota_{t_0}^*(dt)=0 for any t_0\in[0,1], we know that both terms on the left hand side of the chain homotopy equation is zero. Meanwhile, we calculate

\displaystyle\begin{aligned}{}[I(d\omega)](p)&=\left[I\left(-\sum\limits_{\alpha=1}^n\frac{\partial f}{\partial\bar{x}^\alpha}dt\wedge d\bar{x}^\alpha\wedge d\bar{x}^I\right)\right](p)\\&=-\sum\limits_\alpha\left(\int\limits_0^1\frac{\partial f}{\partial\bar{x}^\alpha}\bigg\vert_{(p,t)}\,dt\right)dx^\alpha\wedge dx^I\end{aligned}


\displaystyle\begin{aligned}{}[d(I\omega)](p)&=d\left(\left(\int\limits_0^1f(p,t)\,dt\right)dx^I\right)\\&=\sum\limits_\alpha\frac{\partial}{\partial x^\alpha}\left(\int\limits_0^1f(p,t)\,t\right)dx^\alpha\wedge dx^I\end{aligned}

so I(d\omega)+d(I\omega)=0 as well, just as asserted.

December 3, 2011 Posted by | Differential Topology, Topology | Leave a comment



Get every new post delivered to your Inbox.

Join 366 other followers