# The Unapologetic Mathematician

## 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 $\displaystyle\iota_1^*\omega-\iota_0^*\omega=d(I\omega)+I(d\omega)$

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 $\displaystyle[[I\omega](p)](v_1,\dots,v_{k-1})=\int\limits_0^1[\eta(p,t)](\iota_{t*}v_1,\dots,\iota_{t*}v_{k-1})\,dt$

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}

and \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