# proof of closed differential forms on a simple connected domain

###### lemma 1.

Let $\gamma_{0}$ and $\gamma_{1}$ be two regular homotopic curves in $D$ with the same end-points. Let $\sigma\colon[0,1]\times[0,1]\to D$ be the homotopy between $\gamma_{0}$ and $\gamma_{1}$ i.e.

 $\sigma(0,t)=\gamma_{0}(t),\qquad\sigma(1,t)=\gamma_{1}(t).$

Notice that we may (and shall) suppose that $\sigma$ is regular too. In fact $\sigma([0,1]\times[0,1])$ is a compact subset of $D$. Being $D$ open this compact set has positive distance from the boundary $\partial D$. So we could regularize $\sigma$ by mollification leaving its image in $D$.

Let $\omega(x,y)=a(x,y)\,dx+b(x,y)\,dy$ be our closed differential form and let $\sigma(s,t)=(x(s,t),y(s,t))$. Define

 $F(s)=\int_{0}^{1}a(x(s,t),y(s,t))x_{t}(s,t)+b(x(s,t),y(s,t))y_{t}(s,t)\,dt;$

we only have to prove that $F(1)=F(0)$.

We have

 $F^{\prime}(s)=\frac{d}{ds}\int_{0}^{1}ax_{t}+by_{t}\,dt$
 $=\int_{0}^{1}a_{x}x_{s}x_{t}+a_{y}y_{s}x_{t}+ax_{ts}+b_{x}x_{s}y_{t}+b_{y}y_{s% }y_{t}+by_{ts}\,dt.$

Notice now that being $a_{y}=b_{x}$ we have

 $\frac{d}{dt}\left[ax_{s}+by_{s}\right]=a_{x}x_{t}x_{s}+a_{y}y_{t}x_{s}+ax_{st}% +b_{x}x_{t}y_{s}+b_{y}y_{t}y_{s}+by_{st}$
 $=a_{x}x_{s}x_{t}+b_{x}x_{s}y_{t}+ax_{ts}+a_{y}y_{s}x_{t}+b_{y}y_{s}y_{t}+by_{ts}$

hence

 $F^{\prime}(s)=\int_{0}^{1}\frac{d}{dt}\left[ax_{s}+by_{s}\right]\,dt=\left[ax_% {s}+by_{s}\right]_{0}^{1}.$

Notice, however, that $\sigma(s,0)$ and $\sigma(s,1)$ are constant hence $x_{s}=0$ and $y_{s}=0$ for $t=0,1$. So $F^{\prime}(s)=0$ for all $s$ and $F(1)=F(0)$. ∎

###### Lemma 2.

Let us fix a point $(x_{0},y_{0})\in D$ and define a function $F\colon D\to\mathbb{R}$ by letting $F(x,y)$ be the integral of $\omega$ on any curve joining $(x_{0},y_{0})$ with $(x,y)$. The hypothesis assures that $F$ is well defined. Let $\omega=a(x,y)\,dx+b(x,y)\,dy$. We only have to prove that $\partial F/\partial x=a$ and $\partial F/\partial y=b$.

Let $(x,y)\in D$ and suppose that $h\in\mathbb{R}$ is so small that for all $t\in[0,h]$ also $(x+t,y)\in D$. Consider the increment $F(x+h,y)-F(x,y)$. From the definition of $F$ we know that $F(x+h,y)$ is equal to the integral of $\omega$ on a curve which starts from $(x_{0},y_{0})$ goes to $(x,y)$ and then goes to $(x+h,y)$ along the straight segment $(x+t,y)$ with $t\in[0,h]$. So we understand that

 $F(x+h,y)-F(x,y)=\int_{0}^{h}a(x+t,y)dt.$

For the integral mean value theorem we know that the last integral is equal to $ha(x+\xi,y)$ for some $\xi\in[0,h]$ and hence letting $h\to 0$ we have

 $\frac{F(x+h,y)-F(x,y)}{h}=a(x+\xi,y)\to a(x,y)\quad h\to 0$

that is $\partial F(x,y)/\partial x=a(x,y)$. With a similar argument (exchange $x$ with $y$) we prove that also $\partial F/\partial y=b(x,y)$. ∎

###### Theorem.

Just notice that if $D$ is simply connected, then any two curves in $D$ with the same end points are homotopic. Hence we can apply Lemma 1 and then Lemma 2 to obtain the desired result. ∎

Title proof of closed differential forms on a simple connected domain ProofOfClosedDifferentialFormsOnASimpleConnectedDomain 2013-03-22 13:32:49 2013-03-22 13:32:49 paolini (1187) paolini (1187) 9 paolini (1187) Proof msc 53-00 SubstitutionNotation