Seventh assignment - Solutions

1. Spivak, no. 4-23:

   For $R>0,$ and $n$ an integer, define the singular $1$-cube $c_{R,n}:[0,1]\longrightarrow \mathbb{R}^2\setminus\{0\}$ by

   $$c_{R,n}(t)=(R\cos{2\pi nt}, R\sin{2\pi nt}).$$

   
   
   Show that there is a singular $2$-cube $c:[0,1]^2\longrightarrow \mathbb{R}^2\setminus\{0\}$ such that $c_{R_1,n}-c_{R_2,n}=\partial c.$

   Solution: Define $c:[0,1]^2\longrightarrow \mathbb{R}^2$ by

   $$c(t,s)=((sR_1+(1-s)R_2)\cos{2\pi nt}, (sR_1+(1-s)R_2)\sin{2\pi nt}).$$

   
   
   This looks simpler in complex notation for $\mathbb{R}^2=\mathbb{C}$ as

   $$c(t,s)=(sR_1+(1-s)R_2)e^{2\pi int}.$$

   
   
   Its boundary consists of four 1-cubes, up to sign. Namely:

   $$\partial c(t)=c(1,t)-c(0,t)-c(t,1)+c(t,0)=c_{R_1,n}(t)-c_{R_2,n}(t)$$

   
   
   since the last two 1-cubes are actually the same.

2. Spivak, no. 4-24:

   If $c$ is a singular $1$-cube in $\mathbb{R}\setminus\{0\}$ with $c(0)=c(1),$ show that there is an integer $n$ such that $c-c_{1,n}=\partial c^{(2)}$ for some $2$-chain $c^{(2)}.$ Hint: First partition $[0,1]$ so that each $c([t_{i-1},t_i])$ is contained on one side of some line through $0.$

   Solution: We define a continuous function $\theta:[0,1]\longrightarrow \mathbb{R}$ so that

   $$c(t)=(\vert c(t)\vert\cos(2\pi \theta (t)),\vert c(t)\vert\sin(2\pi \theta (t))) \forall t\in[0,1]. \in\mathbb{R}^2\setminus\{0\}.$$ (1)

   
   
   To do so, let $L$ be the line through the origin and $c(0),$ then let $L^\pm$ be the two halves of $L\setminus\{0\}$ so that $c(0)\in L^+.$ As the hint suggest we can choose $t_0=0$ and then choose $t_1$ to be the first point at which $c(t)\in L^-,$ $t_2$ the first later point at which $c(t)\in L^+$ and so on. At the last step, either $t_{2N}=1$ or else $t_{2N}<1$ and there is no later point at which $c(t)\in L^-;$ in this case set $t_{2N+1}=1.$ This gives a partition of $[0,1]$ so that on $[t_{2i},t_{2i+1})$ $c(t)$ does not hit $L^-$ whereas on $[t_{2i+1},t_{2i+2})$ it does not hit $L^+.$ Set

   $$c(t_0)=(\vert c(t_0)\vert\cos(2\pi \theta_0),\vert c(t_0)\vert\sin(2\pi \theta_0))$$

   
   
   for some $\theta _0\in[0,1).$ Now for each $i$ there are unique continuous functions $\tau_{i}:[t_{2i},t_{2i+1})\longrightarrow [-\frac12,\frac12]$ and $\tau'_{i}:[t_{2i+1},t_{2i+2})\longrightarrow (0,1)$ so that

   $$c(t)=(\vert c(t)\vert\cos(2\pi(\theta_0+\tau _i(t))),\vert c(t_0)\vert\sin(2\pi (\theta_0+\tau _i(t)))) t\in[t_{2i},t_{2i+1}),$$

   $$c(t)=(\vert c(t)\vert\cos(2\pi (\theta_0+\tau' _i(t))),\vert c(t_0)\vert\sin(2\pi (\theta_0+\tau '_i(t)))) t\in[t_{2i+1},t_{2i+2}).$$

   
   
   It follows that

   $$\lim_{t\uparrow t_{2i+1}}=\tau'_i(t_{2i+1})+n_i \lim_{t\uparrow t_{2i+1}}=\tau_{i+1}(t_{2(i+1)})+n_i$$

   $$n_i, n'_i\in\mathbb{Z}.$$

   
   
   Now the function $\theta:[0,1]\longrightarrow \mathbb{R}$ can be defined by insisting that $\theta =\theta_0+\tau_0$ on $[0,t_1)$ and that only later intervals $\theta(t)=\theta_0+\tau _i+q_i$ or $\theta (t)=\theta_0+\tau _i+q'_i$ for some integers $q_i,$ $q;_i$ and that it be be continuous at each $t_i.$ Necessarily $\theta(1)=\theta(0)+n,$ $n\in\mathbb{Z}.$ Clearly (2) holds.

   Now define a 2-cube by

   $$\begin{multline*} d(t,s)=((s\vert c(t)\vert+(1-s))\cos(2\pi(s\theta(t)+(1-s)nt))... ...1-s))\sin(2\pi(s\theta(t)+(1-s)nt))), (s,t)\in[0,1]\times[0,1]. \end{multline*}$$
   
   
   This $2$-cube takes values in $\mathbb{R}^2\setminus\{0\}$ and

   $$\partial d=c_{1,n}(t)-c(t)$$

   
   
   since the two terms from $t=0$ and $t=1$ cancel. Thus $c^{(2)}=-d$ has the desired property

   [Notice this actually gives a cube $d(t,-s),$ with boundary $c-c_{1,n};$ you were only asked for a chain $c^{(2)}$ so you could do it in pieces.]

3. Spivak, no. 4-25:

   (Independence of parameterization.) Let $c$ be a singular $k$-cube and $p:[0,1]^k\longrightarrow [0,1]^k$ a $1$-$1$ function such that $p([0,1]^k)=[0,1]^k$ and $\det p'(x)\ge0$ for all $x\in[0,1]^k.$ If $\omega$ is a $k$-form, show that

   $$\int_c\omega =\int_{c\circ p}\omega .$$

   
   

   Solution: By definition

   $$\int_c\omega=\int_{[0,1]^k}c^*\omega=\int{[0,1]^k}f, c^*\omega=f(x)dx^1\wedge\cdots\wedge dx^k$$

   
   
   and similarly
   $$\begin{multline*} \int_{c\circ p}\omega=\int_{[0,1]^k}(c\circ p)^*\omega=\int_{[... ...f(x)dx^1\wedge\cdots\wedge dx^k) =gdx^1\wedge\cdots\wedge dx^k. \end{multline*}$$
   
   
   Thus

   $$g=(f\circ p)\vert\det p'\vert=(f\circ p)\det p'$$

   
   
   since $\det p'\ge0$ by assumption. Now, since $p$ is $1$-$1$ it follows from the change of variable formula that

   $$\int_{[0,1]^k}g=\int_{[0,1]^k}(f\circ p)\vert\det p'\vert=\int_{[0,1]^k}f.$$

   
   

4. Spivak, no. 4-26:

   Show that $\int{c_{R,n}}d\theta=2\pi n$ and use Stokes' theorem to conclude that $c_{R,n}\not=\partial c$ for any $2$-chain in $\mathbb{R}^2\setminus\{0\}.$ Here

   $$d\theta=\frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy$$

   
   
   is a smooth $1$-form on $\mathbb{R}\setminus\{0\}$ which, despite the notation, is not exact.

   Solution: If we substitute the definition of $c_{R,n},$ $x=R\cos{2\pi nt},$ $y=R\sin{2\pi nt}$ we find that

   $$c_{R,n}^*d\theta=-\sin(2\pi nt)\frac{d\cos{2\pi nt}}{dt}+ \cos(2\pi nt)\frac{d\sin{2\pi nt}}{dt}=2\pi ndt$$

   
   
   Thus

   $$\int_{c_{R,n}}d\theta=\int_{[0,1]}c_{R,n}^*d\theta=\int_0^12\pi ndt=2\pi n.$$

   
   

   It follows that $c_{R,n}\not=\partial c$ for any $2$-chain $c$ in $\mathbb{R}^2\setminus\{0\},$ if $n\not=0,$ since by Stokes' theorem this would imply

   $$\int_{c_{R,n}}d\theta=\int_{\partial c}d\theta=\int_{c}d(d\theta)=0$$

   
   
   since $d\theta$ is a well-defined closed form.

5. Spivak, no. 4-27:

   Show that the integer $n$ of Problem 4-24 is unique. This integer is called the winding number of $c$ around $0.$

   Solution: If $c$ is a closed $1$-cube in $\mathbb{R}^2$ then with $n$ from Problem 4-24 above, using Stokes' theorem

   $$\int_cd\theta=\int_{c_{1,n}}d\theta+\int_{\partial c^{(2)}}d\theta=2\pi n$$

   
   
   which shows that $n$ is unique, since it is determined by integration over $c.$

6. Spivak, no. 4-28:

   Recall that the set of complex number, $\mathbb{C},$ is simply $\mathbb{R}^2$ with $(a,b)=a+bi.$ If $a_1,\dots,a_n\in\mathbb{C}$ let $f:\mathbb{C}\longrightarrow \mathbb{C}$ be

   $$f(z)=z^n+a_1z^{n-1}+\cdots+a_n.$$

   
   
   Define the singular $1$-cube $c_{R,f}:[0,1]\longrightarrow \mathbb{C}\setminus\{0\}$ by $c_{R,f}=f\circ c_{R,1}$ and the singular $2$-cube $c$ by

   $$c(s,t)=t\cdot c_{R,n}(s)+(1-t)c_{R,f}(s).$$

   
   
   1. Show that $\partial c=c_{R,f}-c_{R,n}$ and that

      $$c([0,1]\times[0,1])\subset\mathbb{C}\setminus\{0\}$$

      
      
      if $R$ is large enough.
   2. Using Problem 4-26 above, prove the Fundamental theorem of algebra: Every polynomal $=z^n+a_1z^{n-1}+\cdots+a_n$ with $a_i\in\mathbb{C}$ has a root in $\mathbb{C}.$

   Solution

   1. By definition $\partial c$ is the sum of four $1$-cubes

      $$\partial c=c(s,0)-c(s,1)-c(0,s)+c(1,s).$$

      
      
      The last two are the same $1$-cube with opposite signs, since both $c_{R,n}$ and $c_{R,f}$ are closed.
   2. We can estimate $f$ by
      $$\begin{multline} \vert f(z)-z^n\vert=\vert a_1z^{n-1}+\cdots+a_n\vert\\ \le\ver... ...ert a_n\vert \le C\vert z\vert^{n-1},\text{ in }\vert z\vert\ge1. \end{multline}$$
      
      
      This shows that $f(z)\not=0$ on $c_{R,1}$ for $R$ large enough. Moreover for each $s\in[0,1]$ consider
      $$\begin{multline*} t\longmapsto tc_{R,f}(s)+(1-t)c_{R,n}(s)\\ =tf(Re^{2\pi is})+(1-t)Re^{2\pi ins} =tR^n(e^{2\pi ins}+\tau)+(1-t)Re^{2\pi ins}. \end{multline*}$$
      
      
      Here $\vert\tau\vert\le C/R$ is very small by (2). Thus the straight line given by varying $t\in[0,1]$ does not pass through the origin, so $c([0,1]\times[0,1])\subset\mathbb{C}\setminus\{0\}.$
   3. Since $c_{R,f}$ and $c_{R,n}$ differ by an exact $2$-chain in $\mathbb{C}\setminus\{0\}$ it follows from Problem 4-16 that

      $$\int{c_{R,f}}d\theta=2\pi n.$$

      
      

      Now suppose, to get a contradiction, that $f(z)\not=0$ for all $z\in\mathbb{C}.$ In particular, $a_n\not=0$ since otherwise $f(0)=0.$ Then

      $$d(s,t)=f\circ c_{Rs,1}(t)\in\mathbb{C}\setminus\{0\} \forall (t,s)\in[0,1]\times[0,1].$$

      
      
      Since $\partial d=c_{R,f}-c'$ where $c'(t)=a_n$ is constant it follows from Stokes' theorem that $2\pi n=\int_{c'}d\theta=0$ which is a contradiction. This proves the Fundamental Theorem of Algebra, that $f(z)$ must have a zero.