Tuesday, August 28, 2012

On the Banchoff-Chmutov surfaces

This discusses a  question posed on MathOverflow   by Leon Lampret.

Denote by $T_n$ the  Chebysev polynomial of the first kind and degree $n$ uniquely determined by the  equality $\newcommand{\bR}{\mathbb{R}}$

$$ T_n(\cos t)=\cos nt,\;\;\forall t\in\bR. $$

Denote by $U_n$ the  Chebysev polynomial of degree $n$ and of the second kind uniquely determined  by the equality

$$ U_n(\cos t)=\frac{\sin (n+1) t}{\sin t},\;\;t\in\bR. $$

They are related by two equalities

$$T_n'= n U_{n-1}, \;\; T_n(x)^2 -(x^2-1)U_{n-1}(x)^2=1. \tag{1}$$

The polynomial $T_n$ has $n$ distinct real zeros located in $(-1,1)$ and thus, by Rolle's theorem, all its critical points are located in $(-1,1)$.

The polynomial $T_n$ is a solution of the second order linear differential equation

$$ (1-x^2)y''-xy'+n^2 y=0, $$

which shows that all the critical points of $T_n$ are nondegenerate.

The  Banchoff-Chmutov surface   $Z_n$ is defined by

$$Z_n:=\Bigl\lbrace (x,y,z)\in\bR^3;\;\; \underbrace{T_n(x)+T_n(y)+ T_n(z)}_{=: f_n(x,y,z)} =0\;\bigr\rbrace. $$

Remark.  (a)   $Z_n$ is a smooth submanifold of $\bR^3$. To see this, we  rely on the implicit function theorem. Observe that if  $df_n(x_0,y_0,z_0)=0$, then

$$ T_n'(x_0)=T_n'(y_0)=T_n'(z_0) =0. $$

In particular (1) implies

$$U_{n-1}(x_0)=U_{n-1}(y_0)=U_{n-1}(z_0)=0. $$

Invoking (1) again we deduce


This  shows that $(x_0,y_0,z_0)\not\in Z_n$.

(b) If $n$ is even, then $Z_n$ is compact. Indeed, in this case $T_n$ is an even polynomial and

$$\lim_{|x|\to\infty} T_n(x)=\infty. $$

This implies that $Z_n$ is  bounded, thus compact since it is obviously closed.

Assume that $n$ is   even  and consider the function

$$f: Z_n\to \bR,\;\; h(x,y,z)= z. $$

The critical points of  $h$.  A point $(x,y,z)$ on $Z_n$ is critical for $h$ iff

$$  T'_n(x)= T_n'(y)=0, \;\; T_n(z)=-T_n(x)-T_n(y) $$

Now the critical points  of $T_n$ are all located in the interval $[-1,1]$ and can be  easily determined from the defining equality

$$ T_n( \cos t) = \cos nt, \;\;t\in [0,\pi], \tag{A} $$

so that

$$ T_n'(\cos t) = n\frac{\sin nt}{\sin t} $$

This  nails the critical points  of $T_n$ to

$$u_k = \cos \frac{k\pi}{n},\;\; k=1,\dotsc, n-1.$$

Note that

$$ T_n(u_k)= \cos k\pi=(-1)^k $$

so that the critical points of $h$ on the surface $Z_n$  are

$$\bigl\lbrace (u_j,u_k,z);\;\; T_n(z)+(-1)^j+(-1)^k=0,\;\;j,k=1,\dotsc, n-1 \bigr\rbrace.  $$

Now we need to count the solutions of the equations

$$T_n(x)=0,\;\pm 2. $$

The equation $T_n(x)=0$ has $n$ solutions, all situated  in $[-1,1]$.

On the interval $[-1,1]$  we deduce from  (A) that $|T_n|\leq 1$.  The polynomial $T_n$ is even and is increasing on $[1,\infty)$. We conclude that the equation $T_n(x)=-2$ has no solutions, while the equality $T_n(x)=2$ has two solutions.   Thus the critical set of $h$  splits  into three parts

$$ C_0= \lbrace (u_j,u_k,z);\;\;j+k\in 2\mathbb{Z}+1,\;\;T_n(z)=0\rbrace, $$

$$ C_2^+= \lbrace (u_j,u_k,z);\;\;j,k\in 2\mathbb{Z}+1,\;\;T_n(z)=2, z>1\rbrace, $$

$$ C_2^-= \lbrace (u_j,u_k,z);\;\;j,k\in 2\mathbb{Z}+1,\;\;T_n(z)=2, z<-1\rbrace. $$

From the above discussion we deduce  that the points in $C_2^-$ are minima and the points in $C_2^+$ are maxima.

The function $h$ is a Morse function.         Note that $h$  is defined implicitly, by solving for $z$ in the  nonlinear equation

$$  T_n(x)+ T_n(y) + T_n(z)=0. \tag{2} $$

Suppose that $(x_0,y_0,z_0)$ is a critical point of $h$,  $x_0=u_j$, $y_0=u_k$. Differentiating (2) near this critical point we deduce $\newcommand{\pa}{\partial}$

$$ \frac{\pa z}{\pa x}T_n'(z) +T_n'(x) =0,\;\;  \frac{\pa z}{\pa y} T_n'(z) + T_n'(y)=0. $$

 Differentiating the above agian we deduce that

$$\frac{\pa^2 z}{\pa x\pa y}|_{(u_j,u_k)}=0,  \tag{3}$$

$$\frac{\pa^2 z}{\pa x^2}|_{(u_j,u_k)} T_n'(z_0) +T_n''(u_j)= 0,\;\; \frac{\pa^2 z}{\pa y^2}|_{(u_j,u_k)} T_n'(z_0) +T_n''(u_k)= 0. $$

Now let observe that $T_n(z_0)=0$ or $T_n(z_0)=2$.  In the first case $T_n'(z_0)\neq 0$ bcauise $T_n$ has only simple zeros. In the second case $T_n'(z_0)\neq 0$ because  in this case $|z_0|>1$ and $T_n$ has no critical points outside $(-1,1)$.  Hence

$$\frac{\pa ^2 z}{\pa x^2}|_{(u_j,u_k)}= -\frac{T_n''(u_j)}{T_n'(z_0)},\;\;\frac{\pa ^2 z}{\pa y^2}|_{(u_j,u_k)}= -\frac{T_n''(u_k)}{T_n'(z_0)}. \tag{4} $$

The point $u_k$ is a local minimum for $T_n$ if $k$ is odd and a local maximum if $k$ is even.  Moreover, these are nondegenerate critical points  of $T_n$.

This proves that all the critical points of $h$ are nondegenerate. Moreover if $(u_j, u_k)\in C_0$ then the numbers  $T_n''(u_j)$ and $T_n''(u_k)$ have opposite signs  and invoking (3) and (4) we deduce that  this point is a saddle point.

Thus  the Euler characteristic of $Z_n$ is

$$ \chi(Z_n)= {\rm card}\; C_2^+ +{\rm card}\; C_2^- -{\rm card}\; C_0. $$

Now  observe that

$$ {\rm card}\; C_2^\pm = \Bigl(\;{\rm card}\; [1,n-1]\cap (2\mathbb{Z}+1) \;\Bigr)^2= \frac{n^2}{4},$$

$$ {\rm card} \; C_0 = n\Bigl( \frac{n(n-2)}{4}+\frac{n(n-2)}{4}\Bigr)=\frac{n^2(n-2)}{2}. $$

Thus the Euler characteristic of $Z_n$  is

$$\chi(Z_n)=\frac{n^2(3-n)}{2}. \tag{E} $$

Here are  images of $Z_2, Z_4, Z_6$, courtesy of  StackExchange (hat tip to Igor Rivin)


The above computations do not  explain whether $Z_n$ is connected or not.  To check that it suffices to look at the critical values  of the above  function corresponding to saddle points. These critical values are the zeros $\zeta_1<\dotsc <\zeta_n$ of $T_n$. The level set

$$ Z_n\cap \lbrace z=\zeta_k\rbrace $$

is the algebraic curve

$$ T_n(x)+T_n(y)=0. \tag{C} $$

This forces $|x|,|y|\leq 1$ because $T_n(x)> 1$ for $|x|> 1$ and $|T_n(x)\leq 1$ for $|x|\leq 1$.  We can use the  *homeomorphism*

$$[0,\pi]\ni t\mapsto x=\cos t\in [-1,1] $$

to give an alternate description to (C).  It is  the singular curve  inside  the square $[0,\pi]\times [0,\pi]$ with coordinates $(s,t)$ described by

$$\cos ns+ \cos nt =0.$$

This can be easily visualized as the intersection of the square with the  grid

$$ s\pm t\in (2\mathbb{Z}+1)\frac{\pi}{n} $$

which is  connected. Now it is not very difficult to conclude using the Morse theoretic data on $h$ that $Z_n$ is connected.

Example.    The equality (E) predicts that $\chi(Z_6)=-54$. Let us  verify this directly.    Here is a more detailed image of $Z_6$.


We can give an alternate description of $Z_6$ as follows. Consider the $1$-dimensional simplicial complex $C\subset \bR^3$ depicted below

 The surface $Z_6$ is homeomorphic to the boundary of a thin tubular neighborhoof $T$ of $C$ in $\bR^3$ and thus

$$\chi(Z_6)=\chi(\pa T)=2\chi(T)=2\chi(C). $$

Thus formula (E) predicts that

$$\chi(C)= -27.$$

Let us verify this directly.   The vertex set of $C$  consists of

  • 8 Green vertices of degree 3.
  • 12 Red vertices of degree 4.
  • 6 Blue vertices of degree 5.
  • 1 Black vertex of degree 6.
The number $V$ of vertices of $C$ is thus

$$V= 8+12+6+1=27. $$

The total number $E$ of edges of $C$   is half the sum of degrees of vertices so that

$$ E=\frac{1}{2}( 3\times 8+ 4\times 12+ 5\times 6+ 6\times 1)= \frac{1}{2} (24+48+30+6)=54. $$


$$\chi(C)= 27-54 =-27. $$

Post a Comment