This discusses a question posed on MathOverflow by Leon Lampret.
Denote by $T_n$ the Chebyshev 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 Chebyshev 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
$$T_n(x_0)=T_n(y_0)=T_n(z_0)=1.$$
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 the gradient of $f_n$ points in the $z$-direction, i.e.,
$$ T'_n(x)= T_n'(y)=0, \;\; T_n(z)=-T_n(x)-T_n(y) $$
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 $[1,\infty)$. (It is increasing since $T_n'(x)\neq 0$ for $x\geq 1$.)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
$$ C_0=\big\lbrace\, (u_j,u_k,z);\;\;1\leq j\leq n-1,\;j+k\in 2\mathbb{Z}+1,\;\;T_n(z)=0\,\big\rbrace, $$
$$ C_2^+=\big\lbrace \,(u_j,u_k,z);\;\; j,k\in [1,n-1]\cap 2\mathbb{Z}+1,\;\;T_n(z)=2, z>1\,\big\rbrace, $$
$$ C_2^-= \lbrace (u_j,u_k,z);\;\;j,k\in [1,n-1]\cap 2\mathbb{Z}+1,\;\;T_n(z)=2, z<-1\rbrace. $$
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 again 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 us observe that $T_n(z_0)=0$ or $T_n(z_0)=2$. In the first case $T_n'(z_0)\neq 0$ because $T_n$ 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)
 |
| Z_2 |
 |
| Z_6 |
 |
| Z_4 |
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 $Z_6$.
 |
| 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.
$$V= 8+12+6+1=27. $$
The total number $E$ of edges of $C$ is half the sum of degrees of vertices so that
$$ E=\frac{1}{2}( 3\times 8+ 4\times 12+ 5\times 6+ 6\times 1)= \frac{1}{2} (24+48+30+6)=54. $$
Hence
$$\chi(C)= 27-54 =-27. $$
No comments:
Post a Comment