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

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


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

Wednesday, August 22, 2012

Bill Thurston has passed away

I have found out that Bill Thurston has passed away last night after a fight with melanoma.            This is  a premature closing of an amazing and transformative  chapter in the history of mathematics.      Like all the greats,    there won't  be anyone like  Thurston.     Rest in peace!

Thursday, August 16, 2012

Invariants of symmetric matrices

$\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\eS}{\mathscr{S}}$ $\DeclareMathOperator{SO}{SO}$  $\newcommand{\bZ}{\mathbb{Z}}$ $\newcommand{\eQ}{\mathscr{Q}}$ $\DeclareMathOperator{\tr}{tr}$

Fix an integer $m>2$, and denote by $\eS_m$  the  space of real, symmetric   $m\times m$ matrices. The group  $\SO(m)$ of  orthogonal transformations  of $\bR^m$ acts on $\eS_m$ by conjugation. Fix a unit vector  $\eta\in\bR^m$. We obtain a subgroup   $\SO(m-1)$ of $\SO(m)$ consisting of transformations that fix the vector $\eta$.   Denote by $\eQ_m$ the space  of   $\SO(m-1)$-invariant  homogeneous quadratic polynomials  on $\eS_m$.   Then  $\eQ_m$ is spanned by the quadratic polynomials

$$   A\mapsto \tr A^2,\;\; A\mapsto (\tr A)^2, \;\;A\mapsto |A\eta|^2, $$

$$ A\mapsto (A\eta,\eta)^2,\;\; A\mapsto (\tr A)(A\eta,\eta), $$

where   $(-,-)$ denotes the canonical inner product on $\bR^m$.

There is   a simple proof  of this fact due to Robert Bryant.

Tuesday, August 14, 2012

A simple homotopic trick

Here is a simple fact, which seems to be well known to   homotopy theorists.  It might come in handy.

 Suppose that  we have two fibrations

$$ f:Y\to B,\;\; \pi:E\to B, $$

such that  $E$ is contractible.  The pullback to $Y$ of the fibration $\pi: E\to B$  via the map $f$ is a new fibration $g: X\to Y$.   Then the homotopy fiber  $Z$ of $Y\to B$ is homotopic to  $X$.

 Indeed we have a fibration $F:X\to E$ with t homotopy fiber $Z$.  Since $E$ is contractible we  deduce that $X$  is homotopic to $Z$.

This is particularly useful when $X\to Y$ is a principal  $G$-bundle classified by a map $f: Y\to BG$.  The map $f$ can be assumed to be a fibration. The homotopy fiber of $f$ is then $X$.   From the Leray-Serre spectral sequence we obtain a spectral sequnce  converging to the cohomology of $Y$ with $E_2$-term

$$E_2^{p,q}= H^p\bigl(\;BG, \tilde{H}^q(X)\;\bigr), $$

where $\tilde{H}^q(X)$ denotes a local system on $BG$ with stalk $H^q(X)$.

Monday, August 13, 2012

A wonderful math video archive

CUNY has a wonderful video archive going back to the 80s.     What a great idea!  You get to see some of the classics that are no longer among  the living. Hat tip to Richard Hind!