## 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

$$T_n(x_0)=T_n(y_0)=T_n(z_0)=1.$$

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

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

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)

 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.
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

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

Hence

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

## 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


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!