Let $T_n$ denote the $n$-th Chebyshev polynomial and define $$f_n(x,y,z):=T_n(x)+T_n(y)+T_n(z)\;\;\;\text{ and}$$ $$Z_n:=\mathcal{Z}(f_n) \subseteq \mathbb{R}^3,$$ the Banchoff-Chmutov surface, where in general, $\mathcal{Z}(f_1,\ldots,f_k)$ denotes the zero set of polynomials $f_1,\ldots,f_k$, i.e. $\{(x,y,z) \in \mathbb{R}^3; f_1(x,y,z)=\ldots=f_k(x,y,z)=0\}$.

Let us prove, that this is a surface. By the implicit function theorem, it suffices to prove that the points, where $[D_x{f_n},D_y{f_n},D_z{f_n}]$ is zero, do not lie in $Z_n$ (here $D_x$ is just the partial derivative). This is quivalent to showing that the set $$\mathcal{Z}(f_n,D_xf_n,D_yf_n,D_zf_n)=\mathcal{Z}(T_n(x) + T_n(y) + T_n(z),D_xT_n(x),D_yT_n(y),D_zT_n(z))$$ is empty. This can be done by using (from wiki page) $D_xT_n(x) = nU_{n-1}(x)$ and Pell's equation $T_n(x)^2 - (x^2 - 1)U_{n-1}(x)^2 = 1$, to obtain $\mathcal{Z}(1 + 1 + 1) = \emptyset$.

Let us observe the height function $Z_n \rightarrow \mathbb{R}$, $(x,y,z) \mapsto ax + by + cz = [a,b,c][x,y,z]^t$. It is linear, so its derivative is $[a,b,c] :T_pZ_n \rightarrow T_p\mathbb{R} = \mathbb{R}$. Its critical points are therefore those, where the tangent plane $T_pZ_n$ has normal $[a,b,c]$. But the tangent plane of $\mathcal{Z}(f)$ always has normal $[D_xf,D_yf,D_zf]$. Thus the critical points of our height function are those $x,y,z$ where $[D_xf_n,D_yf_n,D_zf_n]=[a,b,c]$, i.e. the critical points are $$\mathcal{Z}(f_n,T_n(x) - a,T_n(y) - b,T_n(z) - c).$$ Now I don't know how to check if these critical points are nondegenerate. I don't even have local parametrizations to work with.

Question: Can one calculate the homology $H_\ast(Z_n)$ by using the elementary methods from Morse theory (i.e. structural theorem, handle decomposition, Morse inequalities, Morse complex)?

  • 5
    $\begingroup$ Check out math.stackexchange.com/questions/46212/… for lots of cool pictures and code to generate these. $\endgroup$
    – Igor Rivin
    Aug 28, 2012 at 0:24
  • 2
    $\begingroup$ Shouldn't the formula be $[Dxf_n,D_yf_n,D_zf_n]=[a,b,c]$ up to multiplication by a constant? Also, $D_xf_n=nU_{n-1}(x)$, not $T_n(x)$, so the big equation is wrong. Finding a set of local coordinates should not be hard: Assuming without loss of generality that $c\neq 0$, then $x$ and $y$ will form a set of local coordinates. If you can find these critical points and check that they are nondegenerate and such, the only difficulty in computing the homology will be the compactness of the surface. If I understand correctly it will be compact for $n$ even and not for $n$ odd. $\endgroup$
    – Will Sawin
    Aug 28, 2012 at 0:30
  • 1
    $\begingroup$ @Igor WOW! WOW! $\endgroup$ Aug 28, 2012 at 14:02

1 Answer 1


The function $h(x,y,z)=z$, corresponding to $a=b=0$ will do the trick. Assume $n$ is even. Using a bit of Morse theory I will show that

$$ \chi(Z_n)= \frac{n^2(3-n)}{2}. \tag{1} $$

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

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

Note that

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

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

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

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

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

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

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

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 and the saddle points are exactly the points in $C_0$; for a proof, click here.

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

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

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

(To explain the above equality note that there are $n$ independent possible choices for $z$, the zeros of $T_n$. Then we need to choose integers $(j,k)$ in $[1,n-1]\times [1,n-1]$ so that exactly one of them is odd. The number of pairs $(j,k)$ with $j$ odd, $k$ even and $1\leq j,k\leq n-1$ is $\frac{n}{2}\times \frac{n-2}{2}$. We have an equal number of pairs $(j,k)$, $1\leq j,k\leq n-1$ with $j$ even and $k$ odd.)

We conclude that the Euler characteristic of $Z_n$ is

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

For $n=2$ we get that $Z_2$ is a sphere. This agrees with the pictures on the site indicated by I. Rivin.

Update. 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 zet

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

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.

Update To better explain my answer to Leon, below is a rendition of $Z_6$ where one can see three layers, yellow, green and blue.

alt text

The equality (1) predicts

$$\chi(Z_6)=\frac{6^2(3-6)}{2}=-54. $$

One can verify this directly as follows. Consider the $1$-dimensional simplicial complex $C$ embedded in $\mathbb{R}^3$ depicted below

alt text

The surface $Z_6$ is homeomorphic to the boundary of a thin tubular neighborhood $T$ of this set in $\mathbb{R}^3$. (Think of the edges as thin spaghetti.) For this reason

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

Let me give an alternate proof of the equality

$$\chi(C)=-27. \tag{E} $$

The complex $C$ has $8$ Green vertices of degree $3$, $12$ Red vertices of degree $4$, $6$ Blue vertices of degree $5$ and a unique Black vertex of degree $6$. Thus the number $V$ of vertices of this complex is

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

The number $E$ of edges is half the sum of degrees of vertices. Thus

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


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

  • $\begingroup$ Hmm, judging from the pictures, if I'm not mistaken, for $n=2(k+1)$, we have a cube (sphere) with $k^2$ holes drilled in each direction $x,y,z$, so I would say that $Z_n$ has genus $k^2+k^2(k+1)+k^2(k+1)=k^2(2k+3)$, for$k\in\mathbb{N}$. For $n=2,4,6,8$, this gives genus $0,5,28,81$. Thus the Euler characteristic is $\chi=2-2g=2-2k^2(2k+3)$, i.e. for $n=2,4,6,8$, we get $2,-8,-54,-160$. But $n^2-\frac{n^3}{4}$ gives $2,0,-18,-64,-150$. Where did I go wrong? $\endgroup$
    – Leo
    Aug 29, 2012 at 13:58
  • $\begingroup$ The picture is more complicated than that. For $n=2(k+1)$ there are $(k+1)$-layers the look like the top face of the cube. $\endgroup$ Aug 29, 2012 at 15:24
  • $\begingroup$ Umm, $Z_4$ has just one hole drilled in each of the three directions, and hopefully, we agree that it has genus $5$, hence $\chi=-8$. However, $n^2-\frac{n^3}{4}\neq-8$ for $n=4$. Something must be wrong. $\endgroup$
    – Leo
    Aug 29, 2012 at 16:17
  • $\begingroup$ Thanks for your comment. There is an errorin $C_0$. The answer is $n^2(3-n)/2$. $\endgroup$ Aug 29, 2012 at 17:51
  • $\begingroup$ If $n=4$ we obtain according to the corrected formula that $\chi(Z_4)= -8$ so that the genus of $Z_4$ is $5$. $\endgroup$ Aug 29, 2012 at 18:11

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