Morse theory and homology of an algebraic surface (example) 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)?
 A: 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.

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

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
$$ E=\frac{1}{2}( 3\times 8 + 4\times 12 + 5\times 6+ 6\times 1)=\frac{1}{2} (24+48+30+6)=54. $$
Hence 
$$\chi(C)= 27=54=-27.  $$
