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Let $T_n$ denote the $n$-th Chebyshev polynomial and define $f_n(x,y,z)\!:=\!T_n(x)\!+\!T_n(y)\!+\!T_n(z)$ and $$Z_n:=\mathcal{Z}(f_n) \subseteq \mathbb{R}^3,$$ the Bachoff-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: How can one calculate the Euler characteristic $\chi(Z_n)$ by using elementary methods from Morse theory (i.e. handle decomposition, Morse inequalities, Morse complex)?

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A nice solution, due to L. Nicolaescu, can be found here and here. – Leon Sep 2 '12 at 21:06

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