Morse theory and homology of an algebraic surface (example) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T21:47:33Z http://mathoverflow.net/feeds/question/105664 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/105664/morse-theory-and-homology-of-an-algebraic-surface-example Morse theory and homology of an algebraic surface (example) Leon Lampret 2012-08-28T00:21:10Z 2012-08-30T14:29:20Z <p>Let $T_n$ denote the $n$-th <a href="http://en.wikipedia.org/wiki/Chebyshev_polynomial" rel="nofollow"><em>Chebyshev polynomial</em></a> 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 <a href="http://mathworld.wolfram.com/ChmutovSurface.html" rel="nofollow"><em>Banchoff-Chmutov surface</em></a>, where in general, $\mathcal{Z}(f_1,\ldots,f_k)$ denotes the zero set of polynomials $f_1,\ldots,f_k$, i.e. ${(x,y,z) \in \mathbb{R}^3; f_1(x,y,z)=\ldots=f_k(x,y,z)=0}$.</p> <p>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 $\mathcal{Z}(1 + 1 + 1) = \emptyset$.</p> <p>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. </p> <p><strong>Question:</strong> 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)?</p> http://mathoverflow.net/questions/105664/morse-theory-and-homology-of-an-algebraic-surface-example/105706#105706 Answer by Liviu Nicolaescu for Morse theory and homology of an algebraic surface (example) Liviu Nicolaescu 2012-08-28T11:27:26Z 2012-08-30T14:29:20Z <p>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</p> <p>$$\chi(Z_n)= \frac{n^2(3-n)}{2}. \tag{1}$$ </p> <p>A point $(x,y,z)$ on $Z_n$ is critical for $h$ iff</p> <p>$$T'_n(x)= T_n'(y)=0, \;\; T_n(z)=-T_n(x)-T_n(y)$$</p> <p>Now the critical points of $T_n$ are all located in the interval $[-1,1]$ and can be easily determined from the defining equality</p> <p>$$T_n( \cos t) = \cos nt, \;\;t\in [0,\pi], \tag{A}$$</p> <p>so that</p> <p>$$T_n'(\cos t) = n\frac{\sin nt}{\sin t}$$</p> <p>This nails the critical points of $T_n$ to </p> <p>$$x_k = \cos \frac{k\pi}{n},\;\; k=1,\dotsc, n-1.$$</p> <p>Note that</p> <p>$$T_n(x_k)= \cos k\pi=(-1)^k$$</p> <p>so that the critical points of $h$ on the surface $Z_n$ are </p> <p>$$\bigl\lbrace (x_j,x_k,z);\;\; T_n(z)+(-1)^j+(-1)^k=0,\;\;j,k=1,\dotsc, n-1 \bigr\rbrace.$$</p> <p>Now we need to count the solutions of the equations</p> <p>$$T_n(x)=0,\;\pm 2.$$</p> <p>The equation $T_n(x)=0$ has $n$ solutions, all situated in $[-1,1]$. </p> <p>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</p> <p>$$C_0= \lbrace (x_j,x_k,z);\;\;j+k\in 2\mathbb{Z}+1,\;\;T_n(z)=0\rbrace,$$</p> <p>$$C_2^+= \lbrace (x_j,x_k,z);\;\;j,k\in 2\mathbb{Z}+1,\;\;T_n(z)=2, z>1\rbrace,$$</p> <p>$$C_2^-= \lbrace (x_j,x_k,z);\;\;j,k\in 2\mathbb{Z}+1,\;\;T_n(z)=2, z&lt;-1\rbrace.$$</p> <p>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 <em>Morse</em> function and the saddle points are exactly the points in $C_0$; for a proof, <a href="http://liviusmathblog.blogspot.com/2012/08/on-banchoff-cmutov-surfaces.html" rel="nofollow">click here.</a></p> <p>Thus the Euler characteristic of $Z_n$ is</p> <p>$$\chi(Z_n)={\rm card}\; C_2^+ +{\rm card}\; C_2^- -{\rm card}\; C_0.$$</p> <p>Now observe that </p> <p>$${\rm card}\; C_2^\pm = \Bigl(\;{\rm card}\; [1,n-1]\cap (2\mathbb{Z}+1) \;\Bigr)^2= \frac{n^2}{4},$$</p> <p>$${\rm card} \; C_0 = n\times \Bigl( \frac{n(n-2)}{4}+ \frac{n(n-2)}{4}\Bigr)= \frac{n^2(n-2)}{2}.$$</p> <p>(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.) </p> <p>We conclude that the Euler characteristic of $Z_n$ is</p> <p>$$\chi(Z_n)= \frac{n^2}{2}- \frac{n^2(n-2)}{2}=\frac{n^2(3-n)}{2}.$$ </p> <p>For $n=2$ we get that $Z_2$ is a sphere. This agrees with the pictures on the site indicated by I. Rivin.</p> <p><strong>Update.</strong> 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 $\zeta_1&lt;\dotsc &lt;\zeta_n$ of $T_n$. The level zet</p> <p>$$Z_n\cap \lbrace z=\zeta_k\rbrace$$</p> <p>is the algebraic curve </p> <p>$$T_n(x)+T_n(y)=0. \tag{C}$$</p> <p>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 <em>homeomorphism</em></p> <p>$$[0,\pi]\ni t\mapsto x=\cos t\in [-1,1]$$</p> <p>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</p> <p>$$\cos ns+ \cos nt =0.$$</p> <p>This can be easily visualized as the intersection of the square with the grid</p> <p>$$s\pm t\in (2\mathbb{Z}+1)\frac{\pi}{n}$$</p> <p>which is connected. Now it is not very difficult to conclude using the Morse theoretic data on $h$ that $Z_n$ is connected.</p> <p><strong>Update</strong> To better explain my answer to Leon, below is a rendition of $Z_6$ where one can see three layers, yellow, green and blue.</p> <p><img src="http://www.nd.edu/~lnicolae/Z6.jpg" alt="alt text"></p> <p>The equality (1) predicts </p> <p>$$\chi(Z_6)=\frac{6^2(3-6)}{2}=-54.$$</p> <p>One can verify this directly as follows. Consider the $1$-dimensional simplicial complex $C$ embedded in $\mathbb{R}^3$ depicted below</p> <p><img src="http://www.nd.edu/~lnicolae/Z6a.jpg" alt="alt text"></p> <p>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</p> <p>$$\chi(Z_6)= 2\chi(T)= 2\chi(C).$$</p> <p>Let me give an alternate proof of the equality</p> <p>$$\chi(C)=-27. \tag{E}$$</p> <p>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 </p> <p>$$V= 8+12+6+1=27.$$</p> <p>The number $E$ of edges is half the sum of degrees of vertices. Thus</p> <p>$$E=\frac{1}{2}( 3\times 8 + 4\times 12 + 5\times 6+ 6\times 1)=\frac{1}{2} (24+48+30+6)=54.$$</p> <p>Hence </p> <p>$$\chi(C)= 27=54=-27.$$</p>