\documentclass[12pt]{article} \usepackage{amsfonts} %\documentstyle[12pt,amsfonts]{article} %\documentstyle{article} \setlength{\topmargin}{-.5in} \setlength{\oddsidemargin}{0 in} \setlength{\evensidemargin}{0 in} \setlength{\textwidth}{6.5truein} \setlength{\textheight}{8.5truein} %\input ../basicmath/basicmathmac.tex % %\input ../adgeomcs/lamacb.tex \input ../adgeomcs/mac.tex \input ../adgeomcs/mathmac.tex \def\fseq#1#2{(#1_{#2})_{#2\geq 1}} \def\fsseq#1#2#3{(#1_{#3(#2)})_{#2\geq 1}} \def\qleq{\sqsubseteq} % \begin{document} \begin{center} \fbox{{\Large\bf Spring, 2005 \hspace*{0.4cm} CIS 610}}\\ \vspace{1cm} {\Large\bf Advanced Geometric Methods in Computer Science\\ Jean Gallier \\ \vspace{0.5cm} Homework 3}\$10pt] March 16, 2005; Due April 6, 2005\\ Note: New due date! \end{center} A problems'' are for practice only, and should not be turned in. \vspace {0.25cm}\noindent {\bf Problem A1.} Let B_r = \{x = (x_1, \ldots, x_{n})\in \reals^n \mid x_1^2 + \cdots + x_n^2 < r\} be the open ball of radius r (centered at the origin) in \reals^n (where r > 0). Prove that the map \[ x \mapsto \frac{rx}{\sqrt{r^2 - (x_1^2 + \cdots + x_n^2)}}$ is a diffeomorphism of $B_r$ onto $\reals^n$ (where $x = (x_1, \ldots, x_{n}$)). \medskip\noindent {\it Hint\/}. Compute explicity the inverse of this map. \vspace {0.25cm}\noindent {\bf Problem A2.} A smooth bijective map of manifolds need not be a diffeomorphism. For example, show that $\mapdef{f}{\reals}{\reals}$ given by $f(x) = x^3$ is not a diffeomorphism. \vspace {0.25cm}\noindent {\bf Problem A3.} (a) Let $X\subseteq \reals^M$ and $Y\subseteq \reals^N$ be two smooth manifolds of dimension $m$ and $n$ respectively. We can make $X\times Y\subseteq \reals^{M+ N}$ into a smooth manifold of dimension $m + n$ as follows: for any $(p, q)\in X\times Y$, if $\mapdef{\varphi}{\Omega_1}{U}$ and $\mapdef{\psi}{\Omega_2}{V}$ are parametrizations at $p\in U\subseteq X$ and $q\in V\subseteq Y$ respectively, then show that $\mapdef{\varphi\times \psi}{\Omega_1\times \Omega_2}{U\times V}$ is indeed a parametrization at $(p, q)\in X\times Y$. As the $U\times V$'s cover $X\times Y$, these parametrizations make $X\times Y$ into a manifold. \medskip Check that $T_{(p, q)} (X\times Y) = T_p X\times T_q Y$. \medskip (b) Given a set, $X$, let $\Delta = \{(x, x)\mid x\in X\} \subseteq X\times X$, called the {\it diagonal of $X$\/}. If $X$ is a manifold, then prove that $\Delta$ is a manifold diffeomorphic to $X$. \medskip (c) The {\it graph\/} of a function, $\mapdef{f}{X}{Y}$, is the subset of $X\times Y$ given by $\mathrm{graph}(f) = \{(x, f(x)) \mid x\in X\}.$ Define $\mapdef{F}{X}{\mathrm{graph}(f)}$ by $F(x) = (x, f(x))$. Prove that if $X$ and $Y$ are smooth manifolds and if $f$ is smooth, then $F$ is a diffeomorphism and thus, $\mathrm{graph}(f)$ is a manifold diffeomorphic to $X$. \medskip (d) Given any (smooth) map, $\mapdef{f}{X}{X}$, some $x\in X$ is a {\it fixed point\/} of $f$ iff $f(x) = x$. Prove that $f$ has a fixed point iff $\mathrm{graph}(f) \cap \Delta \not= \emptyset$ (where $\Delta$ is the diagonal in $X\times X$). \vspace {0.5cm} B problems'' must be turned in. \vspace {0.25cm}\noindent {\bf Problem B1 (60 pts).} Recall from Homework 1, Problem B6, the Cayley parametrization of rotation matrices in $\mathbf{SO}(n)$ given by $C(B) = (I - B)(I + B)^{-1},$ where $B$ is any $n \times n$ skew symmetric matrix. In that problem, it was shown that $C(B)$ is a rotation matrix that does not admit $-1$ as an eigenvalue and that every such rotation matrix is of the form $C(B)$. \medskip (a) If you have not already done so, prove that the map $B \mapsto C(B)$ is injective. \medskip (b) Prove that $dC(B)(A) = D_A((I - B)(I + B)^{-1}) = -[I + (I - B)(I + B)^{-1}]A(I + B)^{-1}.$ {\it Hint\/}. First, show that $D_A(B^{-1}) = - B^{-1}AB^{-1}$ (where $B$ is invertible) and that \\ $D_A (f(B)g(B)) = (D_Af(B))g(B) + f(B)(D_Ag(B))$, where $f$ and $g$ are differentiable matrix functions. \medskip Deduce that $dC(B)$ is injective, for every skew-symmetric matrix, $B$. If we identify the space of $n\times n$ skew symmetric matrices with $\reals^{n(n-1)/2}$, show that the Cayley map, $\mapdef{C}{\reals^{n(n-1)/2}}{\mathbf{SO}(n)}$, is a parametrization of $\mathbf{SO}(n)$. \medskip (c) Now, consider $n = 3$, i.e., $\mathbf{SO}(3)$. Let $E_1$, $E_2$ and $E_3$ be the rotations about the $x$-axis, $y$-axis, and $z$-axis, respectively, by the angle $\pi$, i.e., $E_1 = \mattc{1}{0}{0} {0}{-1}{0} {0}{0}{-1}, \quad E_2 = \mattc{-1}{0}{0} {0}{1}{0} {0}{0}{-1}, \quad E_3 = \mattc{-1}{0}{0} {0}{-1}{0} {0}{0}{1}.$ Prove that the four maps \begin{eqnarray*} B & \mapsto & C(B) \\ B & \mapsto & E_1C(B) \\ B & \mapsto & E_2C(B) \\ B & \mapsto & E_3C(B) \end{eqnarray*} where $B$ is skew symmetric, are parametrizations of $\mathbf{SO}(3)$ and that the union of the images of $C$, $E_1C$, $E_2C$ and $E_3C$ covers $\mathbf{SO}(3)$, so that $\mathbf{SO}(3)$ is a manifold. \medskip (d) Let $A$ be {\it any\/} matrix (not necessarily invertible). Prove that there is some diagonal matrix, $E$, with entries $+1$ or $-1$, so that $EA + I$ is invertible. \medskip (e) Prove that every rotation matrix, $A\in \mathbf{SO}(n)$, is of the form $A = E(I - B)(I + B)^{-1},$ for some skew symmetric matrix, $B$, and some diagonal matrix, $E$, with entries $+1$ and $-1$, and where the number of $-1$ is even. Moreover, prove that every orthogonal matrix $A\in \mathbf{O}(n)$ is of the form $A = E(I - B)(I + B)^{-1},$ for some skew symmetric matrix, $B$, and some diagonal matrix, $E$, with entries $+1$ and $-1$. The above provide parametrizations for $\mathbf{SO}(n)$ (resp. $\mathbf{O}(n)$) that show that $\mathbf{SO}(n)$ and $\mathbf{O}(n)$ are manifolds. However, observe that the number of these charts grows exponentially with $n$. \vspace {0.25cm}\noindent {\bf Problem B2 (20 pts).} (1) For every symmetric, positive, definite matrix, $S$, and for every invertible matrix, $A$, prove that $A S\transpos{A}$ is symmetric, positive, definite. \medskip (2) Prove that for any symmetric, positive, definite matrix, $S$, there is some symmetric, positive, definite matrix, $S_1$, so that $S = S_1^2 = S_1\transpos{S_1}$. \medskip (3) Use (2) to prove that given any two symmetric, positive, definite matrices, $S$ and $S'$, there is some invertible matrix, $A$, so that $A S\transpos{A} = S'.$ Conclude that the action of $\mathbf{GL}(n, \reals)$ on $\mathbf{SPD}(n)$ given by $A\cdot S = A S \transpos{A}$ is well-defined and transitive. \vspace {0.25cm}\noindent {\bf Problem B3 (100 pts).} Consider the action of the group $\mathbf{SL}(2, \reals)$ on the upper half-plane, $H = \{z = x + iy \in \complex \mid y > 0\}$, given by $\matta{a}{b}{c}{d}\cdot z = \frac{az + b}{cz + d}.$ \medskip (a) Check that for any $g\in \mathbf{SL}(2, \reals)$, $\Immag(g\cdot z) = \frac{\Immag(z)}{|cz + d|^2},$ and conclude that if $z\in H$, then $g\cdot z \in H$, so that the action of $\mathbf{SL}(2, \reals)$ on $H$ is indeed well-defined (Recall, $\Re(z) = x$ and $\Immag(z) = y$, where $z = x + i y$.) \medskip (b) Check that if $c \not= 0$, then $\frac{az + b}{cz + d} = \frac{-1}{c^2 z + cd} + \frac{a}{c}.$ Prove that the group of M\"obius transformations induced by $\mathbf{SL}(2, \reals)$ is generated by M\"obius transformations of the form \begin{enumerate} \item $z \mapsto z + b$, \item $z \mapsto k z$, \item $z \mapsto -1/z$, \end{enumerate} where $b\in \reals$ and $k\in \reals$, with $k > 0$. Deduce from the above that the action of $\mathbf{SL}(2, \reals)$ on $H$ is transitive and that transformations of type (1) and (2) suffice for transitivity. \medskip (c) Now, consider the action of the discrete group $\mathbf{SL}(2, \integs)$ on $H$, where $\mathbf{SL}(2, \integs)$ consists of all matrices $\matta{a}{b}{c}{d}, \quad ad - bc = 1,\quad a, b, c, d \in \integs.$ Why is this action not transitive? Consider the two transformations $S\co z \mapsto -1/z$ associated with $\matta{0}{-1}{1}{0}$ and $T\co z \mapsto z + 1$ associated with $\matta{1}{1}{0}{1}$. \medskip Define the subset, $D$, of $H$, as the set of points, $z = x + iy$, such that $-1/2 \leq x \leq -1/2$ and $x^2 + y^2 \geq 1$. Observe that $D$ contains the three special points, $i$, $\rho = e^{2\pi i/3}$ and $-\overline{\rho} = e^{\pi i/3}$. \medskip Draw a picture of this set, known as a {\it fundamental domain\/} of the action of $G = \mathbf{SL}(2, \integs)$ on $H$. You will now prove the following result first established by Gauss: \medskip\noindent {\bf Theorem I\/}. Let $G'$ be the subgroup of $G = \mathbf{SL}(2, \integs)$ generated by $S$ and $T$. \begin{enumerate} \item[(1)] For every point, $z\in H$, there is some $g\in G'$ so that $g\cdot z \in D$. \item[(2)] If two distinct points $z, z'\in D$ are in the same orbit under the action of $G = \mathbf{SL}(2, \integs)$, then either $\Re(z) = \pm 1/2$ and $z = z' \pm 1$ or $|z| = 1$ and $z' = -1/z$. \item[(3)] Let $z\in D$ and consider the stabilizer, $G_z$ of $z$ (under the action of $G$). Then, $G_z = \{1\}$, unless \begin{enumerate} \item[(i)] $z = i$, in which case, $G_z$ is the group of order $2$ generated by $S$ (note, $S^2 = 1$) \item[(ii)] $z = \rho = e^{2\pi i/3}$, in which case $G_{z}$ is the group of order $3$ generated by $ST$ (composition is written from right to left, as usual, and note that $(ST)^3 = 1$) \item[(iii)] $z = -\overline{\rho} = e^{\pi i/3}$, in which case $G_{z}$ is the group of order $3$ generated by $TS$ (note that $(TS)^3 = 1$). \end{enumerate} \end{enumerate} Deduce from Theorem I that the natural map $D \longrightarrow H/G$ is surjective and that its restriction to the interior of $D$ is injective. \medskip\noindent {\it Hints\/}. Observe that since $c, d$ are integers, for a fixed $z$, there are finitely many pairs $(c, d)$ so that $|cz + d| < K$, for any fixed $K > 0$. Thus, there is some $g\in G'$ so that $\Immag(g\cdot z)$ is maximum. Next, show that there is some $n$ so that the real part of $T^ngz$ is between $-1/2$ and $1/2$. Show that this element, $z' = T^ngz$, is actually in $D$. \medskip For (2) and (3), show that it may be assumed that $\Immag(g\cdot z) \geq \Immag(z)$, i.e., $|cz + d| \leq 1$. \medskip (d) Use Theorem I to prove \medskip\noindent {\bf Theorem II\/}. The group $G = \mathbf{SL}(2, \integs)$ is generated by $S$ and $T$, i.e., $G' = G$. \medskip (e) In view of Theorem I, as every point in the interior of $D$ corresponds to a unique orbit and every orbit has some representative in $D$, by applying all elements of $G = \mathbf{SL}(2, \integs)$ to $D$, we get a {\it tesselation\/} of $H$, i.e., we get $H = \bigcup_{g\in G} g\cdot D,$ where the interiors of $g\cdot D$ and $g' \cdot D$ are disjoint whenever $g\cdot D$ and $g' \cdot D$ are distinct. By Theorem II, we get all $g\cdot D$'s by applying $S$ and $T$ to $D$. Draw the picture obtained by applying $1, T, TS, ST^{-1}S, ST^{-1}, S, ST, STS, T^{-1}S, T^{-1}.$ \vspace {0.25cm}\noindent {\bf Problem B4 (30 pts).} Let $J$ be the $2\times 2$ matrix $J = \matta{1}{0}{0}{-1}$ and let $\mathbf{SU}(1, 1)$ be the set of $2\times 2$ complex matrices $\mathbf{SU}(1, 1) = \{A \mid A^* J A = J,\quad \det(A) = 1\},$ where $A^*$ is the conjugate transpose of $A$. \medskip (a) Prove that $\mathbf{SU}(1, 1)$ is the group of matrices of the form $A = \matta{a}{b}{\overline{b}}{\overline{a}}, \quad\hbox{with}\quad a\overline{a} - b\overline{b} = 1.$ If $g = \matta{1}{-i}{1}{i}$ prove that the map from $\mathbf{SL}(2, \reals)$ to $\mathbf{SU}(1, 1)$ given by $A \mapsto gAg^{-1}$ is a group isomorphism. \medskip (b) Prove that the M\"obius transformation associated with $g$, $z \mapsto \frac{z - i}{z + i}$ is a bijection between the upper half-plane, $H$, and the unit open disk, $D = \{z \in \complex \mid |z| < 1\}$. Prove that the map from $\mathbf{SU}(1, 1)$ to $S^1\times D$ given by $\matta{a}{b}{\overline{b}}{\overline{a}} \mapsto (a/|a|, b/a)$ is a continuous bijection (in fact, a homeomorphism). Conclude that $\mathbf{SU}(1, 1)$ is topologically an open solid torus. \medskip (c) Check that $\mathbf{SU}(1, 1)$ acts transitively on $D$ by $\matta{a}{b}{\overline{b}}{\overline{a}}\cdot z = \frac{az + b}{\overline{b}z + \overline{a}}.$ Find the stabilizer of $z = 0$ and conclude that $\mathbf{SU}(1, 1)/\mathbf{SO}(2) \cong D.$ \vspace {0.25cm}\noindent {\bf Problem B5 (100 pts).} (a) Let $U\subseteq \reals^m$ be an open subset of $\reals^m$ and pick some $a\in U$. If $\mapdef{f}{U}{\reals^n}$ is a submersion at $a$, i.e., $df_a$ is surjective (so, $m\geq n$), prove that there is an open set, $W \subseteq U \subseteq \reals^m$, with $a\in W$ and a diffeomorphism, $\psi$, with domain $V \subseteq \reals^m$, so that $\psi(V) = W$ and $f(\psi(x_1, \ldots, x_m)) = (x_1, \ldots, x_n),$ for all $(x_1, \ldots, x_m)\in V$. \medskip\noindent {\it Hint\/}. Since $df_a$ is surjective, the rank of the Jacobian matrix, $(\partial f_i/\partial x_j(a))$ ($1\leq i \leq n$, $1\leq j \leq m$), is $n$ and after some permutation of $\reals^m$, we may assume that the square matrix, $B = (\partial f_i/\partial x_j(a))$ ($1\leq i, j \leq n$), is invertible. Define the map, $\mapdef{h}{U}{\reals^m}$, by $h(x) = (f_1(x), \ldots, f_n(x), x_{n+1}, \ldots, x_{m}),$ where $x = (x_1, \ldots, x_m)$. Check that the Jacobian matrix of $h$ at $a$ is invertible. Then, apply the inverse function theorem and finish up. \medskip (b) Let $\mapdef{f}{M}{N}$ be a map of smooth manifolds. A point, $p\in M$, is called a {\it critical point (of $f$)\/} iff $df_p$ is {\it not\/} surjective and a point $q\in N$ is called a {\it critical value (of $f$)\/} iff $q = f(p)$, for some critical point, $p \in M$. A point $p\in M$ is a {\it regular point (of $f$)\/} iff $p$ is not critical, i.e., $df_p$ is surjective, and a point $q\in N$ is a {\it regular value (of $f$)\/} iff it is not a critical value. In particular, any $q\in N - f(M)$ is a regular value and $q\in f(M)$ is a regular value iff {\it every\/} $p\in f^{-1}(q)$ is a regular point (but, in contrast, $q$ is a critical value iff {\it some\/} $p \in f^{-1}(q)$ is critical). \medskip Prove that for every regular value, $q\in f(M)$, the preimage $Z = f^{-1}(q)$ is a manifold of dimension $\dimm(M) - \dimm(N)$. \medskip\noindent {\it Hint\/}. Pick any $p\in f^{-1}(q)$ and some parametrizations, $\varphi$ at $p$ and $\psi$ at $q$, with $\varphi(0) = p$ and $\psi(0) = q$ and consider $h = \psi^{-1}\circ f\circ \varphi$. Prove that $dh_0$ is surjective and then apply (a). \medskip (c) Under the same assumptions as (b), prove that for every point $p\in Z = f^{-1}(q)$, the tangent space, $T_p Z$, is the kernel of $\mapdef{df_p}{T_p M}{T_q N}$. \medskip (d) If $X, Z\subseteq \reals^N$ are manifolds and $Z\subseteq X$, we say that {\it $Z$ is a submanifold of $X$\/}. Assume there is a smooth function, $\mapdef{g}{X}{\reals^k}$, and that $0\in \reals^k$ is a regular value of $g$. Then, by (b), $Z = g^{-1}(0)$ is a submanifold of $X$ of dimension $\dimm(X) - k$. Let $g = (g_1, \ldots, g_k)$, with each $g_i$ a function, $\mapdef{g_i}{X}{\reals}$. Prove that for any $p\in X$, $dg_p$ is surjective iff the linear forms, $\mapdef{(dg_i)_p}{T_p X}{\reals}$, are linearly independent. In this case, we say that $g_1, \ldots, g_k$ are {\it independent at $p$\/}. We also say that $Z$ is {\it cut out\/} by $g_1, \ldots, g_k$ when $Z = \{p\in X \mid g_1(p) = 0, \ldots, g_k(p) = 0\}$ with $g_1, \ldots, g_k$ independent for all $p\in Z$. \medskip Let $\mapdef{f}{X}{Y}$ be a smooth maps of manifolds and let $q\in f(X)$ be a regular value. Prove that $Z = f^{-1}(q)$ is a submanifold of $X$ cut out by $k = \dimm(X) - \dimm(Y)$ independent functions. \medskip\noindent {\it Hint\/}. Pick some parametrization, $\psi$, at $q$, so that $\psi(0) = q$ and check that $0$ is a regular value of $g = \psi^{-1}\circ f$, so that $g_1, \ldots, g_k$ work. \medskip (e) Let $U\subseteq \reals^m$ be an open subset of $\reals^m$ and pick some $a\in U$. If $\mapdef{f}{U}{\reals^n}$ is an immersion at $a$, i.e., $df_a$ is injective (so, $m\leq n$), prove that there is an open set, $V \subseteq \reals^n$, with $f(a)\in V$, an open subset, $U'\subseteq U$, with $a\in U'$ and $f(U') \subseteq V$ and a diffeomorphism, $\varphi$, with domain $V$, so that $\varphi(f(x_1, \ldots, x_m)) = (x_1, \ldots, x_m, 0, \ldots, 0),$ for all $(x_1, \ldots, x_m)\in U'$. \medskip\noindent {\it Hint\/}. Since $df_a$ is injective, the rank of the Jacobian matrix, $(\partial f_i/\partial x_j(a))$ ($1\leq i \leq n$, $1\leq j \leq m$), is $m$ and after some permutation of $\reals^n$, we may assume that the square matrix, $B = (\partial f_i/\partial x_j(a))$ ($1\leq i, j \leq m$), is invertible. Define the map, $\mapdef{g}{U\times \reals^{n-m}}{\reals^n}$, by $g(x,y) = (f_1(x), \ldots, f_m(x), y_{1} + f_{m+1}(x), \ldots, y_{n-m} + f_n(x)),$ where $x = (x_1, \ldots, x_m)$ and $y = (y_1, \ldots, y_{n - m})$. Check that the Jacobian matrix of $g$ at $(a, 0)$ is invertible. Then, apply the inverse function theorem and finish up. \medskip Now, assume $Z$ is a submanifold of $X$. Prove that locally, $Z$ is cut out by independent functions. This means that if $k = \dimm(X) - \dimm(Z)$, the {\it codimension\/} of $Z$ in $X$, then for every $z\in Z$, there are $k$ independent functions, $g_1, \ldots, g_k$, defined on some open subset, $W\subseteq X$, with $z\in W$, so that $Z\cap W$ is the common zero set of the $g_i$'s. \medskip (f) We would like to generalize our result in (b) to the more general situation where we have a smooth map, $\mapdef{f}{X}{Y}$, but this time, we have a submanifold, $Z\subseteq Y$ and we are investigating whether $f^{-1}(Z)$ is a submanifold of $X$. In particular, if $X$ is also a submanifold of $Y$ and $f$ is the inclusion of $X$ into $Y$, then $f^{-1}(Z) = X\cap Z$. \medskip Convince yourself that, in general, the intersection of two submanifolds is {\it not\/} a submanifold. Try examples involving curves and surfaces and you will see how bad the situation can be. What is needed is a notion generalizing that of a regular value, and this turns out to be the notion of transversality. \medskip We say that {\it $f$ is transveral to $Z$\/} iff $df_p(T_pX) + T_{f(p)} Z = T_{f(p)} Y,$ for all $p\in f^{-1}(Z)$. (Recall, if $U$ and $V$ are subspaces of a vector space, $E$, then $U + V$ is the subspace $U + V = \{u + v \in E \mid u\in U,\, v\in V\}$). In particular, if $f$ is the inclusion of $X$ into $Y$, the transversality condition is $T_{p} X + T_p Z = T_p Y,$ for all $p \in X \cap Z$. \medskip Draw several examples of transversal intersections to understand better this concept. Prove that if $f$ is transversal to $Z$, then $f^{-1}(Z)$ is a submanifold of $X$ of codimension equal to $\dimm(Y) - \dimm(Z)$. \medskip\noindent {\it Hint\/}. The set $f^{-1}(Z)$ is a manifold iff for every $p\in f^{-1}(Z)$, there is some open subset, $U \subseteq X$, with $p\in U$, and $f^{-1}(Z) \cap U$ is a manifold. First, use (e) to assert that locally near $q = f(p)$, $Z$ is cut out by $k = \dimm(Y) - \dimm(Z)$ independent functions, $g_1, \ldots, g_k$, so that locally near $p$, the preimage $f^{-1}(Z)$ is cut out by $g_1\circ f, \ldots, g_k\circ f$. If we let $g = (g_1, \ldots, g_k)$, it is an immersion and the issue is to prove that $0$ is a regular value of $g\circ f$ in order to apply (b). Show that transversality is just what's needed to show that $0$ is a regular value of $g\circ f$. \medskip (g) With the same assumptions as in (f) ($f$ is transversal to $Z$), if $W = f^{-1}(Z)$, prove that for every $p\in W$, $T_p W = (df_p)^{-1}(T_{f(p)} Z),$ the preimage of $T_{f(p)} Z$ by $\mapdef{df_p}{T_p X}{T_{f(p)} Y}$. In particular, if $f$ is the inclusion of $X$ into $Y$, then $T_p (X \cap Z) = T_p X\cap T_p Z.$ \medskip (h) Let $X, Z\subseteq Y$ be two submanifolds of $Y$, with $X$ compact, $Z$ closed, $\dimm(X) + \dimm(Z) = \dimm(Y)$ and $X$ transversal to $Z$. Prove that $X\cap Z$ consists of a finite set of points. \vspace{0.5cm}\noindent {\bf TOTAL: 310 points.} \end{document}