Friday, November 18, 2016

Loose ends of smooth manifolds

 Preliminary exam prep

Here we round up some theorems that have escaped previous roundings-up. Let $X,Y$ be smooth manifolds and $f:X\to Y$ a smooth map.

Theorem: (Inverse function theorem) If $df_p$ is invertible for some $p\in M$, then there exist $U\owns p$ and $V\owns f(p)$ connected such that $f|_U:U\to V$ is a diffeomorphism.

Corollary: (Stack of records theorem) If $\dim(X)=\dim(Y)$, then every regular value $y\in Y$ has a neighborhood $V\owns y$ such that $f^{-1}(Y)=U_1\sqcup \cdots \sqcup U_k$, where $f|_{U_i}:U_i\to V$ is a diffeomorphism.

Proof: Since $y\in Y$ is a regular value, $df_x$ is surjective for all $x\in f^{-1}(y)$. Since $\dim(X)=\dim(Y)$ and $df_x$ is linear, $df_x$ is an isomorphism, hence invertible. By the inverse function theorem, there exist $U\owns x$ and $V\owns y$ connected such that $f|_U:U\to V$ is a diffeomorphism. Before we actually apply this, we need to show that $f^{-1}(y)$ is a finite set.

First we note that by the preimage theorem, since $y$ is a regular value, $f^{-1}(y)$ is a submanifold of $X$ of dimension $\dim(X)-\dim(Y)=0$. Next, if $f^{-1}(y) = \{x_i\}$ were infinite, since $X$ is compact, there would be some limit point $p\in X$ of $\{x_i\}$. But then by continuity,
\[
y = \lim_{i\to \infty}\left[f(x_i)\right] = f\left(\lim_{i\to\infty}\left[x_i\right]\right) = f(p),\]so $p\in f^{-1}(y)$. But then either $p$ cannot be separated from other elements of $f^{-1}(y)$, meaning $f^{-1}(y)$ is not a manifold, or the sequence $\{x_i\}$ is finite in length. Hence $f^{-1}(y) = \{x_1,\dots,x_k\}$. Let $U_i\owns x_i$ and $V_i\owns y$ be the sets asserted to exist by the inverse function theorem (the $U_i$ may be assumed to be disjoint without loss of generality). Let $V = \bigcap_{i=1}^k V_i$ and $U_i' = f^{-1}(V)\cap U_i$, for which we still have $f|_{U_i'}:U_i'\to V$ a diffeomorphism. $\square$

Theorem: (Classification of manifolds) Up to diffeomorphism,
  • the only 0-dimensional manifolds are collections of points,
  • the only 1-dimensional manifolds are $S^1$ and $\R$, and
  • the only 2-dimensional compact manifolds are $S^2\# (T^2)^{\#n}$ or $S^2\#(\R\P^2)^{\#n}$, for any $n\geqslant 0$.
Compact 2-manifolds are homeomorphic iff they are both (non)-orientable and have the same Euler characteristic. Note that
\[
\chi\left(S^2\# (T^2)^{\#n}\right) = 2-2n,
\hspace{1cm}
\chi\left(S^2\#(\R\P^2)^{\#n}\right) = 2-n.
\]
These surfaces are called orientable (on the left) and non-orientable (on the right) surfaces of genus $n$.

Theorem: (Stokes' theorem) For $X$ oriented and $\omega\in \Omega^{n-1}_X$, $\int_X d\omega = \int_{\dy X} \omega$.

Proposition: The tangent bundle $TX$ is always orientable.

Proof: Let $U,V\subset X$ with $\vp:U\to \R^n$ and $\psi:V\to \R^n$ trivializing maps, and $\psi\circ \vp^{-1}:\R^n\to \R^n$ the transition function. To show that $TX$ is always orientable, we need to show the Jacobian of the induced transition function (determinant of the derivative) on $TX$ is always non-negative. On $TU$ and $TV$, we have trivializing maps $(\varphi,d\varphi)$ and $(\psi,d\psi)$, giving a transition function
\[
(\psi\circ \varphi^{-1}, d\psi \circ d\varphi^{-1}) = (\psi\circ \varphi^{-1}, d(\psi \circ \varphi^{-1})). 
\]
The Jacobian of this is
\[
\det(d(\psi\circ \varphi^{-1}, d(\psi \circ \varphi^{-1}))) = \det(d(\psi\circ \varphi^{-1}), d(\psi \circ \varphi^{-1}))) = \det(d(\psi\circ \varphi^{-1}))\cdot \det(d(\psi \circ \varphi^{-1})) \>0,
\]
and since $d(\psi\circ \varphi^{-1})\neq 0$ (as $\psi \circ \varphi^{-1}$ is a diffeomorphism, its derivative is an isomorphism), the result is always positive. $\square$

References: Lee (Introduction to smooth manifolds, Chapter 4), Guillemin and Pollack (Differential topology, Chapter 1

Sunday, November 13, 2016

Covering spaces

 Preliminary exam prep

Let $X,Y$ be topological spaces.

Definition: A space $\widetilde X$ and a map $p:\widetilde X\to X$ are called a covering space of $X$ if either of two equivalent conditions hold:
  • There is a cover $\{U_\alpha\}_{\alpha\in A}$ of $X$ such that $p^{-1}(U_\alpha)\cong \bigsqcup_{\beta\in B_\alpha} U_\beta$.
  • Every point $x\in X$ has a neighborhood $U\subset X$ such that $p^{-1}(U) \cong \bigsqcup_{\beta\in B}U_\beta$.
We also demand that every $U_\beta$ is carried homeomorphically onto $U_\alpha$ (or $U$) by $p$, and the $U_\alpha$ (or $U$) are called evenly covered.

Some definitions require that $p$ be surjective. A universal cover of $X$ is a covering space that is universal with respect to this property, in that it covers all other covering spaces. Moreover, a cover that is simply connected is immediately a universal cover.

Remark:
Every path connected (pc), locally path connected (lpc), and semi locally simply connected (slsc) space has a universal cover.

Theorem: (Lifting criterion) Let $Y$ be pc and lpc, and $\widetilde X$ a covering space for $X$. A map $f:Y\to X$ lifts to a map $\widetilde f:Y\to \widetilde X$ iff $f_*(\pi_1(Y))\subset p_*(\pi_1(\widetilde X))$.

Further, if the initial map $f_0$ in a homotopy $f_t:Y\to X$ lifts to $\widetilde f_0:Y\to \widetilde X$, then $f_t$ lifts uniquely to $\widetilde X$. This is called the homotopy lifting property. Next, we will see that path connected covers of $X$ may be classified via a correspondence through the fundamental group.

Theorem:
Let $X$ be pc, lpc, and slsc. There is a bijection (up to isomorphism) between pc covers $p:\widetilde X\to X$ and subgroups of $\pi_1(X)$, described by $p_*(\pi_1(\widetilde X))$.

Example:
Let $X=T^2$, the torus, with fundamental group $\Z\oplus \Z$. Below are some covering spaces of $p:\widetilde X\to X$ with the corresponding subgroups $p_*(\pi_1(\Z\oplus\Z))$.
Definition: Given a covering space $p:\widetilde X\to X$, an isomorphism $g$ of $\widetilde X$ for which $\id_X\circ p = p\circ g$, is called a deck transformation, the collection of which form a group $G(\widetilde X)$ under composition. Further, $\widetilde X$ is called normal (or regular) if for every $x\in X$ and every $\widetilde x_1,\widetilde x_2\in p^{-1}(x)$, there exists $g\in G(\widetilde X)$ such that $g(\widetilde x_1)=\widetilde x_2$.

For path connected covering spaces over path connected and locally path connected bases, being normal is equivalent to $p_*(\pi_1(\widetilde X))\leqslant  \pi_1(X)$ being normal. In this case, $G(\widetilde X)\cong \pi_1(X)/p_*(\pi_1(\widetilde X))$. This simplifies even more for $\widetilde X$ a universal cover, as $\pi_1(\widetilde X)=0$ then.

Theorem: Let $G$ be a group, and suppose that every $x\in X$ has a neighborhood $U\owns x$ such that $g(U)\cap h(U) = \emptyset$ whenever $g\neq h\in G$. Then:
  • The quotient map $q:X\to X/G$ describes a normal cover of $X/G$.
  • If $X$ is pc, then $G = G(X)$.
A group action satisfying the hypothesis of the previous theorem is called a covering space action.

Proposition: For any $n$-sheeted covering space $\widetilde X\to X$ of a finite CW complex, $\chi(\widetilde X) = n\chi(X)$.

References: Hatcher (Algebraic Topology, Chapter 1)

Thursday, November 10, 2016

Differential 1-forms are closed if and only if they are exact

 Preliminary exam prep

The title refers to 1-forms in Euclidean $n$-space $\R^n$, for $n\geqslant 2$. This theorem is instructive to do in the case $n=2$, but we present it in general. We will use several facts, most importantly that the integral of a function $f:X\to Y$ over a curve $\gamma:[a,b]\to X$ is given by
\[
\int_\gamma f\ dx_1\wedge \cdots \wedge dx_k = \int_a^b (f\circ \gamma)\ d(x_1\circ \gamma)\wedge \cdots \wedge d(x_n\circ \gamma),
\]
where $x_1,\dots,x_n$ is some local frame on $X$. We will also use the fundamental theorem of calculus and one of its consequences, namely
\[
\int_a^b \frac{\dy f}{\dy t}(t)\ dt = f(b)-f(a).
\]

Theorem:
A 1-form on $\R^n$ is closed if and only if it is exact, for $n\geqslant 2$.

Proof: Let $\omega = a_1dx_1+\cdots a_ndx_n\in \Omega^1_{\R^n}$ be a 1-form on $\R^n$. If there exists $\eta\in \Omega^0_{\R^n}$ such that $d\eta = \omega$, then $d\omega = d^2\eta = 0$, so the reverse direction is clear. For the forward direction, since $\omega$ is closed, we have
\[
0 = d\omega = \sum_{i=1}^n \frac{\dy a_1}{\dy x_i}dx_i \wedge dx_1 + \cdots + \sum_{i=1}^n \frac{\dy a_n}{\dy x_n} dx_i\wedge dx_n
\ \ \ \implies\ \ \
\frac{\dy a_i}{\dy x_j} = \frac{\dy a_j}{\dy x_i}\ \forall\ i\neq j.
\]
Now fix some $(\textbf{x}_1,\dots,\textbf{x}_n)\in \R^n$, and define $f\in \Omega^0_{\R^n}$ by
\[
f(\textbf x_1,\dots,\textbf x_n) = \int_{\gamma(\textbf x_1,\dots,\textbf x_n)}\omega,
\]
for $\gamma$ the composition of the paths
\[
\begin{array}{r c l}
\gamma_1\ :\ [0,\textbf x_1] & \to & \R^n, \\
t & \mapsto & (t,0,\dots,0),
\end{array}
\hspace{5pt}
\begin{array}{r c l}
\gamma_2\ :\ [0,\textbf x_2] & \to & \R^n, \\
t & \mapsto & (\textbf x_1,t,0,\dots,0),
\end{array}
\hspace{5pt}\cdots\hspace{5pt}
\begin{array}{r c l}
\gamma_n\ :\ [0,\textbf x_n] & \to & \R^n, \\
t & \mapsto & (\textbf x_1,\dots,\textbf x_{n-1},t).
\end{array}
\]
By applying the definition of a pullback and the change of variables formula (use $s=\gamma_i(t)$ for every $i$),
\begin{align*}
\int_{\gamma(\textbf x_1,\dots,\textbf x_n)}\omega  & = \sum_{i=1}^n \int_{\gamma_i} a_1 dx_1 + \cdots + \sum_{i=1}^n \int_{\gamma_i}a_n dx_n \\
& = \sum_{i=1}^n \int_{\gamma_i} a_1(x_1,\dots,x_n)\ dx_1 + \cdots + \sum_{i=1}^n \int_{\gamma_i}a_n(x_1,\dots,x_n)\ dx_n \\
& = \sum_{i=1}^n \int_0^{\textbf x_i} a_1(\gamma_i(t))\ d(x_1\circ \gamma_i)(t) + \cdots + \sum_{i=1}^n \int_0^{\textbf x_i}a_n(\gamma_i(t))\ d(x_n\circ \gamma_i)(t) \\
& = \int_0^{\textbf x_1} a_1(\gamma_1(t))\gamma'_1(t)\ dt + \cdots + \int_0^{\textbf x_n}a_n(\gamma_n(t))\gamma_n'(t)\ dt \\
& = \int_{(0,\dots,0)}^{(\textbf x_1,0,\dots,0)} a_1(s)\ ds + \cdots + \int_{(\textbf x_1,\dots,\textbf x_{n-1},0)}^{(\textbf x_1,\dots,\textbf x_n)}a_n(s)\ ds \\
& = \int_0^{\textbf x_1} a_1(s,0,\dots,0)\ ds + \cdots + \int_0^{\textbf x_n}a_n(\textbf x_1,\dots,\textbf x_{n-1}, s)\ ds.
\end{align*}
To take the derivative of this, we consider the partial derivatives first. In the last variable, we have
\[
\frac{\dy f}{\dy \textbf x_n} = \frac\dy{\dy \textbf x_n}\int_0^{\textbf x_n}a_n(\textbf x_1,\dots,\textbf x_{n-1}, s)\ ds = a_n(\textbf x_1,\dots,\textbf x_n) = a_n.
\]
In the second-last variable, applying one of the identities from $\omega$ being closed, we have
\begin{align*}
\frac{\dy f}{\dy \textbf x_{n-1}}  & = \frac\dy{\dy \textbf x_{n-1}}\int_0^{\textbf x_{n-1}}a_{n-1}(\textbf x_1,\dots,\textbf x_{n-2}, s,0)\ ds  +  \frac\dy{\dy \textbf  x_{n-1}}\int_0^{\textbf x_n}a_n(\textbf x_1,\dots,\textbf x_{n-1}, s)\ ds \\
& = a_{n-1}(\textbf x_1,\dots,\textbf x_{n-1},0) + \int_0^{\textbf x_n}\frac{\dy a_n}{\dy \textbf x_{n-1}}(\textbf x_1,\dots,\textbf x_{n-1}, s)\ ds \\
& = a_{n-1}(\textbf x_1,\dots,\textbf x_{n-1},0) + \int_0^{\textbf x_n}\frac{\dy a_{n-1}}{\dy s}(\textbf x_1,\dots,\textbf x_{n-1}, s)\ ds \\
& = a_{n-1}(\textbf x_1,\dots,\textbf x_{n-1},0) + a_{n-1}(\textbf x_1,\dots,\textbf x_n) - a_{n-1}(\textbf x_1,\dots,\textbf x_{n-1}, 0) \\
& = a_{n-1}(\textbf x_1,\dots, \textbf x_n) \\
& = a_{n-1}.
\end{align*}
This pattern continues. For the other variables we have telescoping sums, and we compute the partial derivative in the first variable as an example:
\begin{align*}
\frac{\dy f}{\dy \textbf x_1}  & = \frac{\dy}{\dy \textbf x_1}\int_0^{\textbf x_1}a_1(s,0,\dots,0)\ ds + \sum_{i=2}^n\frac\dy{\dy \textbf x_1}\int_0^{\textbf x_i}a_i(\textbf x_1,\dots,\textbf x_{i-1}, s,0,\dots,0)\ ds \\
& = a_1(\textbf x_1,0,\dots,0) + \sum_{i=2}^n \int_0^{\textbf x_i} \frac {\dy a_i}{\dy \textbf x_1} (\textbf x_1,\dots,\textbf x_{i-1}, s,0,\dots,0)\ ds \\
& = a_1(\textbf x_1,0,\dots,0) + \sum_{i=2}^n \int_0^{\textbf x_i} \frac {\dy a_1}{\dy s} (\textbf x_1,\dots,\textbf x_{i-1}, s,0,\dots,0)\ ds \\
& = a_1(\textbf x_1,0,\dots,0) + \sum_{i=2}^n \left(a_1(\textbf x_1,\dots,\textbf x_i, 0,\dots,0) - a_1(\textbf x_1,\dots,\textbf x_{i-1},0,\dots,0)\right) \\
& = a_1(\textbf x_1,\dots,\textbf x_n) \\
& = a_1.
\end{align*}
Hence we get that
\[
df = \frac{\dy f}{\dy x_1} dx_1 + \cdots + \frac{\dy f}{\dy x_n}dx_n = a_1dx_1 + \cdots + a_ndx_n = \omega,
\]
so $\omega$ is exact. $\square$

References: Lee (Introduction to smooth manifolds, Chapter 11)

Tuesday, November 8, 2016

More (co)homological constructions

 Preliminary exam prep

Recall a previous post (2016-09-16, "Complexes and their homology") that focused on constructing topological spaces in different ways and recovering the homology. Here we complete that task, introducing cellular homology. Recall a cell complex (or CW complex) $X$ was a sequence of skeleta $X_k$ for $k=0,\dots,\dim(X)$ consisting of $k$-cells $e^k_i$ and their attaching maps to the $(k-1)$-skeleton.

Cellular homology


Definition: The long exact sequence in relative homology for the pair $X_k,X_{k-1}$ shares terms with the long exact sequence for the pair $X_{k+1},X_k$, as well as $X_{k-1},X_{k-2}$. By letting $d_k$ be the composition of maps in different long exact sequences, for $k>1$, that make the diagram
commute, we get a complex of equivalence classes of chains
\[
\cdots \to H_{k+1}(X_{k+1},X_k) \tov{d_{k+1}} H_k(X_k,X_{k-1})\tov{d_k} H_{k-1}(X_{k-1},X_{k-2})\to \cdots \to H_1(X_1,X_0)\tov{d_1} H_0(X_0) \tov{d_0} 0,
\]
whose homology $H_k^{CW}(X) = \ker(d_k)/\text{im}(d_{k-1})$ is called the cellular homology of $X$. The map $d_1$ is the connecting map in the long exact sequence of the pair $X_1,X_0$, and $d_0=0$.

This seems quite a roundabout way of defining homology groups, but it turns out to be very useful. Note that for $k=1$, the map $d_1$ is the same as for a simplicial complex, hence

Theorem:
In the context above,
  1. for $k\>0$, $H^{CW}_k(X)\cong H_k(X)$;
  2. for $k\>1$, $H_k(X_k,X_{k-1})=\Z^\ell$, where $\ell$ is the number of $k$-cells in $X$; and
  3. for $k\>2$, $d_k(e^k_i) = \displaystyle\sum_j\deg(\underbrace{\dy e^k_i}_{S^{k-1}}\tov{f_{k,i}} X_{k-1}\tov{\pi} \underbrace{X_{k-1}/X_{k-1}-e^{k-1}_j}_{S^{k-1}})e^{k-1}_j$.
Example: Real projective space $\R\P^n$ has a cell decomposition with one cell in each dimension, and 2-to-1 attaching maps $\dy(e_k) =2X_{k-1}$ for $k>1$. This gives us a construction
\[
X_0 = e_0,
\hspace{1cm}
X_1 = e_1 \bigsqcup_{\dy(e_1)=e_0} X_0,
\hspace{1cm}
X_2 = e_2 \bigsqcup_{\dy(e_2)=2e_1} X_1,
\hspace{1cm}
X_3 = e_3 \bigsqcup_{\dy(e_3)=2e_2} X_2, \dots
\]It is immediate that $d_0=d_1=0$, and for higher degrees, we have
\[
d_k(e^k) = \deg(S^{k-1}\to \R\P^{k-1}\to S^{k-1})e^{k-1}.\]
Since this is a map between spheres, we may apply local degree calculations. The first part is the 2-to-1 cover, where every point in $\R\P^{k-1}$ is covered by two points from $S^{k-1}$, one in each hemisphere. One covers it via the identity, the other via the antipodal map. As long as we choose a point not in $\R\P^{k-2}\subset \R\P^{k-1}$, the second step doesn't affect these degree calculations. The antipodal map $S^{k-1}\to S^{k-1}$ has degree $(-1)^k$, hence for $a$ the antipodal map, the composition has degree
\[
\deg(S^{k-1}\to \R\P^{k-1}\to S^{k-1}) = \deg(\id_{S^{k-1}}) + \deg(a_{S^{k-1}}) = 1+(-1)^k = \begin{cases} 2 & k\text{ even}, \\ 0 & k \text{ odd.}\end{cases}
\]

Products in (co)homology


Recall that an $n$-chain on $X$ is a map $\sigma:\Delta^n\to X$, where $\Delta^n=[v_0,\dots,v_n]$ is an $n$-simplex. These form the group $C_n$ of $n$-chains. An $n$-cochain is an element of $C^n = \Hom(C_n,\Z)$, though the coefficient group does not need to be $\Z$ necessarily.

Definition: The diagonal map $X\to X\times X$ induces a map on cohomology $H^*(X\times X)\to H^*(X)$, and by Kunneth, this gives a map $H^*(X)\otimes H^*(X)\to H^*(X)$, and is called the cup product.

For $a\in H^p(X)$ and $b\in H^q(X)$, representatives of the class $a$ are in $\Hom(C_p,\Z)$ and representatives of the class $b$ are in $\Hom(C_q,\Z)$, though we will conflate the notation for the class with that of a representative. Hence for a $(p+q)$-chain $\sigma$ the cup product of $a$ and $b$ acts as
\[
(a\smile b)\sigma = a\left(\sigma|_{[v_0,\dots,v_p]}\right)\cdot b\left(\sigma|_{[v_p,\dots,v_{p+q}]}\right).
\]
Definition: The cap product combines $p$-cochains with $q$-chains to give $(q-p)$-chains, by
\[
\begin{array}{r c l}
\frown\ :\ H^p(X) \times H_q(X) & \to & H_{q-p}(X), \\\
(a, \sigma) & \mapsto & a\left(\sigma|_{[v_0,\dots,v_p]}\right)\cdot \sigma|_{[v_p,\dots,v_q]}.
\end{array}\]
The cap product with the orientation form of an orientable manifold $X$ gives the isomorphism of Poincare duality.

Remark: Given a map $f:X\to Y$, the cup and cap products satisfy certain identities via the induced map on cohomology groups. Let $a,b\in H^*(Y)$ and $c\in H_*(X)$ be cochain and chain classes, for which
\[f^*(a\smile b) = f^*(a)\smile f^*(b),
\hspace{1cm}
a\frown f_*c = f_*(f^*a\frown c).\]
The first identity asserts that $f^*$ is a ring homomorphism and the second describes the commutativity of an appropriate diagram. The cup and cap products are related by the equation
\[a(b\frown \sigma) = (a\smile b)\sigma,\]for $a\in H^p$, $b\in H^q$ and $\sigma\in C_{p+q}$.

References: Hatcher (Algebraic topology, Chapter 2.2), Prasolov (Elements of homology theory, Chapter 2)

Monday, November 7, 2016

Images of manifolds and transversality

 Preliminary exam prep

Let $X,Y$ be manifolds embedded in $\R^n$, and $f:X\to Y$ a map, with $df_x:T_xX\to T_{f(x)}Y$ the induced map on tangent spaces.

Definition: The map $f$ is a
  • homeomorphism if it is continuous and has a continuous inverse, 
  • diffeomorphism if it is smooth and has a smooth inverse,
  • injection if $f(a)=f(b)$ implies $a=b$,
  • immersion if $df_x$ is injective for all $x\in X$,
  • embedding if it is an immersion and $df_x$ is a homeomorphism onto its image,
  • submersion if $df_x$ is surjective for all $x\in X$.
Transversality is a mathematical relic whose only practical use is, perhaps, in classical algebraic geometry.

Definition: The manifolds $X$ and $Y$ are transverse if $T_pX\oplus T_pY \cong \R^n$ for every $p\in X\cap Y$. The map $f$ and $Y$ are transverse if $\text{im}(f)$ and $Y$ are transverse.

Note that being transverse (or transversal) is a symmetric, but not a reflexive, nor a transitive relation. Recall that a regular value of $f$ is $y\in Y$ such that $df_x:T_xX \to T_{f(x)}Y$ is surjective for all $x\in f^{-1}(y)$. If $y$ is not in the image of $f$, then $f^{-1}(y)$ is empty, so $y$ is trivially a regular value. Every value that is not a regular value is a critical value.

Theorem: (Preimage theorem) For every regular value $y$ of $f$, the subset $f^{-1}(y)\subset X$ is a submanifold of $X$ of dimension $\dim(X)-\dim(Y)$.

Now let $M$ be a submanifold of $Y$.

Corollary: If $f$ is transverse to $M$, then $f^{-1}(M)$ is a manifold, with $\codim_Y(M)=\codim_X(f^{-1}(M))$.

Theorem: (Transversality theorem) Let $\{g_s:X\to Y\ |\ s\in S\}$ be a smooth family of maps. If $g:X\times S\to Y$ is transverse to $M$, then for almost every $s\in S$ the map $g_s$ is transverse to $M$.

If we replace $f$ with $df$, and ask that it be transverse to $M$, then $df|_s$ is also transverse to $M$.

Example: Consider the map $g_s:X\to \R^n$ given by $g_s(X)=i(X)+s=X+s$, where $i$ is the embedding of $X$ into $\R^n$. Since $g(X\times \R^n)=\R^n$ and $g$ varies smoothly in both variables, we have that $g$ is transverse to $X$. Hence by the transversality theorem, $X$ is transverse to its translates $X+s$ for almost all $s\in \R^n$.

Theorem: (Sard) For $f$ smooth and $\dy Y=\emptyset$, almost every $y\in Y$ is a regular value of $f$ and $f|_{\dy X}$. Equivalently, the set of critical values of $f$ has measure zero.

Resources: Guillemin and Pollack (Differential topology, Chapters 1, 2), Lee (Introduction to smooth manifolds, Chapter 6)

Friday, November 4, 2016

Tools of homotopy

 Preliminary exam prep

Let $X,Y$ be topological spaces and $A$ a subspace of $X$. Recall that a path in $X$ is a continuous map $\gamma:I\to X$, and it is closed (or a loop), if $\gamma(0)=\gamma(1)$. When $X$ is pointed at $x_0$, we often require $\gamma(0)=x_0$, and call such paths (and similarly loops) based.

Definitions


Definition:
  • $X$ is connected if it is not the union of two disjoint nonempty open sets.
  • $X$ is path connected if any two points in $X$ have a path connecting them, or equivalently, if $\pi_0(X)=0$.
  • $X$ is simply connected if every loop is contractible, or equivalently, if $\pi_1(X)=0$.
  • $X$ is semi-locally simply connected if every point has a neighborhood whose inclusion into $X$ is $\pi_1$-trivial.
Path connectedness and simply connectedness have local variants. That is, for $P$ either of those properties, a space is locally $P$ if for every point $x$ and every neighborhood $U\owns x$, there is a subset $V\subset U$ on which $P$ is satisfied.

Remark: In general, $X$ is $n$-connected whenever $\pi_r(X)=0$ for all $r\leqslant n$. Note that 0-connected is path connected and 1-connected is simply connected and connected. Also observe that the suspension of path connected space is simply connected.

Definition:
  • A retraction (or retract) from $X$ to $A$ is a map $r:X\to A$ such that $r|_A = \id_A$.
  • A deformation retraction (or deformation retract) from $X$ to $A$ is a family of maps $f_t:X\to X$ continuous in $t,X$ such that $f_0 = \id_X$, $f_1(X) = A$, and $f_t|_A = \id_A$ for all $t$.
  • A homotopy from $X$ to $Y$ is a family of maps $f_t:X\to Y$ continuous in $t,X$.
  • A homotopy equivalence from $X$ to $Y$ is a map $f:X\to Y$ and a map $g:Y\to X$ such that $g\circ f \simeq \id_X$ and $f\circ g \simeq \id_Y$.
Definition: A pair $(X,A)$, where $A\subset X$ is a closed subspace, is a good pair, or has the homotopy extension property (HEP), if any of the following equivalent properties hold:
  • there exists a neighborhood $U\subset X$ of $A$ such that $U$ deformation retracts onto $A$,
  • $X\times \{0\}\cup A\times I$ is a retract of $X\times I$, or
  • the inclusion $i:A\hookrightarrow X$ is a cofibration.
In some texts such a pair $(X,A)$ is called a neighborhood deformation retract pair, and HEP is reserved for any map $A\to X$, not necessarily the inclusion, that is a cofibration. For more on cofibrations, see a previous blog post (2016-07-31, "(Co)fibrations, suspensions, and loop spaces").

Definition: There is a functor $\pi_1:\text{Top}_*\to \text{Grp}$ called the fundamental group, that takes a pointed topological space $X$ to the space of all pointed loops on $X$, modulo path homotopy.

This may be generalized to $\pi_n$, which takes $X$ to the space of all pointed embeddings of $S^n$.

Definition: Let $G,H$ be groups. The free product of $G$ and $H$ is the group
\[
G*H = \{a_1\cdots a_n\ :\ n\in \Z_{\>0}, a_i\in G\text{ or }H, a_i\in G(H)\implies a_{i+1}\in H(G)\},
\]
with group operation concatenation, and identity element the empty string $\emptyset$. We also assume $e_Ge_H=e_He_G=e_G=e_H=\emptyset$, for $e_G$ ($e_H$) the identity element of $G$ ($H$).

The above construction may be generalized to a collection of groups $G_1*\cdots*G_m$, where the index may be uncountable. If every $G_\alpha=\Z$ (equivalently, has one generator), then $*_{\alpha\in A} G_\alpha$ is called the free group on $|A|$ generators.

Theorems


Theorem: (Borsuk-Ulam) Every continuous map $S^n\to \R^n$ takes a pair of antipodal points to the same value.

Theorem: (Ham Sandwich theorem) Let $U_1,\dots,U_n$ be bounded open sets in $\R^n$. There exists a hyperplane in $\R^n$ that divides each of the open sets $U_i$ into two sets of equal volume.

Volume is taken to be Lebsegue measure. The Ham sandwich theorem is an application of Borsuk-Ulam (see Terry Tao's blog post for more).

Theorem: If $X$ and $Y$ are path-connected, then $\pi_1(X\times Y)\cong \pi_1(X)\times \pi_1(Y)$.

Now suppose that $X = \bigcup_\alpha A_\alpha$ is based at $x_0$ with $x_0\in A_\alpha$ for all $\alpha$. There are natural inclusions $i_\alpha:A_\alpha\to X$ as well as $j_\alpha:A_\alpha\cap A_\beta \to A_\alpha$ and $j_\beta:A_\alpha\cap A_\beta \to A_\beta$.
Both $i_\alpha$ and $j_\alpha$ induce maps on the fundamental group, each (and all) of the $i_{\alpha*}:\pi_1(A_\alpha)\to \pi_1(X)$ extending to a map $\Phi:*_\alpha \pi_1(A_\alpha)\to \pi_1(X)$.

Theorem: (van Kampen)
  • If $A_\alpha\cap A_\beta$ is path-connected, then $\Phi$ is a surjection. 
  • If $A_\alpha\cap A_\beta\cap A_\gamma$ is path connected, then $\ker(\Phi) = \langle j_{\alpha*}(g)(j_{\beta*}(g))^{-1}\ |\ g\in \pi_1(A_\alpha\cap A_\beta,x_0)\rangle$.
As a consequence, if triple intersections are path connected, then $\pi_1(X) \cong *_\alpha A_\alpha /\ker(\Phi)$. Moreover, if all double intersections are contractible, then $\ker(\Phi)=0$ and $\pi_1(X)\cong *_\alpha A_\alpha$.

Proposition: If $\pi_1(X)=0$ and $\widetilde H_n(X)=0$ for all $n$, then $X$ is contractible.

References: Hatcher (Algebraic topology, Chapter 1), Tao (blog post "The Kakeya conjecture and the Ham Sandwich theorem")

Tuesday, November 1, 2016

Explicit pushforwards and pullbacks

 Preliminary exam prep

Here we consider a map $f:M\to N$ between manifolds of dimension $m$ and $n$, respectively, and the maps that it induces. Let $p\in M$ with $x_1,\dots,x_m$ a local chart for $U\owns p$ and $y_1,\dots,y_n$ a local chart for $V\owns f(p)$. Induced from $f$ are the differential (or pushforward) $df$ and the pullback $df^*$, which are duals of each other:
\[
\begin{array}{r c l}
df_p\ :\ T_p M & \to & T_{f(p)}N \\[10pt]
df\ :\ TM & \to & TN \\
\alpha & \mapsto & (\beta\mapsto \alpha(\beta\circ f)) \\[10pt]\\\\
\end{array}
\hspace{1cm}
\begin{array}{r c l}
df^*_p\ :\ T_{f(p)}^* N & \to & T_p^*M\\[10pt]
df^*\ :\ T^*N & \to & T^*M \\
\omega & \mapsto & \omega\circ f\\[10pt]
\bigwedge ^k T^*N & \to & \bigwedge^k T^*M\\
\omega\ dy_1\wedge\cdots \wedge dy_k & \mapsto & (\omega \circ f)\ d(y_1\circ f)\wedge \cdots \wedge d(y_k\circ f)
\end{array}
\]
These maps may be described by the diagram below.
Example: For example, consider the map $f:\R^3\to \R^3$ given by $f(x,y,z) = (x-y,3z^2,xz+yz)$, with the image having coordinates $(u,v,w)$. With elements
\[
2x\frac\dy{\dy x} - 5z\frac\dy{\dy y}\in TM,
\hspace{2cm}
2uv+\sqrt w-5\in C^\infty(N),
\hspace{2cm}
\cos(uv)\in T^*N,
\]
we have
\begin{align*}
df_p\left(2x\frac\dy{\dy x} - 5z\frac\dy{\dy y}\right)(2uv+\sqrt w-5) & = \left(2x\frac\dy{\dy x} - 5z\frac\dy{\dy y}\right)\left(6(x-y)z^2+\sqrt{xz+yz}-5\right)(p),\\
df_p^*\left( \cos(uv)\right) & = \cos((x-y)3z^2), \\
\left(\textstyle\bigwedge^2 df_p^*\right)(\cos(uv)du\wedge dw) & = \cos((x-y)3z^2)d(3z^2)\wedge d(xz+yz) \\
& = \cos((x-y)3z^2)\left(-6z^2\ dx\wedge dz -6z^2\ dy \wedge dz\right).
\end{align*}