Sampling points uniformly on parametrized manifolds

Here I'll describe how to sample points uniformly on a (parametrized) manifold, along with an actual implementation in Python. Let $M$ be a $m$-dimensional manifold embedded in $\R^n$ via $f:\R^m\to \R^n$. Moreover, assume that $f$ is Lipschitz (true if $M$ is compact), injective (true if $M$ is embedded), and is a parameterization, in the sense that there is an $m$-rectangle $A = [a_1,b_1]\times \cdots \times [a_m,b_m]$ such that $f(A) = M$ (the intervals need not be closed). Set $(\widetilde Jf)^2 = \det(Df\cdot Df^T)$ to be the $m$-dimensional Jacobian, and calculate $$c = \int_{a_m}^{b_m}\cdots \int_{a_1}^{b_1} \widetilde J f\ dx_1\cdots dx_m.$$ Recall the brief statistical background presented in a previous blog post ("Reconstructing a manifold from sample data, with noise," 2016-05-26). A uniform or constant probability density function is valued the same at every point on its domain.

Proposition: In the setting above:

• (completely separable) Let $g_1,\dots,g_m$ be probability density functions on $[a_1,b_1],\dots,[a_m,b_m]$, respectively. If $g_1\cdots g_m = \widetilde J f / c$, then the joint probability density function of $g_1,\dots,g_m$ is uniform on $M$ with respect to the metric induced from $\R^n$.
• (non-separable) Let $g$ be a probability density function on $[a_1,b_1]\times\cdots\times[a_m,b_m]$. If $g=\widetilde J f/c$, then $g$ is uniform on $M$ with respect to the metric induced from $\R^n$.

A much more abstract statement and proof are given in [2], Section 3.2.5, but assuming $f$ is injective and $M$ is in $\R^n$, we evade the worst notation. Section 2.2 of [1] gives a brief explanation of how the given statement follows, while Section 2 of [3] goes into more detail of why the above is true.

Example: Let $M = S^2$, the sphere of radius $r$, and $f:[0,2\pi)\times [0,\pi) \to \R^3$ the natural embedding given by
$$(\theta,\varphi) \mapsto (r\cos(\theta)\sin(\varphi),r\sin(\theta)\sin(\varphi), r\cos(\varphi)),$$ with $$\begin{align*} Df & = \begin{bmatrix} -r \sin(\varphi) \sin(\theta) & r \cos(\theta)\sin(\varphi) &  0 \\ r \cos(\varphi) \cos(\theta) & r \cos(\varphi) \sin(\theta) & -r \sin(\varphi) \end{bmatrix}, & \widetilde Jf & = r^2\sin(\varphi), \\ Df\cdot Df^T & = \begin{bmatrix} r^2 \sin^2(\varphi) & 0 \\ 0 & r^2 \end{bmatrix}, & c & = 4\pi r^2. \end{align*}$$
Let $g_1(\theta) = 1/2\pi$ be the uniform distribution over $[0,2\pi)$, meaning that $g_2(\varphi) = \sin(\varphi)/2$ over $[0,\pi)$. Sampling points randomly from these two distributions and applying $f$ will give uniformly sampled points on $S^2$.

Example: Let $M = T^2$, the torus of major radius $R$ and minor radius $r$, and $f:[0,2\pi)\times [0,2\pi) \to \R^3$ the natural embedding given by
$$(\theta,\varphi) \mapsto ((R+r\cos(\theta))\cos(\varphi),(R+r\cos(\theta))\sin(\varphi), r\sin(\theta)),$$ with $$\begin{align*} Df & = \begin{bmatrix} -r \cos(\varphi)\sin(\theta) & -r \sin(\varphi) \sin(\theta) & r \cos(\theta) \\ -(R + r \cos(\theta)) \sin(\varphi) & \cos(\varphi) (R + r \cos(\theta)) & 0 \end{bmatrix}, & \widetilde Jf & = r(R+r\cos(\theta)), \\ Df\cdot Df^T & = \begin{bmatrix} r^2 & 0 \\ 0 & (R + r \cos(\theta))^2 \end{bmatrix}, & c & = 4\pi^2rR. \end{align*}$$
Let $g_2(\varphi) = 1/2\pi$ be the uniform distribution over $[0,2\pi)$, meaning that $g_1(\theta) = (1+r\cos(\theta)/R)/(2\pi)$ over $[0,2\pi)$. Sampling points randomly from these two distributions and applying $f$ will give uniformly sampled points on $T^2$.

Below I give a simple implementation of how to actually sample points, in Python using the SciPy package. The functions $f,g_1,\dots,g_m$ are all assumed to be given.

import scipy.stats as st

class var_g1(st.rv_continuous):
    'Uniform variable 1'
    def _pdf(self, x):
        return g1(x)
...
class var_gm(st.rv_continuous):
    'Uniform variable m'
    def _pdf(self, x):
        return gm(x)

dist_g1 = var_g1(a=a1, b=b1, name='Uniform distribution 1')
...
dist_gm = var_gm(a=am, b=bm, name='Uniform distribution m')

def mfld_sample():
    return f(dist_g1.rvs(),...,dist_gm.rvs())

A further application for this would be to understand how to sample points uniformly on projective manifolds, with a leading example the Grassmannian, embedded via Plucker coordinates.

References:
[1] Diaconis, Holmes, and Shahshahani (Sampling from a manifold, Section 2.2)
[2] Federer (Geometric measure theory, Section 3.2.5)
[3] Rhee, Zhou, and Qiu (An iterative algorithm for sampling from manifolds, Section 2)

The conditioning number of a helix, part 2

Recall the previous attempt to find the conditioning number of a helix (see post "The conditioning number of a helix, part 1," 2016-10-31). Here we complete the approach and although exact solutions are hard to find, we give close approximations.

The setting was a helix $C$ of radius $r$ and stretch $c$, so given as the zero locus of $x-r\cos(z/c)$ and $y-r\sin(z/c)$, and we wanted to find where the normal plane at a point $p\in C$ intersects $C$ again. It may intersect $C$ several times, but we are only interested in the shortest distances. Without loss of generality, assume that $p=(r,0,0)$. The normal plane at $p$ is then given by
$$0 = \det\begin{bmatrix} x-r\cos(p_z/c) & y-r\sin(p_z/c) & z-p_z \\ 1 & 0 & r\sin(p_z/c)/c \\ 0 & 1 & -r\cos(p_z/c)/c \end{bmatrix} = \det\begin{bmatrix} x-r & y & z \\ 1 & 0 & 0 \\ 0 & 1 & -r/c \end{bmatrix} = \frac rc y + z.$$ Since the cylinder on which the helix $C$ lies is $x^2+y^2=r^2$, the curve $C'$ representing the intersection of the plane with the cylinder is given by the zero locus of $\pm r\sqrt{x^2-r^2} + cz$. This allows us to find the intersection with the helix. However, since $C$ is parametrized with $z$ the free variable and $C'$ with $x$ free, its is easier to switch to cylindrical coordinates
$$\left(r = \sqrt{x^2+y^2},\theta = \arctan(y/x), z = z\right).$$ Doing so gives a nice description of $C$ and $C'$ as below. $$\begin{array}{r l c l} C : &  (r\cos(z/c),r\sin(z/c),z) & = & (r,\theta,\theta c) \\ C' : &  (x,-\sqrt{r^2-x^2},r\sqrt{r^2-x^2}/c) & = & (r.\theta,r^2\sin(\theta)/c) \end{array}$$ The switch in coordinates is represented by the diagram below, where we have only used the top half of $C'$.

Finding $C\cap C'$ is equivalent to solving $\frac{c^2}{r^2} = \frac{\sin(\theta)}\theta$ for $\theta$, a task that can not be solved exactly. Instead we take the tangent lines to $C'$ on the unrolled cylinder at its base, and see where those intersect the line $\theta c$. Inspecting the areas of the tangent lines closer and calculating the euclidean distances in $\R^3$ from $p$ to $a$ and $b$, which is, I can't believe I'm saying this, a great exercise for the reader, we get the distances to be
$$d(p,a) = \sqrt{2r^2\left(1+\cos\left(\frac{\pi c^2}{r^2-c^2}\right)\right) + \left(\frac{\pi cr^2}{r^2-c^2}\right)^2}, \hspace{.2cm} d(p,b) = \sqrt{2r^2\left(1-\cos\left(\frac{2\pi c^2}{r^2+c^2}\right)\right) + \left(\frac{2\pi c r^2}{r^2+c^2}\right)^2}.$$ Truthfully, the diagrams are tricky to draw in TikZ and I don't want to simply have a scan of some rough work. More importantly, $d(p,a) = d(p,b)$ implies $c = r/\sqrt 3$, meaning that when the stretch $c$ is larger than $r/\sqrt 3$, the normal planes certainly do not intersect the helix again.

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

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)

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)

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)

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)

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")

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
$$\begin{align*} df_p\left(2x\frac\dy{\dy x} - 5z\frac\dy{\dy y}\right)(2uv+\sqrt w-5) & = \left(2x\frac\dy{\dy x} - 5z\frac\dy{\dy y}\right)\left(6(x-y)z^2+\sqrt{xz+yz}-5\right)(p),\\ df_p^*\left( \cos(uv)\right) & = \cos((x-y)3z^2), \\ \left(\textstyle\bigwedge^2 df_p^*\right)(\cos(uv)du\wedge dw) & = \cos((x-y)3z^2)d(3z^2)\wedge d(xz+yz) \\ & = \cos((x-y)3z^2)\left(-6z^2\ dx\wedge dz -6z^2\ dy \wedge dz\right). \end{align*}$$

The conditioning number of a helix, part 1

Definition: Let $M$ be a smooth $d$-manifold embedded in $\R^n$ and $N^\epsilon_pM = N_pM \cap B(p,\epsilon)$ the natural embedding of the $\epsilon$-normal plane at $p\in M$. The pairwise conditioning number of $p$ and $q$ is
$$\tau_{p,q} = \sup\{\epsilon \ :\ N^\epsilon_pM\cup N^\epsilon_qM\text{ embeds in }\R^n\}.$$

The condition on $\epsilon$ is the same as saying $i(N^\epsilon_pM)\cap i(N^\epsilon_qM) = \emptyset$, where $i$ is induced by the embedding of $M$. It is immediate that $\tau = \inf_{p,q}\{\tau_{p,q}\}$, so we will try to find $\tau_{p,q}$ first. Recall that a helix of radius $r$ and vertical period $2\pi c$ is a 1-dimensional manifold

embedded in $\R^3$ as the zero locus of
$$f(x,y,z) = x-r\cos(z/c), \hspace{2cm} g(x,y,z) = y-r\sin(z/c).$$
We first find the normal plane at two arbitrary points $p_1,p_2$ on the helix, then their intersection (which is a line), and then the distance from $p_1$ and $p_2$ to that line. The smallest of these two distances bounds $\tau_{p_1,p_2}$ from below (and the bound is achieved on pairs of points defining the medial axis). Then take the infimum of this value over all points on the helix. However, this excludes the case when the normal planes are parallel (for instance when the two points have the same $x$- and $y$-values).

Moreover, even just calculating the infimum for points whose normal planes are not parallel yields a result of zero. We describe the process nonetheless. For the first step, we need the equations of the normal planes. Let
$$D^f = \begin{bmatrix} 1 & 0 & r\sin(z/c)/c \end{bmatrix}, \hspace{2cm} D^g = \begin{bmatrix} 0 & 1 & -r\cos(z/c)/c \end{bmatrix}.$$
be the Jacobians of $f$ and $g$. The points $p_1$, $p_2$ are completely described by the $z$-coordinate, so we have two values $z_1$, $z_2$ for $p_1$, $p_2$, respectively. The normal plane at $p_i$ is the zero locus of
$$\det\begin{bmatrix} x-r\cos(z_i/c) & y-r\sin(z_i/c) & z-z_i \\ 1 & 0 & r\sin(z_i/c)/c \\ 0 & 1 & -r\cos(z_i/c)/c \end{bmatrix}  = z - z_i - \frac{xr}c \sin(z_i/c) + \frac{yr}c\cos(z_i/c).$$
We have two equations and three unknowns, so one independent variable. Solving for $x$ and $y$ gives us
$$x = \frac{(z-z_1)\cos(z_2/c) - (z-z_2)\cos(z_1/c)}{r\sin(\frac{z_1-z_2}c)/c}, \hspace{1cm} y = \frac{(z-z_1)\sin(z_2/c) - (z-z_2)\sin(z_1/c)}{r\sin(\frac{z_1-z_2}c)/c}.$$
These are functions of $z$, giving us two new functions
$$h_i(z) = (x(z)-r\cos(z_i/c))^2+(y(z)-r\sin(z_i/c))^2 + (z-z_i)^2,$$
for $i=1,2$, which, when minimized, give a lower bound for the pairwise conditioning number of $p_1$ and $p_2$. Indeed, by slowly increasing the $\epsilon$ until the $\epsilon$-normal planes at $p_1$ and $p_2$ intersect, the first point of intersection will happen on the intersection $N_{p_1}M\cap N_{p_2}M$. Hence finding the shortest distance from $p_1$ and $p_2$ to this line gives a definite lower bound. The functions $h_i$ are quadratic in $z$, and we know the function $az^2+bz+c$, for $a>0$, has minimum at $-b/2a$. The values of $h_1$ and $h_2$ at their minima are the same and equal to
$$h_m := h_i\left(\frac{-b}{2a}\right) = \frac{2(c^2+r^2)\cos^2\left(\frac{z1-z2}{2c}\right)\left(r^2+c(z_1-z_2)\csc\left(\frac{z_1-z_2}c\right)\right)^2}{2c^2r^2+r^4+r^4\cos\left(\frac{z_1-z_2}c\right)}.$$
A natural limit of $h_m$ to consider is $z_2\to z_1$. If any of the factors in the numerator are zero, we also get a minimum, so another limit to look for is $z_2\to cz_1$, which makes the cosine factor zero. These are
$$\lim_{z_2\to z_1}[h_m] = \frac{(c^2+r^2)^2}{r^2}, \hspace{2cm} \lim_{z_2\to z_1+c\pi}[h_m] = \frac{c^2\pi^2(c^2+r^2)}{4r^2},$$
which are finite nonzero for positive values of $c$ and $r$. For the last factor, fix $z_1=0$. Then finding when the factor vanishes is equivalent to finding when $\sin(z_2/c)$ and $-z_2 c/r^2$ intersect. There are values for which this happens, and the other factors in $h_m$ are all finite at these values, so $\inf_{z\in \R}[h_i(z)]=0$. Visual confirmation is given by the cases below.

Hence this is not the best approach to calculate the conditioning number of a curve. The next attempt will be to calculate the actual pairwise conditioning number, rather than trying to bound it from below.

Basic topological constructions

 Preliminary exam prep

Let $X,Y$ be topological spaces based at $x_0,y_0$, respectively, and $I=[0,1]$ the unit interval.
$$\begin{array}{r r c l} \text{cone} & CX & = & X\times I / X\times \{0\} \\[5pt] \text{suspension} & \Sigma X & = & X\times I / X\times \{0\}, X\times \{1\}\\[5pt] \text{reduced suspension} & \widetilde\Sigma X & = & X\times I/X\times\{0\}, X\times \{0\}, \{x_0\}\times I \\[5pt] \text{wedge} & X\vee Y & = & X\sqcup Y / \{x_0\} \sim \{y_0\} \\[5pt] \text{smash} & X\wedge Y & = & X\times Y / X\times \{y_0\}, \{x_0\}\times Y \\[5pt] \text{join} & X * Y & = & X\times Y \times I \left/\begin{array}{l l} X\times \{y\}\times \{0\} & \forall\ y\in Y \\ \{x\}\times Y \times \{1\} & \forall\ x\in X \end{array}\right. \\[5pt] \text{connected sum} & X \# Y & = & (X\setminus D^n_X)\sqcup (Y\setminus D^n_Y) / \partial D^n_X \sim \partial D^n_Y \end{array}$$
In the last description, $X$ and $Y$ are assumed to be $n$-manifolds, with $D^n_X$ a closed $n$-dimensional disk in $X$ (similarly for $Y$). The quotient identification may also be made via some non-trivial map. In fact, only the interior of each $n$-disk is removed from $X$ and $Y$, so that the quotient makes sense.

Remark: Some of the above constructions may be expressed in terms of others, for example
$$X\wedge Y = X\times Y / X\vee Y, \hspace{1cm} X*Y = \Sigma(X\wedge Y).$$
The first is clear by viewing $X = X\times \{y_0\}$ and $Y = \{x_0\}\times Y$ as sitting inside $X\times Y$. The second is clear by letting $X\times \{y\}\times \{0\}$ be identified to $\{x_0\}\times\{y\}\times \{0\}$ for every $y\in Y$, and analogously with $Y$.

Example: Here are some of the constructions above applied to some common spaces.
$$\begin{align*} CX & \simeq \text{pt} & \Sigma S^n & = S^{n+1} & S^n \wedge S^m & = S^{n+m}\\ \Sigma X & = S^1 \wedge X & S^n * S^m & = S^{n+m+1}\end{align*}$$
Remark: We may also calculate the homology of the new spaces in terms of the old ones.
$$\begin{array}{r c l l} \widetilde H_k(CX) & = & 0 & \text{via homotopy} \\ \widetilde H_k(\Sigma X) & = & \widetilde H_{k-1}(X) & \text{via Mayer--Vietoris} \\ \widetilde H_k(X\vee Y) & = & \widetilde H_k(X)\oplus \widetilde H_k(Y) & \text{via Mayer--Vietoris}\\ \widetilde H_k(X\wedge S^\ell) & = & \widetilde H_{k-\ell}(X) & \text{via Kunneth} \\ \widetilde H_k(X\# Y) & = & \widetilde H_k(X) \oplus \widetilde H_k(Y) & \text{via Mayer--Vietoris and relative homology} \end{array}$$
The last equality holds for $k<n-1$, for $M$ and $N$ both $n$-manifolds, and for $k=n-1$ when at least one of them is orientable.

References: Hatcher (Algebraic Topology, Chapters 0, 2)

Tools of (co)homology

 Preliminary exam prep

Let $X,Y$ be topological spaces, $G$ a group, and $R$ a unital commutative ring.

Defining homology groups

Theorem: If $(X,A)$ is a good pair (there exists a neighborhood $U\subset X$ of $A$ such that $U$ deformation retracts onto $A$), then for $i:A\hookrightarrow X$ the inclusion and $q:X\twoheadrightarrow X/A$ the quotient maps, there exists a long exact sequence of reduced homology groups
$$\cdots \to \widetilde H_n(A) \tov{i_*} \widetilde H_n(X) \tov{q_*} \widetilde H_n(X/A) \to \cdots.$$

Theorem: For any pair $(X,A)$, there exists a long exact sequence of homology groups
$$\cdots \to H_n(A) \to H_n(X) \to H_n(X,A) \to \cdots,$$
where the last is called a relative homology group. Hence $H_n(X,A)\cong \widetilde H_n(X/A)$ for a good pair $(X,A)$.

Theorem (Excision): For any triple of spaces $(Z,A,X)$ with $\text{cl}(Z)\subset \text{int}(A)$, there is an isomorphism $H_n(X-Z,A-Z)\cong H_n(X,A)$.

For any $x\in X$, the local homology of $X$ at $x$ is the relative homology groups $H_n(X,X-\{x\})$. By excision, these are isomorphic to $H_n(U,U-\{x\})$ for $U$ any neighborhood of $x$. If $X$ is nice enough around $x$ (that is, if $U\cong \R^k$), then these groups are isomorphic to $H_n(\R^k,\R^k-\{x\})\cong H_n(D^k,\dy D^k) = H_n(S^k)$.

Theorem (Mayer-Vietoris): For $X=A\cup B$, there is a long exact sequence of homology groups
$$\cdots \to H_n(A\cap B) \to H_n(A)\oplus H_n(B) \to H_n(X) \to \cdots,$$

and if $A\cap B$ is non-empty, there is an analogous sequence for reduced homology groups.

Extending with coefficients

Recall the $\Tor$ and $\Ext$ groups, which were, respectively, the left and right derived functors of, respectively, $\otimes$ and $\Hom$ (see post "Exactness and derived functors," 2016-03-20). Here we only need $\Tor_1$ and $\Ext^1$, which are given by, for any groups (that is, $\Z$-modules) $A$, $B$,
$$\begin{array}{r c c c l} \Tor(A,B) & = & H_1(\text{projres}(A)\otimes B) & = & H_1(A\otimes \text{projres}(B)), \\ \Ext(A,B) & = & H^1(\Hom(A,\text{injres}(B))) & = & H^1(\Hom(\text{projres}(A),B)). \end{array}$$
Note that $\Tor$ is symmetric in its arguments, while $\Ext$ is not. Recall that $\Tor_0(A,B)=A\otimes B$ and $\Ext^0(A,B) = \Hom(A,B)$.

Theorem (Universal coefficient theorem): There exist isomorphisms
$$\begin{array}{r c c c l} H_n(X;G) & \cong & \Hom(H^n(X),G)\oplus \Ext(H^{n+1}(X),G) & \cong & H_n(X)\otimes G\ \oplus\ \Tor(H_{n-1}(X),G),  \\ H^n(X;G) & \cong & \Hom(H_n(X),G)\oplus \Ext(H_{n-1}(X),G) & \cong & H^n(X)\otimes G\ \oplus\ \Tor(H^{n+1}(X),G). \end{array}$$

Here are some common $\Hom$, $\Tor$, and $\Ext$ groups:
$$\begin{align*} \Hom(\Z,G) & = G & \Tor(\Z,G) & = 0 & \Ext(\Z,G) & = 0 \\ \Hom(\Z_m,\Z) & = 0 & \Tor(G,\Z) & = 0 & \Ext(\Z_m,\Z) & = \Z_m \\ \Hom(\Z_m,\Z_n) & = \Z_{\gcd(m,n)} & \Tor(\Z_m,\Z_n) & = \Z_{\gcd(m,n)} & \Ext(\Z_m,\Z_n) & = \Z_{\gcd(m,n)} \\ \Hom(\Q,\Z_n) & = 0 & & & \Ext(\Q,\Z_n) & = 0 \\ \Hom(\Q,\Q) & = \Q & & & \Ext(G,\Q) & = 0 \end{align*}$$

Theorem (Künneth formula): For $X,Y$ CW-complexes, $F$ a field, and $H^k(Y;G)$ or $H^k(X;G)$ finitely generated for all $k$, there are isomorphisms, for all $k$,

$$H_k(X\times Y;F) \cong \bigoplus_{i+j=k} H_i(X;F)\otimes_FH_j(Y;F), \hspace{1cm} H^k(X\times Y;G) \cong \bigoplus_{i+j=k} H^i(X;G)\otimes_GH^j(Y;G)$$

Dualities

Theorem (Poincaré duality): For $X$ a closed $n$-manifold (compact, without boundary) that is $R$-orientable (consistent choice of $R$-generator for each local homology group), for $k=0,\dots,n$ there are isomorphisms

$$H^k(X;R)\cong H_{n-k}(X;R).$$

Note that a simply orientable manifold means $\Z$-orientable. A manifold that is not $\Z$-orientable is always $\Z_2$-orientable (in fact all manifolds are $\Z_2$-orientable).

Theorem (Alexander duality): For $X\subsetneq S^n$ a non-empty closed locally contractible subset, for $k=0,\dots,n-1$ there are isomorphisms
$$\widetilde H^k(X) \cong \widetilde H_{n-k-1}(S^n-X).$$

References: Hatcher (Algebraic topology, Chapters 2, 3), Aguilar, Gitler, and Prieto (Algebraic Topology from a Homotopical Viewpoint, Chapter 7)

Vector fields

 Preliminary exam prep

Here we will have an overview of vector fields and all things related to them. Let $M$ be an $n$-dimensional manifold, and $\pi:M\to TM$ its tangent bundle.

Definition: A vector field is a map $X:M\to TM$ such that $\pi\circ X = \id_M$.

A vector field may also be viewed as a section of the tangent bundle, and smooth vector fields as the space of smooth sections $\Gamma(TM)$. Given a chart $(U,\varphi)$ of $M$ near $p$, we have the pushforward $\varphi_*:T_pM\to T_{\varphi(p)}(\R^n) = \R^n$, where we may assume $\varphi(p)=0$. Given the standard basis $\{e_i\}$ of $\R^n$, we get a basis of $T_pM$ given by
$$\left\{\left.\frac{\dy}{\dy x_i}\right|_p = (\varphi_*)^{-1}(e_i)\right\}_{i=1}^n.$$
Recall that $TM$ may be viewed as the space of derivations, or maps $C^\infty(M)\to \R$ satisfying the Leibniz rule. Then for $p\in M$, we have $X(p):C^\infty(M)\to \R$, so we have $X(p)(f) = X_p(f)\in \R$ for all $f\in C^\infty(M)$. Hence $X_p\in T_pM$, and $X(f)\in C^\infty(M)$. Briefly,
$$\begin{array}{r c l} f\ :\ M & \to & \R, \\ X\ :\ M & \to & TM, \end{array} \hspace{2cm} \begin{array}{r c l} Xf\ :\ M & \to & \R, \\ fX\ :\ M & \to & TM. \end{array}$$

Definition: Given a vector field $X\in \Gamma(TM)$, an integral curve of $X$ is a smooth curve $\gamma:\R \to M$ such that $\gamma'(t) = X_{\gamma(t)}$ for all $t\in \R$.

The domain of $\gamma$ need not be all of $\R$, though any integral curve may be extended to a maximal integral curve, for which the domain can not be made larger. A collection of integral curves for a particular vector field is a flow.


Definition: A flow, or a one paramater group of diffeomorphisms, is a smooth map $\psi:\R\times M\to M$ such that

1. $\psi(t,\cdot)$ is a diffeomorphism of $M$, for all $t$,
2. $\psi(0,\cdot) = \id_M$,
3. $\psi(s+t,\cdot) = \psi(s,\cdot)\circ \psi(t,\cdot)$.

For convenience, we write $\psi_t(p) = \psi(t,p)$, Note that fixing $p\in M$, the map $\psi(\cdot,p)$ is a integral curve. Moreover, flows and vector fields are related uniquely by
$$\left.\frac{d f}{d t} \psi_t(p)\right|_{t=0} = X_p(f).$$
Indeed, if we have a flow $\psi$ and an element $f\in \Hom(T^*_pM,\R)$, this gives us a vector field $X\in \Gamma(TM)$. Conversely, if we have a vector field $X$, by the existence and uniqueness of solutions to first order ordinary differential equations (with boundary conditions), we can find a $\psi$ that satisfies this equality.

Definition: Let $X,Y\in \Gamma(TM)$ and $\psi$ be the associated flow of $X$. The Lie derivative of $Y$ in the direction of $X$, or Lie bracket of $X$ and $Y$, is an element of $\Gamma(TM)$ given by
$$\begin{align*} \left(\mathcal L_XY\right)_p(f) & = \left.\frac{df}{dt}\right|_{t=0}\bigg((\psi_t)_*^{-1}(Y_{\psi_t(p)}(f))\bigg) \\ & = [X,Y]_p(f) \\ & = X_p(Y(f)) - Y_p(X(f)) \end{align*}$$

The Lie derivative has some properties, among them $\mathcal L_X(fY) = X(fY) + f(\mathcal L_XY)$ for any $f\in C^\infty(M)$. If we let $Y$ be the map $M\to TM$ given by
$$\begin{array}{r c l} Y\ :\ M & \to & \Hom(T^*M,\R),\\ p & \mapsto & \left(\begin{array}{r c l} f_p\ :\ C^\infty(M) & \to & \R, \\ g & \mapsto & g(p), \end{array}\right), \end{array}$$
then $Yf = f$, so $\mathcal L_XY = X-X = 0$, and we have $\mathcal L_X f = Xf$.

Remark: Vector fields are 1-forms, or elements of $\mathcal A^0_M(TM) = \Gamma(TM\otimes \bigwedge^0T^*M) = \Gamma(TM)$. We may generalize the definition above to consider the Lie derivative $\mathcal L_X\omega$ of a differential $k$-form $\omega$ . Note that a differential $k$-form takes in $k$ vector fields and gives back a smooth function $M\to \R$. With this in mind, we may define new operations on vector fields:
$$\begin{align*} (\mathcal L_X\omega)(Y_1,\dots,Y_k) & = \mathcal L_X(\omega(Y_1,\dots,Y_k)) - \sum_{i=1}^k\omega(Y_1,\dots,\mathcal L_XY_i,\dots,Y_k) \\ (d\omega)(Y_1,\dots,Y_{k+1}) & = \sum_{i=1}^{k+1}(-1)^{i-1}Y_i(\omega(Y_1,..,\widehat{Y_i},..,Y_{k+1})) + \sum_{j>i=1}^{k+1}(-1)^{i+j}\omega([Y_i,Y_j],Y_1,..,\widehat{Y_i},..,\widehat{Y_j},..,Y_{k+1}) \\ (i_X\omega)(Y_1,\dots,Y_{k-1}) & = \omega(X,Y_1,\dots,Y_{k-1}) \end{align*}$$

The last is the interior product. All three are related by Cartan's formula $\mathcal L_X\omega = d(i_X\omega)+i_X(d\omega)$:
$$\begin{align*} (\mathcal L_{Y_1}\omega)(Y_2,\dots,Y_{k+1}) & = Y_1(\omega(Y_2,\dots,Y_{k+1})) - \sum_{i=2}^{k+1}\omega(Y_2,\dots,[Y_1,Y_i],\dots,Y_k) \\ & = Y_1(\omega(Y_2,\dots,Y_{k+1})) - \sum_{i=2}^{k+1}(-1)^i\omega([Y_1,Y_i],Y_2,\dots,\widehat{Y_i},\dots,Y_k) \\ (d(i_{Y_1}\omega))(Y_2,\dots,Y_{k+1}) & =  \sum_{i=2}^{k+1}(-1)^iY_i(\omega(Y_1,..,\widehat{Y_i},..,Y_{k+1})) - \sum_{j>i=2}^{k+1}(-1)^{i+j}\omega([Y_i,Y_j],Y_1,..,\widehat{Y_i},..,\widehat{Y_j},..,Y_{k+1})\\ (i_{Y_1}(d\omega))(Y_2,\dots,Y_{k+1}) & = (d\omega)(Y_1,\dots,Y_{k+1}) \\ & = \sum_{i=1}^{k+1}(-1)^{i-1}Y_i(\omega(Y_1,..,\widehat{Y_i},..,Y_{k+1})) + \sum_{j>i=1}^{k+1}(-1)^{i+j}\omega([Y_i,Y_j],Y_1,..,\widehat{Y_i},..,\widehat{Y_j},..,Y_{k+1}) \end{align*}$$

Remark: The action of a $k$-differential form on a $k$-vector field is given by $$\left(dx_1\wedge \cdots \wedge dx_k\right)\left(\frac\dy{\dy y_1},\dots,\frac\dy{\dy y_p}\right) = \det\begin{bmatrix} dx_1\frac\dy{\dy y_1} & dx_1\frac{\dy}{\dy y_2} & \cdots & dx_1\frac\dy{\dy y_p} \\ dx_2\frac\dy{\dy y_1} & dx_2\frac{\dy}{\dy y_2} & \cdots & dx_2\frac\dy{\dy y_p} \\ \vdots & \vdots & \ddots & \vdots \\ dx_p\frac\dy{\dy y_1} & dx_p\frac{\dy}{\dy y_2} & \cdots & dx_p\frac\dy{\dy y_p} \end{bmatrix} = \det\left(dx_i\frac\dy{\dy y_j}\right).$$ This may be generalized to get a map $\wedge^k T^*M \oplus \Gamma(TM)^{\oplus \ell} \to \bigwedge^{k-\ell}T^*M$, for $\ell\leqslant k$. For example, given a basis $x,y$ of our space $M$, $$(dx\wedge dy)\left(x\frac\dy{\dy x} + y \frac\dy{\dy y}\right) = dx\left(x\frac\dy{\dy x} + y \frac\dy{\dy y}\right)dy - dy\left(x\frac\dy{\dy x} + y \frac\dy{\dy y}\right)dx = x\ dy - y\ dx.$$ When $\ell=1$, this is just the interior product.

References: Lee (Introduction to smooth manifolds, Chapter 8), Hitchin (Differentiable manifolds, Chapter 3)

The tangent space and differentials

 Preliminary exam prep

Let $M,N$ be smooth $n$-manifolds. Here we discuss different definitions of the tangent space and differentials, or pushforwards, of smooth maps $f:M\to N$.

Derivations (Lee)

Definition: A derivation of $M$ at $p\in M$ is a linear map $v:C^\infty(M)\to \R$ such that for all $f,g\in C^\infty(M)$,
$$v(fg) = f(p)v(g) + g(p)v(f).$$
The tangent space $T_pM$ to $M$ at $p$ is the set of all derivations of $M$ at $p$.

Given a smooth map $F:M\to N$ and $p\in M$, define the differential $dF_p:T_pM\to T_{f(p)}N$, which, for $v\in T_pM$ and $f\in C^\infty(N)$ acts as
$$dF_p(v)(f) = v(f\circ F)\in \R.$$

Dual of cotangent (Hitchin)

Definition: Let $Z_p\subset C^\infty(M)$ be the functions whose derivative vanishes at $p\in M$. The cotangent space $T_p^*M$ to $M$ at $P$ is the quotient space $C^\infty(M)/Z_p$. The tangent space to $M$ at $P$ is the dual of the cotangent space $T_pM = (T_p^*M)^* = \Hom(T_p^*M,\R)$.

Given a smooth map $F:M\to N$ and $p\in M$, define the differential
$$\begin{array}{r c l} dF_p\ :\ T_pM & \to & T_{F(p)}N, \\ \left(f:C^\infty(M)/Z_p \to \R\right) & \mapsto & \left(\begin{array}{r c l} g\ :\ C^\infty(N)/Z_{F(p)} & \to & \R, \\ h & \mapsto & f(h\circ F). \end{array}\right) \end{array}$$
This definition makes clear the relation to the first approach. Since $h\not\in Z_{F(p)}$, the derivative of $h$ does not vanish at $F(p)$. Hence the derivative of $h\circ F$ at $p$, which is the derivative of $h$ at $F(p)$ multiplied by the derivative of $F$ at $p$, does not a priori vanish at $p$.

Derivative of chart map (Guillemin and Pollack)

Definition: Let $f:\R^n\to \R^m$ be a smooth map. Then the derivative of $f$ at $x\in \R^n$ in the direction $y\in \R^n$ is defined as
$$df_x(y) = \lim_{h\to 0}\left[\frac{f(x+yh)-f(x)}h\right].$$
Given $x\in M$ and charts $\varphi:\R^n\to M\subset \R^m$, the tangent space to $M$ at $p$ is the image $T_pM = d\varphi_0(\R^n)$, where we assume $\varphi(0)=p$.

Given a smooth map $F:M\to N$ and charts $\varphi:\R^n\to M$, $\psi:\R^n\to N$, with $\varphi(0)=p$ and $\psi(0)=F(p)$, define the differential $dF_p:T_pM\to T_{F(p)}N$ via the diagrams below.

Here $h = \psi^{-1}\circ F\circ \varphi$, so $dh_0$ is well-defined. Hence $dF_p = d\psi_0\circ dh_0\circ d\varphi_0^{-1}$ is also well-defined.

Sometimes the differential is referred to as the pushforward, in which case it is denoted by $(F_*)_p$.

References: Lee (Introduction to Smooth Manifolds, Chapter 3), Hitchin (Differentiable manifolds, Chapter 3.2), Guillemin and Pollack (Differential topology, Chapter 1.2)

Degree and orientation

 Preliminary exam prep

Topology

 Recall that a topological manifold is a Hausdroff space in which every point has a neighborhood homeomorphic to $\R^n$ for some $n$. An orientation on $M$ is a choice of basis of $\R^n$ in each neighborhood such that every path in $M$ keeps the same orientation in each neighborhood. Every manifold $M\owns x$ appears in a long exact sequence (via relative homology) with three terms
$$H_n(M-\{x\}) \tov{f} H_n(M) \tov{g} H_n(M,M-\{x\}).$$
The first term is 0, because removing a point from an $n$-dimensional space leaves only its $(n-1)$-skeleton, which is at most $(n-1)$-dimensional. For $U$ a neighborhood of $x$ in $M$, the last term (via excision) is
$$H_n(M-U^c, M-\{x\}-U^c) = H_n(U, U-\{x\})\cong H_n(\R^n, \R^n-\{x\}) \cong H_n(\R^n, S^{n-1}),$$
which in turn fits into a long exact sequence whose interesting part is
$$H_n(\R^n)\to H_n(\R^n,S^{n-1}) \to H_{n-1}(S^{n-1})\to H_{n-1}(\R^n),$$
and since the first and last terms are zero, $H_n(M,M-\{x\})=\Z$. Since $f$ is zero, $g$ into $\Z$ must be injective, meaning that $H_n(M)=\Z$ or 0.

Theorem: Let $M$ be a connected compact (without boundary) $n$-manifold. Then

1. if $M$ is orientable, $g$ is an isomorphism for all $x\in M$, and
2. if $M$ is not orientable, $g=0$.

Definition: Let $f:M\to N$ be a map of connected, oriented $n$-manifolds. Since $H_n(M)=H_n(N)$ is infinite cyclic, the induced homomorphism $f_*:H_n(M)\to H_n(N)$ must be of the form $x\mapsto dx$. The number $d$ is called the degree of $f$.

In the special case when we are computing the degree for a map $f:S^n\to S^n$, by excision we get
$$\deg(f) = \sum_{x_i\in f^{-1}(y)} \deg\left(H_n(U_i,U_i-x_i)\tov{f_*} H_n(V,V-y)\right),$$ for any $y\in Y$, some neighborhood $V$ of $y$, and preimages $U_i$ of $V$. This is called the local degree of $f$.

Geometry

Let $M$ be a smooth $n$-manifold. Recall $\Omega_M^r$ is the space of $r$-forms on $M$ and $d^r:\Omega^r_M\to \Omega^{r+1}_M$ is the differential map. Also recall the de Rham cohomology groups $H^r(M) = \text{ker}(d^r)/\text{im}(d^{r-1})$.

Definition: An $n$-manifold $M$ is orientable if it has a nowhere-zero $n$-form $\omega\in \Omega^n_M$. A choice of $\omega$ is called an orientation of $M$.

We also have a map $H^n(M)\to \R$, given by $\alpha\mapsto \int_M\alpha$, where the integral is normalized by the volume of $M$, so that integrating 1 across $M$ gives back 1. It is immediate that this doesn't make sense when $M$ is not compact, but when $M$ is compact and orientable, we get that $H^n(M)\neq0$. Indeed, if $\eta\in \Omega^{n-1}_M$ with $d\eta =\omega$, by Stokes' theorem we have
$$\int_M\omega = \int_Md\eta = \int_{\dy M}\eta = \int_\emptyset\eta = 0,$$
as $M$ has no boundary (since it is a manifold). But $\omega$ is nowhere-zero, meaning the first expression on the left cannot be zero. Hence $\omega$ is not exact and is a non-trivial element of $H^n(M)$.

Theorem: Let $M$ be a smooth, compact, orientable manifold of dimension $n$. Then $H^n(M)$ is one-dimensional.

Proof: The above discussion demonstrates that $\dim(H^n(M))\>1$. We can get an upper bound on the dimension by noting that the space of $n$-forms on $M$, given by $\Omega^n_M = \bigwedge^n(TM)^*$, has elements described by $dx_{i_1}\wedge \cdots \wedge dx_{i_n}$, with $\{i_1,\dots,i_n\}\subset \{1,\dots,n\}$. By rearranging the order of the $dx_{i_j}$, every element looks like $\alpha dx_1\wedge \cdots\wedge dx_n$ for some real number $\alpha$. Hence $\dim(\Omega^n_M) \leqslant 1$, so $\dim(H^n(M))$ is either 0 or 1. Therefore $\dim(H^n(M))=1$. $\square$

Definition: Let $f:M\to N$ be a map of smooth, compact, oriented manifolds of dimension $n$. Since $H^n(M)$ and $H^n(N)$ are 1-dimensional, the induced map $f^*:H^n(N)\to H^n(M)$ must be of the form $x\mapsto dx$. The number $d$ is called the degree of $f$. Equivalently, for any $\omega\in \Omega^n_N$,
$$\int_M f^*\omega = d\int_N\omega$$

References: Hatcher (Algebraic Topology, Chapters 2, 3.3), Lee (Introduction to Smooth Manifolds, Chapter 17)

The Grassmannian is a complex manifold

 Lecture topic

Let $Gr(k,\C^n)$ be the space of $k$-dimensional complex subspaces of $\C^n$, also known as the complex Grassmannian. We will show that it is a complex manifold of dimension $k(n-k)$. Thanks to Jinhua Xu and professor Mihai Păun for explaining the details.

To begin, take $P\in Gr(k,\C^n)$ and an $n-k$ subspace $Q$ of $\C^n$, such that $P\cap Q = \{0\}$. Then $P\oplus Q = \C^n$, so we have natural projections

A neighborhood of $P$, depending on $Q$ may be described as $U_Q = \{S\in Gr(k,\C^n)\ :\ S\cap Q = \{0\}\}$. We claim that $U_Q \cong \Hom(P,Q)$. The isomorphism is described by
$$\begin{array}{r c c c l} \Hom(P,Q) & \to & U_Q & \to & \Hom(P,Q), \\ A & \mapsto & \{v+Av\ :\ v\in P\}, \\ & & S & \mapsto & \left(\pi_Q|_S\right) \circ \left(\pi_P|_S\right)^{-1}. \end{array}$$
The map on the right, call it $\varphi_Q$, is also the chart for the manifold structure. The idea of decomposing $\C^n$ into $P$ and $Q$ and constructing a homomorphism from $P$ to $Q$ may be visualized in the following diagram.

Then $\Hom(P,Q) \cong \Hom(\C^k,\C^{n-k})\cong \C^{k(n-k)}$, so $Gr(k,\C^n)$ is locally of complex dimension $k(n-k)$. To show that there is a complex manifold structure, we need to show that the transition functions are holomorphic. Let $P,P'\in Gr(k,\C^n)$ and $Q,Q'\in Gr(n-k,\C^n)$ such that $P\cap Q = P'\cap Q' = \{0\}$. Let $X\in \Hom(P,Q)$ such that $X\in \varphi_Q(U_Q\cap U_{Q'})$, with $\varphi_Q(S)=X$ and $\varphi_{Q'}(S)=X'$ for some $S\in U_Q\cap U_{Q'}$. Define $I_X(v) = v+Xv$, and note the transition map takes $X$ to
$$\begin{align*} X' & = \varphi_{Q'}\circ \varphi_Q^{-1}(X) & (\text{definition}) \\ & = \varphi_{Q'}(S) & (\text{assumption})\\ & = \left(\pi_{Q'}|_S\right)\circ \left(\pi_{P'}|_S\right)^{-1} & (\text{definition}) \\ & = \left(\pi_{Q'}|_S\right)\circ I_X\circ I_X^{-1}\circ \left(\pi_{P'}|_S\right)^{-1} & (\text{creative identity}) \\ & = \left(\pi_{Q'}|_S\circ I_X\right)\circ \left(\pi_{P'}|_S\circ I_X\right)^{-1} & (\text{redistribution}) \\ & = \left(\pi_{Q'}|_P +\pi_{Q'}|_Q\circ X\right) \circ \left(\pi_{P'}|_P +\pi_{P'}|_Q\circ X\right). & (\text{definition}) \end{align*}$$
At this last step we have compositions and sums of homomorphisms and linear maps, which are all holomorphic. Hence the transition functions of $Gr(k,\C^n)$ are holomorphic, so it is a complex manifold.

Complexes and their homology

 Preliminary exam prep

Here I'll present complexes from the most restrictive to the most general. Recall the standard $n$-simplex is
$$\Delta^n = \{x\in \R^{n+1}\ :\ \textstyle\sum x_i = 1, x_i\>0\}.$$

Definition: Let $V$ be a finite set. A simplicial complex $X$ on $V$ is a set of distinct subsets of $V$ such that if $\sigma\in X$, then all the subsets of $\sigma$ are in $X$.

Every $n$-simplex in a simplicial complex is uniquely determined by its vertices, hence no pair of lower dimensional faces of a simplex may be identified with each other.

Definition: Let $A,B$ be two indexing sets. A $\Delta$-complex (or delta complex) $X$ is
$$X = \left.\bigsqcup_{\alpha\in A} \Delta^{n_\alpha}_\alpha \right/\left\{\mathcal F_{\beta}^{k_\beta}\right\}_{\beta\in B}\ , \hspace{1cm} \mathcal F_\beta^{k_\beta} = \{\Delta_1^{k_\beta},\dots,\Delta_{m_\beta}^{k_\beta}\},$$
such that if $\sigma$ appears in the disjoint union, all of its lower dimensional faces also appear. The identification of the $k$-simplices in $\mathcal F^k$ is done in the natural (linear) way, and restricting to identified faces gives the identification of the $\mathcal F^{k-1}$ where the faces appear.

To define simplicial homology of a simplicial or $\Delta$-complex $X$, fix an ordering of the set of 0-simplices (which gives an ordering of every $\sigma\in X$), define $C_k$ to be the free abelian group generated by all $\sigma\in X$ of dimension $k$ (defined by $k+1$ 0-simplices), and define a boundary map
$$\begin{array}{r c l} \partial_k\ :\ C_k & \to & C_{k-1}, \\\ [v_0,\dots,v_k] & \mapsto & \sum_{i=0}^k(-1)^i[v_0,\dots,\widehat{v_i},\dots,v_k]. \end{array}$$
Then $H_k(X):= \text{ker}(\partial_k)/\text{im}(\partial_{k+1})$.

Recall the standard $n$-cell is $e^n = \{x\in \R^n\ :\ | x| \leqslant 1\}$, also known as the $n$-disk or $n$-ball.


Definition: Let $X_0$ be a finite set. A cell complex (or CW complex) is a collection $X_0,X_1,\dots$ where
$$X_k := \left.X_{k-1}\bigsqcup_{\alpha\in A_k} e^k_\alpha \right/\left\{\partial e^k_\alpha\sim f_{k,\alpha}(\dy e^k_\alpha)\right\}_{\alpha\in A_k},$$
where the $f_{k,\alpha}$ describe how to attach $k$-cells to the $(k-1)$-skeleton $X_{k-1}$, for $k\>1$. $X_k$ may also be described by pushing out $e^k\sqcup_{\dy e^k}X_{k-1}$. Note that $\dy e^k = S^{k-1}$, the $(k-1)$-sphere.

To define cellular homology, we need more tools (relative homology and excision) that require a blog post of their own.

References: Hatcher (Algebraic topology, Chapter 2.1)

Classical Lie groups

 Lecture topic

A Lie group $G$ is both a group and a manifold, with a smooth map $G\times G\to G$, given by $(g,h)\mapsto gh^{-1}$. The Lie algebra $\mathfrak g$ of $G$ is the tangent space $T_eG$ of $G$ at the identity.

We distinguish between real and complex Lie groups by saying that the base manifold is either real or complex analytic, respectively.

Example: Here are some examples of classical Lie groups and their dimension:
$$\begin{array}{r c r c l} \text{general linear group} & n^2 & GL(n) & = & \{n\times n\ \text{matrices with non-zero determinant}\} \\ \text{special linear group} & n^2-1 & SL(n) & = & \{M\in GL(n)\ :\ \det(M)=1\} \\ \text{orthogonal group} & n(n-1)/2 & O(n) & = & \{M\in GL(n)\ :\ MM^t = I\} \\ \text{special orthogonal group} & n(n-1)/2 & SO(n) & = & \{M\in O(n)\ :\ \det(M)=1\} \\ \text{unitary group} & n^2 & U(n) & = & \{M\in GL(n,\textbf{C})\ :\ MM^* = I\} \\ \text{special unitary group} & n^2-1 & SU(n) & = & \{M\in U(n)\ :\ \det(M)=1\} \\ \text{symplectic group} & n(2n+1) & Sp(n) & = & \{n\times n\ \text{matrices}:\ \omega(Mx,My)=\omega(x,y)\} \end{array}$$
For the symplectic group, the skew-symmetric bilinear form $\omega$ is defined as
$$\omega(x,y) = \sum_{i=1}^n x_iy_{i+n} - y_ix_{i+n} = \begin{pmatrix} 0 & -I \\ I & 0 \end{pmatrix} x\cdot y,$$
where $\cdot$ is the regular dot product (a symmetric bilinear form). Also note that the unitary group is a real Lie group - real because there is no holomorphic map $G\times G\to G$ as would be necessary, so we view the entries of a matrix in $U(n)$ in terms of its real and imaginary parts. Hence the dimension indicated above is real dimension.

References: Kirillov Jr (An introduction to Lie groups and Lie algebras, Chapter 2)

The real and complex Jacobian

 Lecture topic

Let $f:\C^n\to \C^n$ be a holomorphic function. We will show that the the real Jacobian is the square of the complex Jacobian. Write $f = (f_1,\dots,f_n)$ with $f_i = u_i+\img v_i$, where the $u_i$ are functions of the $z_j = x_j+\img y_j$. By the Cauchy-Riemann equations
$$\frac{\dy u_i}{\dy x_j} = \frac{\dy v_i}{\dy y_j} \hspace{1cm} \text{and} \hspace{1cm} \frac{\dy u_i}{\dy y_j} = -\frac{\dy v_i}{\dy x_j}$$
and expanding, we have that
$$\begin{align*} \frac{\dy f_i}{\dy z_j} & = \frac 12\left(\frac{\dy f_i}{\dy x_j} - \img \frac{\dy f_i}{\dy y_j}\right) \\ & = \frac12\left(\frac{\dy u_i}{\dy x_j} + \img \frac{\dy v_i}{\dy x_j} - \img\left(\frac{\dy u_i}{\dy y_j} + \img \frac{\dy v_i}{\dy y_j}\right)\right) \\ & = \frac12 \left(\frac{\dy u_i}{\dy x_j}+ \frac{\dy v_i}{\dy y_j} + \img \left(\frac{\dy v_i}{\dy x_j} - \frac{\dy u_i}{\dy y_j}\right)\right) \\ & = \frac{\dy u_i}{\dy x_j} + \img \frac{\dy v_i}{\dy x_j}. \end{align*}$$
The complex Jacobian of $f$ is $J_\C f$ (or its determinant), with entries
$$(J_\C f)_{i,j} = \frac{\dy f_i}{\dy z_j},$$
and the real Jacobian of $f$ is $J_\R f$ (or its determinant), with entries
$$\begin{align*} \begin{bmatrix} (J_\R f)_{2i-1,2j-1} & (J_\R f)_{2i-1,2j} \\ (J_\R f)_{2i,2j-1} & (J_\R f)_{2i,2j} \end{bmatrix} & = \begin{bmatrix} \displaystyle \frac{\dy u_i}{\dy x_j} & \displaystyle \frac{\dy u_i}{\dy y_j} \\ \displaystyle \frac{\dy v_i}{\dy x_j} & \displaystyle \frac{\dy v_i}{\dy y_j} \end{bmatrix} \\ & \tov{R_{2i-1} + \img R_{2i} \to R_{2i-1}} \begin{bmatrix} \displaystyle \frac{\dy f_i}{\dy z_j} & \displaystyle \img\frac{\dy f}{\dy z} \\ \displaystyle \frac{\dy v_i}{\dy x_j} & \displaystyle \frac{\dy v_i}{\dy y_j} \end{bmatrix} \\ & \tov{C_{2j} - \img C_{2j-i} \to C_{2j}} \begin{bmatrix} \displaystyle \frac{\dy f_i}{\dy z_j} & 0 \\ \displaystyle \frac{\dy v_i}{\dy x_j} & \displaystyle \overline{\frac{\dy f_i}{\dy z_j}} \end{bmatrix}, \end{align*}$$
where the row and column operations have been performed for all rows $2i$ and all columns $2j$. Moving all the odd-indexed columns to the left and all odd-indexed rows to the top, we get that
$$J_\R f \simeq \begin{bmatrix} A & 0 \\ * & B \end{bmatrix} \hspace{1cm} \text{with} \hspace{1cm} A_{i,j} = \frac{\dy f_i}{\dy z_j},\ \ \ B_{i,j} = \overline{\frac{\dy f_i}{\dy z_j}}.$$
Since the number of operations to switch the columns is the same as the number of operations to switch the rows, the sign of the determinant of $J_\R f$ will not change. That is,
$$\det(J_\R f) = \det(A)\det(B) = \det(J_\C f) \overline{\det( J_\C f)} = |\det(J_\C f)|^2.$$

Equations on Riemann surfaces

Recall that a Riemann surface is a complex 1-manifold $M$ with a complex structure $\Sigma$ (a class of analytically equivalent atlases on $X$). Here we consider equations that relate connections and Higgs fields with solutions on Riemann surfaces. Let $G=SU(2)$ (complex 2-matrices with determinant 1) or $SO(3)$ (real orthogonal 3-matrices with determinant 1), $\theta$ a Higgs field over $M$, and  $P$ a principal $G$-bundle over $M$.

Definition: The curvature of a principal $G$-bundle $P$ is the map
$$\begin{array}{r c l} F_\nabla\ :\ \mathcal A^0_M(P) & \to & \mathcal A^2_M(P), \\ \omega s & \mapsto & (d_\nabla \circ \nabla)(\omega s), \end{array}$$
where the extension $d_\nabla:\mathcal A^k_M(P)\to \mathcal A^{k+1}_M(P)$ is defined by the Leibniz rule, that is $d_\nabla (\omega\otimes s) = (d\omega)\otimes s +(-1)^k\omega \wedge \nabla s$, for $\omega$ a $k$-form and $s$ a smooth section of $P$.

Since we may write $\mathcal A^1 = \mathcal A^{1,0}\oplus\mathcal  A^{0,1}$ as the sum of its holomorphic and anti-holomorphic parts, respectively (see post "Smooth projective varieties as Kähler manifolds," 2016-06-16), we may consider the restriction of $d_\nabla$ to either of these summands.

Definition: For a vector space $V$, define the Hodge star $*$ by
$$\begin{array}{r c l} *\ :\ \bigwedge^k(V^*) & \to & \bigwedge^{n-k}(V^*), \\ e^{i_1}\wedge \cdots \wedge e^{i_k} & \mapsto & e^{j_1}\wedge\cdots \wedge e^{j_{n-k}}, \end{array}$$
so that $e^{i_1}\wedge\cdots \wedge e^{i_k}\wedge e^{j_1}\wedge\cdots \wedge e^{j_{n-k}} = e^1\wedge \cdots \wedge e^n$. Extend by linearity from the chosen basis.

The dual of the generalized connection $d_\nabla$ is written $d_\nabla^* = (-1)^{m+mk+1}*d_\nabla *$, where $\dim(M)=m$ and the argument of $d_\nabla^*$ is in $\mathcal A^k_M$ (this holds for manifolds $M$ that are not necessarily Riemann surfaces as well).

Now we may understand some equations on Riemann surfaces. They all deal with the connection $\nabla$, its generalization $d_\nabla$, its curvature $F_\nabla$, and the Higgs field $\theta$. Below we indicate their names and where they are mentioned (and described in further detail).
$$\begin{align*} \text{Hitchin equations} && \left.d_\nabla\right|_{A^{0,1}}\theta & = 0 && [2],\ \text{Introduction}\\ && F_\nabla + [\theta,\theta^*] & = 0\\[10pt] \text{Yang-Mills equations} && d^*_\nabla d_\nabla \theta + *[*F_\nabla,\theta] & = 0 && [1],\ \text{Section 4} \\ && d_\nabla^*\theta & = 0 \\[10pt] \text{self-dual Yang-Mills equation} && F_\nabla  - *F_\nabla & = 0 && [2],\ \text{Section 1}\\[10pt] \text{Yang-Mills-Higgs equations} && d_\nabla *F_\nabla + [\theta,d_\nabla \theta] & = 0 && [4],\ \text{equation (1)} \\ && d_\nabla * d_\nabla \theta & = 0 \end{align*}$$

Recall the definitions of $\theta$ and $\theta*$ from a previous post ("Higgs fields of principal bundles," 2016-08-24). Now we look at these equations in more detail. The first of the Hitchin equations says that $\theta$ has no anti-holomorphic component, or in other words, that $\theta$ is holomorphic. In the second equation, the Lie bracket $[\cdot,\cdot]$ of the two 1-forms is
$$\begin{align*} [\theta,\theta^*] & = \left[\textstyle\frac12f(dz+i\ dy), \frac12\bar f(dz - i\ dy) \right] \\ & = \textstyle -\frac i4f\bar f\ dx\wedge dy + \frac i4 f\bar f\ dy \wedge dx -\frac i4 f\bar f\ dx\wedge dy +\frac i4 f\bar f\ dy\wedge dx \\ & = -i|f|^2\ dx\wedge dy. \end{align*}$$
In the Yang-Mills and Yang-Mills-Higgs equations, we can simplify some parts by noting that, for a section $s$ of the complexification of $P\times_\ad \mathfrak g$,
$$\begin{align*} d_\nabla (\theta\otimes s) & = \textstyle \frac12d_\nabla (fdx\otimes s) + \frac i2 d_\nabla (fdy \otimes s) \\ & = \textstyle \frac12 (df\wedge dx \otimes s - fdx \wedge \nabla s) +\frac i2 (df\wedge dy - fdy \wedge \nabla s) \\ & = \left(\frac i2\frac{\dy f}{\dy x} - \frac 12 \frac{\dy f}{\dy y}\right)dx\wedge dy \otimes s - \underbrace{\textstyle \frac12f(dx+idy)}_{\theta}\wedge \nabla s. \end{align*}$$
The Hodge star of $\theta$ is $*\theta = \frac 12f(dy -idx)$, so
$$\begin{align*} d_\nabla *(\theta\otimes s) & = \textstyle \frac12d_\nabla (fdy\otimes s) - \frac i2 d_\nabla (fdx \otimes s) \\ & = \textstyle \frac12 (df\wedge dy \otimes s - fdy \wedge \nabla s) -\frac i2 (df\wedge dx - fdx \wedge \nabla s) \\ & = \left(\frac 12\frac{\dy f}{\dy x} + \frac i2 \frac{\dy f}{\dy y}\right)dx\wedge dy \otimes s + \underbrace{\textstyle \frac12f(idx-dy)}_{i\theta}\wedge \nabla s. \end{align*}$$
We could express $\nabla s = (s_1dx + s_2dy)\otimes s^1$, but that would not be too enlightening. Next, note the self-dual Yang-Mills equation only makes sense over a (real) 4-dimensional space, since the degrees of the forms have to match up. In that case, with a basis $dz_1=dx_1+idy_1, dz_2 = dx_2+idy_2$ of $\mathcal A^1$, we have
$$\begin{align*} F_\nabla & = F_{12} dx_1\wedge dy_1 + F_{13} dx_1 \wedge dx_2 + F_{14} dx_1\wedge dy_2 + F_{23} dy_1\wedge dx_2 + F_{24} dy_1\wedge dy_2 + F_{34} dx_2\wedge dy_2, \\ *F_\nabla & = F_{12} dx_2\wedge dy_2 - F_{13} dy_1 \wedge dy_2 + F_{14} dy_1\wedge dx_2 + F_{23} dx_1\wedge dy_2 - F_{24} dx_1\wedge dx_2 + F_{34} dx_1\wedge dy_1. \end{align*}$$
Then the self-dual equation simply claims that
$$F_{12} = F_{34} \hspace{1cm},\hspace{1cm} F_{13} = -F_{24} \hspace{1cm},\hspace{1cm} F_{14} = F_{23}.$$

Remark: This title of this post promises to talk about equations on Riemann surfaces, yet all the differential forms are valued in a principal $G$-bundle over $\R^2$ (or $\R^4$). However, since the given equations are conformally invariant (this is not obvious), and as a Riemann surface locally looks like $\R^2$, we may consider the solutions to the equations as living on a Riemann surface.

References:
[1] Atiyah and Bott (The Yang-Mills equations over Riemann surfaces)
[2] Hitchin (Self-duality equations on a Riemann surface)
[3] Huybrechts (Complex Geometry, Chapter 4.3)
[4] Taubes (On the Yang-Mills-Higgs equations)

Higgs fields of principal bundles

The goal here is to understand the setting of Higgs fields on Riemannian manifolds, in the manner of Hitchin. First we consider general topological spaces $X$ and groups $G$.

Definition: Let $X$ be a topological space and $G$ a group. A principal bundle (or principal $G$-bundle) $P$ over $X$ is a fiber bundle $\pi:P\to X$ together with a continuous, free, and transitive right action $P\times G\to P$ that preserves the fibers. That is, if $p\in \pi^{-1}(x)$, then $pg\in \pi^{-1}(x)$ for all $g\in G$ and $x\in X$.

Now suppose we have a principal bundle $\pi:P\to X$, a representation $\rho$ of $G$, and another space $Y$ on which $G$ acts on the left. Define an equivalence relation $(p,y)\sim (p',y')$ on $P\times Y$ iff there is some $g\in G$ for which $p'=pg$ and $y'=\rho(g^{-1})y$. This is an equivalence relation. We will be interested in the adjoint representation (induced by conjugation).

Proposition: The projection map $\pi':P\times_\rho Y := (P\times Y)/\sim\ \to X$, where $\pi'([p,y]) = \pi(p)$, defines a vector bundle over $X$, called the associated bundle of $P$.

Recall a Lie group $G$ is a group that is also a topological space, in the sense that there is a continuous map $G\times G\to G$, given by $(g,h)\mapsto gh^{-1}$. The Lie algebra $\mathfrak g$ of the Lie group $G$ is the tangent space $T_eG$ of $G$ at the identity $e$. We will be interested in principal $G$-bundles $P\to \R^2$ and associated bundles $P\times_\ad \mathfrak g\to \R^2$, where $\ad$ is the adjoint representation of $G$.

Next, recall we had the space $\mathcal A^k_M$ of $k$-differential forms on $M$ (see post "Smooth projective varieties as Kähler manifiolds," 2016-06-16), defined in terms of wedge products of elements in the cotangent bundle $(TM)^* = T^*M$ of $M$. Now we generalize this to get differential forms over arbitrary vector bundles.

Definition: Let $E\to M$ be a vector bundle. Let
$$\begin{align*} \mathcal A^k_M(E) & := \Gamma(E\otimes \textstyle\bigwedge^k T^*M) = \Gamma(E)\otimes_{\mathcal A^0_M}\mathcal A^k_M, \\ \mathcal A^{p,q}_M(E) & := \Gamma(E\otimes \textstyle\bigwedge^p (T^{1,0}M)^*\otimes \bigwedge^q (T^{0,1}M)^*) = \Gamma(E)\otimes_{\mathcal A^0_M}\mathcal A^{p,q}_M \end{align*}$$
be the spaces of $k$- and $(p,q)$-differential forms, respectively, over $M$ with values in $E$.

Equality above follows by functoriality. Now we are close to understanding where exactly the Higgs field lives, in Hitchin's context.

Definition: Given a function $f:\C\to \C$, the conjugate of $f$ is $\bar f$, defined by $\bar f(z) = \overline{f(\bar z)}$.

Hitchin denotes this as $f^*$, but we will stick to $\bar f$. Finally, let $P$ be a $G$-principal bundle over $\R^2$ and $P\times_\ad \mathfrak g$ the associated bundle of $P$. Given $f\in \mathcal A^0_{\R^2}( (P\times_\ad \mathfrak g)\otimes \C)$, set
$$\begin{align*} \theta & = \textstyle \frac12 f(dx+i\ dy) \in \mathcal A^{1,0}_{\R^2}((P\times_\ad\mathfrak g)\otimes \C) ,\\ \theta^* & = \textstyle \frac12 \bar f(dx-i\ dy) \in \mathcal A^{0,1}_{\R^2}((P\times_\ad\mathfrak g)\otimes \C), \end{align*}$$
called a Higgs field over $\R^2$ and (presumably) a dual (or conjugate) Higgs field over $\R^2$. Note this agrees with the definition in a previous post ("Connections, curvature, and Higgs bundles," 2016-07-25).

References: Hitchin (Self-duality equations on a Riemann surface), Wikipedia (article on associated bundles, article on vector-valued differential forms)

What is a stack?

 Conference topic

This is from discussions at the 2016 West Coast Algebraic Topology Summer School (WCATSS) at The University of Oregon. Thanks to Piotr Pstragowski for explaining the material.

Definition: A groupoid is a category where all the morphisms are invertible. Alternatively, a groupoid is a set of objects $A$, a set of morphisms $\Gamma$, and a collection of maps as described by the diagram below.

To describe stacks, we compare them with sheaves. Both start out with a space $X$ and a topology on it, so that we may consider open sets $U$.

In addition to these conditions, there is a triple intersection condition for stacks that does not have an analogous one in sheaves. It is given by:

for every $U_i,U_j,U_k$ and $s_i,s_j,s_k\in \widehat{\mathcal F}(U_i), \widehat{\mathcal F}(U_j), \widehat{\mathcal F}(U_k)$, respectively, such that there exist isomorphisms $\varphi_{ij}:s_i|_{U_i\cap U_j}\to s_j|_{U_i\cap U_j}$, $\varphi_{jk}:s_j|_{U_j\cap U_k}\to s_k|_{U_j\cap U_k}$, and $\varphi_{ik}:s_i|_{U_i\cap U_k}\to s_k|_{U_i\cap U_k}$, the diagram below commutes:

Example: A Hopf algebroid may be viewed as a functor into groupoids, so that with the appropriate topology, it becomes a stack. Indeed, by definition a Hopf algebroid is a pair of $k$-algebras $(A,\Gamma)$ such that $(\Spec(A),\Spec(\Gamma))$ is a groupoid object in affine schemes, or in other words, is a functor from affine schemes into groupoids.

References: nLab (article on groupoids)

Morphisms of schemes

 Conference topic

This is from discussions at the 2016 West Coast Algebraic Topology Summer School (WCATSS) at The University of Oregon. Thanks to Zijian Yao for explaining the material.

Consider a morphism of schemes $\varphi:S'\to S$ and coherent sheaves $\mathcal F,\mathcal G$ over $S$. Consider also a map of sheaves $f:\mathcal F\to \mathcal G$ and a map $f'$ between the pullbacks of $\mathcal F$ and $\mathcal G$, as described by the diagram below.

There are two natural questions to ask.

1. When is $f' = \varphi^*f$?
2. If we start with $\mathcal G'$ over $S'$, when is $\mathcal G' = \varphi^*\mathcal G$?

To answer these questions, consider fiber products of schemes and projections from them, as given below.

Remark: If 1. is true, then $p_1^*(f') = p_2^*(f')$. If the previous statement is an equivalence, then $\varphi$ is a morphism of descent.

Remark: If 2. is true, then there exists $\alpha:p_1^*(\mathcal G') \to p_2^*(\mathcal G')$ such that $\pi_{32}^*(\alpha)\pi_{21}^*(\alpha) = \pi_{31}^*(\alpha)$ and $\pi^*(\Delta) = \alpha$. If the previous statement is an equivalence, then $\varphi$ is effective.