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)