quaternionic projective line$\,\mathbb{H}P^1$
topology (point-set topology, point-free topology)
see also differential topology, algebraic topology, functional analysis and topological homotopy theory
Basic concepts
fiber space, space attachment
Extra stuff, structure, properties
Kolmogorov space, Hausdorff space, regular space, normal space
sequentially compact, countably compact, locally compact, sigma-compact, paracompact, countably paracompact, strongly compact
Examples
Basic statements
closed subspaces of compact Hausdorff spaces are equivalently compact subspaces
open subspaces of compact Hausdorff spaces are locally compact
compact spaces equivalently have converging subnet of every net
continuous metric space valued function on compact metric space is uniformly continuous
paracompact Hausdorff spaces equivalently admit subordinate partitions of unity
injective proper maps to locally compact spaces are equivalently the closed embeddings
locally compact and second-countable spaces are sigma-compact
Theorems
Analysis Theorems
manifolds and cobordisms
cobordism theory, Introduction
Stereographic projection is the name for a specific homeomorphism (for any $n \in \mathbb{N}$) form the n-sphere $S^n$ with one point $p \in S^n$ removed to the Euclidean space $\mathbb{R}^n$
One thinks of both the $n$-sphere as well as the Euclidean space $\mathbb{R}^n$ as topological subspaces of $\mathbb{R}^{n+1}$ in the standard way, such that they intersect in the equator of the $n$-sphere.
For $p \in S^n$ one of the corresponding poles, the stereorgraphic projection is the map which sends a point $x \in S^{n}\backslash \{p\}$ along the line connecting it with $p$ to the equatorial plane.
If one applies stereographic projection to both possible poles $p_+, p_- \in S^n$ of the sphere given a fixed equatorial plane, then one obtains two different homeomorphisms
The set of these two projections constitutes an atlas that exhibits the n-sphere as a topological manifold, in fact a differentiable manifold and in fact as a smooth manifold.
For $n = 2$ and with $\mathbb{R}^2 \simeq \mathbb{C}$ regarded as the complex plane, then this atlas realizes the 2-sphere as a complex manifold: the Riemann sphere.
We consider the ambient canonical coordinates of $\mathbb{R}^{n+1}$, in terms of which the $n$-sphere is the topological subspace whose underlying subset is presented as follows:
Without restriction, we may identify the given pole point in the coordinates with
hence the corresponding equatorial hyperplane with
(standard stereographic projection)
The function which sends a point $x \in S^{n} \backslash p \subset \mathbb{R}^{n+1}$ to the intersection of the line through $p$ and $x$ with the equatorial hyperplane is a homeomorphism which is given in terms of ambient coordinates by
First consider more generally the stereographic projection
of the entire ambient space minus the point $p$ onto the equatorial plane, still given by mapping a point $x$ to the unique point $y$ on the equatorial hyperplane such that the points $p$, $x$ any $y$ sit on the same straight line.
This condition means that there exists $d \in \mathbb{R}$ such that
Since the only condition on $y$ is that $y_1 = 0$ this implies that
This equation has a unique solution for $d$ given by
and hence it follow that
Since rational functions are continuous, this function $\sigma$ is continuous and since the topology on $S^n\backslash p$ is the subspace topology under the canonical embedding $S^n \backslash p \subset \mathbb{R}^{n+1} \backslash p$ it follows that the restriction
is itself a continuous function (because its pre-images are the restrictions of the pre-images of $\sigma$ to $S^n\backslash p$).
To see that $\sigma \vert_{S^n \backslash p}$ is a bijection of the underlying sets we need to show that for every
there is a unique $(x_1, \cdots , x_{n+1})$ satisfying
$(x_1, \cdots, x_{n+1}) \in S^{n} \backslash \{p\}$, hence
$x_1 \lt 1$;
$\underoverset{i = 1}{n+1}{\sum} (x_i)^2 = 1$;
$\underset{i \in \{2, \cdots, n+1\}}{\forall} \left(y_i = \frac{x_i}{1-x_1} \right)$.
The last condition uniquely fixes the $x_{i \geq 2}$ in terms of the given $y_{i \geq 2}$ and the remaining $x_1$, as
With this, the second condition says that
hence equivalently that
By the quadratic formula the solutions of this equation are
The solution $\frac{ 2 r^2 + 2 }{ 2 r^2 + 2 } = 1$ violates the first condition above, while the solution $\frac{ 2 r^2 - 2 }{ 2 r^2 + 2 } \lt 1$ satisfies it.
Therefore we have a unique solution, given by
In particular therefore also an inverse function to the stereographic projection exists and is a rational function, hence continuous. So we have exhibited a homeomorphism as required.
Of course the construction in prop. does not really depend on the specific coordinates chosen there.
More generally, for any point $p\in S^n$, considered as a vector in $\mathbb{R}^{n+1}$, let $W\subset \mathbb{R}^{n+1}$ be the linear subspace that is the orthogonal complement to the linear span $span\{v\}$. Then there is a homeomorphism
given by sending a point $p\in S^n\setminus \{v\}$ to the point in $W$ that is the intersection of the line through $v$ and $p$ with $W$.
More generally, one may take for $W$ any affine subspace of $\mathbb{R}^{n+1}$ normal to $span\{v\}$ and not containing $v$. For instance one option is to take $W$ to be the tangent space to $S^n$ at $-v$, embedded as a subspace of $\mathbb{R}^{n+1}$.
(see e.g. Cook-Crabb 93, section 1)
The inverse map $\sigma^{-1}$ exhibits $S^n$ as the one-point compactification of $W$.
In most cases of interest one is doing geometry using stereographic projection, so the sphere and the subspace $W$ are equipped with extra structure. Taking the explicit case above ($v=(1,0,\ldots,0)$), as the formula is so simple, some structures are automatically preserved, for instance:
orthogonalgroup action fixing $v,-v$ pointwise (note that these two points are sent to β$\infty$β and $0$, respectively, under $\sigma$)
Stereographic projection is a general geometric technique which can be applied to other commutative rings $k$ besides $k = \mathbb{R}$. We give a brief taste of this for the case of fields. For simplicity, we focus on conic sections, i.e., solutions sets in the projective plane $\mathbb{P}^2(k)$ to homogeneous polynomials of degree $2$.
Via stereographic projection, all pointed nonsingular conic sections $C \subset \mathbb{P}^2(k)$ are isomorphic and can be identified explicitly with a projective line $\mathbb{P}^1(k)$ by means of a stereographic projection.
Geometrically, if $p$ is the chosen basepoint of $C$ and $L \subset \mathbb{P}^2(k)$ is a line not incident to $p$, then for any other point $q$ of $C$ the unique line $L(p, q)$ incident to $p$ and $q$ intersects $L$ in exactly one point, denoted $\phi(q)$. (Here $\phi(p)$ is taken to be the intersection of the tangent to $p$ at $C$ with $L$; this can be considered the basepoint of $L$.) In the opposite direction, to each point $x$ of $L$, the line $L(p, x)$ intersects $C$ in $p$ and (since a quadratic with one root will also have another root) another point $q$ (which might be the same as $p$; this happens precisely when $L(p, x)$ is the tangent to $C$ at $p$); this gives the inverse $\phi^{-1}(x) = q$. In this way we obtain an isomorphism $\phi: C \to L$ of subvarieties.
For the case $k = \mathbb{Q}$ and the rational conic $C = \{(x, y) \in \mathbb{Q} \times \mathbb{Q}: x^2 + y^2 = 1\}$, stereographic projection from the point $(-1, 0)$ on $C$ to the rational line $x = 0$ defines an isomorphism of algebraic varieties $C \to \mathbb{P}^1(\mathbb{Q})$. As the calculations above show, its inverse takes $(0, t)$ (for $t \in \mathbb{Q}$) to $(\frac{1 - t^2}{1 + t^2}, \frac{2 t}{1 + t^2})$.
A Pythagorean triple is a triple of integers $(a, b, c)$ such that $a^2 + b^2 = c^2$, so that $(\frac{a}{c}, \frac{b}{c})$ lies on $C$. The isomorphism indicates that for each such point on $C$ (written as a pair of fractions in reduced form), there is a rational number $t = \frac{p}{q}$ (again in reduced form) so that
If we assume $a, b, c$ are coprime, and further (re)arrange the triple so that $a$ is odd and $b$ is even, one may quickly conclude from the first of this pair of equations that $p, q$ must have opposite parity. A little further argumentation then shows $q^2 - p^2, 2 p q, q^2 + p^2$ must be mutually coprime, so that in fact $a = q^2 - p^2, b = 2 p q$, and $c = q^2 + p^2$. Thus we arrive at the classical parametrization of Pythagorean triples, essentially on the basis of the geometry of stereographic projection!
Textbook accounts:
See also:
Wikipedia, Stereographic projection
A. L. Cook, M.C. Crabb, Fiberwise Hopf structures on sphere bundles, J. London Math. Soc. (2) 48 (1993) 365-384 (pdf)
Last revised on December 1, 2021 at 08:16:27. See the history of this page for a list of all contributions to it.