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On Euclidean space, every Killing vector field is a constant translation generator plus a constant infinitesimal rotation generator:Here $A$ is a constant vector and $B$ is an antisymmetric matrix. The second covariant derivative of a Killing vector vanishes for the Euclidean metric, so the components are affine functions of Cartesian coordinates; the Killing equation makes the linear coefficient antisymmetric.

In the Euclidean plane with metric tensor $g=d{x}^2+d{y}^2$, two useful examples areThe notation ${\partial }_x$ denotes a coordinate basis vector; $dx$ denotes a differential one-form. The corresponding flow of a vector field maps areFor translation, all arrows have the same direction and length. For rotation, arrows are tangent to circles, have length $\sqrt{x^2+y^2}$, and vanish at the origin. The rotation flow keeps $x^2+y^2$ fixed, and its nonstationary integral curves of a vector field are circles. Both flows preserve the Euclidean metric. See Carroll's original notes, section 5, equations (5.45)-(5.46).

The circular integral curves of a vector field are symmetry orbits, whereas geodesics of the Euclidean plane are straight lines. A Killing vector field does not require its own orbits to be geodesics. The geodesic conserved quantity from a Killing vector pairs the field with the velocity of a separate geodesic; it does not require the velocity to equal the Killing vector field.

Build a closed annulus from two nested $n$-sided polygons, $n\ge 3$, with $n$ radial edges joining corresponding vertices. There are $2n$ vertices, $3n$ edges and $n$ quadrilateral topological cells, giving $\chi =2n-3n+n=0$. The central missing region and the exterior are not faces of the annulus. A deformation retraction onto a circle gives $(b_0,b_1,b_2)=(1,1,0)$, agreeing with the cell count. This is a useful reduction: the annulus is two-dimensional, yet its homology is that of a one-dimensional cycle graph.

Exoplanet transit photometry measures the decrease in a star's observed radiative flux during an exoplanet transit. Define $F_0$ as the unobscured star's observed radiative flux and $F(t)$ as the value at time $t$. For a dark, opaque exoplanet crossing a uniformly bright circular stellar disc, a fully superimposed planetary disc blocks exactly the fractionHere $R_p$ and $R_{\star }$ are the planet and star radii. This exoplanet transit depth determines a radius ratio, $R_p/R_{\star }=\sqrt{\delta }$; an absolute planet radius requires an independently estimated stellar radius. For $R_p/R_{\star }=0.10$, $\delta =0.010$: a one-percent dip, with $F_{\mathrm{full}}/F_0=0.990$. Halving the radius ratio to $0.05$ reduces the exoplanet transit depth to $0.0025$, a quarter as much.

For the central exoplanet transit illustrated below, the transit impact parameter is $b=0$. Write $k=R_p/R_{\star }<1$ and let $z$ be the projected centre-to-centre separation in units of $R_{\star }$. The light curve is $F/F_0=1-A(z)/\pi$, where $A(z)$ is the overlap area of circles of radii $1$ and $k$. There is no overlap when $z\ge 1+k$; the planetary disc is fully inside when $z\le 1-k$, giving $A=\pi k^2$. Between those limits, the overlap area grows or shrinks continuously. Thus even a uniform star has curved ingress and egress, with a flat bottom only during full overlap. The illustration uses approximately constant transverse velocity over the short crossing, so $z=|ut|/R_{\star }$ with mid-transit at $t=0$. Analytic models also handle limb darkening. ^[ref].

The simple exoplanet transit depth formula assumes constant stellar brightness, no limb darkening or starspots, negligible light from the exoplanet, and no transit dilution by other light. It describes full overlap in a non-grazing exoplanet transit, requiring $b<1-k$ for a full-overlap interval. A grazing exoplanet transit never hides the whole planetary disc against the star, and limb darkening weights different blocked areas differently. Repeating dips are valuable evidence, but eclipsing binary stars and starspots can imitate parts of the signal, so confirmation and joint modelling matter.

For the upper half disk $|z|<2$ slit along $z=iy$, $0<y\le 1$, squaring maps onto $|w|<4$ minus $[-1,4)$. After scaling the radius-four disc, the disk automorphism $v=4(w+1)/(w+16)$ sends this slit to $[0,1)$. Its square root with argument in $(0,\pi )$ maps onto the upper half unit disc. There $u+u^{-1}$ maps bijectively onto the lower half-plane: its imaginary part is $\mathrm{Im}u(1-|u{|}^{-2})<0$, and the two roots of its inverse equation are reciprocal, with exactly one inside the unit circle. The final Möbius map gives the displayed conformal disk bijection.

Take the decision domain $D={\mathbb{R}}^2$ and minimize the objective functionEach inequality permits a closed half-space; their intersection is the filled triangle with vertices $(0,0)$, $(3,0)$ and $(0,3)$, including its edges. The feasible region is a compact set and a convex set. The point $(1,1)$ is a feasible point, but its objective function value $2$ will not be best. Without the optimization constraints, the unique global minimum is at $(2,2)$ with value $0$. That point is infeasible because $2+2>3$.

The objective function is the squared Euclidean distance from $(2,2)$. Its positive level sets are circles centered there. Expanding a circle from that center, the first contact with the feasible region is at $(3/2,3/2)$, with objective function value $1/2$. Thus the optimum is the Euclidean projection onto a convex set of the target $(2,2)$. Squaring a nonnegative Euclidean distance preserves which point is nearest.^[ref]

Here is a direct certificate, without relying on the picture. Put $s=x+y$ and $d=x-y$. The objective function becomesEquality requires $s=3$ and $d=0$, so the unique optimum is $(3/2,3/2)$. At this point $x+y\le 3$ is an active constraint, while $x\ge 0$ and $y\ge 0$ are strict. An active constraint identifies a tight restriction at a specified point; it alone does not establish optimality. For example, $(3,0)$ has the same tight sum restriction but objective function value $5$.

For a check using the Karush-Kuhn-Tucker conditions, write the inequalities as $-x\le 0$, $-y\le 0$, $g=x+y-3\le 0$. Their Lagrange multipliers at the optimum are $(0,0,1)$. The gradient equation isFeasibility, nonnegative Lagrange multipliers and complementary slackness also hold. With this strictly convex function and affine optimization constraints, the Karush-Kuhn-Tucker conditions certify the unique global minimum.^[ref]

The fundamental group ${\pi }_1(X,x_0)$ consists of based loops modulo based homotopy. Multiplication follows one based loop and then the other; the constant based loop represents the identity, and path reversal gives the inverse. Passing to homotopy classes makes this operation associative, as explained by well-defined loop concatenation.

Identify the circle with $S^1=\{z\in \mathbb{C}:|z|=1\}$ and choose basepoint $1$. Its based loopshave winding number $n$: positive $n$ counts counterclockwise turns, negative $n$ counts clockwise turns, and $n=0$ gives the constant based loop. Every based loop on the circle is joined by a based homotopy to exactly one ${\gamma }_n$. ConsequentlyThis is Hatcher's Theorem 1.7 in Algebraic Topology, §1.1, pp. 29–30. For example, two counterclockwise turns followed by one clockwise turn represent $2+(-1)=1$, rather than three turns. The winding graph in the figure above makes the signed arithmetic visible.

A proof outline uses the covering map $q:\mathbb{R}\to S^1$, $q(u)=e^{2\pi iu}$. By the path lifting theorem, a based loop has a unique lift starting at zero and ending at an integer $n$. Interpolating its lift to $s\mapsto ns$ produces a based homotopy to ${\gamma }_n$. The homotopy lifting theorem for a covering map keeps that endpoint unchanged during a based homotopy, proving uniqueness.

In the filled disk, every based loop $\gamma$ at $x_0$ has a based homotopy to the constant based loop:Both endpoints remain $x_0$, and the convex set property of the disk keeps every intermediate point inside. Thus ${\pi }_1(D^2,x_0)$ is the trivial group. Together with the deformation retraction above, this computes three concrete answers: the filled disk has trivial fundamental group, while the circle and the annulus have infinite cyclic groups.

Maps have computable algebraic shadows too. The continuous map $p_n:S^1\to S^1$, $p_n(z)=z^n$, fixes $1$ and sends ${\gamma }_k$ to ${\gamma }_{nk}$. Its induced group homomorphism is therefore $k\mapsto nk$, and its mapping degree is $n$. For $n=2$ the image consists of even integers, so this map cannot be a homotopy equivalence. Composing $p_2$ with $p_{-3}$ gives $p_{-6}$: map composition multiplies degrees, whereas concatenating based loops adds winding numbers. Distinct degrees distinguish these maps up to homotopy; this calculation does not classify arbitrary topological spaces.

Historically, Poincaré's 1895 Analysis Situs introduced the fundamental group; its original §12 explicitly names it. The CNRS transcription of Poincaré's §12, Groupe fondamental provides a direct view of this early formulation.

Each Killing vector field supplies a conserved quantity for freely moving test particles: it turns a geometric symmetry into a first integral of the geodesic equation. Let $\gamma (\lambda )$ be a geodesic with affine parameter $\lambda$, tangent vector $U^{\mu }=d{x}^{\mu }/d\lambda$, and ${\nabla }_{U}U=0$. Wherever a Killing vector field $K$ is defined along the geodesic,The statement applies to Riemannian manifolds and to timelike geodesics, null geodesics and spacelike geodesics in a Lorentzian manifold. It requires an affine parameter; for a massive test particle, proper time is one choice. See Tong, section 4.3.2, equation (4.32).

For a worked example in the Euclidean plane, take $\gamma (t)=(1+t,2-t)$ with affine parameter $t$ and velocity $U=(1,-1)$. The Euclidean Killing vector fields for translation and rotation giveThe rotational field along the geodesic is $K_{\mathrm{rot}}(\gamma (t))=(t-2,1+t)$: it changes from $(-2,1)$ at $t=0$ to $(-1,2)$ at $t=1$, yet its pairing with $U$ stays $-3$. With toy mass $m=1$ and $t$ interpreted as time, the first constant is the $x$ component of linear momentum, and the second is the signed $z$ component of angular momentum about the origin. Their numerical values use toy units. The geodesic is straight; the rotation flow orbits are circles.

This conservation law follows from metric compatibility and the Killing equation:The geodesic equation kills the acceleration term, and the symmetric product $U^{\mu }{U}^{\nu }$ selects the vanishing symmetric derivative. This tensor contraction is the whole proof outline; see Carroll, equation (5.43).

In a stationary spacetime, a chosen time-translation Killing vector field $K$ gives Killing energy. With metric signature $(-+++)$, $c=1$, and a massive test particle's four-momentum $p^{\mu }=m{U}^{\mu }$, the physical normalization isHere $\Phi$ generates the axial rotational symmetry when one exists. In coordinates with $K={\partial }_t$ and $\Phi ={\partial }_{\varphi }$, these are $E=-p_t$ and $L_z=p_{\varphi }$. Both are conserved quantities along the geodesic, even if the metric has a time-angle cross term. A future timelike $K$ makes $E$ positive for a future massive particle; the scale of $E$ depends on the normalization of $K$. Not every spacetime has a global timelike Killing vector field, and a stationary field can become spacelike in an ergoregion of a stationary spacetime. Conservation on the field's domain does not imply a universal global positive energy; see Tong, sections 4.3.1-4.3.2 on the timelike-domain qualification and conserved charges.

A geometric locus is the set of points satisfying a specified geometric or algebraic condition in an ambient space. For example, points at a fixed positive distance $r$ from $p$ form a circle in the plane and a sphere in three dimensions. In the complex plane, $|z-z_0|=r$ has Cartesian equation $(x-\mathrm{Re}{z}_0{)}^2+(y-\mathrm{Im}{z}_0{)}^2=r^2$, proving the circle description. A locus can also be a union, such as $x(y-1)=0$, which consists of two lines; domain restrictions in the original equation must be retained when simplifying it.

How much information about a shape do we need to keep? Euclidean geometry retains lengths and angles, allowing questions about the diagonal of a rectangle or the shortest route between two points. Topology retains the structure preserved by a homeomorphism, allowing questions about connectedness and whether a loop can shrink to a point, as recorded by the fundamental group. These are different levels of description of the same object.

For example, stretching a circle horizontally produces an ellipse. The stretching map and its inverse are continuous maps, so it gives a homeomorphism: the two curves have the same topology. It changes some Euclidean distances, however, so it is not an isometry. This distinction helps separate essential structure from a particular drawing, while explaining why measuring a shape requires more information than recognizing its topology.

The choice of geometry matters too. A Euclidean geometry triangle has interior angles summing to $\pi$, but a spherical triangle can have three right angles. Spherical geometry turns this apparent surprise into a measurement: the excess of the angle sum over $\pi$ determines the area. Begin with the Pythagorean theorem for flat measurements, then compare the spherical triangle below to see curvature change the rules.

Choose a fixed support point $O$ and nonrotating axes in an inertial frame. All angular momenta, torques, and moments of inertia below refer to $O$. Let $\hat{\mathbf{e}}$ point along the axle from $O$ toward the centre of mass, and let $\hat{\mathbf{z}}$ point vertically upward. Write $\theta$ for their angle. For a rapidly spinning symmetric top, the dominant angular momentum is ${\mathbf{L}}_O\simeq L_s\hat{\mathbf{e}}$, where $L_s=I_3{\omega }_3$ is signed and $I_3$ is the axial principal moment of inertia. The sign of the axial angular velocity ${\omega }_3$ follows the right-hand rule.

With mass $M$, centre of mass distance $d$, and constant gravitational acceleration $g$, the gravitational force and its torque areAn ideal support applies a force at $O$, so its lever arm and torque about $O$ vanish. At the instant the axle lies in the $xz$ plane with a positive $x$ component, the gravitational torque points along $+y$. It adds $d{\mathbf{L}}_O={\mathbf{\tau }}_{O}dt$ toward $+y$, rather than along the downward force. The dot product identity $d|{\mathbf{L}}_O{|}^2/dt=2{\mathbf{L}}_O\cdot {\mathbf{\tau }}_O$ shows why perpendicular torque changes direction without instantly changing magnitude. This vector balance and the slow-turning relation ${\mathbf{\tau }}_O=\mathbf{\Omega }\times {\mathbf{L}}_O$ are developed in Chapter 20: Rotation in space (sections 20-2 and 20-3, equations 20.13–20.15).

For steady precession about the vertical, $\mathbf{\Omega }=\Omega \hat{\mathbf{z}}$ and $d\hat{\mathbf{e}}/dt=\Omega \hat{\mathbf{z}}\times \hat{\mathbf{e}}$. Equating the two cross products givesThe angular momentum tip traces a circle of radius $|L_s|\sin \theta$: its sideways speed is $|\Omega L_s|\sin \theta$. Both that radius and the gravitational lever arm contain $\sin \theta$, explaining its cancellation for $0<\theta <\pi$. Exactly vertical alignment has zero gravitational torque and no distinguishable azimuthal precession to infer by this cancellation. Positive $L_s$ gives positive precession, counterclockwise viewed from above; reversing the spin reverses the slow precession. ^[ref]

Collapse a separating simple closed curve on a closed orientable surface of genus $g$. If its two sides have genera $g_1,g_2$, the quotient is homeomorphic to ${\Sigma }_{g_1}\vee {\Sigma }_{g_2}$. Its integral homology is $\mathbb{Z}$ in degree zero, ${\mathbb{Z}}^{2g}$ in degree one, ${\mathbb{Z}}^2$ in degree two, and zero above degree two. In relative homology, the circle maps to zero in $H_1({\Sigma }_g)$, and the second relative group fits into a split sequence $0\to \mathbb{Z}\to H_2({\Sigma }_g,A)\to \mathbb{Z}\to 0$. The two collapsed sides retain independent fundamental classes.

A homotopy between continuous maps $f,g:X\to Y$ is a continuous map $H:X\times [0,1]\to Y$ satisfying $H(x,0)=f(x)$ and $H(x,1)=g(x)$. Think of $t$ as time: continuity is required jointly in position and time. The target $Y$ specifies where every intermediate point is allowed to go. See Hatcher, Algebraic Topology, Chapter 0, pp. 3–4.

For the filled disk $D^2=\{x\in {\mathbb{R}}^2:\Vert x\Vert \le 1\}$, the formulais a contraction of a topological space: it starts at the identity map and ends at the constant map to zero. The Euclidean norm obeys $\Vert H(x,t)\Vert =(1-t)\Vert x\Vert \le 1$, so the formula never leaves the disk. For $x=(3/5,4/5)$, the midpoint of the deformation is $(3/10,2/5)$, whose Euclidean norm is $1/2$.

On the circle $S^1=\{x:\Vert x\Vert =1\}$, the same formula fails: at $t=1/2$ every image has Euclidean norm $1/2$, so lies outside $S^1$. This checks one failed candidate; the nonzero winding number of a once-around loop proves that no contraction within the circle exists. The distinction is between allowing the filling and requiring the deformation to stay on its boundary.

A horocycle in the Poincare disc model is a Euclidean circle internally tangent to the boundary circle. In the upper half-plane model, a horocycle is a Euclidean circle tangent to the real axis from above, or a horizontal line $y=h>0$. The common geometric feature is a single ideal centre of a horocycle.