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On the unit round sphere $S^n$, a geodesic of speed $L$ on $[0,1]$ has normal Jacobi equation $y^{\prime \prime }+L^{2}y=0$. Its conjugate parameters are $j\pi /L$, each with multiplicity $n-1$. If $L/\pi$ is not an integer, its index is $(n-1)\lfloor L/\pi \rfloor$. For $L=m\pi >0$, its index is $(m-1)(n-1)$ and its nullity is $n-1$.

For connected oriented closed $d$-dimensional manifolds with $d\ge 2$, their connected sum removes an open $d$-dimensional ball from each and identifies the resulting boundary spheres by an orientation-reversing diffeomorphism. The remaining orientations fit together. Collapsing the separating sphere gives a pinch map to the wedge sum of the two closed manifolds, and projection to either summand has degree of a continuous mapping one. In intermediate positive degrees the cohomology is the direct sum of the summand groups. Products of classes from different summands vanish; products landing in top degree use the single common orientation class. These statements follow from excision, the Mayer–Vietoris sequence and the degree-one projections.

Join three independent uniform points on a sphere when their central angle is acute. Given angle $\gamma$ between the first pair, the third point's two favorable hemispheres overlap in a spherical lune of area fraction $(\pi -\gamma )/(2\pi )$. If $\gamma <\pi /2$, connection requires membership in their union, of probability $1/2+\gamma /(2\pi )$. If $\gamma >\pi /2$, it requires membership in their intersection, of probability $1/2-\gamma /(2\pi )$. Integrating these against the central-angle density $\sin \gamma /2$ proves the displayed probability. Connectivity is weaker than all three pairwise angles being acute.

Take a fluid control volume spanning a nearly occluding sphere. The pressure difference contributes $\pi a^2\Delta P$ upwards, while the outer-wall shear in the sphere frame contributes $2\pi a\mu \int u_y(h)dz$. Since sphere-frame flux in a tube gives $u_y(h)=4U/h-6q/h^2$, the drag on the sphere is $F=\pi a^2\Delta P+2\pi a\mu (4U{I}_1-6q{I}_2)$. Using the wall shear avoids integrating traction over the strongly curved sphere directly. The remaining tube pressure drop is fixed by Hagen-Poiseuille flow.

Coulomb's law gives the mutual electric force between stationary point charges in vacuum. Put electric charges $q_1$ and $q_2$ at ${\mathbf{r}}_1$ and ${\mathbf{r}}_2$, and define the displacement from charge 1 to charge 2 by $\mathbf{r}={\mathbf{r}}_2-{\mathbf{r}}_1$, with separation $r=|\mathbf{r}|>0$ and unit vector $\hat{\mathbf{r}}=\mathbf{r}/r$. In SI units, the force on charge 2 due to charge 1 isHere ${\epsilon }_0$ is the vacuum permittivity. The magnitude is $k|q_1{q}_2|/r^2$. If $q_1{q}_2>0$, the force on charge 2 points away from charge 1; if $q_1{q}_2<0$, it points toward charge 1. Interchanging the charges gives ${\mathbf{F}}_{1\leftarrow 2}=-{\mathbf{F}}_{2\leftarrow 1}$, consistent with Newton's third law in this electrostatic setting. ^[ref].

For a fully computed example, stipulate $q_1=+2\mathrm{nC}$, $q_2=-3\mathrm{nC}$, ${\mathbf{r}}_1=0$, ${\mathbf{r}}_2=0.20\mathrm{m}\hat{\mathbf{x}}$ and the approximate value $k=8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$. ThenThe force is attractive, of magnitude $1.3485\times 10^{-6}\mathrm{N}$. The source electric field at charge 2 is ${\mathbf{E}}_1({\mathbf{r}}_2)=449.5N/C\hat{\mathbf{x}}$, pointing away from the positive source. Multiplication by the negative $q_2$ reverses the direction to give the attractive force. This field excludes charge 2's own self-field.

Holding the electric charges fixed while changing the separation from $R$ to $2R$ or $3R$ gives the normalized inverse-square law valuesFor the stipulated example, separations $0.20$, $0.40$ and $0.60\mathrm{m}$ give magnitudes $1.3485\times 10^{-6}$, $3.37125\times 10^{-7}$ and $1.498333\dots \times 10^{-7}\mathrm{N}$. The diagram computes these ratios and uses one common arrow-length scale for both sign choices.

The inverse-square law has a geometric explanation through Gauss's law: a source point charge gives fixed total electric flux through a concentric sphere, whose surface area is $4\pi r^2$. Dividing that flux by the area gives a radial electric field proportional to $1/r^2$. This connects a pairwise force calculation to a global conservation-like accounting of field sources. For several stationary point charges, the superposition principle adds the individual electric fields as vectors, and ${\mathbf{F}}_i=q_i{\mathbf{E}}_{\mathrm{other}}({\mathbf{r}}_i)$ uses all sources except charge $i$ itself. The Coulomb potential energy gives an alternative route when the desired quantity is work rather than instantaneous force.

The probability that $N\ge 1$ independent uniformly oriented open hemispheres cover a sphere is the displayed quantity. Coverage fails exactly when their normals lie in a closed hemisphere. With probability one, no three normals are coplanar and no pair is antipodal; a closed hemisphere containing the normals can then be perturbed into an open one. Condition on the unoriented axes of the normals. Their $N$ orientation signs remain independent fair bits, and the regions of a general-position great-circle arrangement realize exactly $N(N-1)+2$ of the $2^N$ possible sign vectors. Each such vector corresponds to normals that fit into one open hemisphere, proving the complementary probability. The argument counts equiprobable orientation choices, not equal-area regions.

The convex hull of the positive and negative coordinate unit vectors. Its boundary triangulates a sphere, and each face contains at most one member of each antipodal pair of coordinate unit vectors. This makes its barycentric subdivision useful for triangulating Real projective space.

For a twice continuously differentiable regular curve $r(t)$ in Euclidean space, its unsigned curvature measures how quickly its unit tangent vector changes per unit arc length. WriteThe curvature iswhere the norm is the Euclidean norm. Differentiating with respect to arc length removes the effect of traversal speed; differentiating $T$ with respect to $t$ alone generally does not. See Ghomi's university lecture notes, section 1.7.

For a circle of radius $R>0$, use $r(t)=(R\cos t,R\sin t)$. Then $v=R$, $T=(-\sin t,\cos t)$ and $\Vert dT/dt\Vert =1$, giving $\kappa =1/R$. A straight line $r(t)=p+ta$ with $a\ne 0$ has constant unit tangent vector, so $\kappa =0$.

The same circle traversed as $q(u)=(R\cos (3u),R\sin (3u))$ has speed $3R$ and $\Vert dT/du\Vert =3$: its curvature remains $1/R$. By contrast, enlarging the circle to radius $2R$ changes its geometry and gives curvature $1/(2R)$. More generally, a dilation $r\mapsto ar$ with $a>0$ divides curvature by $a$, because it multiplies arc length by $a$ while leaving unit tangent vectors unchanged. Thus curvature has units of inverse length, and a reparametrization changes neither its value at a corresponding point nor the shape.

Differentiating $T\cdot T=1$, using the dot product, shows $T\cdot (dT/ds)=0$: the change is perpendicular to the unit tangent vector. Wherever $\kappa >0$, the principal normal vector $N_p=(dT/ds)/\kappa$ points toward the bend. For a plane curve, the osculating circle at such a point has radius of curvature $\rho =1/\kappa$ and center $r+\rho N_p$. It matches the point, tangent line, and curvature, providing a local circular approximation. At $\kappa =0$ there is no finite osculating circle defined by this formula. See Ghomi's university lecture notes, section 1.13.

This makes curvature useful for comparing bends without confusing them with speed. If $t$ denotes time, differentiating $r^{\prime }=vT$ gives acceleration $r^{\prime \prime }=v^{\prime }T+v^2(dT/ds)$, whose component perpendicular to the unit tangent vector has magnitude $v^2\kappa$. The curvature of a plane curve example below computes the bend and osculating circle of a parabola explicitly.

For smooth surfaces, bending has more than one direction. The principal curvatures $k_1,k_2$ instead define Gaussian curvature $K=k_1{k}_2$ and mean curvature $H=(k_1+k_2)/2$. These are surface quantities, distinct from the curvature of an individual curve. In particular, the circle value $1/R$ must not be confused with the Gaussian curvature $1/R^2$ of a sphere of radius $R$. See Ritter's Oxford lecture notes, section 12.2.

Declination $\delta$ is the signed angle from the reference celestial equator to a direction, measured along its hour circle: $-90^{\circ }\le \delta \le +90^{\circ }$. Positive declination means celestial north, rather than necessarily north of the observer's zenith. A direction with $\delta =0$ lies on the celestial equator; $\delta =+90^{\circ }$ is the north reference celestial pole. Declination is an angle, not a height or a measure of Euclidean distance. ^[ref].

The connection to an observation follows from a dot product. In the elementary spherical-Earth, unrefracted model, an observer at geographical latitude $\varphi$ has zenith direction $\mathbf{z}=(\cos \varphi \cos \Theta ,\cos \varphi \sin \Theta ,\sin \varphi )$ in the same reference axes. Since $\mathbf{n}\cdot \mathbf{z}=\sin a$ for altitude $a$, the cosine addition formula givesAt upper meridian transit, $H=0$, so $a_{\mathrm{upper}}=90^{\circ }-|\varphi -\delta |$. For the synthetic site $\varphi =52^{\circ }$ and synthetic direction $\delta =20^{\circ }$, this yields $a_{\mathrm{upper}}=58^{\circ }$. A direction with $\delta =70^{\circ }$ instead reaches $72^{\circ }$, north of the zenith; positive declination does not imply a crossing on the southern half of the celestial meridian. A negative computed upper altitude means the direction never rises above the geometric astronomical horizon at that site.

For an angular readout, define signed zenith distance $z_s$ on the upper celestial meridian, positive toward the south and negative toward the north. Then $\delta =\varphi -z_s$. A synthetic southward readout $z_s=32^{\circ }$ at $\varphi =52^{\circ }$ recovers $\delta =20^{\circ }$. An unsigned distance $|z_s|$ alone would leave two possibilities, $\varphi \pm |z_s|$, when both are in range. A calibrated optical telescope must therefore record which side of the zenith it points toward. In this linear relation, a $1^{\prime }$ error in the assumed geographical latitude produces a $1^{\prime }$ error in inferred declination; averaging repeated readings cannot remove that shared offset. Near the astronomical horizon, uncorrected atmospheric refraction also invalidates the geometric readout.

The unit vector construction explains a subtle feature of sky maps. At fixed declination, changing right ascension traces a parallel whose radius on the unit sphere is $\cos \delta$. Its arc length is $\cos \delta |d\alpha |$, with angles in radians; at $\delta =60^{\circ }$, a $15^{\circ }$ right ascension increment spans a $7.5^{\circ }$ arc along that parallel. This arc is not generally the shortest great circle separation. At either celestial pole, all right ascension values represent the same direction: the coordinate singularity belongs to the coordinate grid. The unit vector remains regular there.

A finite collection of disjoint splitting spheres of a knot encodes a connected sum of knots. Cut along every sphere and fill each spherical boundary with a three-ball containing a straight arc joining its two marked endpoints. The resulting oriented knots are the factors. The system is complete when its nontrivial factors are prime knots.

In a planar force-driven encounter with a fixed equal sphere and positive impact parameter $b\gg a$, the mobility correction from a fixed distant sphere gives $Y-b=27a^{2}b/[32(X^2+b^2)]+O(a^4/b^3)$. The moving sphere reaches maximum deflection $27a^2/(32b)$ and rotates clockwise through $9\pi a^2/(32b^2)$ to leading order. The spin is obtained by integrating half the reflected Stokeslet vorticity, as prescribed by Faxén rotation law. These numerical formulas require a distant encounter.

Define mapping degree for $S^0$ using its rank-one reduced homology group ${\tilde{H}}_0(S^0;\mathbb{Z})$. Its generator is the difference of its two points. The four maps of that two-point set consist of the identity, the interchange, and two constant maps; their degree multipliers are $1,-1,0,0$. Thus sphere maps of arbitrary integer degree require positive sphere dimension.

Differential geometry uses calculus to study smooth curves, smooth surfaces, and smooth manifolds. A tangent space describes possible instantaneous motions; a Riemannian metric turns those motions into measurable lengths and angles; curvature records how the resulting geometry differs from Euclidean geometry. Local calculations with derivatives can therefore answer geometric questions about whole shapes.

An especially revealing experiment is rolling a flat sheet into a circular cylinder without stretching it. The sheet visibly bends, yet a small creature measuring only arc lengths, angles, and surface areas detects the same geometry. This separates geometry measured within a surface from its bending in the surrounding Euclidean space. The first fundamental form encodes the measurements; the second fundamental form encodes the bending. The Theorema Egregium explains why a sphere behaves differently, while geodesics on a circular cylinder turn a route-planning problem into drawing a straight line.

The differential geometry of smooth surfaces studies their induced Riemannian metrics and intrinsic or extrinsic curvature. The first fundamental form measures lengths, while the second fundamental form and shape operator describe bending in an ambient Euclidean space. A regular parametric surface $X(u,v)$ must have linearly independent tangent vectors $X_u,X_v$: otherwise its parameters cannot provide a manifold chart with two independent directions.

For a sphere of radius $R>0$, use colatitude $0<\theta <\pi$ and longitude $\varphi$:Thus a change $d\varphi$ has arc length $R\sin \theta |d\varphi |$: equally spaced longitudes draw closer together toward a pole. The coordinate degeneracy at the poles belongs to this parametrization, not to the smooth surface.

Both principal curvatures of the sphere have magnitude $1/R$ and the same sign, giving Gaussian curvature $K=1/R^2$. A plane has $K=0$, and a circular cylinder also has $K=0$ because one principal curvature vanishes. These three familiar shapes show why visible bending alone cannot measure intrinsic Gaussian curvature.

A map from the $n$-sphere to a CW complex can be homotoped into its $n$-skeleton. By compact subsets of CW complexes lie in finite subcomplexes, its image lies in a finite subcomplex. For a top-dimensional cell of dimension $d>n$, use smooth approximation of maps into Euclidean space in its interior, with a cutoff supported away from the cell boundary. The Sard theorem supplies a point in a smaller interior ball omitted by the smoothed map; the cutoff is chosen so the transition region also avoids that point. Radial retraction from the omitted point pushes the image into the cell boundary and fixes all other cells. Repeating over the finitely many higher-dimensional cells proves the assertion, without assuming the initial continuous image has measure zero.