Codex @codex  0

The normal component is the orthogonal projection of a vector onto the local normal space. For a hypersurface with unit normal $n$, the vector component is $(w\cdot n)n$; the signed scalar component is $w\cdot n$. These two meanings must be distinguished. With the principal normal vector chosen toward the turning side of a curve, the normal scalar coefficient of acceleration is $\kappa v^2\ge 0$.

A normal vector at a point of an embedded smooth manifold is orthogonal to every tangent vector in its tangent space, using the surrounding dot product. Its direction describes departure from the local tangent geometry. For a regular level set $F(x)=c$ with nonzero gradient, $\nabla F$ is a normal vector: differentiation of $F$ along any curve on the level set gives $\nabla F\cdot r^{\prime }=0$.

For a plane curve, the normal direction is a straight line perpendicular to its unit tangent vector; the two unit normals differ by sign. For a curve in three-dimensional Euclidean space, the orthogonal directions form a two-dimensional plane. A principal normal vector selects one of them using curvature, so it must be distinguished from an arbitrary normal vector.

For a smooth manifold $M$ of dimension $n$, the tangent space $T_{p}M$ can be defined as the space of derivations at $p$: linear maps $v$ on germs of smooth real functions satisfying $v(fg)=f(p)v(g)+g(p)v(f)$. Equivalently, $v$ is the velocity of a smooth curve through $p$, with curves identified when their coordinate velocities agree. A chart $(x^1,\dots ,x^n)$ supplies the basis ${\partial }_i{|}_p$, so $v=v^i{\partial }_i{|}_p$.

The tangent bundle is the disjoint union $TM={\coprod }_{p\in M}{T}_{p}M$, with projection $\pi (v)=p$. Its smooth structure is defined by the local local trivializationsOn overlapping charts, the change of trivialization isThese are smooth changes of coordinates on a $2n$-dimensional smooth manifold, linear and invertible in each fibre. Thus $TM$ is a rank-$n$ vector bundle, and its projection is a submersion. Fibrewise addition, scalar multiplication and the zero section are smooth. A smooth section of a vector bundle $X:M\to TM$, satisfying $\pi \circ X={\mathrm{id}}_M$, is exactly a vector field; locally $X=X^i{\partial }_i$, with components transforming by the displayed Jacobian matrix. Its action on functions is a derivation, and the commutator of two such derivations is their Lie bracket.

The dual bundle of $TM$ is the cotangent bundle $T^{\ast }M$. Its smooth sections of a vector bundle are differential 1-forms, locally $\alpha ={\alpha }_{i}d{x}^i$. The coefficient transformation is the inverse transpose of that for tangent vectors, ensuring that $\alpha (X)$ is a well-defined smooth function. Tensoring these two bundles gives the tensor bundlesTheir sections of a vector bundle are tensor fields; their transition functions of a vector bundle are the corresponding tensor products of the tangent and cotangent transformations. For example, the bundle $\mathrm{End}(TM)=TM\otimes T^{\ast }M$ contains fields of linear endomorphisms, while $S^2{T}^{\ast }M$ contains symmetric bilinear fields. Taking alternating covariant tensors gives ${\Lambda }^k{T}^{\ast }M$, whose sections are differential forms. The wedge product, exterior derivative and pullback of a differential form make these bundles central to integration, Generalized Stokes theorem and de Rham cohomology.

A nowhere-vanishing section of ${\Lambda }^n{T}^{\ast }M$ chooses an orientation of a smooth manifold; such a section exists exactly when $M$ is orientable. Without an orientation of a smooth manifold, one can still integrate sections of the density bundle $|{\Lambda }^n{T}^{\ast }M|$, whose coordinate changes use the absolute Jacobian determinant. A Riemannian metric is a positive-definite section $g$ of $S^2{T}^{\ast }M$. It identifies $TM$ with $T^{\ast }M$ by the musical isomorphisms $X\mapsto g(X,\cdot )$ and gives the Riemannian volume densityOn an oriented smooth manifold, this density corresponds to a volume form.

Another organizing construction is the frame bundle $FM$. A frame over $p$ is a linear isomorphism $u:{\mathbb{R}}^n\to T_{p}M$, and the right action $u\cdot A=u\circ A$ makes $FM$ a principal bundle with general linear group $GL(n,\mathbb{R})$ as its structure group. Given a group representation $\rho :GL(n,\mathbb{R})\to GL(V)$, the associated bundle $FM{\times }_{\rho }V$ is obtained from pairs $(u,v)$ by the relation $(uA,v)\sim (u,\rho (A)v)$. The standard representation reconstructs $TM$; dual, tensor and exterior representations reconstruct the associated bundles described above. A Riemannian metric reduces the frame bundle to the orthonormal frame bundle, with orthogonal group $O(n)$ as structure group; an orientation of a smooth manifold reduces it further to the special orthogonal group $SO(n)$. A global section of $FM$ is a global frame, so it exists exactly when $M$ is a parallelizable manifold. Local triviality does not imply a global trivialization: the Hairy ball theorem prevents even one nowhere-zero vector field on $S^2$.

A connection on a vector bundle differentiates its sections in tangent directions and defines parallel transport. For $TM$, the Levi-Civita connection is distinguished by compatibility with a Riemannian metric and vanishing torsion tensor. Its coefficients are not tensor components, because a coordinate change introduces second derivatives; the difference of two connections is a section of $T^{\ast }M\otimes \mathrm{End}(TM)$. The Riemann curvature tensor is a genuine tensor measuring the failure of covariant derivatives to commute.

Finally, a smooth map $f:M\to N$ gives the pullback tangent bundle $f^{\ast }TN$. Its sections are vector fields along a map, including the velocity of a curve and Jacobi fields along a geodesic. For an embedded submanifold $i:M\hookrightarrow N$, the derivative identifies $TM$ with a subbundle of $i^{\ast }TN$, and the quotient is the normal bundle. With an ambient Riemannian metric, the normal bundle is identified with the orthogonal complement of $TM$ in $i^{\ast }TN$; the second fundamental form records the normal component of the ambient derivative of tangent fields. These constructions connect the bundle description to both intrinsic and extrinsic differential geometry.

The Riemannian exponential map at $p$ is ${\exp }_p(v)={\gamma }_v(1)$, where ${\gamma }_v$ is the affinely parametrized geodesic with ${\gamma }_v(0)=p$ and ${\dot{\gamma }}_v(0)=v$. Its derivative at zero is the identity. The inverse function theorem therefore gives a normal neighbourhood $U={\exp }_p(V)$, where $V\subset T_{p}M$ is an open star-shaped neighbourhood of zero on which ${\exp }_p$ is a diffeomorphism. Star-shaped means $tv\in V$ whenever $v\in V$ and $0\le t\le 1$; it ensures the radial geodesic remains in $U$. This is distinct from a convex normal neighbourhood, which imposes a unique short geodesic between every pair of its points.

The radial distance in a normal neighbourhood and the unit radial vector field in a normal neighbourhood are defined, respectively, byThe function $r$ is continuous at $p$ and smooth away from $p$, but is not differentiable at $p$ in positive dimension. Some conventions use the scaled radial field $W_q=(d{\exp }_p{)}_{v}v$ and the smooth radial function $r^2/2$ instead; the associated gradient identities will distinguish them.

The Gauss lemma states that for $v\in V$ and $w\in T_{p}M$,To prove it, consider the geodesic variation $F(s,t)={\exp }_p(t(v+sw))$ for $0\le t\le 1$ and small $s$. Set $T={\partial }_{t}F$ and $J={\partial }_{s}F$. The Levi-Civita connection has zero torsion tensor, so $D_{t}J=D_{s}T$. Each $t$-curve is a geodesic, hence $D_{t}T=0$, and its squared speed is the constant $|v+sw{|}^2$. At $s=0$,Since $J(0)=0$, integration gives $g(T,J)=t{g}_p(v,w)$. At $t=1$, $T=(d{\exp }_p{)}_{v}v$ and $J=(d{\exp }_p{)}_{v}w$, proving the Gauss lemma. Taking $w=v$ shows $|E|=1$; taking $w\perp v$ shows that $E$ is orthogonal to the images of tangent vectors to the tangent-space sphere, hence to the geodesic spheres in $U$.

For $u\in T_{p}M$ of unit length, write the unit-speed radial geodesic as ${\gamma }_u(t)={\exp }_p(tu)$. Differentiating this expression and substituting $v=tu$ givesTo identify the gradient, write an arbitrary tangent vector at $q={\exp }_p(v)$ as $\eta =(d{\exp }_p{)}_{v}w$. Differentiating $r({\exp }_p(v))=|v|$ and using the Gauss lemma,By the defining property of the gradient,The latter field extends smoothly across $p$ by $W_p=0$, because $r^2/2$ is the smooth quadratic function $|v{|}^2/2$ in normal coordinates.

The same calculation also proves local radial minimization: for a piecewise smooth curve in the normal neighbourhood, $|dr(\dot{\alpha })|\le |\dot{\alpha }|$, so its length from $p$ to $q$ is at least $r(q)$, attained by the radial geodesic. Take a small tangent ball with closure inside a larger normal ball. Any path leaving the small ball before reaching a point inside it has already accumulated length at least its radius. Therefore, for points in the small ball, $r(q)=d(p,q)$ even when competitors may leave the neighbourhood. This local argument does not assert that $r$ equals global Riemannian distance throughout every arbitrarily large star-shaped normal neighbourhood.

Choose homogeneous polynomials $F,G$ defining the two projective plane curves. Their absence of a common irreducible component makes them coprime. For a point $O=[1:0:0]$ outside both curves, they have degrees $n,m$ in $X$ and nonzero constant leading coefficients. The resultantis therefore nonzero. Indeed, a vanishing resultant over $k(Y,Z)$ would give a common factor there, and Gauss's lemma would give a common component of the original curves.

The homogeneity of $F,G$ makes $R$ a homogeneous polynomial of degree $nm$: scaling $(Y,Z)$ by $a$ scales their $X$-roots by $a$, and the product of the $nm$ differences of roots scales by $a^{nm}$. Every point in the intersection projects from $O$ to a zero of $R$ in the projective line. Each such projection line contains only finitely many points of either curve, since it passes through $O$ outside them. Thus the intersection is finite already.

We may now choose $O$ also outside the finitely many joining lines of distinct intersection points. Projection is then injective on the intersection. A nonzero degree-$nm$ homogeneous polynomial on the projective line has at most $nm$ distinct zeros, so the resultant bound for intersections of plane curves givesThis derives the required cardinality bound without assuming the full intersection-multiplicity statement of Bézout's theorem.

Fix an explicit stationary-grid convention. For subdivision arity $m>1$, write the univariate rule asA column of the subdivision matrix is one translated subdivision mask; the next column is translated by $m$ fine-grid rows. Thus the column displacement is the arity. If $a_k$ is active from $k=\alpha$ to $k=\beta$, the subdivision mask width is $w=\beta -\alpha +1$ fine-grid positions, after removing zero padding. Equivalently the mask span is $\beta -\alpha$ fine intervals. This is different from counting old control points in a row stencil: that count measures how many controls influence one newly created point. Stating the convention avoids confusing these two widths.

For a convergent subdivision curve scheme, its support is the influence region of one original control point, equivalently the support of a function of the basic limit $\varphi$ obtained from data $p_j={\delta }_{j0}$. Translation gives the representation $P(t)={\sum }_j{p}_j\varphi (t-j)$. Descendant indices of index zero have parameter positionsThus the limiting support enclosure is $[\alpha /(m-1),\beta /(m-1)]$, as in the support of a stationary subdivision scheme. Endpoint activity gives equality for the ordinary positive convergent masks under discussion.

The functional precision set of a subdivision scheme consists of functions exactly recovered from their uniformly sampled data using the scheme's consistent parameter placement. Polynomial precision records the exactly reproduced polynomial space. Sampling-grid spacing and origin must be specified; the usual arbitrary-grid precision requirement uses every spacing and shift, rather than merely membership in the fixed-grid reconstruction space.

Center the given mask at indices $-1,0,1$. Its binary linear interpolatory subdivision rule isEach new control point is either an old point or an edge midpoint. Hence every refined control polygon is the same curve with extra vertices. Starting from scalar impulse data, the limit is explicitlyThe mask width is three positions, its mask span is two, its largest row stencil uses two old control points, and its subdivision arity is two. Translating the mask indexing to $0,1,2$ translates the basic support to $[0,2]$; the support width remains two original grid intervals.

On $[j,j+1]$, reconstruction is $(j+1-t)p_j+(t-j)p_{j+1}$. Constant and affine samples are therefore recovered exactly. A non-affine polynomial cannot agree with a linear expression on a whole open interval. Thus its polynomial functional precision set of a subdivision scheme isFor continuous functions reproduced at every translated sampling grid, the same conclusion holds: agreement with all chord interpolants forces the affine interpolation identity, hence affinity. At one fixed grid, the larger exactly representable class is the space of linear splines with those spline knots; this is not a higher polynomial precision claim.

For arbitrary data, the one-sided derivatives at $j$ are $p_j-p_{j-1}$ and $p_{j+1}-p_j$. They need not coincide. Therefore the generic limit is $C^0$, not $C^1$. Its first derivative is piecewise constant; exceptional affine data, or individual matching neighboring slopes, can be smoother. This conclusion follows directly from the limit formula, not just from a heuristic mask-width test.

A geodesic is an affinely parametrized curve satisfying ${\nabla }_{\dot{\gamma }}\dot{\gamma }=0$ for the Levi-Civita connection of the Riemannian metric. In coordinates, write ${\nabla }_{{\partial }_i}{\partial }_j={\Gamma }_{ij}^k{\partial }_k$. For a vector field $V(t)=V^k(t){\partial }_k$ along the curve,Putting $V=\dot{\gamma }$ gives the geodesic equationThis is a smooth second-order ordinary differential system, so the Picard-Lindelöf theorem gives local existence and uniqueness for prescribed position and velocity.

To prove local extension of geodesic velocity, first suppose $\dot{\gamma }(0)=0$. The constant curve through $\gamma (0)$ solves this initial value problem, so uniqueness makes $\gamma$ constant locally. The zero vector field is the desired extension.

Otherwise choose a chart about $\gamma (0)$ in which the first component of the coordinate velocity is nonzero. Write the coordinate curve as $c(t)\in {\mathbb{R}}^n$ and defineAt $(0,0)$ its derivative has columns $c^{\prime }(0),e_2,\dots ,e_n$ and is invertible, because the first component of $c^{\prime }(0)$ is nonzero. The inverse function theorem therefore supplies coordinates $(t,z_2,\dots ,z_n)$ on an open neighbourhood after shrinking the parameter interval to $|t|<\delta$. Along the curve these coordinates are $(t,0,\dots ,0)$. The smooth coordinate field $\partial /\partial t$, pushed back through $\Psi$ and the original chart, agrees with $\dot{\gamma }(t)$. This proves the extension without assuming that a long geodesic has no self-intersections.

By metric compatibility, differentiation along the curve givesThe local extension above permits the ordinary vector-field version of the metric identity to be used along the arc, or the same identity follows directly from the coordinate derivative. Hence an affinely parametrized geodesic has constant speed, including the zero-speed case.

Now let $S^n$ be the unit sphere in ${\mathbb{R}}^{n+1}$. Its induced Levi-Civita derivative is the tangential projection of the ordinary ambient derivative: projection preserves the metric identity, and the commuting ambient coordinate derivatives give zero torsion. By uniqueness of the Levi-Civita connection, proved in Question 5, this is the induced connection.

Thus a curve on the sphere is a geodesic exactly when its ambient acceleration has no tangential component. The normal space is spanned by $\gamma$, so ${\gamma }^{\prime \prime }=a(t)\gamma$. Differentiating $\gamma \cdot \gamma =1$ twice gives $\gamma \cdot {\gamma }^{\prime }=0$ andwhere $c\ge 0$ is the constant speed. Therefore ${\gamma }^{\prime \prime }+c^2\gamma =0$. With initial data $\gamma (0)=p$ and ${\gamma }^{\prime }(0)=v$, all solutions areand, when $v=0$, $\boxed{\gamma (t)=p}$. The orthogonality and length conditions show directly that the displayed curve remains on the sphere, and its acceleration is $-c^2\gamma$, so it is indeed a geodesic. The nonconstant geodesics are the great circles traversed at any constant speed; arbitrary parameter intervals give their restrictions, and shifting the initial parameter simply shifts the formula. There are no other geodesics.

For the unit-speed curve, the unit tangent vector, curvature of a space curve, principal normal vector and binormal vector areDifferentiating $t\cdot t=1$ gives $t\cdot {\gamma }^{\prime \prime }=0$, so $t,p,b$ form a positively oriented orthonormal basis. The hypothesis ${\gamma }^{\prime \prime }\ne 0$ makes the principal normal vector well-defined.

To determine the sign of the torsion of a space curve without relying on a pictorial handedness convention, take $R>0$, $h\ne 0$, $L=\sqrt{R^2+h^2}$ and use the unit-speed circular helixWriting $u=s/L$, direct differentiation gives its Frenet frame:ConsequentlyAt $s=0$ the marked point is $(R,0,0)$, with $t=(0,R/L,h/L)$, $p=(-1,0,0)$, $b=(0,-h/L,R/L)$ and $b^{\prime }=(h/L^2,0,0)$. For $h>0$, $b^{\prime }$ points opposite to $p$ and the torsion of a space curve is positive. Replacing $h$ by $-h$ reflects the circular helix across the horizontal plane and makes the torsion of a space curve negative; then $b^{\prime }$ points in the same direction as $p$. The direction of increasing $s$ is fixed in both panels.

For a Riemannian metric $g$, write $s(t)=\sqrt{g_{\gamma (t)}(\dot{\gamma }(t),\dot{\gamma }(t))}$. Use the energy of a curve normalization compatible with the requested inequality:The Cauchy-Schwarz inequality here has the direct variance formEquality holds exactly when $s$ is constant almost everywhere, hence everywhere for a smooth curve. With the alternative energy of a curve convention containing a factor $1/2$, the same assertion reads $L^2\le 2E$.

In the upper half-plane write $\gamma =u+iv$, $v>0$, and use $d{s}^2=(d{u}^2+d{v}^2)/v^2$. Let the endpoints be $P=ip$, $Q=iq$, $p,q>0$. ThenThe first equality requires $\dot{u}=0$ throughout. Endpoint values force $u=0$. The second equality requires the derivative of $\log v$ to have one sign, so $v$ is monotone. Conversely every smooth monotone positive $v$ with these endpoints attains this bound. Consequently the absolute length minimizers are precisely the vertical curves with monotone height, including a constant curve when $P=Q$, and the distance is $|\log (q/p)|$.

For curves stationary for the energy of a curve, fixed-endpoint variations give the Euler–Lagrange equations for $F=({\dot{u}}^2+{\dot{v}}^2)/v^2$:Thus $\dot{u}=c{v}^2$ for a constant $c$. Since $u(1)-u(0)=0$, integration gives $c{\int }_0^1{v}^{2}dt=0$, hence $c=0$. The second equation then becomes $(\dot{v}/v{)}^{\prime }=0$. ThereforeIt has constant hyperbolic speed, is one of the absolute length minimizers, and also has minimum energy of a curve $E=[\log (q/p){]}^2$.