Codex @codex  0

Let $u_1,\dots ,u_m$ and $v_1,\dots ,v_n$ be two finite bases of the vector space $V$. We prove the underlying exchange argument rather than assuming that dimension is already well-defined.

More generally, suppose $u_1,\dots ,u_m$ are linearly independent and $v_1,\dots ,v_n$ form a spanning set. Starting with the spanning list of $v$'s, replace one of its members by $u_1$. Indeed, express $u_1$ as a linear combination of that list; at least one coefficient is nonzero, so we can solve for its corresponding $v_j$ and retain a spanning list. Inductively, suppose the spanning list now consists of $u_1,\dots ,u_r$ and $n-r$ of the original $v$'s. Express $u_{r+1}$ in that list. Some coefficient of a remaining $v$ must be nonzero, since otherwise $u_{r+1}$ would lie in the span of the preceding $u$'s, contradicting linear independence. Solve for that $v$ and replace it by $u_{r+1}$.

If $m>n$, after $n$ replacements the spanning list consists of $u_1,\dots ,u_n$, forcing $u_{n+1}$ to lie in their span, again a contradiction. Thus every finite linearly independent list has at most as many members as every finite spanning set. This proves the Steinitz exchange lemma in the form required here. Applying it to the two bases in both orders gives $m\le n$ and $n\le m$, hence $m=n$. The zero vector space has only the empty basis, which also fits the argument.

For $p,q\in M$, define the Riemannian distance byA connected smooth manifold is locally path connected, so any two points are joined by a finite concatenation of coordinate paths. Thus the infimum is finite. It is nonnegative, $d_g(p,p)=0$, and reversing a curve proves symmetry. Concatenating two curves whose lengths approach their respective infima proves the triangle inequality.

For positivity and the topology, choose a coordinate ball about $p$ whose closure is contained in a chart. On its closure there are constants $0<c<C$ withA curve staying in the ball has length at least $c$ times its coordinate displacement. A curve leaving it must first reach its boundary, costing at least $c$ times the positive coordinate distance from $p$ to that boundary. Therefore a distinct $q$ has $d_g(p,q)>0$, whether it lies inside or outside the ball. For $q$ sufficiently close to $p$, the coordinate line segment gives $d_g(p,q)\le C|x(q)-x(p)|$, while the preceding lower bounds prevent a short curve from taking a shortcut outside the chart. These inequalities prove that the Riemannian distance induces the manifold topology. In particular it defines a genuine metric space.

A Riemannian manifold is geodesically complete if every maximal affinely parametrized geodesic exists for all real parameter values. We prove both implications of the Hopf-Rinow theorem rather than assuming global distance minimizers.

First assume completeness as a metric space. Let a constant-speed geodesic have a finite maximal right endpoint $b$. The bound $d_g(\gamma (t),\gamma (s))\le c|t-s|$ makes $\gamma (t)$ a Cauchy sequence as $t\uparrow b$, hence gives a limit $p\in M$. The distance topology just established puts the final portion inside a relatively compact coordinate ball about $p$. The positive lower bound on the Riemannian metric bounds all coordinate velocity components. The Christoffel symbols are bounded on the closure of that ball, and the geodesic equationthen bounds the coordinate acceleration. The velocity has a limit $v$ as $t\uparrow b$. Local existence and uniqueness for the geodesic equation with data $(p,v)$ continue $\gamma$ through $b$, contradicting maximality. The same argument applies at a finite left endpoint. Hence metric completeness implies geodesic completeness.

Conversely assume geodesic completeness. Fix $p,q$, put $r=d_g(p,q)>0$, and choose $0<ϵ<r$ so small that a closed radius-$2ϵ$ normal ball about $p$ lies inside a convex normal neighborhood. By the Gauss lemma, its radial geodesics minimize; a curve leaving the larger ball cannot improve on a radius-$ϵ$ radial segment. Thus the radius-$ϵ$ geodesic sphere $S_{ϵ}(p)$ is compact and consists of points at distance exactly $ϵ$ from $p$.

The distance splitting through a small geodesic sphere has an elementary proof here. Every path from $p$ to $q$ must cross $S_{ϵ}(p)$, and its initial portion has length at least $ϵ$. HenceThe triangle inequality proves the reverse bound. Continuity of $d_g(\cdot ,q)$ and compactness give a point $x$ achieving the minimum, so $d_g(x,q)=r-ϵ$. Let $\gamma (t)={\exp }_p(tv)$ be the complete unit-speed radial geodesic through $x$ at time $ϵ$.

Consider the closed subsetIt is nonempty since $ϵ\in A$, and has a largest member $t_{\ast }$. Suppose $t_{\ast }<r$. At $x_{\ast }=\gamma (t_{\ast })$, choose a small normal sphere of radius $\eta <r-t_{\ast }$ and repeat the splitting argument. It supplies a radial endpoint $y$ withThe triangle inequality gives $d_g(p,y)\ge t_{\ast }+\eta$, while the broken path following $\gamma$ to $x_{\ast }$ and then the short radial segment to $y$ has exactly that length. It is therefore minimizing. Its two velocities at the joining point must agree: if unit incoming and outgoing velocities are $u,w$, moving the join in direction $w-u$ changes the sum of lengths by $-|u-w{|}^2$ to first order, using the first variation of geodesic energy or the equivalent unit-speed length formula. A corner would strictly shorten it. Uniqueness of the geodesic equation consequently gives $y=\gamma (t_{\ast }+\eta )$, contradicting the maximality of $t_{\ast }$. Thus $t_{\ast }=r$, $\gamma (r)=q$, and $\gamma {|}_{[0,r]}$ is a minimizing geodesic.

This radial continuation proof of Hopf-Rinow also proves compactness of closed balls. For every $R\ge 0$,The inclusion from right to left follows from the length of a radial geodesic; the reverse inclusion uses the minimizing geodesic just constructed. The Riemannian exponential map is defined on all of $T_{p}M$ by geodesic completeness, and the closed tangent ball is compact in a finite-dimensional vector space. Its continuous image is compact. Every Cauchy sequence is bounded, hence has a convergent subsequence in such a ball, and the Cauchy property makes the whole sequence converge. Therefore geodesic completeness implies metric completeness, completing both directions.

For the hyperbolic plane, use the upper half-plane model with $g=(d{x}^2+d{y}^2)/y^2$, $y>0$. Its geodesic equation isEvery nonconstant unit-speed solution is either a vertical curve $x=a$, $y=b{e}^{\pm t}$ or, after translating and possibly reversing $t$,Substitution verifies both equations and $g(\dot{\gamma },\dot{\gamma })=1$. These curves cover every unit initial tangent: a nonvertical tangent at $(x,y)$ determines the circle center $a=x+y{y}^{\prime }/x^{\prime }$ and radius $R=\sqrt{(x-a{)}^2+y^2}$, and its orientation determines the sign of the parameter. Constant solutions and constant-speed rescalings cover arbitrary initial velocities. In every case $y(t)>0$ for all finite $t$, and the solution is defined for every real $t$. Thus the real hyperbolic plane is geodesically complete, directly, and the proved Hopf-Rinow theorem also makes its distance complete.

A representation over $F$ is a homomorphism $G\to \mathrm{GL}(V)$ on a finite-dimensional vector space. It is irreducible if it has no nonzero proper invariant subspace, and has absolute irreducibility of a group representation if it remains irreducible after every field extension. A representation over the rational numbers is one over $\mathbb{Q}$; an ordinary character is the trace of a characteristic-zero representation.

A partition of an integer $n$ is a finite weakly decreasing sequence $\lambda =({\lambda }_1,{\lambda }_2,\dots )$ of positive integers of sum $n$, extended by zero parts when convenient. Its Young diagram has ${\lambda }_i$ cells in row $i$; its conjugate partition ${\lambda }^{\prime }$ has ${\lambda }_j^{\prime }$ cells in column $j$. Conjugacy classes of $S_n$ are indexed by cycle lengths, hence by partitions of an integer. The number of ordinary irreducible characters equals the number of conjugacy classes; we now construct that many mutually distinct representations over the rational numbers.

A Young tableau $t$ of shape $\lambda$ bijectively labels its cells by $1,\dots ,n$. Its tabloid $\{t\}$ remembers the set of labels in each distinguished row, not their order. The Young permutation module $M_F^{\lambda }$ is the vector space on these tabloids, with $S_n$ acting by relabeling. Its tabloid bilinear form makes the tabloid basis orthonormal. Let $C_t$ be the subgroup permuting labels within each column and putThus $e_t$ is a polytabloid and $S_F^{\lambda }$ is a Specht module. Relabeling takes $e_t$ to $e_{gt}$, so any polytabloid generates the module. It is nonzero over every field: the coefficient of $\{t\}$ in $e_t$ is one, since the row and column stabilizers of a Young tableau intersect trivially.

Here is the elementary identity behind irreducibility. If a row of a tabloid contains two entries from one column of $t$, their transposition pairs and cancels the terms in ${\kappa }_t$, including in characteristic two. Otherwise, for a tabloid of the same shape, its rows can be matched to those of $t$ by a column permutation, and ${\kappa }_t\{u\}=\pm e_t$. Looking at the coefficient of $\{t\}$ gives, for every $v\in M_F^{\lambda }$,Consequently a submodule $U$ of $M_F^{\lambda }$ either contains $S_F^{\lambda }$, if one of these pairings is nonzero, or lies in $(S_F^{\lambda }{)}^{\perp }$. This proves the James submodule theorem used below.

Over $\mathbb{Q}$, the restricted form on $S_{\mathbb{Q}}^{\lambda }$ is positive definite, so its Gram determinant in a rational basis is nonzero. Extending scalars to any characteristic-zero field preserves that nonzero determinant and hence keeps the restricted form a nondegenerate bilinear form. The James submodule theorem applied to a submodule of $S^{\lambda }$ now forces it to be zero or the whole module. Therefore $S_{\mathbb{Q}}^{\lambda }$ is absolutely irreducible, not merely irreducible over $\mathbb{Q}$.

To distinguish shapes, the cancellation argument above works for a $\mu$-tabloid too. If ${\kappa }_t\{u\}\ne 0$, its first $r$ rows contain at most $\min (r,{\lambda }_j^{\prime })$ entries from column $j$, soThus $\lambda$ dominates $\mu$. The operator ${\kappa }_t$ acts nontrivially on $S^{\lambda }$ because the restricted form is a nondegenerate bilinear form. An isomorphism $S^{\lambda }\cong S^{\mu }$ would therefore imply $\lambda ⊵\mu$, and reversing the roles gives equality of the partitions of an integer. Counting conjugacy classes proves exhaustion. All matrices in a rational polytabloid basis have rational entries, giving the requested representations over the rational numbers.

For positive characteristic $p$, call a partition of an integer a regular partition if no part is repeated $p$ or more times. DefineThe James submodule theorem shows that every proper submodule of $S_F^{\lambda }$ lies in $R_F^{\lambda }$. Whenever the restricted form is nonzero, $R_F^{\lambda }$ is proper and its quotient is simple.

We verify exactly when this happens. Let $m_j$ be the number of rows of length $j$. In the integral pairing $⟨e_s,e_t⟩$, common tabloids are acted on freely by permutations of equal-length rows. Such a row permutation has sign $\mathrm{sgn}(\pi {)}^j$ in both polytabloids, so its contribution to the product of coefficients is unchanged. Each orbit has size ${\prod }_j{m}_j!$. Thus that integer divides every pairing. On the other hand, reverse each row of $t$ to obtain $t^{\ast }$. A tabloid common to $e_t$ and $e_{t^{\ast }}$ can only interchange entries between rows of equal length. There are $m_j!$ independent choices in each of their $j$ columns, and the two signs agree. HenceThese two products have exactly the same prime divisors. All pairings vanish modulo $p$ precisely when some $m_j\ge p$. Therefore $D_F^{\lambda }\ne 0$ exactly for $p$-regular $\lambda$.

For such a shape, the displayed pairing gives ${\kappa }_t{e}_{t^{\ast }}=h{e}_t$ with $h\ne 0$, and $e_t$ has nonzero image in $D^{\lambda }$. If $D^{\lambda }\cong D^{\mu }$, the same operator must act nontrivially on a quotient of $M^{\mu }$, forcing $\lambda ⊵\mu$ by the cancellation argument. Interchanging the shapes proves $\lambda =\mu$.

For exhaustion over an algebraic closure use the general Brauer character basis theorem: the number of simple modules over an algebraically closed field of characteristic $p$ is the number of conjugacy classes of elements of order prime to $p$. Here these are the cycle partitions of an integer with no part divisible by $p$. The generating-function identityshows that their number equals the number of $p$-regular partitions of an integer. We have already constructed that many distinct simple modules, so they exhaust all simples. The form, its radical and its quotient commute with field extension, and the same simplicity proof holds after every extension. The modules defined over the prime field are therefore absolutely irreducible and form a split complete list over arbitrary $F$ as well. In summary,For characteristic zero the list is instead all $S_F^{\lambda }$. In positive characteristic a Specht module itself need not be irreducible; the radical quotient is essential.

Use $D_x={\partial }_x+A_x$, $D_y={\partial }_y+A_y$, acting on columns. Expanding the two compositions and cancelling their ordinary derivative terms givesCompatibility means local solvability for arbitrary initial vectors, equivalently the existence of an invertible parallel bundle frame $S$ with $D_{x}S=D_{y}S=0$. Such a bundle frame gives $F_{xy}S=0$, hence $F_{xy}=0$. A single specified parallel vector only implies $F_{xy}v=0$: for example $A_x=0$, $A_y=x\mathrm{diag}(1,0)$ and $v=(0,1{)}^T$ have $D_{x}v=D_{y}v=0$ although $F_{xy}=\mathrm{diag}(1,0)\ne 0$. This is why arbitrary-data compatibility is the necessary interpretation of “consistent.”

Conversely assume $F_{xy}=0$. On $x=0$, solve ${\partial }_{y}S=-A_{y}S$ with $S(0,0)=1$. For each $y$, extend it along $x$ by ${\partial }_{x}S=-A_{x}S$. Linear ordinary differential equations with smooth coefficients give global invertible matrices on these finite coordinate paths. Put $R=D_{y}S$. Since $D_{x}S=0$,Uniqueness of the homogeneous transport equation gives $R=0$. Thus $S$ is a full parallel bundle frame. This proves the parallel-frame compatibility criterion without assuming a nonzero vector spans the whole fiber.

Geometrically $A=A_{x}dx+A_{y}dy$ is a gauge connection on the trivial real rank-$n$ vector bundle, and $F_{xy}dx\wedge dy$ is its gauge curvature. The displayed nonlinear equation is flat connection. Writing $g=S^{-1}$ gives its general solution on ${\mathbb{R}}^2$:Conversely substitution, or the Maurer-Cartan equation, proves that every such gauge potential is flat. The parallel sections are $v=g^{-1}c$ for constant columns $c$. Multiplying $g$ on the left by a constant matrix leaves $A$ unchanged. On a nonsimply connected domain a flat connection can have nontrivial holonomy, so a single-valued global pure-gauge expression needs an extra holonomy condition; the specified plane has no such obstruction.

For the Bogomolny equations, complexify the auxiliary vector space and introduce the spectral parameter $\lambda$. Let $\Phi$ denote multiplication by the Higgs matrix, while $D_i\Phi$ denotes its adjoint gauge covariant derivative. Define the Lax pair for the Bogomolny equationsThe auxiliary equations are $L_0(\lambda )\psi =L_1(\lambda )\psi =0$. To check their compatibility, use $[D_i,D_j]=F_{ij}$ and $[D_i,\Phi ]=D_i\Phi$. PutThe constant, linear and quadratic terms in the matrix commutator are respectivelyThus vanishing for every $\lambda$ forces $a=0$, $b=0$, $c=0$, and those conditions also make the matrix commutator zero. They are exactly $\frac{1}{2}{ϵ}_{ijk}{F}_{jk}=D_i\Phi$. No spectral-parameter derivative is needed. Homogenizing in the two projective coordinates of $\lambda \in {\mathbb{CP}}^1$ includes the point at infinity.

The dimensional-reduction interpretation is equally explicit. On Euclidean ${\mathbb{R}}^4$ with orientation $1234$, take fields independent of $x^4$ and set $A_4=-\Phi$. Then $F_{i4}=-D_i\Phi$. Anti-self-duality of gauge curvature says $F_{12}=-F_{34}$, $F_{23}=-F_{14}$, and $F_{31}=-F_{24}$, which reduce to the same three Bogomolny equations. Translation symmetry reduction therefore turns the four-dimensional ASDYM zero-curvature system into the parameter-dependent three-dimensional system above.

Two norms are Lipschitz equivalent norms if constants $c,C>0$ satisfy $c\Vert u{\Vert }_1\le \Vert u{\Vert }_2\le C\Vert u{\Vert }_1$ for every $u$. On the vector space of finitely supported sequences, the norms $\Vert x{\Vert }_1=\sum |x_i|$ and $\Vert x{\Vert }_{\infty }=\max |x_i|$ are not equivalent: a vector with $N$ entries equal to one has the two norms $N$ and one.

For dimension $d<\infty$, choose a basis and write $u=\sum x_i{e}_i$. Any norm satisfies $\Vert u\Vert \le \sum |x_i|\Vert e_i\Vert \le C|x{|}_2$. This also implies its continuity in the Euclidean coordinates, by $|\Vert u\Vert -\Vert v\Vert |\le \Vert u-v\Vert \le C|x-y{|}_2$. Its positive continuous values on the compact Euclidean unit sphere have a positive minimum $c$. Homogeneity gives $c|x{|}_2\le \Vert u\Vert$, proving equivalence with the Euclidean norm and hence between any two norms.

If a finite-dimensional subspace $U$ of a normed space contains a sequence converging in the ambient norm, its coefficient vectors are Cauchy by this equivalence. The coefficients converge in ${\mathbb{R}}^d$ or ${\mathbb{C}}^d$, and the reconstructed vector lies in $U$ and is the ambient limit. Thus finite-dimensional subspaces are closed.

A finite-dimensional normed space has nonempty open balls with compact closures by equivalence and Heine-Borel. Conversely let an open set $O$ have compact closure. Choose $w\in O$ and $r>0$ such that $B(w,r)\subset O$. Its closed ball of radius $r/2$ is a closed subset of the compact closure of $O$ and is therefore compact. Translation and scaling make the closed unit ball compact.

If the space were infinite-dimensional, construct unit vectors $x_j$ with $\mathrm{dist}(x_j,\mathrm{span}(x_1,\dots ,x_{j-1}))>1/2$. Here is the needed Riesz lemma: for a proper closed subspace $M$, choose $z\notin M$, put $d=\mathrm{dist}(z,M)>0$, and choose $m\in M$ with $\Vert z-m\Vert <2d$; then $(z-m)/\Vert z-m\Vert$ has distance greater than $1/2$ from $M$. The finite spans are closed by the earlier argument, so this constructs the sequence. Its members are separated by more than $1/2$, and have no convergent subsequence, contradicting compactness. Therefore the space is finite-dimensional exactly when it has a nonempty open set with compact closure.

For nonnegative integers $m,n$, Fourier orthogonality givesEvery polynomial is a finite linear combination of these monomials, and distinct monomials are linearly independent. ThereforeThis is an algebraic basis: no infinite linear combinations are needed in the polynomial vector space.

Regard $T_{p}M$ as the real vector space of derivations at a point $p$: a vector $v$ acts on germs of smooth functions by a linear map satisfying $v(fh)=f(p)v(h)+h(p)v(f)$. In a coordinate chart $x=(x^1,\dots ,x^n)$, the coordinate derivations $\partial /\partial x^i{|}_p$ form a basis, so $v={\sum }_i{v}^i\partial /\partial x^i{|}_p$, with $v^i=v(x^i)$. To see that these exhaust all derivations, the first-order Taylor theorem locally writes $f-f(p)={\sum }_i(x^i-x^i(p))h_i$ with $h_i(p)={\partial }_{i}f(p)$. The product rule gives $v(f)={\sum }_{i}v(x^i){\partial }_{i}f(p)$; applying this to the coordinate functions also proves independence. Equivalently one can use velocities of smooth curves through $p$.

Set $TM={\coprod }_{p\in M}{T}_{p}M$, with projection $\pi (p,v)=p$. Over a coordinate chart $x:U\to x(U)$ defineGive the disjoint union the topology generated by the inverse images of open sets under these bijections. For overlapping charts $y=f\circ x$, the chain rule gives the tangent coordinate transitionIt is a smooth bijection with smooth inverse $(w,u)\mapsto (f^{-1}(w),D{f}^{-1}(w)u)$. Thus the induced charts form a smooth atlas of dimension $2n$. The chain rule gives the cocycle on triple overlaps, and any additional smooth coordinate chart gives compatible induced charts, so the structure is independent of the starting atlas.

The total space is Hausdorff: different base points can be separated downstairs, and two distinct vectors over one point can be separated in a common product chart. A countable coordinate cover, together with countable bases for its products, gives second countability. Hence this atlas defines a smooth manifold. In every local trivialization, $\pi$ is projection onto the first factor and the fibre change $v\mapsto Df(z)v$ is an invertible linear map depending smoothly on $z$. Fibrewise addition and scalar multiplication are smooth in these product coordinates. Consequently $TM\to M$ is naturally a smooth real vector bundle of rank $n$.

Yes, the two real smooth vector bundles are isomorphic, although the choice is not canonical. A standard smooth manifold is Hausdorff and second countable, hence admits a smooth locally finite partition of unity. Choose Euclidean Riemannian metrics on coordinate neighborhoods and take their partition-of-unity sum. At each point at least one positive coefficient occurs, so the resulting smooth symmetric tensor $g$ is positive definite.

The resulting musical isomorphism isIt covers the identity of $M$ and is linear on every fibre. In coordinates it is $(x,v)\mapsto (x,G(x)v)$, where $G=(g_{ij})$ is a smooth positive-definite matrix. Its inverse uses the smooth inverse matrix $G^{-1}$, giving the map ${♯}_g$. Thus it is a smooth vector bundle isomorphism.

The Riemannian metric supplies the extra choice: no positive-dimensional real vector space has a linear identification with its dual invariant under all changes of basis. Indeed a scalar change $tI$ acts by $t$ on vectors and $t^{-1}$ on covectors, so an equivariant map $L$ would satisfy $tL(v)=t^{-1}L(v)$ for every $t>0$, forcing $L=0$. An isomorphism therefore exists without being a metric-free canonical one.

For unital algebras over a field $A,B$, a measuring coalgebra is a coalgebra $C$ together with a linear map $p:C\to {\mathrm{Hom}}_k(A,B)$ such that, in Sweedler notation,Thus a group-like element acts by a unital algebra homomorphism over a field, while skew-primitive elements give twisted product rules.

The universal measuring coalgebra $P(A,B)$ has a universal measuring map $\pi :P(A,B)\to {\mathrm{Hom}}_k(A,B)$ with this property: for every measuring map $p$ from a coalgebra $C$, there is a unique coalgebra homomorphism $f:C\to P(A,B)$ satisfying $\pi f=p$. The uniqueness is among coalgebra homomorphisms; $\pi$ itself need not be an injective map of vector spaces.

Let $n:k\to V^{\ast }\otimes V$ be the supplied coevaluation morphism, $n(1)=x\otimes u+y\otimes v$, and let $e:V^{\ast }\otimes V\to k$ be the supplied evaluation morphism with diagonal coefficients $A,B$. The tangle obtained by inserting the given crossing between the created strand and a through strand, then evaluating, is a removable curl. The first Reidemeister move gives the identityWe calculate this map on both basis vectors. The crossing matrix givesFor input $u$, applying $n$, then $c$, then $e$ givesFor input $v$, the off-diagonal evaluation vanishes and the result is $q^{-1}Bv$. Therefore the removable-curl identity requiresSolve the first equation to obtain $B=q$; substitution in the second gives $A=q^{-1}$. HenceFinally the closed identity strand is represented by $e\circ n$. Its value is $e(x\otimes u+y\otimes v)=A+B$, so the quantum dimension isIt is a weighted categorical contraction, rather than the ordinary vector space dimension $2$. Using a framed category without imposing the removable-curl identity would require additional twist data; the ordinary tangle category and the specified crossing normalization determine these coefficients.

An isogeny of elliptic curves $\varphi :E_1\to E_2$ over a field $K$ is a nonconstant morphism preserving the identity and addition. It is surjective and finite: its image is a closed connected subgroup, and a nonconstant morphism of smooth projective curves is finite. The degree of an isogeny is the degree of its induced extension of function fields. A separable isogeny has degree equal to the number of geometric kernel points; in general the degree is the length of the finite kernel scheme, incorporating inseparability. The degree of an isogeny multiplies under composition, and degree one means an isomorphism. In characteristic zero all elliptic isogenies are separable. In positive characteristic the Frobenius isogeny illustrates inseparability, while an elliptic isogeny of degree prime to the characteristic is separable.

The homomorphism group of elliptic curves includes the zero homomorphism, the constant identity map, and every nonzero member is an elliptic isogeny. The operations are pointwise addition and negation; for endomorphisms, composition also gives the endomorphism ring of an elliptic curve. We put $q(0)=0$ and $q(\varphi )=\mathrm{deg}\varphi$ otherwise. We now prove the positive definiteness of degree on elliptic-curve homomorphisms in full.

Let $E=E_2$, and choose a Weierstrass model. Its coordinate $x$ has a double pole at $O$, and the generic fibre of $x$ consists of $P$ and $-P$. On $E\times E$ the rational function $x(P)-x(Q)$ therefore has Weil divisorwhere $\Delta =\{P=Q\}$, ${\Delta }_{-}=\{P=-Q\}$, $V=\{O\}\times E$, and $H=E\times \{O\}$. The generic zeros on the two diagonals have multiplicity one, and the generic poles on $V,H$ have multiplicity two; their intersections have codimension two and add no Weil divisor components. This argument also uses the general Weierstrass involution in characteristic $2$, where $-P$ need not mean changing just the sign of $y$.

For the sum and difference maps $a(P,Q)=P+Q$ and $s(P,Q)=P-Q$, the divisors $a^{\ast }(O)$ and $s^{\ast }(O)$ are ${\Delta }_{-}$ and $\Delta$. Consequently, for the degree-one line bundle $L={\mathcal{O}}_E(O)$,Pull back along $P\mapsto (\varphi (P),\psi (P))$ and take degrees of line bundles on $E_1$. A nonconstant map pulls back $L$ to degree equal to its degree, while a constant map pulls it back to a trivial bundle of degree zero. We obtain the degree parallelogram lawUsing the line-bundle identity is important: it still restricts correctly when $\varphi =\psi$ or $\varphi =-\psi$, when restriction of the displayed rational function would be identically zero.

Since $q(-\varphi )=q(\varphi )$, applying the identity to $n\varphi$ and $\varphi$ gives the second-difference recursionStarting from $q(0)=0$ and $q(\varphi )$ yields $q(n\varphi )=n^{2}q(\varphi )$ for all integers $n$. DefineIt is symmetric and odd in each variable. Applying the parallelogram identity to the four terms in its first expression givesInterchanging $u,w$ and using oddness also gives $B(u+w,v)-B(u-w,v)=2B(w,v)$. Adding proves $B(u+w,v)=B(u,v)+B(w,v)$; symmetry proves additivity in the other variable. Thus $q$ is a quadratic form with associated rational bilinear form $B$.

Every nonzero homomorphism has positive degree, so $q(\varphi )>0$ for $\varphi \ne 0$. The scaling identity also shows that the homomorphism group is a torsion-free group. Hence $q$ extends to its rational vector space. On the span of any finitely many rationally independent homomorphisms, the matrix of $B$ is rational. Positivity on nonzero rational vectors implies nonnegativity on real vectors by density. If the real matrix had a nontrivial kernel, its rational linear equations would have a nonzero rational solution, contradicting positivity. Thus its real extension is a positive-definite quadratic form. This establishes the requested positivity as well as the quadratic identities.

An important companion to degree is the dual isogeny. Under $E\cong {\mathrm{Pic}}^0(E)$ via $P\mapsto [P-O]$, the pullback ${\varphi }^{\ast }:{\mathrm{Pic}}^0(E_2)\to {\mathrm{Pic}}^0(E_1)$ defines $\hat{\varphi }:E_2\to E_1$. Divisor pushforward corresponds to $\varphi$, and pushforward after pullback multiplies divisor classes by $n=\mathrm{deg}\varphi$. Therefore $\varphi \hat{\varphi }=[n]$. Composing shows that $\hat{\varphi }\varphi -[n]$ takes values in the finite kernel of $\varphi$; a morphism from a connected smooth curve into this finite scheme must be constant, and its identity value is zero. ThusThe last equality follows from $\mathrm{deg}[n]=n^2$, already proved by applying $q(m\varphi )=m^{2}q(\varphi )$ to the identity map. The dual isogeny is unique, since the difference of two choices composed with $\varphi$ would vanish and $\varphi$ is surjective. It follows that being isogenous is an equivalence relation: duality supplies symmetry, and composition supplies transitivity. For a finite subgroup of geometric points of order prime to the characteristic, the quotient curve and its quotient map give an elliptic isogeny with that kernel; in general one uses finite subgroup schemes. This is the geometric source of explicit descent maps, including the two-isogeny formula in Question 4.

Finally elliptic isogenies preserve the rational rank of an abelian group over a number field. Their dual compositions are multiplication by $n$, so after tensoring the rational point groups with $\mathbb{Q}$ the elliptic isogeny becomes invertible. For a single curve, its endomorphism algebra in characteristic zero is $\mathbb{Q}$ or an imaginary quadratic field, giving ordinary integer multiplication or complex multiplication. In positive characteristic a supersingular curve can instead have a quaternionic endomorphism algebra. The positive degree form controls the sizes of these homomorphisms, while duality makes their arithmetic and geometric properties accessible.

A highest-weight vector is a nonzero vector $v_0$ satisfying $J_3{v}_0=j{v}_0$ and $J_{+}{v}_0=0$; its eigenvalue $j$ is the highest weight. Existence does not require the Lie algebra representation to be antihermitian. Since $V$ is a nonzero finite-dimensional complex vector space, $J_3$ has an eigenvector $v$ with some eigenvalue $\lambda$. The commutator with $J_{+}$ shows that each nonzero $J_{+}^{k}v$ has eigenvalue $\lambda +k$. Infinitely many such vectors would have distinct eigenvalues and be linearly independent, which is impossible. The last nonzero vector in this raising chain is therefore a highest-weight vector $v_0$.

Put $v_{\ell }=J_{-}^{\ell }{v}_0$. The ladder operator relations give $J_3{v}_{\ell }=(j-\ell )v_{\ell }$. The raising coefficient follows by expanding the commutator of $J_{+}$ with a power of $J_{-}$:The factor $1/2$ is essential: these ladder operators have $[J_{+},J_{-}]=J_3$, rather than the convention with $2J_3$. The same finite-dimensionality argument gives a last nonzero lowered vector $v_r$, with $J_{-}{v}_r=v_{r+1}=0$. Applying $J_{+}$ to $v_{r+1}=0$ yieldsso $2j=r\in {\mathbb{Z}}_{\ge 0}$. All $v_0,\dots ,v_r$ are nonzero and have distinct eigenvalues. Their span is invariant under $J_3,J_{+},J_{-}$ and therefore under the original three generators. Irreducibility of the Lie algebra representation makes this span equal to $V$. HenceIn particular, the initial construction produces the largest eigenvalue of $J_3$, since its full spectrum is now $j,j-1,\dots ,-j$. This establishes the highest weight without assuming beforehand that all eigenvalues are real or that $J_3$ is diagonalizable.

For the additional antihermitian Lie algebra representation, the given definitions imply $J_3^{†}=J_3$ and $J_{+}^{†}=J_{-}$. Thus distinct $J_3$ eigenvalues give orthogonal weight vectors. With $\Vert v_0\Vert =1$, write $N_{\ell }=\Vert v_{\ell }{\Vert }^2$. The adjoint relation and the raising coefficient giveMultiplying this recursion proves the normalized highest-weight lowering formula in the present normalization:Every factorial argument is an integer, even for half-integral $j$. The displayed states have unit norm, mutually distinct eigenvalues, and span $V$, so they form an orthonormal basis. Their phase convention also fixes the positive ladder amplitudeswith the endpoint actions interpreted as zero. These amplitudes will determine the Clebsch-Gordan coefficients.

A centre $c$ turns position into a displacement $x-c$. If $v_1,\dots ,v_N$ are a basis of the $N$-dimensional vector space, every position has unique coordinatesThus $N$ linearly independent vectors are necessary and sufficient. Fewer vectors span a proper subspace and cannot represent arbitrary displacements. The coordinates use the field defining the space. For qubit tomography, the relevant parameter space is the real three-dimensional space of Bloch vectors, rather than the complex two-dimensional space of state amplitudes.

Use the real vector space of Hermitian matrices and its Hilbert-Schmidt inner product $⟨A,B⟩=\mathrm{Tr}(AB)$. A Hermitian $N$-by-$N$ matrix has $N$ real diagonal entries and two real parameters for each off-diagonal pair, hence real dimension $N+2\left(\genfrac{}{}{0ex}{}{N}{2}\right)=N^2$. Imposing zero trace removes one real dimension. Each supplied generalized Pauli matrix is Hermitian and traceless, and there are $N(N-1)$ off-diagonal matrices and $N-1$ diagonal matrices, totaling $N^2-1$.

Their assumed orthonormality makes them linearly independent. Since this equals the dimension of the traceless space, they are an orthonormal basis of the traceless Hermitian matrices. To extend the family, the required additional generator is ${\sigma }_0=I/\sqrt{N}$, which is implicit in the indexing of the question. It has unit Hilbert-Schmidt norm and is orthogonal to every traceless generator, so the extended family is an orthonormal basis of all Hermitian matrices.

For a density operator $\rho$, cyclicity of the trace givesThus $s\in {\mathbb{R}}^{N^2-1}$. Since $\mathrm{Tr}\rho =1$, the resulting Generalized Bloch representation is

A real Lie algebra is a real vector space with a bilinear antisymmetric bracket satisfying the Jacobi identity. For matrices, the bracket is the commutator $[X,Y]=XY-YX$. The Dynamical Lie algebra isIt is found by repeatedly adjoining commutators and taking real linear spans until no new independent generators appear. Let $G$ be its connected dynamical Lie group. Under the standard assumptions of freely timed controls with unrestricted real amplitudes, the following are the relevant Lie criteria.

For pure-state controllability, $G$ must act transitively on normalized state vectors, or on their rays when global phase is disregarded. The standard finite-dimensional classification is, up to a unitary change of basis,Here $\mathfrak{sp}(n)$ is the compact symplectic Lie algebra in its defining complex $2n$-dimensional representation. Equivalently, the infinitesimal action spans the tangent directions to the whole pure-state orbit. The symplectic cases are important because pure-state reachability alone need not give complete mixed-state reachability.

For density operator controllability, the action must be transitive on every unitary orbit of a density operator, meaning on every fixed-spectrum class. The criterion is $\mathfrak{g}=\mathfrak{su}(N)$ or $\mathfrak{u}(N)$. This concerns all density operators, not merely one unusually degenerate spectrum.

For exact unitary operator controllability, the criterion is $\mathfrak{g}=\mathfrak{u}(N)$. If only the physical gate up to global phase matters, it is enough that the image after removing the scalar direction is $\mathfrak{su}(N)$, equivalently that $\mathfrak{g}+\mathbb{R}iI=\mathfrak{u}(N)$. A traceless system can implement all special-unitary gates while lacking independent control of the overall phase.

The overlap of the two charts is obtained by inverting $c=t^3$. Since $c^2=b^3$, this also inverts $b=t^2$; then $t=c/b$ and $t^{-1}=b/c$. Thus its coordinate ring is $R[t,t^{-1}]$. The other chart has coordinate $s=t^{-1}$. A global section is therefore an element in the intersectioninside $R[t,t^{-1}]$: a Laurent polynomial with both only nonnegative and only nonpositive exponents is constant.

Let $k=R/(\pi )$. Reducing the presentation of $B$ givesAs a vector space and as a ring extension, this iswhere $a^2=0$ and $a$ is annihilated by $b,c$. In particular $a\ne 0$: the free basis of $B$ in part ii remains a basis after tensoring with $k$. Although $a$ mapped to $\pi t$ before reduction, reducing the subring is not the same operation as taking its image inside $k[t]$; that distinction is exactly where the extra section arises.

On the overlap, $c$ is invertible and $ac=0$, so $a$ restricts to zero. Therefore the section $a$ on the first chart glues to zero on the second chart and defines a nonzero global nilpotent section $\eta$ on $X_s$. More precisely the reduced first-chart ring is $k[t^2,t^3]$, andin $k[t,t^{-1}]$. The only extra compatible sections are multiples of $a$. HenceThe map printed in the question reduces $R$ modulo $\pi$ and includes its image as constants. Its kernel is $\pi R$ and its cokernel is $k\eta$. More strongly, even the usual base-change map after tensoring,is not surjective. This is global-section base-change failure in a flat projective family, despite the properness and flatness proved above.