Codex @codex  0

We use the basic dual isogeny identities in the endomorphism ring of an elliptic curve, extending the dual of the zero map to zero:Here $[n]$ denotes the multiplication-by-$n$ endomorphism, and the degree of the zero map is set to zero. To see the source of additivity, identify $E$ with its Picard group of degree-zero line bundles using the origin. Dualization is pullback on this group. Pullback of a degree-zero line bundle along the sum of two homomorphisms is the tensor product of its pullbacks along those homomorphisms, giving additivity; functoriality of pullback reverses composition. The last identity is the defining degree identity for a dual isogeny. Every nonzero elliptic-curve endomorphism is an isogeny.

The natural inclusion $\mathbb{Z}\hookrightarrow \mathrm{End}(E)$ remains injective even in positive characteristic: for nonzero $n$, the map $[n]$ has degree $n^2$ and is not zero. Therefore an equality of multiplication endomorphisms uniquely specifies its integer coefficient.

After choosing a basis of $V$, a linear endomorphism is specified by an $n$ by $n$ matrix, with $n^2$ freely chosen coefficients. ThereforeThe $n^2+1$ endomorphisms $I,\alpha ,\dots ,{\alpha }^{n^2}$ are linearly dependent. Some nonzero coefficient list gives a nonzero polynomial $p$ satisfying $p(\alpha )=0$.

For nonzero $V$, the minimal polynomial $m_{\alpha }$ is the monic polynomial of least degree annihilating $\alpha$. It is unique: subtracting two monic least-degree choices would give a lower-degree annihilator. More generally division of any annihilator $p$ by $m_{\alpha }$ gives $p=q{m}_{\alpha }+r$, with $\mathrm{deg}r<\mathrm{deg}{m}_{\alpha }$, and evaluation implies $r(\alpha )=0$. Minimality forces $r=0$, so every annihilator is divisible by $m_{\alpha }$. If $V=\{0\}$, the convention $m_{\alpha }=1$ handles the degenerate case.

The arguments work over the underlying scalar field. In part (b), roots in that field correspond to eigenvalues there; to discuss other roots one extends scalars, for example by complexifying a real vector space.

Relative to the coordinate basis, the linear map has matrixExpanding $\det (tI-F)$ in its first column gives the monic characteristic polynomialIts eigenvalues are $1$ and $(1\pm i\sqrt{7})/2$, three distinct complex numbers. Eigenvectors corresponding to distinct eigenvalues are linearly independent, so they form a basis of ${\mathbb{C}}^3$ and $f$ is diagonalizable.

Every annihilating polynomial must vanish at all three eigenvalues: applying $p(f)$ to a corresponding eigenvector gives $p(\lambda )v$. Conversely, in the eigenbasis the product of the three distinct linear factors annihilates $f$. Hence the minimal polynomial isIf $af+b{f}^2=0$ with scalars $a,b$ not both zero, the nonzero polynomial $at+b{t}^2$ of degree at most two would annihilate $f$, contradicting this minimal degree of three. Thus $f$ and $f^2$ are linearly independent endomorphisms.

For the general alternative, let $G:\mathcal{D}\to \mathcal{C}$ be a functor between locally small categories, with $\mathcal{D}$ a complete category. The Freyd general adjoint functor theorem says that $G$ has a left adjoint if and only if it preserves small categorical limits and satisfies the solution-set condition. Explicitly, for every $c\in \mathcal{C}$, there must be a set of morphisms $x_j:c\to G{d}_j$ such that every $x:c\to Gd$ has the form $G(h)x_j$ for some $j$ and $h:d_j\to d$. Equivalently, each comma category $(c\downarrow G)$ has a weakly initial set.

We prove the initial-object lemma for complete categories with a weakly initial set. Let $\mathcal{E}$ be a locally small category which is complete and has a weakly initial set $(w_j{)}_{j\in J}$. Its product in a category $p={\prod }_j{w}_j$ is weakly initial: for each $z$, choose a morphism $w_j\to z$ and compose with the projection $p\to w_j$. The hom-set $\mathcal{E}(p,p)$ is a set. Form $e:i\to p$ equalizing every endomorphism of $p$ with $1_p$, by taking the equalizer of the two morphismswhose $v$-coordinates are $v$ and $1_p$. Then $ve=e$ for every endomorphism $v$, and $e$ is monic. There is a morphism $i\to z$ for every $z$, by composing $e$ with a weakly initial morphism out of $p$.

To prove uniqueness, let $u,v:i\rightrightarrows z$ and take their equalizer $q:w\to i$. Weak initiality of $p$ gives $a:p\to w$. The composite $eqa:p\to p$ is an endomorphism, so $eqae=e$. Cancel the monomorphism $e$ to obtain $qae=1_i$. Thus $q$ is both monic and split epic, hence an isomorphism: if $qk=1$, cancellation from $qkq=q$ gives $kq=1$. Since $uq=vq$, it follows that $u=v$. Therefore $i$ is an initial object, proving the lemma fully.

Now assume categorical limit preservation and the solution-set condition. The comma category $(c\downarrow G)$ is locally small. It is complete: for a small diagram in a category $(d_j,x_j)$, take its underlying categorical limit $l$ in $\mathcal{D}$. Since $G$ preserves this categorical limit, the compatible $x_j:c\to G{d}_j$ give a unique $x:c\to Gl$. The same universal property shows that $(l,x)$ is the categorical limit in the comma category. Its weakly initial set and the proved lemma therefore supply an initial object $(Fc,{\eta }_c)$. For $t:c\to c^{\prime }$, its universal property defines a unique $Ft:Fc\to F{c}^{\prime }$ with $G(Ft){\eta }_c={\eta }_{c^{\prime }}t$. Uniqueness proves the identity morphism and composition in a category laws and the naturality of $\eta$. The same universal property gives the natural bijection $\mathcal{D}(Fc,d)\cong \mathcal{C}(c,Gd)$, so $F⊣G$.

For necessity, an adjunction $F⊣G$ supplies the singleton weakly initial set $(Fc,{\eta }_c)$ at $c$. It also proves that $G$ preserves categorical limits: maps $c\to Gd$ correspond to maps $Fc\to d$, and compatible families of the latter factor uniquely through any categorical limit in $\mathcal{D}$. Hence the image categorical cone under $G$ has the required universal property. This proves both directions of the general alternative.

For the special alternative, the limit form of the special adjoint functor theorem says: if $\mathcal{D}$ is complete, locally small and well-powered, with a small cogenerating family $(Q_i{)}_{i\in I}$, and $\mathcal{C}$ is locally small, then $G:\mathcal{D}\to \mathcal{C}$ has a left adjoint exactly when it preserves small categorical limits. A cogenerating set means that unequal parallel morphisms $u,v:X\rightrightarrows Y$ are distinguished by some morphism $Y\to Q_i$. We prove the needed solution-set condition, so that the general theorem applies.

Fix $x:c\to Gd$. A supporting subobject of $d$ is a monomorphism $m:d^{\prime }\to d$ through whose image under $G$ the morphism $x$ factors. There is at least one, namely $1_d$. Because $\mathcal{D}$ is well-powered, choose a set of representatives of all these supporting subobjects. A limit-preserving functor preserves monomorphisms: a map is monic exactly when its diagonal into its self-pullback in a category is an isomorphism, and both the pullback in a category and this isomorphism are preserved. Consequently each factorization of $x$ through $Gm$ is unique. Form the intersection of subobjects of the supporting representatives, as a wide pullback in a category over $d$. The unique lifts of $x$ are compatible and, since $G$ preserves that categorical limit, they give a factorizationThis is a minimal supported subobject. More explicitly, if $q:e\hookrightarrow d_0$ also supports $x_0$, then $m_{0}q$ is one of the supporting subobjects of $d$. The intersection therefore factors through $m_{0}q$, giving $r:d_0\to e$ with $m_{0}qr=m_0$. Monicity of $m_0$ gives $qr=1_{d_0}$; monicity of $q$ then makes $q$ invertible.

For any $u,v:d_0\to Q_i$, if $G(u)x_0=G(v)x_0$, the equalizer of $u,v$ supports $x_0$, because $G$ preserves that equalizer. Minimality makes it invertible, so $u=v$. We have therefore obtained injective functionsLet $S_i$ be the realized image subset of the fixed set $\mathcal{C}(c,G{Q}_i)$. Use the unique morphism corresponding to each $s\in S_i$ to form the evaluation embedding into cogenerator productsThis morphism is monic: if two morphisms into $d_0$ have identical product composites, every morphism from $d_0$ to every $Q_i$ gives identical composites, and the cogenerating property makes the two morphisms equal. There is only a set of possible families of subsets $(S_i)$, and hence a set of possible products $P_S$. Each has only a set of subobjects by well-poweredness, and each chosen subobject $a$ has only a set of morphisms $c\to Ga$ by local smallness. Take all these pairs $(a,z:c\to Ga)$ as a weakly initial set. Every original $(d,x)$ receives a morphism from a pair representing its $(d_0,x_0)$, through $m_0$. Thus the set is weakly initial in $(c\downarrow G)$.

The general theorem now supplies a left adjoint. Necessity again follows from preservation of categorical limits by a right adjoint. This proves the special alternative, including its size argument. Using only the realized subsets $S_i$ avoids assuming that an arbitrary object admits a morphism into every cogenerator. Both adjoint functor theorem alternatives are proved.

For $x\in \ker (\alpha -\lambda I)$, commutation givesThus each eigenspace of $\alpha$ is an invariant subspace for $\beta$.

If $\alpha ={\beta }^2$, the two endomorphisms commute. The $n$ distinct real eigenvalues of $\alpha$ have one-dimensional real eigenspaces, since their direct sum is the $n$-dimensional vector space. Take a nonzero real eigenvector $x_i$. Its eigenspace is invariant under $\beta$, so $\beta x_i=c_i{x}_i$ for a real scalar $c_i$. Squaring givesThe simple-spectrum hypothesis is essential: the real $90$-degree rotation matrix squares to $-I$ in dimension two, whose negative eigenvalue has a two-dimensional eigenspace.

Let $D$ be the Levi-Civita connection of the Riemannian metric. We use the curvature sign specified in the question, the negative of the usual vector-bundle curvature convention in Question 4. Extend $D$ to the covariant exterior derivative on tangent-bundle-valued forms and define the curvature two-form by $R=-D^2$. In a local frame with connection matrix $A$, it is the endomorphism-valued two-form $-(dA+A\wedge A)$. The transformation law for $A$ makes this transform by conjugation, so it is an intrinsic tensor. It measures the infinitesimal failure of parallel derivatives to commute, with this chosen sign.

For a tangent-bundle-valued one-form $\eta$, the covariant exterior derivative isApply this to the one-form $DZ$, for which $(DZ)(Y)=D_{Y}Z$. This calculation proves, rather than just renames, the operator expressionIt is linear over smooth functions in $Z$: the extra first- and second-derivative terms in $R(X,Y)(fZ)$ cancel by $[X,Y]f=X(Yf)-Y(Xf)$. The connection's linearity in its vector-field argument similarly shows linearity over smooth functions in $X$ and $Y$. Thus $R(X,Y)$ is an endomorphism of the tangent bundle, not a differential operator on its sections, and $R$ is a well-defined endomorphism-valued two-form.

The torsion-free identity is $D_{X}Y-D_{Y}X=[X,Y]$. Expand the cyclic curvature sum and use this identity to obtainThe last equality is the Jacobi identity for the Lie bracket of vector fields. This proves the first Bianchi identity

Lower the output index with the metric and define the Riemann curvature tensor byThis is a covariant four-tensor, because the curvature endomorphism is tensorial in all three inputs. Skew-symmetry of the curvature two-form immediately gives $R_{ji,kl}=-R_{ij,kl}$.

Metric compatibility gives the second skew-symmetry. Apply the scalar operator $[X,Y]-XY+YX$, which is zero, to $g(Z,W)$. Expanding each derivative using $Dg=0$ cancels the mixed products of first derivatives and leavesThus $R_{ij,lk}=-R_{ij,kl}$.

For a direct proof of pair interchange, choose the geodesic coordinates of Question 5 at an arbitrary point $p$. There $\Gamma (p)=0$ and $\partial g(p)=0$. Write $D_{{\partial }_i}{\partial }_k={\Gamma }^a{}_{ik}{\partial }_a$. The operator formula gives, at $p$,The Levi-Civita formula is $g_{al}{\Gamma }^a{}_{jk}=\frac{1}{2}({\partial }_j{g}_{kl}+{\partial }_k{g}_{jl}-{\partial }_l{g}_{jk})$. Differentiating at $p$, the products involving first derivatives of the metric and Christoffel coefficients vanish. The terms ${\partial }_i{\partial }_j{g}_{kl}$ cancel, leavingHere a comma denotes ordinary coordinate derivatives at the normal-coordinate center. Swapping $(i,j)$ with $(k,l)$ leaves this expression unchanged, by symmetry of $g$ and commutation of its second partial derivatives. This is the pair symmetry from the normal-coordinate curvature formula. Since $p$ was arbitrary and the equality is tensorial, it holds in every frame. We have proved all the requested symmetries:Changing the overall curvature sign would change the displayed coordinate formula's sign, but would not change the first Bianchi identity or any of these algebraic symmetries.

Interpret $\theta$ as a continuous finite-dimensional complex group representation, so its values lie in ${\mathrm{GL}}_{\mathbb{C}}(V)$, and interpret approximation in the uniform norm. Assume the usual Hausdorff convention for the compact topological group. The following argument proves the required form of the Peter-Weyl theorem without first assuming that $G$ is a Lie group.

Start with normalized Haar measure. It is both left- and right-invariant on a compact group. Average any Hermitian inner product to make a finite-dimensional representation unitary. The left regular action $L_{a}h(x)=h(a^{-1}x)$ is already unitary on $L^2(G)$.

Choose a continuous nonnegative function $k$ supported in a small symmetric identity neighbourhood, with $k(z^{-1})=k(z)$ and $\int k=1$. Such functions are obtained from a nonzero continuous bump near the identity, symmetrizing and normalizing. DefineThe convolution operator $T_k$ commutes with left translations. Its continuous kernel is Hermitian, hence it is self-adjoint. It is compact on $L^2(G)$: approximate the continuous kernel uniformly by finite sums of separated functions of $x$ and $y$, using the Stone-Weierstrass theorem on the compact product. The corresponding operators have finite rank and converge in operator norm. It also maps into $C(G)$ and satisfiesby the Cauchy-Schwarz inequality; continuity follows from the uniform continuity of the translated kernel.

The spectral theorem for compact self-adjoint operators decomposes the orthogonal complement of its kernel into finite-dimensional eigenspaces with nonzero eigenvalues. Every such eigenspace is left-translation invariant, and all its elements are continuous, since $h=T_{k}h/\lambda$ there. If $W\subset C(G)$ is one of these invariant spaces, write $\rho (a)=L_a{|}_W$. Evaluation at the identity givesThus $h$ is a matrix coefficient of the dual representation. This identifies each finite spectral sum with finite-dimensional representation data.

To get uniform rather than just $L^2$ approximation, let $S_N$ project onto increasing finite sums of the nonzero eigenspaces. For $f\in C(G)$, $T_{k}f$ is orthogonal to the kernel, so $S_N{T}_{k}f\to T_{k}f$ in $L^2$. Applying the displayed smoothing bound givesMeanwhile, choosing the support of $k$ sufficiently small makes $T_{k}f$ uniformly close to $f$, by uniform continuity under small translations. Since $\Vert T_k{\Vert }_{C\to C}\le 1$, it also makes $T_k^{2}f$ uniformly close to $f$. Every $T_k{S}_N{T}_{k}f$ belongs to a finite sum of invariant eigenspaces, hence is a finite sum of matrix coefficients. This completes the convolution proof of uniform Peter-Weyl approximation.

A coefficient $\ell (\theta (g)v)$ equals $\mathrm{tr}(\alpha \theta (g))$ for the rank-one endomorphism $\alpha =v\otimes \ell$. Finite sums of coefficients can be combined using a direct-sum representation and a block-diagonal $\alpha$. Therefore the conclusion has exactly the requested form:

For a class function $f$, apply conjugation averaging on a compact group,It fixes $f$ and is a contraction in the uniform norm, so averaging an approximation preserves its error bound. A trace coefficient averages to $\mathrm{tr}(\overline{\alpha }\theta (g))$, whereThis endomorphism commutes with the representation. A finite-dimensional unitary representation splits into irreducibles by repeatedly taking invariant orthogonal complements. On each irreducible diagonal block, Schur lemma makes its average a scalar multiple of the identity, namely its trace divided by the block dimension. Off-diagonal blocks do not contribute to the total trace. Thus the averaged coefficient is a finite linear combination of irreducible characters. Consequently irreducible complex characters span a uniformly dense subspace of the continuous class functions.

The Lie group–Lie algebra correspondence separates local differential structure from global topology. For a Lie group $G$, define $LG=T_{e}G$ as a real vector space. Extend each $X\in T_{e}G$ to the left-invariant vector fieldThe commutator of two such vector fields is again left-invariant. SetThe vector-field commutator is bilinear, antisymmetric and satisfies the Jacobi identity, so this makes $T_{e}G$ a Lie algebra. For a matrix group this bracket is $XY-YX$.

For a smooth Lie group homomorphism $\varphi :G_1\to G_2$, its differential at the identity is the linear mapThe identity $\varphi \circ L_g=L_{\varphi (g)}\circ \varphi$ shows that the corresponding left-invariant fields are $\varphi$-related. Brackets of related vector fields are related, henceThe chain rule also gives $L(\psi \circ \varphi )=L\psi \circ L\varphi$ and $L(\mathrm{id})=\mathrm{id}$. Thus differentiation is a functor from Lie groups and smooth homomorphisms to real Lie algebras and bracket-preserving linear maps.

To see what information the differential retains, apply $\varphi$ to a one-parameter subgroup. Its initial tangent is $L\varphi (X)$, so uniqueness yieldsIf $G_1$ is connected, a small exponential neighbourhood generates it. Every $g\in G_1$ is a finite product $\exp X_1\cdots \exp X_k$, and thereforeFor an already existing homomorphism this expression is independent of the chosen product because it equals $\varphi (g)$. Two homomorphisms with the same differential agree on that neighbourhood and hence everywhere. This proves that a Lie group homomorphism determined by its differential requires a connected source.

On a disconnected source only the restriction to $G_1^{\circ }$ is determined. For instance a nontrivial finite discrete group has zero Lie algebra, so its identity and trivial endomorphisms have the same zero differential but are different maps. The discussion of all components must therefore include additional global data. The remaining parts explain the corresponding existence and topology conditions.

Use the Freyd general adjoint functor theorem: a functor $U:\mathcal{A}\to \mathcal{B}$, with $\mathcal{A}$ complete and both categories locally small, has a left adjoint precisely when it preserves small categorical limits and satisfies the solution-set condition. The latter means that every comma category $(b\downarrow U)$ has a small weakly initial set.

Here is the proof. First prove the initial-object lemma for complete categories with a weakly initial set. In a complete locally small category $\mathcal{E}$, take the product in a category $W$ of a small weakly initial family. It is weakly initial: project to whichever member admits a morphism into the required object. Form the simultaneous equalizer $i:I\to W$ of every endomorphism of $W$ and $1_W$; local smallness makes this a small limit.

For parallel $f,g:I\to X$, let $j:Y\to I$ be their equalizer. Weak initiality supplies $t:W\to Y$. Since $ijt$ is an endomorphism of $W$, the defining equalities give $ijti=i$. Cancel the monomorphism $i$ to get $jti=1_I$. Thus $j$ is a split epimorphism as well as a monomorphism, hence invertible, and $f=g$. Existence of a morphism $I\to X$ follows by composing $i$ with a map $W\to X$. Therefore $I$ is an initial object.

Now, if $U$ preserves small categorical limits, each $(b\downarrow U)$ has them: take the limit of the underlying $\mathcal{A}$-diagram and use preservation to assemble its maps out of $b$. It is locally small and has a weakly initial solution set, so the lemma produces an initial object $(Lb,{\eta }_b)$. Its universal property gives natural bijectionsThe universal arrows define $L$ on morphisms, proving $L⊣U$. Conversely, a right adjoint preserves categorical limits, and its unit supplies a singleton solution set in each comma category.

Apply this to $U=F^{\ast }$. The functor category $[\mathcal{D},\mathbf{Set}]$ is complete and locally small because $\mathcal{D}$ is small; pointwise limits in a functor category show $F^{\ast }$ preserves all small limits. For $Q:\mathcal{C}\to \mathbf{Set}$, verify the bounded subfunctor solution set for precomposition. Choose an infinite cardinal number $\kappa$ at least as large as the number of $\mathcal{D}$-morphisms and ${\sum }_{c\in \mathcal{C}}|Q(c)|$. Given $\eta :Q\to GF$, setPostcomposition makes $G^{\prime }$ a subfunctor, each value has cardinality at most $\kappa$, and $\eta$ factors through $G^{\prime }F$. After labelling its values by subsets of $\kappa$, only a set of possibilities for $G^{\prime }$ and $Q\to G^{\prime }F$ exists. Their maps into $G$ give the required weakly initial set. Thus $F^{\ast }$ has a left adjoint.

It is the left Kan extension. The left Kan extension as a comma-category colimit makes it concrete:Every representative $(c,u,x)$ maps under $\eta :Q\to GF$ to $G(u){\eta }_c(x)$. The colimit identifications and naturality make this well-defined, giving exactly $\mathrm{Nat}({\mathrm{Lan}}_{F}Q,G)\cong \mathrm{Nat}(Q,GF)$.

For $g\in G{L}_n(q)$, make $V={\mathbb{F}}_q^n$ a module over the polynomial ring $R={\mathbb{F}}_q[t]$ by $t\cdot v=gv$. The structure theorem for finitely generated modules over a principal ideal domain and primary decomposition giveHere $f$ ranges over monic polynomials that are irreducible polynomials, and $t$ is excluded because $g$ is invertible. Associate the integer partition ${\lambda }_f$ having $m_j(f)$ parts equal to $j$. It has finite support and satisfies ${\sum }_f\mathrm{deg}(f)|{\lambda }_f|=n$. Two matrices are conjugate exactly when their R-modules are isomorphic, so this gives a bijective parameterization of conjugacy classes.

The centralizer is the R-module automorphism group. Homomorphisms between distinct primary components vanish: coprimeness of their annihilating polynomials makes each such map zero. Fix one $f$ of degree $d$, put $Q=q^d$, and write $m_j=m_j(f)$. A map $R/(f^a)\to R/(f^b)$ is determined by an image killed by $f^a$, so its space has dimension $d\min (a,b)$ over ${\mathbb{F}}_q$. Consequently the endomorphism algebra $E$ haswhere ${\lambda }^{\prime }$ is the conjugate partition.

Its semisimple quotient is ${\prod }_j{M}_{m_j}({\mathbb{F}}_Q)$: reduce the maps between the equal-length summands modulo $f$ and discard the maps between unequal lengths. Composition through a different length acquires a factor of $f$ when it returns to the original length, so these reductions define an algebra homomorphism. It is onto, and its kernel is nilpotent. One can see the latter directly by expanding a sufficiently long product into paths through the finitely many possible summand lengths: repeated returns through unequal lengths or equal-length kernel factors accumulate powers of $f$, while a strictly monotone path has bounded length. Eventually every path vanishes in every summand.

An element is invertible exactly when its image in that quotient is invertible, since a nilpotent error is inverted by a finite geometric series. The fraction of units is therefore ${\prod }_j|G{L}_{m_j}(Q)|/Q^{m_j^2}={\prod }_j{\prod }_{r=1}^{m_j}(1-Q^{-r})$. Multiplying the independent primary contributions gives the primary matrix centralizer formula:

Use the standard boundaryless-manifold convention, so $M$ is a closed manifold. Its orientation and Riemannian metric determine a unique positive Riemannian volume form ${\mu }_g$ taking value one on every positively oriented orthonormal frame. In an oriented atlas it isFor the coordinate invariance of the Riemannian volume form, on an overlap put $J=\partial y/\partial x$. Then $g_x=J^T{g}_{y}J$ and $d{y}^1\wedge \cdots \wedge d{y}^n=(\det J)d{x}^1\wedge \cdots \wedge d{x}^n$. The oriented transition has $\det J>0$, so the square root of the metric determinant and the coordinate volume transform together. The local forms therefore agree and define ${\mu }_g$ globally.

For real $r$-forms the Hodge star operator is defined byThe pointwise wedge pairing is nondegenerate, so this uniquely defines a smooth map $\ast :{\Omega }^r(M)\to {\Omega }^{n-r}(M)$. On an oriented orthonormal coframe, it sends a basis wedge to the signed complementary wedge. Applying it twice gives

For Hodge star eigenvalues on graded differential forms, extend $\ast$ complex-linearly to the direct sum of all form degrees. This interpretation matters: on a fixed degree it is an endomorphism only in middle degree. In positive odd dimension, $r(n-r)$ is even for every $r$, so ${\ast }^2=I$ and only $\pm 1$ can occur. Both occur, with eigenforms $1+{\mu }_g$ and $1-{\mu }_g$.

In positive even dimension, ${\ast }^2$ is $+I$ on even-degree forms and $-I$ on odd-degree forms. Hence ${\ast }^4=I$ and the only possible eigenvalues are $\pm 1,\pm i$. The first two occur as above. For the latter, choose a smooth local orthonormal frame and multiply its first dual differential one-form by a nonzero smooth bump function supported in the chart. This gives a form $\eta$ for which $\eta$ and $\ast \eta$ are independent. For $\lambda =\pm i$,Thus all four occur. Consequently, for $n\ge 1$,On degree $m$ in dimension $2m$, the Hodge star has eigenvalues $\pm 1$ for even $m$ and $\pm i$ for odd $m$; signed complementary middle-degree wedges show both occur for $m\ge 1$. In zero dimensions it is multiplication by the signed unit volume at each point, so only the signs actually present occur.

For Hodge integration by parts in arbitrary degree, let $\alpha \in {\Omega }^{r-1}(M)$ and $\beta \in {\Omega }^r(M)$. The graded Leibniz rule and Stokes theorem on the closed manifold giveUsing the given codifferential and the square of the Hodge star,Thereforewhich proves that $\delta$ is the formal adjoint of $d$. The complex version inserts conjugation in the usual Hermitian inner product.

The Hodge decomposition theorem gives an $L^2$-orthogonal decompositionwhere ${\mathcal{H}}^k(M)=\ker \Delta$ is finite dimensional and consists of the harmonic differential forms. Each de Rham cohomology class has a unique harmonic representative. Equivalently, with harmonic projection $P_{\mathcal{H}}$ there is a smooth Green operator of the Hodge Laplacian $G$ satisfying $\Delta G=G\Delta =I-P_{\mathcal{H}}$.

The solvability condition for the Hodge Poisson equation isNecessity follows from $⟨\Delta \alpha ,h⟩=⟨\alpha ,\Delta h⟩=0$. Conversely the orthogonality condition gives $P_{\mathcal{H}}\beta =0$, so ${\alpha }_0=G\beta$ solves the Poisson equation for differential forms.

The nonnegativity of the Hodge Laplacian identityidentifies its kernel with closed and coclosed forms. Thus the affine space of solutions of the Hodge Poisson equation isFor completeness, if a closed form is decomposed as $\omega =h+d\xi +\delta \zeta$, then $d\delta \zeta =0$, and $\Vert \delta \zeta {\Vert }^2=⟨\zeta ,d\delta \zeta ⟩=0$. Hence its cohomology class is represented by $h$. An exact harmonic form $h=d\xi$ must vanish, since $\Vert h{\Vert }^2=⟨\delta h,\xi ⟩=0$. This proves the canonical isomorphism ${\mathcal{H}}^k(M)\cong H_{\mathrm{dR}}^k(M)$, so the solution set is an affine space whose translation vector space is isomorphic to the requested de Rham cohomology.

For a linear map $\theta :U\to V$, its rank of a linear map is $r(\theta )=\dim \mathrm{im}\theta$, and its nullity is $n(\theta )=\dim \ker \theta$. Choose a basis $e_1,\dots ,e_s$ of its kernel and extend it to a basis $e_1,\dots ,e_d$ of $U$. Then $\theta e_{s+1},\dots ,\theta e_d$ span its image. They are linearly independent: a linear combination mapped to zero would belong to the kernel, contradicting independence of the extended basis unless all its coefficients vanish. Hence $r(\theta )=d-s$, provingFor endomorphisms of $U$, define $(\theta +\varphi )(u)=\theta (u)+\varphi (u)$ and $(\theta \varphi )(u)=\theta (\varphi (u))$. The image of the sum lies in the sum of the two images. ThusTo prove the product inequality, restrict $\varphi$ to $K=\ker (\theta \varphi )$. This map has kernel $\ker \varphi$ and image $\mathrm{im}\varphi \cap \ker \theta$: every vector in the intersection has a preimage in $K$. Applying the rank-nullity theorem to the restricted map givesNow put $d=\dim U$ and $s=r(\theta )+r(\varphi )$. If both equalities hold, $r(\theta +\varphi )=s\le d$, while $n(\theta \varphi )=2d-s\le d$ forces $s\ge d$. Therefore $s=d$, the sum has full rank, and the product has nullity $d$. The sum is an isomorphism and the product is zero.

Conversely, if $\theta \varphi =0$, then $\mathrm{im}\varphi \subseteq \ker \theta$, giving $s\le d$. If the sum is also an isomorphism, its rank is $d\le s$. Thus $s=d$, and the two inequalities become equalities. This proves the equality in the rank-sum and nullity-product inequalities in both directions.

Let $S=R[\varphi ]$ be the commutative subring of endomorphisms generated by scalar multiplication and $\varphi$. Commutativity follows because the scalar maps commute with the $R$-linear map $\varphi$, and every element of $S$ is a polynomial in $\varphi$ with scalar coefficients. Thus determinants and the adjugate identity apply to matrices over $S$.

With the convention for the coefficients in the question, the column of generators satisfiesMultiply by $\mathrm{adj}(C)$ over $S$. Since $\mathrm{adj}(C)C=(\det C)I_n$, the endomorphism $\det C$ kills every generator $m_i$. It therefore kills every element $\sum r_i{m}_i$ of the module, by $R$-linearity. Hence it is the zero endomorphism, andThe argument does not require the generating set to be a basis; relations among the generators cause no difficulty. This is the determinant trick.

Write $N=\#E({\mathbb{F}}_q)$. The Hasse bound isWe prove it using isogeny degrees of endomorphisms, with no assumption about a pre-existing Frobenius eigenvalue estimate. The standard elliptic isogeny facts we use are these: isogeny degree is multiplicative under composition; $\mathrm{deg}[m]=m^2$; assigning isogeny degree zero to the zero homomorphism makes isogeny degree a quadratic form, satisfying the degree parallelogram law; and the isogeny degree of a separable isogeny equals the number of its geometric kernel of an isogeny points.

Let $\varphi$ be the $q$-power Frobenius isogeny. It has isogeny degree $q$ and zero differential. Therefore $1-\varphi$ has differential the identity and is a separable isogeny. Its kernel of an isogeny consists exactly of the points fixed by Frobenius, including the identity, soPut $t=q+1-N$. Polarization of the isogeny degree quadratic form gives, for all integers $m,n$,The cross coefficient is fixed by the case $m=n=1$, where the isogeny degree is $N=1-t+q$; the coefficients of $m^2,n^2$ are $\mathrm{deg}1=1$ and $\mathrm{deg}\varphi =q$.

Every isogeny degree is nonnegative. For $n\ne 0$, divide by $n^2$ to obtain $u^2-tu+q\ge 0$ for every rational $u=m/n$. Continuity and density of the rationals imply the same inequality for real $u$. Its minimum, attained at $u=t/2$, is $q-t^2/4$, so $t^2\le 4q$. This is precisely the stated bound. Equality is allowed; a zero value of the quadratic form can occur when the two endomorphisms involved are rationally dependent.

Choose free group generators $a_1,\dots ,a_n$. Construct the Schreier coset graph $\Gamma$ with vertices the right cosets $Hg$ and a positively oriented edge labelled $a_k$ from $Hg$ to $Hg{a}_k$. This is well defined and connected. Its inverse edge carries $a_k^{-1}$. A word in the group generators determines a unique path from the base vertex $H$; that path closes exactly when the word represents an element of $H$.

Deleting a consecutive edge followed by its inverse corresponds exactly to free cancellation in the word. Since reduced words are unique in a free group, the resulting map from based loops modulo these cancellations to $H$ is injective as well as surjective, and respects concatenation. Thus $H$ is the fundamental group of $\Gamma$. More explicitly, choose a spanning tree $T$ and, for each unoriented edge outside $T$, choose one orientation. The tree path from the base vertex to that edge, followed by the edge and the returning tree path, defines an element of $H$. Every loop is a product of these elements and their inverses: insert the appropriate tree paths between successive edges, and tree edges contribute trivial factors. Collapsing the tree leaves one loop for each remaining edge. A product of the chosen basis loops maps to the corresponding word in these loop labels. If the original path could cancel to the constant path, its image could do so as well, since collapsing tree edges preserves every backtracking cancellation. A nonempty reduced word in the remaining labels cannot cancel to the empty word, proving that there are no further relations. These elements therefore form a free basis. This proves that $H$ is a free group without assuming the subgroup theorem.

If $i=[F_n:H]$ is finite, the graph has $i$ vertices and $ni$ unoriented edges, counted by their positive labels; loops and parallel edges count separately. A spanning tree has $i-1$ edges. The Nielsen–Schreier formula follows:For a sequence of finite indices $i_j\to \infty$, this gives the rank-to-index limit for subgroups of a free group

Finally, let $\theta :F_n\to F_n$ be an endomorphism, and put $L=\theta (F_n)$. It is generated by the $n$ images $\theta (a_k)$. If its index $d$ is finite, it is free of rank $r=1+d(n-1)$. A rank-$r$ free group needs at least $r$ group generators: its abelianization is ${\mathbb{Z}}^r$, and reduction modulo $2$ gives the $r$-dimensional vector space $(\mathbb{Z}/2\mathbb{Z}{)}^r$, which cannot be generated by fewer than $r$ vectors. ThereforeSince $n\ge 2$, this forces $d\le 1$, hence $L=F_n$. This proves that a finite-index image of a free-group endomorphism is surjective: