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We regard $N$ as positive and invertible in $k$; $N=1$ gives the trivial group and the same arguments apply. Put $A={\mathrm{Pic}}^0(C)$, an abelian variety of dimension $g$.

First compute the prime-to-characteristic torsion of an abelian variety. The differential of $[N]:A\to A$ at the origin is $N\mathrm{id}$, and translation gives the same assertion everywhere. Consequently $[N]$ is an étale morphism, its fibres have dimension zero, and properness makes it finite. Its image is closed of dimension $g$, hence all of $A$. Choose an ample line bundle $L_0$ on $A$ and put $L=L_0\otimes [-1{]}^{\ast }{L}_0$, which is symmetric and ample. The Theorem of the Cube gives the recurrenceAfter fixing a trivialization at the origin, induction from $n=0,1$ gives $[n{]}^{\ast }L\cong L^{\otimes n^2}$. Comparing top intersection products under the finite map $[N]$ givesThe intersection number is positive because $L$ is ample, so $\mathrm{deg}[N]=N^{2g}$. Since $[N]$ is étale and $k$ is algebraically closed field, its kernel has precisely $N^{2g}$ points.

For a prime power ${\ell }^r\mid N$, write the finite abelian group $A[{\ell }^r]$ as a product of cyclic groups of orders ${\ell }^{a_j}$, with $1\le a_j\le r$. Its subgroup killed by $\ell$ is $A[\ell ]$, so the number of factors is $2g$. Its total order is ${\ell }^{2gr}$, so $\sum a_j=2gr$ forces every $a_j=r$. Combining the prime-power factors by the Chinese remainder theorem provesThe same reasoning works for the dual abelian variety $\hat{A}$, also of dimension $g$.

To define the Weil pairing for an abelian variety on two copies of $A[N]$, we first justify the required theta identification with $\hat{A}$. Identify $X={\mathrm{Pic}}^{g-1}(C)$ with an $A$-torsor and use ordinary translation $T_q(x)=x+q$. The Theorem of the square, a consequence of the Theorem of the Cube, makesa group homomorphism $A\to \hat{A}$. Fix the negative Abel map of an algebraic curve $j:C\to A$, $P\mapsto [P_0-P]$, and let $\rho :\hat{A}\to {\mathrm{Pic}}^0(C)=A$ be pullback by $j$. Pullback by a translate of $j$ gives the same map: algebraically trivial line bundles on an abelian variety are translation-invariant as Picard classes, by the Theorem of the square.

For any $q\in A$, choose a degree-$g$ class $D$ such that both $D$ and $D+q$ have one-dimensional section spaces. This is possible because the nonspecial degree-$g$ classes form a nonempty open subset of the irreducible Picard scheme of a curve, and its translate has a nonempty intersection with it. Since $T_q{\varphi }_D={\varphi }_{D+q}$, the restriction of the theta divisor to an Abel curve proved above givesEach ${\varphi }_D$ differs from $j$ by a translation. Thus $\rho \lambda ={\mathrm{id}}_A$. This equality on closed points is equality of morphisms, since $A$ is reduced and the target is separated. In particular $\lambda$ is a closed immersion, being a section of the separated morphism $\rho$. Its image has dimension $g=\dim \hat{A}$, so it is all of the reduced irreducible abelian variety $\hat{A}$. Therefore $\lambda$ is an isomorphism of schemes. This proves the needed theta autoduality of a Jacobian, including possible infinitesimal issues in positive characteristic.

Here is a definition of the Weil pairing for an abelian variety that also proves nondegeneracy. For $M\in \hat{A}[N]$, the Theorem of the Cube gives $[N{]}^{\ast }M\cong M^{\otimes N}\cong {\mathcal{O}}_A$. Choose a trivialization of $[N{]}^{\ast }M$. The finite étale cover $[N]:A\to A$ has deck group $G=A[N]$. Its natural descent action on $[N{]}^{\ast }M$ becomes multiplication by scalars under this trivialization: these scalars are constant because every invertible regular function on the proper connected abelian variety $A$ is constant. They form a characterwhose values lie in ${\mu }_N$, since $NP=0$ for every $P\in G$. Rescaling the trivialization changes none of these scalars. Translation gives multiplicativity in $P$, and tensor product gives ${\chi }_{M\otimes M^{\prime }}={\chi }_M{\chi }_{M^{\prime }}$. DefineThis fixes a convention for the Weil pairing for an abelian variety; reversing the convention for the descent action inverts all values and changes no nondegeneracy assertion.

If ${\chi }_M=1$, the trivialization is equivariant for the deck group. Descent then makes $M$ the trivial line bundle on the target. Thusis injective. The source and target both have $N^{2g}$ elements, since $k$ contains all $N$ distinct roots of unity and $A[N]\cong (\mathbb{Z}/N{)}^{2g}$. Hence it is an isomorphism. Characters of this group separate its points, so the Weil pairing for an abelian variety is nondegenerate in both variables. Composing with the isomorphism $\lambda$ proves the pairing on $A[N]$ is perfect.

An algebraic line bundle over $A$ is an invertible module: a finitely generated projective module $L$ such that $L_{\mathfrak{p}}\cong A_{\mathfrak{p}}$ for every prime ideal $\mathfrak{p}$. Define the Picard group of a ring as the set of its isomorphism classes, withThe tensor product of modules of two finite projective modules is finite projective: write each as a direct summand of a finite free module and tensor the two decompositions. Localization gives $(L{\otimes }_{A}K{)}_{\mathfrak{p}}\cong A_{\mathfrak{p}}$, so this multiplication stays within the set of invertible modules. It is well defined on isomorphism classes. The associativity and symmetry isomorphisms of the tensor product give associativity and commutativity, and $[A]$ is the identity.

Let $L^{\ast }={\mathrm{Hom}}_A(L,A)$ be the dual module. A finite direct-summand presentation shows that $L^{\ast }$ is finite projective and that localization commutes with this dual. Since $L_{\mathfrak{p}}$ is free of rank one, so is $(L^{\ast }{)}_{\mathfrak{p}}$. The evaluation homomorphismis an isomorphism after every prime localization. Its kernel and cokernel are therefore zero, by exactness of localization and localization detects zero elements for modules. Thus $[L{]}^{-1}=[L^{\ast }]$, proving that $\mathrm{Pic}(A)$ is an abelian group.

For a ring homomorphism $f:A\to B$, tensoring a split presentation $L\oplus L^{\prime }\cong A^r$ gives $(B{\otimes }_{A}L)\oplus (B{\otimes }_A{L}^{\prime })\cong B^r$, so $B{\otimes }_{A}L$ is finite projective. For $\mathfrak{q}\in \mathrm{Spec}B$ put $\mathfrak{p}=f^{-1}(\mathfrak{q})$. Every element of $A\setminus \mathfrak{p}$ becomes a unit in $B_{\mathfrak{q}}$, andThis proves base change of invertible modules. Moreover, the canonical mapis an isomorphism, with inverse $b\otimes (\ell \otimes k)\mapsto (b\otimes \ell )\otimes (1\otimes k)$. It shows that $[L]\mapsto [B{\otimes }_{A}L]$ is a group homomorphism. Identity maps induce the identity, and for $A\to B\to C$, the associativity isomorphism $C{\otimes }_B(B{\otimes }_{A}L)\cong C{\otimes }_{A}L$ proves composition compatibility. Therefore $\mathrm{Pic}$ is a covariant functor to abelian groups.

The polynomial ring $\mathbb{C}[x]$ is a principal ideal domain by division with remainder. A submodule $N$ of a finite free module over a principal ideal domain is free: induct on the ambient rank, project onto the first coordinate, and note that the image is a principal ideal, hence either zero or a free rank-one module. This gives a split exact sequence with that image as quotient; the kernel lies in a free module of one smaller rank, completing the induction. A finite projective module is such a submodule, and so is free. An invertible module has rank one after passing to the fraction field, so its free rank is exactly one. Hence every algebraic line bundle over $\mathbb{C}[x]$ is trivial and

Let $A$ be the known set of affected qubits, with $|A|\le d-1$. Expand arbitrary operators on $A$ in tensor products of $I,X,Y,Z$. These Pauli operators span the complete error space on $A$.

By the distance of a quantum error-correcting code, every Pauli operator $F$ of weight below $d$ has scalar compression $PFP=c_{F}P$. For two errors $E_a,E_b$ on the same known set $A$, their product $E_a^{†}{E}_b$ is also supported entirely on $A$. Its weight is therefore at most $|A|$, rather than twice $|A|$. Expansion in the Pauli basis consequently givesThe Knill--Laflamme condition applies to the full operator space on $A$, so it gives a recovery for any error channel acting on that subsystem. Every set of at most $d-1$ known error locations is correctable. This is quantum erasure correction; the ability to use the locations is what improves the guarantee over unknown-location errors.

If $\rho$ affords an irreducible character $\chi$, entrywise complex conjugation gives a representation with character $\overline{\chi }$. An invariant complex subspace for the conjugate representation conjugates back to one for $\rho$, so irreducibility is preserved. If representations afford $\chi ,\psi$, their tensor product has trace $\chi (g)\psi (g)$, proving that their product is a character.

With the usual character inner product,The character orthogonality and the first assertion therefore make this equal to one if $\chi =\overline{\psi }$ and zero otherwise. This also counts the invariant vectors in the tensor-product representation.

We first prove the simple Artinian case of the Artin–Wedderburn theorem. This part does not need finite dimension over $k$. Choose a minimal nonzero left ideal $L$, using the descending chain condition. Simplicity gives $LA=A$. Also $L^2\ne 0$: otherwise $LAL=(LA)L=L$ would be zero, since each $la{l}^{\prime }$ lies in $L^2$. Choose $x\in L$ for which right multiplication $r_x:L\to L$ is nonzero. By simplicity of $L$, it is an automorphism. Choose $e\in L$ with $ex=x$. Then $(e^2-e)x=0$, and injectivity of $r_x$ gives $e^2=e$. Thus $L=Ae$.

Right multiplication by $e$ splits the left regular module as $A=Ae\oplus A(1-e)$. Repeat within the remaining left ideal: if it contains a minimal left ideal $Af$ generated by an idempotent, the map $v\mapsto vf$ splits off $Af$. Each nonzero remaining complement is a strictly smaller left ideal, so the Artinian condition terminates the process. Therefore the regular module is a finite direct sum of minimal left ideals.

All these simple summands are isomorphic. For primitive idempotents $e,f$, simplicity gives $AeA=A$, so $fAe\ne 0$; otherwise $fAeA=fA$ would be zero. A nonzero element of $fAe$ defines a nonzero homomorphism $Af\to Ae$, which is an isomorphism between simple modules. Thus $A\cong L^{\oplus n}$ as a left module. Put $\Delta ={\mathrm{End}}_A(L)$, a division algebra by the Schur lemma. Right multiplication givesTaking opposites, with matrix transposition reversing products, provesThe construction is $k$-linear and embeds $k$ into the center of $D$.

A central simple algebra is a finite-dimensional simple $k$-algebra whose center is exactly $k$. The Brauer group consists of such algebras modulo matrix stabilization: $A,B$ represent the same class if $M_r(A)\cong M_s(B)$ for some positive integers $r,s$. Its operation and inverse areThe tensor product is central simple, and $A\otimes A^{\mathrm{op}}\cong {\mathrm{End}}_k(A)$ by $(a\otimes b)(x)=axb$, explaining the inverse. Both facts can be checked after a splitting field of a central simple algebra extension, where they are matrix-algebra identities, and descended by faithful scalar extension. Every class has a unique central division-algebra representative, by the regular-module description above.

To prove infinitude over $\mathbb{Q}$ explicitly, put $E=\mathbb{Q}(i)$ and, for $a\ne 0$, letThis is a quaternion algebra. It is central simple: the center is $\mathbb{Q}$, and after adjoining $i$ its generators have the faithful matrix modelThese generate $M_2(E)$; consequently a nonzero proper two-sided ideal over $\mathbb{Q}$ is impossible. The quaternion conjugation givesA nonzero element has an inverse exactly when this norm is nonzero. Thus $A_a$ is a division algebra when $a$ is not an $E/\mathbb{Q}$ norm; if $a$ is a norm it has a nonzero zero divisor and is $M_2(\mathbb{Q})$ by the classification just proved.

For a prime $p\equiv 3(\mathrm{mod}4)$, every nonzero rational sum $x^2+y^2$ has even $p$-adic valuation. Divide out the minimum valuation of $x,y$; cancellation modulo $p$ is impossible because $-1$ is not a square modulo $p$. Therefore $p$ is not an $E/\mathbb{Q}$ norm, and $A_p$ is a nontrivial division-algebra Brauer class. This is the quaternion norm obstruction at primes congruent to three modulo four.

We also prove the product rule needed to distinguish classes. In $A_a\otimes A_b$, write the generators as $i_1,j_1,i_2,j_2$. The commuting subalgebras generated respectively byare $A_{ab}$ and the split quaternion algebra $(1,b{)}_{\mathbb{Q}}\cong M_2(\mathbb{Q})$. The latter has the matrix model $\mathrm{diag}(1,-1)$ and $\left(\begin{array}{cc}0& b\\ 1& 0\end{array}\right)$. Together the two subalgebras generate the whole tensor product, since they recover $i_2$ and $j_1$. Dimensions are sixteen on both sides, givingQuaternion conjugation identifies $A_a$ with its opposite, so these classes have order at most two. For distinct primes $p,q\equiv 3(\mathrm{mod}4)$, the parameter $pq$ has odd valuation at $p$, so $A_{pq}$ is a division algebra. Thus $[A_p]\ne [A_q]$. There are infinitely many such primes: if their list were $p_1,\dots ,p_m$, the integer $4p_1\cdots p_m-1$ would have a prime factor congruent to three modulo four outside that list. We concludeIndeed the same valuation argument for any nonempty finite product gives independent quaternion classes over rational numbers.

For $\mathbb{R}$, we prove the required division-algebra classification. Let $D$ be finite-dimensional and central over $\mathbb{R}$. If $D\ne \mathbb{R}$, choose a nonreal element; its irreducible real minimal polynomial is quadratic, so translating and rescaling it gives $i\in D$ with $i^2=-1$. The centralizer $D_{+}$ of $i$ is a finite-dimensional division algebra over $\mathbb{C}=\mathbb{R}(i)$, with that field central. Every element has a split minimal polynomial over $\mathbb{C}$, hence belongs to $\mathbb{C}$, so $D_{+}=\mathbb{C}$.

Conjugation by $i$ is an involution, giving $D=D_{+}\oplus D_{-}$, where $D_{-}=\{x:ix=-xi\}$. If $D_{-}=0$, then $D=\mathbb{C}$, contrary to its center being $\mathbb{R}$. Choose $j\ne 0$ in $D_{-}$. For any $x\in D_{-}$, $x{j}^{-1}$ commutes with $i$, so $D_{-}=\mathbb{C}j$. Also $j^2\in \mathbb{C}$ and commutes with $j$, forcing $j^2\in \mathbb{R}$. It cannot be positive, since $(j-\sqrt{j^2})(j+\sqrt{j^2})=0$, or zero, since $D$ is a division algebra. Rescale to $j^2=-1$. ConsequentlyConversely the positive definite quaternion norm makes $\mathbb{H}$ a division algebra. The two possible central division representatives are therefore $\mathbb{R}$ and $\mathbb{H}$. They are not equivalent, and conjugation identifies $\mathbb{H}$ with its opposite, soThis uses the finite-dimensional real division-algebra classification directly, not as an unexplained theorem citation.

Finally let $k_s$ be a separable closure and $\Gamma =\mathrm{Gal}(k_s/k)$. The cohomological description isHere $k_s^{\times }$ has the discrete topology and its natural Galois action. For a finite Galois splitting field $L$, identify $A{\otimes }_{k}L$ with a matrix algebra. Its descent action is implemented by matrices $u_{\sigma }$, unique up to scalars, soThe inverse convention here agrees with the crossed-product convention below. The semilinear operators $u_{\sigma }\sigma$ commute with the descended algebra $A$; their crossed product has class $[A{]}^{-1}$ by the commuting-factor decomposition of the full endomorphism algebra. Inverting its scalar cocycle therefore gives the class $[A]$. Associativity gives the two-cocycle identity, and changing the representatives changes $c$ by a coboundary. Conversely a two-cocycle defines the crossed-product algebra of a Galois extension, with $u_{\sigma }a=\sigma (a)u_{\sigma }$ and $u_{\sigma }{u}_{\tau }=c(\sigma ,\tau )u_{\sigma \tau }$. It is central simple and split by $L$. Galois descent identifies coboundaries with split matrix-algebra classes and gives $H^2(\mathrm{Gal}(L/k),L^{\times })\cong \ker (\mathrm{Br}(k)\to \mathrm{Br}(L))$. Every central simple algebra has a finite separable, hence finite Galois, splitting field, yielding the displayed limit.

The canonical commutation relations are $[\hat{x},\hat{p}]=i\hslash I$ (in several dimensions $[x_i,p_j]=i\hslash {\delta }_{ij}I$, with the position-position and momentum-momentum commutators zero). For Hermitian $x,p$, the commutator is anti-Hermitian, just as $i\hslash I$ is; there is no conflict with their self-adjointness. The annihilation operator and its adjoint obeyThus the quantum harmonic oscillator Hamiltonian is $\boxed{H=\hslash (N+1/2)}$. The commutators $[N,a]=-a$, $[N,a^{†}]=a^{†}$ and the normalized ground state constructNormalization follows inductively from $a{a}^{†}=N+1$. Positivity of $N$ and repeated lowering exclude a noninteger eigenvalue: lowering it until it becomes negative would produce a nonzero vector of negative $N$ eigenvalue. Every eigenstate therefore lowers to the unique ground state, so these normalized ladder states exhaust the oscillator eigenstates.

For two independent oscillators, the tensor products $|n_1,n_2⟩$ have $E=\hslash (n_1+n_2+1)$. The level $n=n_1+n_2$ has degeneracy $n+1$. In its degenerate subspace the perturbation $\lambda x_1{x}_2$ reduces tosince terms that raise or lower both numbers leave the subspace. For $n=0$ its matrix is zero. For $n=1$, in $|1,0⟩,|0,1⟩$, it is $(\lambda \hslash /2)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. For $n=2$, in $|2,0⟩,|1,1⟩,|0,2⟩$, it is $(\lambda \hslash /2)\left(\begin{array}{ccc}0& \sqrt{2}& 0\\ \sqrt{2}& 0& \sqrt{2}\\ 0& \sqrt{2}& 0\end{array}\right)$. Their eigenvalues giveThe $n=1$ states are the symmetric and antisymmetric combinations. At $n=2$, the shifted states are $(|2,0⟩\pm \sqrt{2}|1,1⟩+|0,2⟩)/2$, and the zero-shift state is $(|2,0⟩-|0,2⟩)/\sqrt{2}$.

Take the boundary-fitted grid $h=1/(M+1)$ and assume the standard consistency of a numerical method hypothesis $u\in C^4(\overline{\Omega })$ with bounded fourth partial derivatives. Taylor expansion along each coordinate, including a bounded fourth-order remainder, gives the five-point Laplacian truncation errorThe numerical equation and exact grid values therefore imply $A_h\mathbf{e}=\mathbf{\tau }$, where $A_h=-{\Delta }_h$ with zero boundary values. Its discrete sine transform eigenvectors are $\sin (\pi rih)\sin (\pi sjh)$, $1\le r,s\le M$, and its eigenvalues areThe one-dimensional sine vectors form an orthogonal basis, hence so do their tensor products. The least eigenvalue occurs at $r=s=1$. Since $\sin (\pi h/2)\ge h$ for $0<h\le 1$, ${\lambda }_{11}\ge 8$, so the Euclidean operator norm of $A_h^{-1}$ is at most $1/8$. There are $M^2$ residual components, and $Mh<1$, givingThe factor $M$ matters: pointwise second-order consistency of a numerical method becomes first order in the unweighted Euclidean grid norm.

The PDF does not explicitly impose regularity on the exact solution. Without such a hypothesis its unrestricted error claim is false. Here is a counterexample, with a classical solution smooth in the open square and continuous with zero boundary values:The sampled numerical solution separates as $v_i\sin (\pi jh)$. Its error $e_i=v_i-g(ih)$ solves $Ae=\tau$, where $A=-{\Delta }_{h,x}+{\lambda }_{y}I$ and ${\lambda }_y=4h^{-2}{\sin }^2(\pi h/2)$. The one-dimensional residual is ${\tau }_i={\Delta }_{h,x}g(ih)-g^{\prime \prime }(ih)+({\pi }^2-{\lambda }_y)g(ih)$. At the first node,The leading coefficient is strictly negative. Moreover $g^{\prime \prime \prime \prime }<0$ on $(0,1)$, so the triangular-average formula for the second difference and concavity of $g^{\prime \prime }$ give ${\Delta }_{h,x}g\le g^{\prime \prime }$. Thus the positive part of every residual is at most $C{h}^2$. The matrix $A$ has a nonnegative inverse, and its first diagonal inverse entry is at least $1/(2h^{-2}+{\lambda }_y)$, as follows from the nonnegative Neumann expansion of its tridiagonal matrix. The discrete comparison principle bounds the inverse image of the residual's positive part by $C{h}^2$, while the negative first residual gives $e_1\le -c\sqrt{h}+C{h}^2$. Hence $|e_1|\ge c^{\prime }\sqrt{h}$ for small $h$. Since ${\sum }_{j=1}^M{\sin }^2(\pi jh)=1/(2h)$, the full Euclidean error is bounded below by a positive constant. It cannot be $O(h)$. Thus the boxed estimate holds for the intended smooth solution class; a regularity condition is genuinely necessary.

On the tensor product state space, the total angular momentum is $\mathbf{J}={\mathbf{J}}^{(1)}\otimes I+I\otimes {\mathbf{J}}^{(2)}$, conventionally abbreviated ${\mathbf{J}}^{(1)}+{\mathbf{J}}^{(2)}$. Operators on different factors commute. Define $J_{\pm }^{(i)}=J_1^{(i)}\pm i{J}_2^{(i)}$. Expanding the square givesThe first three terms commute with $J_3=J_3^{(1)}+J_3^{(2)}$. In a mixed term one subsystem's magnetic quantum number increases and the other's decreases, so its commutator with total $J_3$ is zero. Hence $\boxed{[{\mathbf{J}}^2,J_3]=0}$.

Use the standard identities $[J_3,J_{\pm }]=\pm J_{\pm }$ and $J_{\pm }|j,m⟩=\sqrt{(j\mp m)(j\pm m+1)}|j,m\pm 1⟩$. The PDF has the commutator in the reversed order with unchanged sign and prints $m+1$ also for lowering; these are sign/label slips. The standard identities are consistent with the meaning of raising and lowering and are used here.

Put $s=j_1+j_2$. The unique product state of magnetic quantum number $-s$ isFor $j_1,j_2>0$, define the orthonormal product states $|a⟩=|j_1,-j_1+1⟩|j_2,-j_2⟩$ and $|b⟩=|j_1,-j_1⟩|j_2,-j_2+1⟩$. Raising the preceding state givesThe normalized orthogonal combination is annihilated by $J_{-}$, because its two lowering amplitudes cancel. It is consequently the lowest-weight state with $J=s-1$:up to an overall phase. For $j_1=3,j_2=1$, the outcome $m_1=-3,m_2=0$ is the $|b⟩$ component. The Born rule therefore gives

Put $a=\hslash /2$. Operators on distinct tensor product factors commute. Acting with ${\sigma }_x\otimes {\sigma }_y\otimes {\sigma }_y$ on $|\uparrow \uparrow \uparrow ⟩$ gives $-|\downarrow \downarrow \downarrow ⟩$, while acting on $|\downarrow \downarrow \downarrow ⟩$ gives $-|\uparrow \uparrow \uparrow ⟩$. Therefore their difference is an eigenvector with eigenvalue $+1$. The same computation applies to the permutations $yxy$ and $yyx$. The product $xxx$ flips the two basis tensors without a phase and so sends their difference to its negative.

Restoring the three factors of $a$, the GHZ state hasThese are certain measurement outcomes for the four product observables in this state.

The connection difference as an endomorphism-valued one-form follows because the graded Leibniz rule terms cancel:Thus $a=D_1-D$ is a global endomorphism-valued one-form. To define the endomorphism bundle connection, requireThe right side is $C^{\infty }$-linear in $s$, so it is indeed an endomorphism-valued one-form, and replacing $\varphi$ by $f\varphi$ gives $df\otimes \varphi +f\tilde{D}\varphi$. In a frame with connection matrix $A$, its extension to an endomorphism-valued $k$-form $b$ isThe endomorphism-valued exterior product combines the wedge of differential forms with composition in $\mathrm{End}E$. In particular,Expanding $d(A+a)+(A+a)\wedge (A+a)$ proves the curvature difference formula:

The Chern connection is characterized by metric compatibility and $D^{0,1}={\bar{\partial }}_E$. Take Hermitian inner products linear in the first variable. In a holomorphic local frame write $⟨eu,ev⟩=v^{†}Hu$, with $H$ a positive Hermitian matrix. The second condition forces $A^{0,1}=0$, and metric compatibility isIts $(1,0)$ component determines $A$ uniquely:Conversely this formula and its conjugate transpose satisfy the metric equation. Under a holomorphic local frame change, $H^{\prime }=g^{†}Hg$ and $\partial g^{†}=0$, so the displayed formula transforms by $A^{\prime }=g^{-1}Ag+g^{-1}\partial g$. Hence the local matrices patch, proving existence and uniqueness.

Differentiating $H^{-1}$ givesTherefore the $(2,0)$ vector-bundle curvature is zero. There is no $(0,2)$ component, andThe last equality is ${\bar{\partial }}^{2}A=0$ in holomorphic local frames, so is intrinsic. This is Dolbeault closedness of Chern curvature. The induced endomorphism bundle connection is itself the Chern connection for the Hilbert-Schmidt metric on $\mathrm{End}E$: the tensor product of a metric-compatible connection and its dual preserves the induced metric, and its $(0,1)$ part is the induced holomorphic Dolbeault operator. These two properties and uniqueness suffice; no further matrix calculation is needed.

For two metrics, both Chern connections have the same $(0,1)$ part, so $a\in A^{1,0}(\mathrm{End}E)$. The $(1,1)$ part of the curvature difference formula is just ${\bar{\partial }}_{\mathrm{End}E}a$; the other two terms have type $(2,0)$, which must cancel because both vector-bundle curvatures have type $(1,1)$. ThusBy Dolbeault theorem, this proves that the curvature Dolbeault classis independent of the metric.

Endomorphism composition and wedge product of differential forms define the powers in $A^{k,k}(\mathrm{End}E)$, rather than in a tensor power of the endomorphism bundle. The Dolbeault operator is a derivation, so $\bar{\partial }({\Theta }^k)=0$. The noncommutative telescoping identity gives the stronger explicit independence statementAll powers of vector-bundle curvature have even total degree, so the derivation introduces no extra signs before $a$. Consequentlyis metric independent. For $k>n$, the form space and the resulting class are zero. This proves metric independence of curvature Dolbeault powers without assuming that endomorphism-valued forms commute.

Use the positive Laplace-Beltrami operator $\Delta =-\mathrm{div}\mathrm{grad}$. A Riemannian submersion is a surjective smooth submersion $\pi :(M,g)\to (N,h)$ for which, at every $p$, the restriction of $d{\pi }_p$ to $H_p=(\ker d{\pi }_p{)}^{\perp }$ is a linear isometry onto $T_{\pi (p)}N$. The spaces $V_p=\ker d{\pi }_p$ and $H_p$ are its vertical and horizontal spaces. Its fibres are totally geodesic submanifolds precisely when ${\nabla }_{U}V$ is vertical for vertical vector fields $U,V$: their second fundamental form vanishes. Equivalently, a geodesic initially tangent to a fibre remains in that fibre while defined.

For a smooth $f:N\to \mathbb{R}$, its basic function $F=f\circ \pi$ is constant along each fibre. The Riemannian gradient of $F$ is the horizontal lift of a vector field through a submersion of ${\mathrm{grad}}_{N}f$, sincefor horizontal $X$, and $dF(U)=0$ for vertical $U$. In particular no derivative of $f$ in a vertical direction occurs.

Here is the needed connection fact, which also follows directly from the Koszul formula: for horizontal lifts $X,Y$ of vector fields $\bar{X},\bar{Y}$ on $N$, the horizontal component of ${\nabla }_{X}Y$ projects to ${\nabla }_{\bar{X}}^N\bar{Y}$. To see this, pair the Koszul formula with a third horizontal lift $Z$. The horizontal inner products are pulled back from $N$, and the horizontal components of their Lie brackets of vector fields project to the brackets on $N$. Thus all six terms are the pullbacks of the corresponding terms on $N$.

Choose an adapted Riemannian orthonormal frame $X_1,\dots ,X_n,U_1,\dots ,U_r$, with the $X_a$ horizontal lifts. Using $\mathrm{Hess}F(A,A)=A(AF)-({\nabla }_{A}A)F$, the connection fact givesFor a vertical $U_b$, both $U_{b}F=0$ and $({\nabla }_{U_b}{U}_b)F=0$, the latter because the fibres are totally geodesic submanifolds. Taking the negative metric trace of the Riemannian Hessian therefore proves the basic-function Laplacian identityThis identity is local and does not require compactness. Vanishing mean curvature of the fibres would already suffice; total geodesicity makes each vertical summand vanish separately.

For the discrete eigenspace assertion, assume the two Riemannian manifolds are closed manifolds. Without a discrete spectral realization, an unrestricted noncompact version need not have an eigenbasis. The projections of the Riemannian product $M\times N$ are Riemannian submersions with totally geodesic submanifolds as fibres. Its Levi-Civita connection splits into the two factor connections. Consequently its positive Laplace-Beltrami operator isThe cross term in the product rule for the positive Laplace-Beltrami operator is zero because the two factor Riemannian gradients are orthogonal.

We use the standard compact elliptic compact elliptic spectral theorem: the positive Laplace-Beltrami operator on a closed manifold is self-adjoint, has compact resolvent, and has a complete orthonormal eigenbasis of smooth eigenfunctions, with finite-dimensional eigenspaces and eigenvalues tending to infinity. Let ${\Delta }_M{u}_i={\mu }_i{u}_i$ and ${\Delta }_N{v}_j={\nu }_j{v}_j$. Fubini's theorem and completeness on each factor show that $u_i(x)v_j(y)$ form a complete orthonormal basis of $L^2(M\times N)$. For example, a function orthogonal to all these products has, for each $i$, zero $u_i$-coefficient as an $L^2(N)$ function, hence is zero.

The displayed operator identity makes $u_i{v}_j$ an eigenfunction with eigenvalue ${\mu }_i+{\nu }_j$. Conversely, if $\Delta w=\lambda w$, self-adjointness of the positive Laplace-Beltrami operator givesAll other coefficients vanish. Only finitely many pairs can have ${\mu }_i+{\nu }_j=\lambda$, since both spectra are nonnegative and have finitely many eigenvalues below any fixed bound. Thus the product Laplacian eigenspace decomposition isThe tensor product summands are mutually orthogonal; their elements are actual smooth eigenfunctions, so this is an equality of eigenspaces, not just a formal expansion.

For a diffeomorphism, the differential and its inverse preserve the natural pairing of a vector with a covector. Consequently the mixed tensor pullback commutes with every tensor contraction $C$:Differentiating at zero gives ${\mathcal{L}}_X(CT)=C({\mathcal{L}}_{X}T)$. The mixed tensor pullback also preserves tensor products, soThe ordinary product rule for differentiation therefore yieldsThis proves the contraction and Leibniz rule properties for all smooth generators, without needing straightening coordinates.

The suggested flow-box theorem applies locally only where $X\ne 0$. For example $X=x{\partial }_x$ vanishes at $x=0$, so it cannot equal a coordinate basis vector there. Its local flow is nevertheless ${\varphi }_s(x)=e^{s}x$ and ${\varphi }_s^{\ast }dx=e^{s}dx$, giving ${\mathcal{L}}_{X}dx=dx$ even at that zero. This Lie derivative at a zero of its generator illustrates why the general flow proof is needed to cover every point.

Expand $T=T^{\mu }{}_{\nu }{\partial }_{\mu }\otimes d{x}^{\nu }$. Using the tensor Lie derivative of functions, vectors and covectors and its Leibniz rule givesThe contravariant slot has a minus sign and the covariant slot a plus sign.

For the commutator identity for Lie derivatives, put $D=[{\mathcal{L}}_X,{\mathcal{L}}_Y]-{\mathcal{L}}_{[X,Y]}$. The commutator of two tensor derivations is itself a tensor derivation, and $D$ commutes with tensor contractions. On a function, $Df=XYf-YXf-[X,Y]f=0$. On a vector field $Z$,by the Jacobi identity, which follows here by expanding the commutators of the operators $X,Y,Z$ acting on functions. For a type $(1,1)$ tensor, $T(Z)$ is a vector field, and contraction compatibility givesAs this holds for every $Z$, $DT=0$. ThereforeThe derivation argument also establishes the identity for arbitrary tensor types by applying it to covector–vector pairings and then to tensor products.