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Write $d=\mathrm{deg}A$ for the degree of a central simple algebra, so ${\dim }_{K}A=d^2$. Choose a finite Galois splitting field of a central simple algebra $E/K$ for $A$ and an isomorphism $\varphi :A{\otimes }_{K}E\to M_d(E)$. The reduced norm isThe existence of a finite separable splitting field is a standard structural property of a central simple algebra; passing to its Galois closure supplies $E$.

Every $E$-algebra automorphism of $M_d(E)$ is inner. Here is the matrix-unit argument for the inner automorphisms of a matrix algebra result. For an automorphism $h$, choose $w\ne 0$ in the image of $h(E_{11})$, and set $w_i=h(E_{i1})w$. Then $h(E_{ij})w_k={\delta }_{jk}{w}_i$. The $w_i$ are nonzero and independent, and their number is $d$, so they form a basis. Relative to that basis $h$ acts on each matrix unit in the usual way. Thus $h(T)=PT{P}^{-1}$ for some $P\in {\mathrm{GL}}_d(E)$. Determinants are unchanged by conjugation, proving independence of $\varphi$.

For $\sigma \in \mathrm{Gal}(E/K)$, compare $\varphi$ with the isomorphism obtained by applying $\sigma$ to matrix entries and ${\sigma }^{-1}$ to the scalar factor of $A\otimes E$. Their difference is again inner. For $a\in A$, this shows $\sigma (\det \varphi (a))=\det \varphi (a)$. The determinant therefore belongs to $K$. In fact, for a $K$-basis $a_1,\dots ,a_{d^2}$, the same comparison shows that all coefficients ofare fixed by the Galois group, so the reduced norm is a homogeneous polynomial of degree $d$ over $K$. This coefficient argument also applies over finite fields, where equality merely as functions would not identify polynomials.

Finally, two finite splitting fields embed into a common finite splitting extension. Determinant commutes with scalar extension, and over that common extension the two matrix identifications differ by an inner automorphism. Hence the resulting polynomials agree over $K$. The reduced norm is independent of both the splitting field and the matrix identification. It is multiplicative, and $\mathrm{Nrd}(a)=0$ exactly when $a$ is not invertible: a matrix with nonzero determinant is invertible after scalar extension, and invertibility descends by the invertibility of the $K$-linear multiplication map. In a division algebra the only element of reduced norm zero is zero.

For a nonnegative integer $r$, a Cr field is a field over which every homogeneous polynomial of positive degree $d$ in $N>d^r$ variables has a nontrivial zero. Thus a C1 field uses the bound $N>d$, and a C2 field uses $N>d^2$.

The required field theorems are as follows. The Chevalley-Warning theorem implies that every finite field is $C_1$: the number of zeros of a polynomial with degree smaller than its number of variables is divisible by the characteristic, and for a homogeneous polynomial the origin is already a zero. The Lang-Nagata theorem for Ci fields states that a finitely generated extension of transcendence degree $s$ of a $C_r$ field is $C_{r+s}$. The Tsen theorem states that the function field of a curve over an algebraically closed field is $C_1$. In particular, finite fields, algebraically closed fields and fields such as $\mathbb{C}(t)$ are examples of $C_1$ fields. A function field of one variable over a finite field is $C_2$ by the Lang theorem. These are statements of the theorems; no theorem proof is needed here.

For the requested finite extension stability of C1 fields, let $m=[L:K]$ and choose a $K$-basis $u_1,\dots ,u_m$ of $L$. Given a homogeneous $f\in L[X_1,\dots ,X_N]$ of degree $d$ with $N>d$, substitute $X_i={\sum }_j{x}_{ij}{u}_j$ and take the field norm:The field norm is the determinant of multiplication on the $m$-dimensional $K$-space $L$, hence a homogeneous polynomial of degree $m$ in its coordinates. Consequently $F$ is homogeneous of degree $md$ in $mN>md$ variables over $K$. The $C_1$ property gives a nonzero coordinate vector $(x_{ij})$ with $F=0$. Its corresponding vector $(X_i)\in L^N$ is nonzero, since the $u_j$ are a basis. The norm of a field element vanishes only for the zero element, so $f(X_1,\dots ,X_N)=0$. Every finite extension of a $C_1$ field is $C_1$. This proof includes inseparable finite extensions, because the determinant definition of the field norm requires no separability.

Let $\pi :T^{\ast }M\to M$ be the cotangent bundle projection. Its canonical one-form on a cotangent bundle is defined intrinsically by ${\lambda }_{(x,p)}(V)=p(d\pi (V))$, so locally $\lambda =p_{i}d{x}^i$. Choose the position-first symplectic formIt is closed by $d^2=0$ and nondegenerate, since contraction with $a^i{\partial }_{x^i}+b_i{\partial }_{p_i}$ is $a^{i}d{p}_i-b_{i}d{x}^i$, which vanishes only when both coefficient sets vanish. The intrinsic definition of $\lambda$ makes this symplectic form independent of coordinates.

Use the convention ${\iota }_{X_H}\omega =dH$. Then the Hamiltonian vector field and the Poisson bracket areso $X_H(F)=\{F,H\}$. Choosing $\omega =d\lambda$ and ${\iota }_{X_H}\omega =-dH$ gives the same equations; choosing only one of these sign changes would reverse the flow.

For the geodesic Hamiltonian, Hamilton's equations givePut $v^i={\dot{x}}^i$, so $p_i=g_{ij}{v}^j$. Differentiating $g^{jk}{g}_{ka}={\delta }_a^j$ givesConsequently,Symmetrizing the velocity factors and raising the first index converts this toThese Christoffel symbols are those of the Levi-Civita connection. Thus a Hamiltonian integral curve of a vector field projects to an affinely parametrized geodesic. Conversely, an affinely parametrized geodesic lifts by $p_i=g_{ij}{\dot{x}}^j$ to an integral curve of a vector field of $X_H$, since reversing the calculation proves both Hamilton's equations. This is the geodesic flow on the cotangent bundle. The conserved Hamiltonian is half the squared speed, and the zero-energy case gives the constant geodesics.

A quadratic homogeneous polynomial depends only on the symmetric part of its coefficient matrix. Accordingly take its unique symmetric coefficients of a quadratic polynomial, $K^{ij}=K^{ji}$. This is the standard implicit convention in identifying such polynomials with symmetric tensors. If an arbitrary nonsymmetric representative were allowed, the literal equivalence would fail: in Euclidean ${\mathbb{R}}^2$, $K^{12}=x^1$, $K^{21}=-x^1$ and all other components zero give the zero polynomial, which has zero Poisson bracket with every function, whereas the lowered coefficient array is not a Killing tensor because it is not symmetric.

Lower the indices of the symmetric coefficient tensor using the Riemannian metric. Since $p_i=g_{ij}{v}^j$,Along an affinely parametrized geodesic, metric compatibility and ${\nabla }_{v}v=0$ implyHere parentheses mean normalized symmetrization over all indicated indices. The left side is $\{K,H\}$ by the Hamiltonian vector field convention. Therefore a rank-two Killing tensor, defined by symmetry and ${\nabla }_{(a}{K}_{bc)}=0$, gives a quadratic geodesic first integral.

Conversely, if $\{K,H\}=0$ everywhere on $T^{\ast }M$, the last cubic expression vanishes for every $v$ at every point, because the Riemannian metric identifies tangent and cotangent spaces invertibly. A symmetric trilinear form is determined by its diagonal cubic polynomial: equivalently, compare its coefficients, or polarize the cubic. Hence ${\nabla }_{(a}{K}_{bc)}=0$. This proves both directions:with symmetry understood on the coefficient representative from the outset. The equivalence is local and does not require geodesic completeness.

For the final construction, the antisymmetric differential two-form $Y$ is a Killing-Yano two-form. Its defining equation says ${\nabla }_a{Y}_{bc}=-{\nabla }_b{Y}_{ac}$. Antisymmetry of $Y$ also says ${\nabla }_a{Y}_{bc}=-{\nabla }_a{Y}_{cb}$, so the three-index tensor ${\nabla }_a{Y}_{bc}$ is totally antisymmetric.

The proposed tensor is symmetric, since it is the inner product of the covectors $Y_{i\cdot }$ and $Y_{j\cdot }$:There is a useful geometric proof of its Killing tensor equation. Along any affinely parametrized geodesic, define $w_k=Y_{ik}{v}^i$. ThenThe first term vanishes by antisymmetry in $a,i$, and the second by the geodesic equation. Thus $w$ is carried by parallel transport. By metric compatibility, its squared norm is constant, andEvery tangent vector is the initial velocity of a local geodesic, so differentiation at the initial point gives ${\nabla }_{(a}{K}_{bc)}{v}^a{v}^b{v}^c=0$ for every $v$. The same cubic-coefficient argument provesThis proves that the square of a Killing-Yano two-form is a rank-two Killing tensor and supplies a nonnegative quadratic geodesic first integral. The argument also explains the conserved quantity: it is the squared norm of a covector that the Killing-Yano two-form makes parallel along every geodesic.

Order the index set of the open cover $\mathcal{U}=(U_a)$. The Čech cochain groups areand their differential is the alternating sum of restrictions:Terms obtained by deleting two indices cancel in pairs, so ${\delta }^2=0$. The Čech cohomology is ${\check{H}}^p=\ker \delta /\mathrm{im}\delta$, with zero incoming differential in degree zero. Its zeroth group is the group of global sections by sheaf gluing.

The acyclic cover theorem applies if every nonempty finite intersection is acyclic for $\mathcal{F}$. For a quasi-coherent sheaf on a variety it is sufficient that every such intersection is affine; on a separated variety, any affine open cover has this property. On ${\mathbb{P}}^n$, take the $n+1$ standard charts $U_i=D_{+}(X_i)$. Every intersection is a principal open in an affine chart, hence affine. The Čech complex has no terms above degree $n$, provingThis is the cohomological dimension bound from an affine cover and requires only quasi-coherence, not finite generation.

For the remaining projective calculations assume $n\ge 1$. There is a necessary zero-dimensional exception to the negative-twist assertion: ${\mathbb{P}}^0$ is a point, every twist is trivial there, and $H^0({\mathbb{P}}^0,\mathcal{O}(m))\cong k$ even for $m<0$.

Put $S=k[X_0,\dots ,X_n]$ with its usual grading. The twisting sheaf on projective space is the sheaf associated with the shifted graded module $S(m)$; on an intersection $U_I={\bigcap }_{i\in I}{U}_i$ its sections areOn $U_i$ a generator is $e_i=X_i^m$, with transition $e_j=(X_j/X_i{)}^m{e}_i$. These are regular units on overlaps and satisfy the cocycle relation, so they glue an invertible sheaf for every integer $m$, including negative $m$.

A global section is a compatible family of these homogeneous fractions, hence a single element of degree $m$ in ${\bigcap }_i{S}_{X_i}\subseteq \mathrm{Frac}S$. For $n\ge 1$, $X_0$ and $X_1$ are relatively prime in the unique factorization domain $S$, so $S_{X_0}\cap S_{X_1}=S$; therefore the full intersection is $S$. It follows thatFor $m\ge 0$ its dimension is $\left(\genfrac{}{}{0ex}{}{n+m}{n}\right)$. The proof uses all-chart compatibility; regularity on one chart alone would permit poles on its complement.

If some $d_i=0$, the ideal $(X_i^{d_i})$ contains $1$, so the requested containment is immediate. Otherwise every $d_i\ge 1$. A monomial $X_0^{a_0}\cdots X_n^{a_n}$ of degree ${\sum }_i{d}_i$ must have $a_i\ge d_i$ for some $i$: if not, its total degree would be at most ${\sum }_i(d_i-1)<{\sum }_i{d}_i$. Thus every degree-${\sum }_i{d}_i$ monomial, and hence every homogeneous polynomial of that degree, belongs to $(X_0^{d_0},\dots ,X_n^{d_n})$. This is monomial containment in an ideal of coordinate powers.

For $n>0$, a top-degree Čech cochain for $\mathcal{O}$ is a degree-zero Laurent polynomial on the full intersection. Write it with a common denominator asThe containment just proved gives $P={\sum }_i{Q}_i{X}_i^{d_i}$, with homogeneous $Q_i$ of degree ${\sum }_j{d}_j-d_i$. ConsequentlyThe $i$th term is regular on the intersection omitting $U_i$, and has degree zero. Give it the sign $(-1{)}^i$ in that component of the preceding Čech cochain. Its coboundary is the original fraction. Every top cochain is thus a coboundary, andThis is top Čech cohomology from missing-denominator monomials.

To obtain the negative-twist bound by induction on dimension, let $H\cong {\mathbb{P}}^{n-1}$ be a hyperplane, with inclusion $j$. Its equation gives the hyperplane exact sequence for twisting sheavesThe associated long exact sequence in sheaf cohomology and sheaf cohomology under a closed inclusion giveFor $n=1$, restriction $H^0({\mathbb{P}}^1,\mathcal{O})\to H^0(H,{\mathcal{O}}_H)$ is the surjection $k\to k$. Exactness and the already proved $H^1({\mathbb{P}}^1,\mathcal{O})=0$ give $H^1({\mathbb{P}}^1,\mathcal{O}(-1))=0$. The displayed surjections then give the same vanishing for every $m\ge 0$. This handles the point hyperplane without making a false negative-twist claim on ${\mathbb{P}}^0$.

For $n\ge 2$, the induction hypothesis in dimension $n-1$ gives $H^{n-1}(H,{\mathcal{O}}_H(m))=0$ whenever $m\ge -(n-1)$. Therefore consecutive top-degree groups are isomorphic for $m\ge 1-n$. Starting with $H^n({\mathbb{P}}^n,\mathcal{O})=0$ and stepping down through $m=0,-1,\dots ,1-n$ reaches $H^n({\mathbb{P}}^n,\mathcal{O}(-n))=0$. Stepping upward proves all positive twists too. We concludeThis is top-twist vanishing by hyperplane induction.

Finally, on $U_0$ write $x_a=X_a/X_0$ and ${\omega }_0=d{x}_1\wedge \cdots \wedge d{x}_n$. On $U_i$, let ${\omega }_i$ be the wedge of $d(X_a/X_i)$ for $a\ne i$, taken in increasing index order. For $i>0$, differentiating $X_0/X_i=x_i^{-1}$ and $X_a/X_i=x_a/x_i$ givesIn the wedge, all terms involving a second copy of $d{x}_i$ disappear; the surviving powers are $x_i^{-2}$ from $d(x_i^{-1})$ and $x_i^{-1}$ from the other $n-1$ differentials. Moving $d{x}_i$ into its original position produces the stated sign. Rescale each local generator by ${\eta }_i=(-1{)}^i{\omega }_i$. Then ${\eta }_j=(X_j/X_i{)}^{-n-1}{\eta }_i$ on every overlap, exactly the transition of the twist with $m=-n-1$. This canonical-form transition on projective space proves the canonical bundle of projective space:For line bundles the Serre duality pairing becomesIn particular $H^n(\mathcal{O}(-r))$ is dual to $H^0(\mathcal{O}(r-n-1))$, which is zero for $r\le n$, precisely the independently obtained bound. The case $r=0$ recovers $H^n(\mathcal{O})=0$, and $H^n(\mathcal{O}(-n-1))\cong k$ matches $H^0(\mathcal{O})\cong k$. Thus the calculated transition functions, global sections and top-degree vanishing agree with Serre duality.