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A line bundle $L$ is ample when some positive tensor power $L^{\otimes m}$ is very ample.

The $k$th exterior power ${\Lambda }^{k}V$ is the quotient vector space of the tensor power $V^{\otimes k}$ that makes every elementary tensor with two equal factors zero. More precisely, take the quotient vector space by the linear span of all such tensors. Its elementary products $v_1\wedge \cdots \wedge v_k$ turn alternating multilinear maps into linear maps: every alternating $k$-linear map on $V$ factors uniquely through ${\Lambda }^{k}V$.^[ref]

If $e_1,\dots ,e_n$ is an ordered basis, the increasing productsform a basis of ${\Lambda }^{k}V$. There is one for each $k$-element choice of basis directions, soRepeated directions vanish; products with a different order can be sorted using their sign. Coordinate determinants distinguish the remaining increasing products, establishing linear independence.^[ref]

For $V={\mathbb{R}}^3$, a basis of ${\Lambda }^{2}V$ is $e_1\wedge e_2,e_1\wedge e_3,e_2\wedge e_3$. This explains the three bivector components in the worked example. The top exterior power ${\Lambda }^{3}V$ has one basis element, $e_1\wedge e_2\wedge e_3$, while ${\Lambda }^{4}V=0$: four independent directions cannot fit into that vector space. By convention ${\Lambda }^{0}V$ is the scalar field, even when $V$ has dimension zero.

For a finite real scalar matrix $(a_{ij})$ and unit vectors $u_i,v_j$ in a real Hilbert space, $|\sum a_{ij}⟨u_i,v_j⟩|\le K_G{\max }_{{\epsilon }_i,{\delta }_j\in \{-1,1\}}|\sum a_{ij}{\epsilon }_i{\delta }_j|$, with a universal constant. One valid bound is $K_G\le \pi /(2\log (1+\sqrt{2}))$. Expand $\sin (ct)$ using odd tensor powers with coefficient magnitudes summing to $\sinh c=1$, then apply the Gaussian sign identity $\mathbb{E}[\mathrm{sgn}⟨G,u⟩\mathrm{sgn}⟨G,v⟩]=(2/\pi )\arcsin ⟨u,v⟩$. Taking $c=\log (1+\sqrt{2})$ converts the expectation back to the original inner product. Complex scalars admit a universal bound twice this real bound by separating real and imaginary parts.

The Kodaira embedding theorem says that a compact complex manifold carrying a positive holomorphic line bundle is projective: sufficiently large tensor powers of the bundle have global holomorphic sections defining a holomorphic embedding into Complex projective space. Conversely, the restriction of $\mathcal{O}(1)$ gives a positive line bundle on every projective complex manifold.

The real Grothendieck inequality asserts that a universal constant $K_G$ satisfiesfor every finite real scalar matrix and vectors of norm at most one in a real Hilbert space. For complex scalars the corresponding scalar supremum is over phases of modulus at most one; a universal complex constant also exists. We prove both versions, without needing the optimal constant.

First, for unit vectors $u,v$ and a standard Gaussian vector $G$ on their finite-dimensional span,To see this, rotational symmetry makes the angle of the Gaussian vector uniform in the plane spanned by $u,v$. The two signs differ with probability $\arccos ⟨u,v⟩/\pi$. Subtracting twice this probability from one gives the identity; collinear cases follow directly. Zero Gaussian coordinates have probability zero.

Put $c=\log (1+\sqrt{2})$, so $\sinh c=1$ and $0<c<\pi /2$. In the Hilbert direct sum of odd tensor powers, formed with the Hilbert tensor product inner product, defineThe Hilbert tensor product satisfies $⟨u^{\otimes m},v^{\otimes m}⟩=⟨u,v{⟩}^m$. For unit $u,v$, both transformed vectors have squared norm ${\sum }_{k\ge 0}{c}^{2k+1}/(2k+1)!=\sinh c=1$, and the power series for sine givesFor any finite collection of these transformed vectors, use a standard Gaussian vector on their finite-dimensional span. The sign identity and $|c⟨u,v⟩|<\pi /2$ giveConsequently, writing $M$ for the scalar sign supremum,Vectors of norm below one are padded to unit vectors in additional orthogonal directions, using separate directions for the two families so their cross inner products remain unchanged. This proves the real Grothendieck inequality with $K_R=\pi /(2\log (1+\sqrt{2}))$.

For the complex Grothendieck inequality, rotate the total scalar sum to have nonnegative real part. Its real part is the sum of the two real bilinear expressions involving $\mathrm{Re}{a}_{ij}\mathrm{Re}⟨u_i,v_j⟩$ and $-\mathrm{Im}{a}_{ij}\mathrm{Im}⟨u_i,v_j⟩$. The real inner products are realized in the underlying real Hilbert space, rotating one family by $i$ for the imaginary part. Each real coefficient matrix has real sign supremum at most the phase supremum of the original complex matrix. Applying the real inequality twice gives the valid constant $K_C=2K_R$. In what follows let $K_G$ denote the appropriate real or complex constant.

Now let $T:L^1(0,1)\to H$ be a bounded linear operator into a Hilbert space, and write $M=\Vert T\Vert$. If $M=0$ there is nothing to prove. Approximate a finite family of functions by simple functions on a common disjoint measurable partition. For such a family, write $f_j={\sum }_k{a}_{jk}{1}_{A_k}$ and $m_k=|A_k|>0$. The vectorshave norm at most $M$. Choose unit vectors $u_j$ with $⟨T{f}_j,u_j⟩=\Vert T{f}_j\Vert$; zero images contribute zero and can be assigned any unit vector. With inner products linear in their first argument, the Grothendieck inequality givesOver the real field, phases here are signs. For any Banach space $E$, operator norm duality givesIndeed, interchange the two suprema and choose each phase to align the scalar values of the functional. Thus the required summing inequality holds for simple functions. Approximation in the Lp space $L^1$ passes it to arbitrary finite families, since both sides vary continuously under finite sums of $L^1$ errors. We obtainFinally let $R$ be the closed span of the Rademacher functions in $L^1(0,1)$. Independence, the scalar second moment, and Q2 give, for each finite scalar family,Completion therefore defines a Banach space isomorphism $S:{\ell }^2\to R$ with $\Vert S\Vert \le 1$ and $\Vert S^{-1}\Vert \le \sqrt{2}$. If a bounded projection $P:L^1\to R$ existed, $S^{-1}P:L^1\to {\ell }^2$ would be an absolutely summing operator by the result just proved. The ideal property ${\pi }_1(BTA)\le \Vert B\Vert {\pi }_1(T)\Vert A\Vert$, obtained directly from the defining weak 1-norm, would make $I_{{\ell }^2}=(S^{-1}P)S$ absolutely summing. But its first $n$ coordinate vectors satisfyA single finite summing constant cannot dominate $\sqrt{n}$ for all $n$. Hence the closed Rademacher span is not complemented in $L^1(0,1)$.

The classification of finite-dimensional sl2 representations identifies $L(n)$ with ${\mathrm{Sym}}^n{\mathbb{C}}^2$. In the defining two-dimensional representation,so $e^2=f^2=0$ andOn the $n$th tensor power, the infinitesimal operator associated with $e$ is the sum of its actions on the individual factors. These actions commute, so exponentiating their sum is the tensor product of the individual exponentials; the same holds for $f$. Restricting to the symmetric power consequently identifies the operator $s$ on $L(n)$ with ${\mathrm{Sym}}^{n}S$. It follows thatIndeed $-I$ contributes one minus sign from each of the $n$ tensor factors. The Weyl reflection lift in an sl2 representation can therefore square to minus the identity, even though its image in the Weyl group is an involution.

Let $\chi$ be the central one-dimensional character of a proposed $W$. Since $Z$ is not cyclic, part (a) makes this one-dimensional character nonfaithful. Choose a nonidentity $z\in \ker \chi$. It acts as identity on every tensor power $W^{\otimes n}$ and therefore on every summand. There is a one-dimensional character $\eta$ of $Z$ with $\eta (z)\ne 1$: decompose the finite abelian group into cyclic factors and choose a primitive one-dimensional character on a factor where $z$ has a nonidentity component. By part (b)(ii), some irreducible representation $V$ of $G$ has central one-dimensional character $\eta$ on $Z$. This $V$ cannot occur in any tensor power of $W$, because $z$ acts nontrivially on it. Therefore no such tensor-generating irreducible $W$ exists.

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.

On $T=V^{\otimes n}$, define the permutation and diagonal actions byThe inverse in the first formula gives a left action. Applying $g$ to every tensor factor and then reordering produces the same tensor as reordering and then applying $g$. Thus the two actions commute. The Schur algebra isIt is the commutant of an operator algebra of the permutation action.

To identify this commutant, use the multilinear isomorphismConjugation by a permutation reorders the factors on the right. Therefore the invariant subspace is the space of symmetric tensors of degree $n$ in $\mathrm{End}(V)$. It is spanned by $a^{\otimes n}$. Indeed, the polarization identityexpresses every symmetrized elementary tensor as a linear combination of pure powers. Those symmetrized elementary tensors span the invariant subspace in characteristic zero.

One may restrict $a$ to invertible endomorphisms without changing the span. For fixed $a$, the vector-valued polynomial $(a+tI{)}^{\otimes n}$ has degree at most $n$. Choose $n+1$ distinct values of $t$ away from the finitely many roots of $\det (a+tI)$. Polynomial interpolation expresses its constant term $a^{\otimes n}$ as a linear combination of those invertible tensor powers. ConsequentlyThe span is an algebra, since products are $(gh{)}^{\otimes n}$.

The image $A_T$ of $\mathbb{C}{S}_n$ is semisimple, as a quotient of a semisimple algebra. The double-centralizer theorem for semisimple operator algebras gives $A_T^{\prime \prime }=A_T$. Since the preceding computation says that $A_T^{\prime }$ is the span of the general linear action, the two actions are mutual commutants. More explicitly, semisimple module decomposition givesHere $D_{\lambda }(V)={\mathrm{Hom}}_{S_n}(S^{\lambda },T)$ is the multiplicity space, with its natural general linear action. The commutant algebra is the product of the full endomorphism algebras of these multiplicity spaces. Hence each nonzero $D_{\lambda }$ is irreducible, and distinct spaces have nonisomorphic general linear representations, because the operators $g^{\otimes n}$ span that commutant. A primitive $e_t$ picks a one-dimensional factor from $S^{\lambda }$, so $e_{t}T$ is an equivalent realization of $D_{\lambda }$ as a Schur module.

The range of shapes is exact. If $\lambda$ has more than $m$ rows, a column antisymmetrizes more than $m$ vectors and its action vanishes by ${\Lambda }^{m+1}V=0$. Conversely, for at most $m$ rows place the $i$th basis vector in every tensor position belonging to row $i$ of $t$. Row symmetrization multiplies this tensor by ${\prod }_i{\lambda }_i!$. Column antisymmetrization is nonzero, because all vectors within each column are distinct, and its different permutations give distinct basis tensors. Thus $h_{t}T\ne 0$. This proves the length bound for a Schur module and the stated Schur–Weyl duality. For $n=0$, the empty partition and $V^{\otimes 0}=\mathbb{C}$ give the trivial version.

The divisor determines the bundle through its local equations. Choose an open cover on which $V$ has a reduced holomorphic defining equation $f_i$, using $f_i=1$ on sets disjoint from $V$. Because $V$ is a complex submanifold of codimension of a submanifold one, its ideal is locally generated by a coordinate; consequently $f_i/f_j$ is a nowhere-zero holomorphic function on each overlap. Inside the sheaf of meromorphic functions on a complex manifold, consider the locally free sheaf of rank oneThese local descriptions agree on overlaps. With $e_i=1/f_i$, its transition functions of a vector bundle are specified byThe ratios satisfy the Čech cocycle condition, and the associated holomorphic line bundle associated to a divisor is $[V]$. The section $1=f_i{e}_i$ is holomorphic and vanishes to order one along $V$. Replacing $f_i$ by $u_i{f}_i$, with $u_i$ a holomorphic unit, replaces the frame by $u_i^{-1}{e}_i$ without changing the subsheaf of meromorphic functions. Refining the cover changes no local sections either. This proves choice independence up to isomorphism rather than merely producing a bundle for one cover.

For a complex manifold of dimension $n$, the canonical bundle isIt is the holomorphic line bundle of holomorphic top-degree differential forms. To compute it on Complex projective space, let $S=\mathcal{O}(-1)$ be the complex tautological line bundle, so that $S^{\ast }=[H]=\mathcal{O}(1)$ is the hyperplane line bundle. At a line $\ell \subset {\mathbb{C}}^{n+1}$, the tangent space is $\mathrm{Hom}(\ell ,{\mathbb{C}}^{n+1}/\ell )$. Tensoring the tautological quotient sequence by $S^{\ast }$ gives the Euler sequence on complex projective spaceTaking the top exterior power gives $\det T^{1,0}{\mathbb{P}}^n=\mathcal{O}(n+1)$; taking its dual bundle yieldsHere a negative tensor power means the corresponding positive tensor power of the dual bundle.

The proposed sections over $U$ form a sheaf: compatible maps into $L$ glue uniquely, and their projections remain $\varphi$. Addition and multiplication by holomorphic functions are defined in the fibres of $L$, making this a sheaf of ${\mathcal{O}}_X$-modules. If $e_i$ is a holomorphic local frame of $L$ on $W_i\subset Y$, every section over $U\subset {\varphi }^{-1}(W_i)$ has the unique expressionThus the sheaf is a locally free sheaf of rank one. Its transition functions of a vector bundle are those of $L$ composed with $\varphi$, so the corresponding pullback vector bundle is precisely ${\varphi }^{\ast }L$. This proves invertibility locally and compatibility globally.

For the local blowup of a complex manifold at a point, work on the projective chart $Y_i\ne 0$ and writeAll the incidence equations reduce to $x_j=u{t}_j$. Therefore $(u,(t_j{)}_{j\ne i})$ gives a holomorphic coordinate chart isomorphic to ${\mathbb{C}}^n$. On an overlap with $Y_{\ell }\ne 0$,which is a biholomorphism where $t_{\ell }\ne 0$. The incidence subset is closed in the product of two Hausdorff spaces and is covered by these charts; hence it is a complex manifold of dimension $n$. In each chart the exceptional fibre is $u=0$, so it is a smooth complex submanifold, with its induced projective coordinates givingAway from the origin the inverse of the projection is $x\mapsto (x,[x])$, which is holomorphic; the projection is therefore a biholomorphism outside $E$.

To find the canonical bundle, pull back the nowhere-zero top form on ${\mathbb{C}}^n$. In the chart above, its Jacobian determinant givesThe sign depends only on the ordering of the coordinates. This is a section of $K_X$ with divisor on a complex manifold exactly $(n-1)E$: it has that vanishing order in every exceptional chart and no zeros elsewhere. A nonzero meromorphic section of a holomorphic line bundle with divisor $D$ identifies its holomorphic line bundle with $[D]$, by sending the local generator $1/f_i$ of $\mathcal{O}(D)$ to the nowhere-zero frame obtained by dividing the section by $f_i$. ConsequentlyThis also includes $n=1$, when the local map is an isomorphism and the exponent is zero.

For a general complex manifold, choose a coordinate neighbourhood $U$ of $P$, with $P$ sent to $0$, and replace $U$ by the inverse image of this coordinate neighbourhood in the local blowup of a complex manifold at a point. Glue this space to $M\setminus \{P\}$ along $U\setminus \{P\}$ using the preceding biholomorphism. The resulting charts give a complex manifold $\tilde{M}$ and a holomorphic map $\varphi$; the exceptional fibre is ${\mathbb{P}}^{n-1}$ and the map is a biholomorphism off that fibre. For completeness, the local projection is proper because its inverse image over a compact subset is closed in that subset times compact Complex projective space. This ensures the gluing is Hausdorff: separate points with distinct images downstairs, and points over $P$ inside the local blowup. The atlas is second countable as well.

The top exterior power of the differential defines a global holomorphic bundle map ${\varphi }^{\ast }{K}_M\to K_{\tilde{M}}$. Equivalently, it is a section of $K_{\tilde{M}}\otimes ({\varphi }^{\ast }{K}_M{)}^{-1}$. The same local Jacobian determinant has vanishing order $n-1$ on $E$, and the differential is invertible elsewhere. Applying the holomorphic line bundle associated to a divisor construction gives the canonical bundle formula for a point blowup

For a Young tableau $t$, let $R_t$ permute the entries within its rows and $C_t$ within its columns. We use the Young symmetrizer conventionReversing the order gives another usual realization of the same irreducible polynomial module. On the tensor power $T=V^{\otimes n}$, the actions areThey commute because applying $g$ to every factor commutes with permuting the factors.

We state the permitted combinatorial input explicitly. A Young symmetrizer satisfies $h_t^2=H_{\lambda }{h}_t$, where $H_{\lambda }$ is the nonzero hook product of a partition. Thus $e_t=H_{\lambda }^{-1}{h}_t$ is a primitive idempotent. The standard-tableau decomposition of the right regular module is $\mathbb{C}{S}_n={⨁}_{\lambda ,t\mathrm{standard}}{e}_t\mathbb{C}{S}_n$. Tensor this right-module direct sum with the left module $T$. The map $e_t\mathbb{C}{S}_n{\otimes }_{\mathbb{C}{S}_n}T\to e_{t}T$ sending $e_{t}a\otimes v$ to $e_{t}av$ is an isomorphism, with inverse $w\mapsto e_t\otimes w$. Therefore the Young-symmetrizer tensor decomposition isThis is a direct sum of $GL(V)$-modules; individual summands need not be $S_n$-invariant.

We also state the allowed Schur algebra result, namely Schur–Weyl duality: the two actions are mutual commutants, andwhere the $D_{\lambda }(V)$ are pairwise nonisomorphic irreducible homogeneous polynomial representations of degree $n$. We also use the standard Schur algebra equivalence between its modules and homogeneous degree-$n$ polynomial representations, so these exhaust the irreducibles in that category. A primitive idempotent $e_t$ has one-dimensional image on $S^{\lambda }$ and zero image on the other simple factors, so $h_{t}T\cong D_{\lambda }(V)$. This identifies the requested irreducibles. They classify the polynomial degree-$n$ representations in this tensor power, not all rational representations of every degree.

The length bound for a Schur module is $D_{\lambda }(V)\ne 0$ if and only if $\ell (\lambda )\le m$. A column of length greater than $m$ antisymmetrizes more than $m$ vectors and gives zero. Conversely, for $\ell (\lambda )\le m$, fill every tensor position in row $i$ with the $i$th basis vector of $V$. Row symmetrization multiplies it by ${\prod }_i{\lambda }_i!$, and column antisymmetrization is nonzero because the vectors within each column are distinct basis vectors.

A rational representation of an algebraic group is a regular morphism into the general linear group of its representation space. For $G{L}_m$, its matrix entries belong to $\mathbb{C}[g_{ij},\det (g{)}^{-1}]$; rational here permits determinant denominators but not arbitrary poles on $G{L}_m$. A one-dimensional rational character of ${\mathbb{C}}^{\times }$ is a Laurent polynomial $p(z)$ with $p(zw)=p(z)p(w)$ and $p(1)=1$. Comparing Laurent coefficients shows that just one monomial occurs and its coefficient is $1$, hence $p(z)=z^r$ for $r\in \mathbb{Z}$.

Restrict a one-dimensional rational character of $G{L}_m$ to its diagonal torus. The same argument in several variables gives ${\prod }_i{x}_i^{r_i}$. Conjugation by permutation matrices makes all $r_i$ equal. On every diagonalizable invertible matrix it consequently agrees with $(\det g{)}^r$. The allowed Zariski-density statement, and equality of regular functions on a dense subset, giveThese are the one-dimensional rational characters of the general linear group.

For complete reducibility of rational GL and SL representations, use the compact-group averaging argument. Average any positive definite Hermitian inner product over $SU(m)$ using normalized Haar measure. The orthogonal complement of an invariant subspace is then $SU(m)$-invariant. Differentiating makes it invariant under $\mathfrak{su}(m)$ and therefore under its complex span ${\mathfrak{sl}}_m(\mathbb{C})$. The elementary unipotent matrices $\exp (t{E}_{ij})$ generate $S{L}_m(\mathbb{C})$, so the complement is $S{L}_m$-invariant. This proves complete reducibility. Averaging over $U(m)$ similarly gives complete reducibility for rational $G{L}_m$ representations.

Here is an explicit rational extension from SL to GL. Decompose the representation space by the finite scalar center of $S{L}_m$ into subspaces $W_k$ on which $\zeta I$ acts as ${\zeta }^k$, with $0\le k<m$ and ${\zeta }^m=1$. These subspaces are invariant. For $g\in G{L}_m$, choose $a$ with $a^m=\det g$, set $s=a^{-1}g\in S{L}_m$, and defineChanging $a$ to $\zeta a$ changes $s$ to ${\zeta }^{-1}s$, so the two factors cancel. It is a homomorphism, since scalar roots multiply up to the same harmless factor, and it restricts to $\rho$ on $S{L}_m$.

It is rational as well. Every matrix coefficient of $\rho {|}_{W_k}$ has a polynomial representative on $S{L}_m$. Averaging that representative over the finite scalar center selects its homogeneous parts $P_d$ with $d\equiv k(\mathrm{mod}m)$. Substitution in the extension gives ${\sum }_d(\det g{)}^{(k-d)/m}{P}_d(g)$, a regular function on $G{L}_m$. Each $S{L}_m$-invariant subspace decomposes into its $W_k$ parts, and the extension acts on each part by a scalar times an $S{L}_m$ action. Thus these subspaces are also invariant under the chosen extension. Consequently $\rho$ is irreducible if and only if this ${\rho }^{\prime }$ is irreducible. For $m=1$, $S{L}_1$ is trivial and the trivial extension supplies the same conclusion.

The effective group is torsion free and every modular cusp is regular in the preceding sense, so orders of a meromorphic weight-$k$ form are integers. At interior points use a local automorphy trivialization; at a modular cusp use the Fourier order of the appropriate slash transform in its cusp width coordinate. Let $C$ be the sum of all modular cusp points, each once, and setA meromorphic function $\phi$ belongs to the Riemann-Roch space $\mathcal{L}(D(f))$ exactly when $\mathrm{div}\phi +D(f)\ge 0$. In the interior this requires $f\phi$ to have no pole; at a modular cusp it requires order at least one. Conversely, the quotient of any weight-$k$ cusp form by $f$ is a meromorphic weight-zero function satisfying precisely those inequalities. This proves the cusp-form divisor presentationTo compute the degree without imposing a valence formula as an extra assumption, use the meromorphic tensor differential $f^{12}(d\tau {)}^{6k}$. Its automorphy factors cancel. Its order at an interior point is $12{\mathrm{ord}}_{P}f$; at a modular cusp it is $12{\mathrm{ord}}_{P}f-6k$, since $d\tau$ is a nonzero constant times $d{q}_c/q_c$. A meromorphic section of the $6k$th tensor power of the canonical bundle has total divisor degree $6k(2g-2)$. The regular-cusp valence formula on a torsion-free modular curve is thereforeFor $k\ge 3$, $\mathrm{deg}D(f)-(2g-2)=(k-2)d/12>0$. The Riemann-Roch theorem says $\ell (D)-\ell (K-D)=\mathrm{deg}D+1-g$, and a divisor of negative degree has no nonzero sections. Thus $\ell (K-D)=0$ andUsing $d=(p^2-1)/2$ and $r_{\infty }=p-1$ givesThe canonical-degree and Riemann-Roch facts used here are general results for compact Riemann surfaces, as permitted.

The Dolbeault theorem identifies $H^2(M,{\mathcal{O}}_M)$ with $H_{\bar{\partial }}^{0,2}(M)$. It vanishes by hypothesis, and complex conjugation in the Hodge decomposition theorem for compact Kähler manifolds makes $H^{2,0}$ vanish too. Thus every real degree-two cohomology class has type $(1,1)$.

Let $\omega$ be a Kähler form. Choose a rational cohomology class $\eta$ sufficiently close to $[\omega ]$. More concretely, using harmonic differential forms as representatives for a fixed Kähler metric, the harmonic form representing $\eta$ is close to $\omega$ in every smooth norm: harmonic differential forms constitute a finite-dimensional vector space. It is a real closed $(1,1)$ form, and is positive if sufficiently close to $\omega$, by compactness. This is the openness of the Kähler cone. Multiplying by a positive integer gives an integral Kähler class $\kappa =N\eta$.

The holomorphic exponential sequence contains the cohomology segmentChoose an integral lift of $\kappa$. Exactness gives a holomorphic line bundle $\mathcal{L}$ having that First Chern class. We must ensure that it has a positive metric, rather than merely a positive representative of its class.

Choose any Hermitian metric $h_0$ and let ${\theta }_0$ be its normalized curvature form of a Hermitian holomorphic line bundle. If $\theta$ is the positive form representing $\kappa$, then $\theta -{\theta }_0$ is an exact real $(1,1)$ form. The ddbar lemma gives a real smooth function $u$ withReplace $h_0$ by $h=e^{-u}{h}_0$. Its normalized curvature is $\theta$, so $\mathcal{L}$ is a positive holomorphic line bundle. The Kodaira embedding theorem now applies: sufficiently many sections of a sufficiently high tensor power define the desired embedding. Therefore

Repeatedly using $V_r\otimes V_1\cong V_{r+1}\oplus V_{r-1}$ shows that the multiplicity of $V_{n-2j}$ in the nth tensor power satisfies Pascal's recursion. Consequentlywhere $\left(\genfrac{}{}{0ex}{}{n}{-1}\right)=0$. The multiplicity is the difference between the numbers of weight vectors of weights $n-2j$ and $n-2j+2$.

A holomorphic line bundle is ample when some positive tensor power is very ample, so its sections define an embedding into Complex projective space. It is a positive holomorphic line bundle when it has a Hermitian metric with positive Chern curvature $i{F}_{\nabla }$. The Kodaira embedding theorem says that a compact complex manifold carrying a positive holomorphic line bundle is projective; sufficiently high tensor powers give a holomorphic embedding into projective space.

Use the B2 root system conventionThus $V=L_{{\omega }_2}$ is the five-dimensional vector representation of the Special orthogonal Lie algebra ${\mathfrak{so}}_5$. Label its weight verticesThe crystal basis is the colored chainbecause each Kashiwara operator ${\tilde{f}}_i$ subtracts ${\alpha }_i$.

For the tensor product of crystals, write $XY$ for $X\otimes Y$. The complete colored-arrow graph is compactly specified byIts three connected highest-weight components start at $AA$, $AB$, and $AE$. Their vertex sets areTheir highest weights and dimensions identify the ten-vertex component with the exterior square and the other two with the symmetric square. Thereforeof dimensions $10$ and $14+1$, respectively.

The module $L=L_{{\omega }_1}$ is the four-dimensional spin representation. Its weights are $(\pm {\epsilon }_1\pm {\epsilon }_2)/2$, and its crystal is

Every weight of $V$ lies in the root lattice, so every weight of every tensor power $V^{\otimes n}$ also lies in that lattice. But ${\omega }_1=({\epsilon }_1+{\epsilon }_2)/2$ represents the nonzero coset in the quotient of the weight lattice by the root lattice. Consequently no irreducible constituent of $V^{\otimes n}$ can have highest weight ${\omega }_1$, and $L$ never occurs.