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Over the Grassmannian ${\mathrm{Gr}}_k({\mathbb{C}}^N)$, the complex tautological bundle on a Grassmannian isIts fibre at $W$ is the k-plane $W$ itself. A neighbourhood of a fixed plane consists of graphs of linear maps $A:W\to W^{\perp }$; $(A,w)\mapsto (\mathrm{graph}A,w+Aw)$ gives a local trivialization. The inclusions of coordinate spaces give the infinite Grassmannian and its tautological rank-k bundle ${\gamma }_k^{\infty }$.

Use the customary compact Hausdorff base convention for the classification of complex vector bundles by a Grassmannian; equivalently the argument uses finite subordinate partitions of unity. Give a rank-k complex vector bundle $E\to X$ a continuous fiber metric given by Hermitian inner products. Choose a finite trivializing cover with local orthonormal coordinates $t_i:E{|}_{U_i}\to U_i\times {\mathbb{C}}^k$, and a subordinate partition of unity ${\rho }_i$ with support contained in $U_i$. Defineextending each component by zero off its support. This is continuous and satisfies $\Vert J_x(v){\Vert }^2={\sum }_i{\rho }_i(x)\Vert v{\Vert }^2=\Vert v{\Vert }^2$, so it is a fibrewise embedding. The image planes vary continuously: in local frames, their orthogonal projection matrix is $J(J^{\ast }J{)}^{-1}{J}^{\ast }$. Thus they define a map $c_E:X\to {\mathrm{Gr}}_k({\mathbb{C}}^{kr})$, and $v\mapsto (x,J_{x}v)$ identifies $E$ with $c_E^{\ast }{\gamma }_k^{\infty }$. This proves that every bundle is a tautological pullback vector bundle, by the finite-dimensional embedding of a complex vector bundle.

We next prove the two independence assertions that turn this construction into a bijection. First, homotopic maps have isomorphic pullback vector bundles. Pulling back along a homotopy gives a bundle over compact Hausdorff $X\times I$; embed that bundle in a finite trivial vector bundle by the construction just given. Its image has a continuous family of rank-k orthogonal projection matrices $P(x,t)$. Compactness gives a subdivision $0=t_0<\cdots <t_s=1$ withRestriction of $P(x,t_{j+1})$ to the image at $t_j$ is injective: if it killed $v$, then $v=(P(x,t_j)-P(x,t_{j+1}))v$, impossible by the strict norm bound unless $v=0$. Equal ranks make it an isomorphism, with continuous local inverses. Composing these bundle maps identifies the endpoint vector bundles. The fact that close orthogonal projections identify their image bundles therefore makes the pullback assignment well-defined on homotopy classes.

Conversely, suppose two classifying maps give isomorphic vector bundles. The infinite Grassmannian has its standard weak CW topology; the compact-subset property of CW complexes puts the compact images of the two maps in finite stages. Their tautological pullback vector bundles therefore give two finite-dimensional embeddings $J_0:E\to {\mathbb{C}}^{N_0}$ and $J_1:E\to {\mathbb{C}}^{N_1}$ of the same bundle, after using the vector bundle isomorphism. In orthogonal coordinate blocks interpolate byAt least one coefficient is nonzero, so every $J_t$ is injective and its image planes give a continuous Grassmannian homotopy. One endpoint uses the first block and the other the second block. Moving the latter coordinate copy to the standard one is a fixed unitary change of coordinates, connected to the identity by a path in the unitary group. Such a path exists by diagonalizing the unitary matrix and continuously varying its eigenvalue phases. Hence the original two maps are homotopic in the infinite Grassmannian. This also proves independence of all embedding and metric choices. Together the two directions establishFor arbitrary non-Hausdorff compact spaces without the required numerability, that usual bundle-classification statement needs additional hypotheses; the construction above specifies precisely the conventional setting being used.

Write $B={\mathrm{Gr}}_k({\mathbb{C}}^{\infty })$. For the assumed fiber bundle $U_k\to \nu \to B$ with contractible total space, the long exact sequence of homotopy groups of a fibration has segmentFor $q\ge 2$ this gives group isomorphisms. For $q=1$, the relevant pointed-set part and connectedness of $U_k$ give ${\pi }_1(B)=0$, corresponding to the single component ${\pi }_0(U_k)$. Thus the claimed formula includes its endpoint:Finally the determinant gives a fiber bundle $S{U}_2\to U_2\to S^1$. Since ${\pi }_3(S^1)={\pi }_4(S^1)=0$, its exact sequence yields ${\pi }_3(U_2)\cong {\pi }_3(S{U}_2)$. The given identification $S{U}_2\cong S^3$ and the degree calculation ${\pi }_3(S^3)=\mathbb{Z}$ therefore giveThe equality of free and based homotopy classes uses the simple connectedness just proved. HenceThe classification of rank-two complex vector bundles over the four-sphere is a bijection of isomorphism classes, labelled by the integer clutching function homotopy class after choosing a generator. It is not an assertion that Whitney sum gives a group operation on bundles of fixed rank two.

Let $C$ be the smooth algebraic curve. A nonzero regular section of a line bundle has an effective zero divisor, so its degree of a line bundle is nonnegative. Degrees add under tensor products and change sign on taking the dual. We use genuine line subbundles, so their quotients in $E$ are locally free. Both bounds can be proved constructively.

For the upper bound, take a rational frame of $E^{\ast }$ over the function field of $C$. Its two members are rational sections with only finitely many poles. Choose an effective divisor $D$ large enough to bound the pole orders of both rational sections. They become global sections of $E^{\ast }\otimes {\mathcal{O}}_C(D)$ and define a sheaf mapwhich is an isomorphism at the generic point. It is therefore injective as a sheaf map: a section in its kernel would vanish generically in a locally free sheaf, which is a torsion-free sheaf. For any line subbundle $L\subset E$, at least one of its two component maps into ${\mathcal{O}}_C(D)$ is nonzero. Such a map is a nonzero global section of ${\mathcal{O}}_C(D)\otimes L^{-1}$. Its zero divisor on an algebraic curve is effective, soThis upper degree bound for line subbundles on a smooth curve depends only on a choice made for $E$, and applies to every one of its line subbundles.

For the lower-bound question we will actually construct subbundles ${\mathcal{O}}_C(-m{p}_0)$ for every sufficiently large integer $m$, where $p_0$ is a fixed point. First prove the globally generated vector bundle property rather than assuming that negative-degree inclusions automatically are subbundles.

Apply the upper-bound argument to the rank-two vector bundle $F=E^{\ast }\otimes K_C$, where $K_C$ is the canonical bundle. Let $B$ be its upper degree bound. Choose $m$ with $m-1>B$. For any point $p\in C$, a nonzero global section of $F(-m{p}_0+p)$ would give a nonzero mapIts image can be saturated to a saturated line subbundle on a smooth curve $M\subset F$. Locally the local ring of a smooth algebraic curve is a discrete valuation ring: dividing out the common power of a uniformizer makes the image a primitive rank-one summand, and its quotient is free. Thus the saturation is indeed a subbundle. The original map gives a nonzero global section of $M\otimes {\mathcal{O}}_C(-m{p}_0+p)$ and consequently $\mathrm{deg}M\ge m-1$, contradicting the upper bound. Hence $H^0(F(-m{p}_0+p))=0$ for every $p$. By Serre duality,The exact sequenceand its long exact sequence in sheaf cohomology show that evaluation $H^0(E(m{p}_0))\to E(m{p}_0){|}_p$ is onto for every $p$. This proves the global generation of positive twists on a smooth curve with one bound on $m$ that works at every point.

Write $W=H^0(E(m{p}_0))$ and $N=\dim W$. The evaluation map $W\otimes {\mathcal{O}}_C\to E(m{p}_0)$ is surjective with kernel bundle of rank $N-2$. Consider the incidence subsetIt is the total space of this kernel bundle, so $\dim Z=1+(N-2)=N-1$. Its projection into $W$ is closed because $C$ is proper, and its image has dimension at most $N-1<N$. Choose $s$ outside this proper closed subset. It vanishes nowhere. In a local trivialization at any point, one component of $s$ is a unit, so it can be completed to a basis. Therefore ${\mathcal{O}}_C\stackrel{s}{\to }E(m{p}_0)$ is a subbundle with a locally free rank-one quotient. Twisting back givesThis proves the unbounded negative degrees of line subbundles. The nowhere-vanishing construction guarantees a subbundle at every point, including the zeros that might otherwise have occurred in a generic meromorphic inclusion. If the curve has several connected components, apply these constructions on each component and add the degrees; the same upper-bound and no-lower-bound conclusions follow.

We prove the required duality pairing explicitly. This also identifies the word “canonical” in the final conclusion: the isomorphism will be the intrinsic residue trace on a smooth projective curve, not a choice of basis for a one-dimensional vector space.

Begin on the projective line. With $s=t^{-1}$, the equation $ds=-t^{-2}dt$ identifies its sheaf of Kähler differentials over a field with $\mathcal{O}(-2)$. The Laurent description in Question 3 pairs $H^1(\mathcal{O}(a))$ with $H^0(\mathcal{O}(-a-2))$ by multiplication and extraction of the coefficient of $t^{-1}dt$. For $a\le -2$ the respective bases areand their pairing matrix has entriesThis is an invertible antidiagonal matrix. For $a\ge -1$ both spaces vanish. The coefficient extraction is well defined on the Čech cohomology quotient: a differential regular on the first chart has no $t^{-1}dt$ term, and one regular on the second chart is a sum of terms $t^{-2-j}dt$ with $j\ge 0$, also with no such term.

To extend this calculation to every vector bundle $E$ on the projective line, we give the splitting argument. Every line bundle on ${\mathbb{P}}^1$ is $\mathcal{O}(a)$: choose a nonzero rational section to obtain a divisor, and use the divisors of $t-c$ to move each finite point to infinity. The denominator-clearing argument in Question 4 makes both $E(n)$ and $E^{\vee }(n)$ globally generated for large $n$. The first assertion ensures $H^0(E(-a))\ne 0$ for some integer $a$; the second gives an embedding $E\hookrightarrow \mathcal{O}(n{)}^{\oplus r}$, so this is impossible for $a>n$. Choose the largest such integer $a$.

A nonzero section gives $\mathcal{O}(a)\to E$. Saturating its image gives a line subbundle; if it were strictly larger, it would be $\mathcal{O}(a+d)$ for a positive integer $d$, recording the section's zeros, contrary to maximality. Thus the quotient $F$ is a vector bundle. Induct on rank and write $F={⨁}_j\mathcal{O}(b_j)$. Twist the sequence by $-a-1$. Since $H^0(E(-a-1))=0$ and $H^1(\mathcal{O}(-1))=0$, exactness forces $H^0(F(-a-1))=0$, whence every $b_j\le a$. Local lifts of a quotient summand differ on overlaps by a Čech cocycle in $\mathcal{O}(a-b_j)$. But $H^1(\mathcal{O}(a-b_j))=0$ by Question 3. Adjusting the lifts by a coboundary makes them glue, splitting the sequence. This proves the Birkhoff–Grothendieck theorem here. The direct-sum Laurent calculation consequently proves the perfect pairingAlthough a splitting proves nondegeneracy, the pairing itself uses only evaluation followed by the coefficient trace on $H^1({\Omega }_{{\mathbb{P}}^1}^1)$, so does not depend on that splitting. This is residue duality on the projective line.

Now choose a rational function $t$ on $C$ with $dt\ne 0$. Such a choice is available over the perfect field $k$: a local parameter at a smooth point has nonzero differential in the rank-one sheaf of Kähler differentials over a field. Since $dt$ spans the differentials of $k(C)/k$, the extension $k(C)/k(t)$ has zero relative differentials and is separable. Question 1 gives a finite generically separable morphism $\pi :C\to {\mathbb{P}}^1$. It is flat: locally its finite module over a discrete valuation ring is torsion-free, hence free. For the invertible sheaf $\mathcal{M}$, its pushforward $E={\pi }_{\ast }\mathcal{M}$ is therefore a vector bundle.

We next prove the differential-valued dual identification needed to transfer the pairing. At a point of the base, complete the local rings. Each branch is of the formIf $e$ is the order of $f$, formal division writes $B=A[u]$ with basis $1,u,\dots ,u^{e-1}$ and a monic degree-$e$ equation $P(u)=0$. In particular this argument does not assume tame ramification or the special equation $s=u^e$. For a polynomial $g$ the elementary interpolation identity isTo see it, pass to a field containing the distinct roots $u_i$ of $P$. Lagrange interpolation writes the remainder as ${\sum }_{i}g(u_i)P(T)/((T-u_i)P^{\prime }(u_i))$. Comparison of its leading coefficient gives the identity, and descent returns it to the original field. The top-coefficient pairing on $A[T]/(P)$ is perfect: in the power basis its matrix has zero entries for $i+j<e-1$ and ones for $i+j=e-1$, so its determinant is a unit. It follows that the trace-dual fractional module is exactly $P^{\prime }(u{)}^{-1}B$.

Formal division of $f(T)-s$ gives $f(T)-s=W(T,s)P(T,s)$ with $W$ a unit. Differentiate in $T$ and evaluate at $u$. This shows that $f^{\prime }(u)$ and $P^{\prime }(u)$ differ by a unit. Since $ds=f^{\prime }(u)du$, we obtainThis is the trace-dual module of a finite curve map identity. Its map sends a differential $\eta$ to the functional $b\mapsto \mathrm{Tr}(b\eta )$, where the trace of $hds$ is $\mathrm{Tr}(h)ds$. For several branches take the direct sum of these maps. Completion is faithfully flat, so their being isomorphisms proves the corresponding isomorphism of the original local modules. A local trivialization of $\mathcal{M}$ now gives, compatibly on overlaps,

The inverse images of the two standard affine charts and their intersection are affine. As in Question 4, the Čech cochain complexes identify $H^1(C,\mathcal{M})$ with $H^1({\mathbb{P}}^1,E)$. Global sections in the last display identify the other factor with $H^0(C,{\mathcal{M}}^{\vee }\otimes {\Omega }_C^1)$. Thus the perfect pairing already calculated on the projective line givesIn particular the stated vanishing of global sections forces $H^1(C,\mathcal{M})=0$. The proof has established the relevant case of Serre duality, rather than invoked it as the requested argument.

To identify its intrinsic trace, use the rational principal-parts resolution on an algebraic curve. The constant sheaf of rational sections and the sheaf of their finite-support principal parts are both flasque sheaves. Consequently a class of $H^1(C,\mathcal{M})$ is represented by finitely many rational principal parts $(m_P)$, modulo the family coming from one global rational section. For a global section $\eta$ of ${\mathcal{M}}^{\vee }\otimes {\Omega }_C^1$, evaluation gives the pairingThe algebraic residue of a rational differential is its $u^{-1}du$ coefficient, independent of the local parameter. Adding a regular representative changes no residue. The local trace calculation above also givesOne can check this directly in the monogenic description: expand a differential in the basis $1,u,\dots ,u^{e-1}$, use the displayed interpolation formula for its trace, and extract the coefficient of $s^{-1}ds$. Formal division of $f(u)-s$ makes that coefficient exactly the coefficient of $u^{-1}du$; for a product of branch fields the trace is the sum. Thus the identity remains valid in positive characteristic, including wild ramification. On ${\mathbb{P}}^1$, partial fractions show that a rational differential's residues at finite points sum to the negative of its residue at infinity. Tracing to ${\mathbb{P}}^1$ therefore proves that the sum of residues of every global rational differential on $C$ is zero. Adding a global rational section to $(m_P)$ consequently changes the pairing by zero. This defines the same pairing as the finite-pushforward calculation, intrinsically, without a chosen map to ${\mathbb{P}}^1$.

Finally take $\mathcal{M}={\Omega }_C^1$. Its differential-valued dual is ${\mathcal{O}}_C$, and Question 4 proved $H^0(C,{\mathcal{O}}_C)=k$. Perfectness identifies $H^1(C,{\Omega }_C^1)$ with the dual of this space. Evaluation at the distinguished constant section $1$ givesThis residue trace on a smooth projective curve uses neither a basis choice nor an auxiliary finite map. It is the canonical isomorphism.

Let $\pi :E\to B$ be an oriented real vector bundle of rank $r$ over a paracompact base. A Thom class is a class $U\in H^r(E,E\setminus B;\mathbb{Z})$ restricting on every oriented fiber pair $({\mathbb{R}}^r,{\mathbb{R}}^r\setminus 0)$ to its positive generator. The Thom isomorphism theorem states that this class exists uniquely with those fiber restrictions and thatHere $B$ is identified with the zero section. Equivalently use the disk/sphere bundle pair. The compact-support version is $H_c^q(B;\mathbb{Z})\cong H_c^{q+r}(E;\mathbb{Z})$, with fiberwise compact support in the Thom construction. If the bundle is not oriented, the integral assertion requires the corresponding orientation local system.

For a closed co-oriented embedded submanifold $Y$ of codimension $r$, co-orientation means an orientation of its normal bundle. A tubular neighborhood identifies a neighborhood of $Y$ with that bundle. Its Thom class, transferred by excision, is a class in $H^r(M,M\setminus Y;\mathbb{Z})$. For compact $Y$ this maps into the direct limit defining compactly supported cohomology, giving $\boxed{{\epsilon }_Y\in H_c^r(M;\mathbb{Z})}$. The construction is independent of the tubular neighborhood and uses the chosen co-orientation. In an oriented ambient manifold it is the Poincare dual of the oriented submanifold. Here “closed submanifold” has the usual compact-without-boundary meaning; if it meant merely a closed subset and $Y$ were noncompact, the supported relative class would still exist but need not have compact support.

We next prove the fixed-point implication without imposing an unprinted orientability assumption on $M$. SetUse a finite triangulation of the closed smooth manifold. The elementary Hopf trace identity says that the alternating trace of a chain endomorphism equals its alternating trace on homology. To verify it, split each finite-dimensional chain space as boundaries, a complement representing homology, and a complement mapping isomorphically onto the preceding boundaries. The traces on these two boundary summands cancel in adjacent degrees, leaving exactly the homology traces. The cohomology traces equal the homology traces because the induced maps are dual over $\mathbb{Q}$.

If $f$ has no fixed points, compactness gives $\delta ={\min }_{x}d(x,f(x))>0$. Choose a triangulation $K$ of mesh less than $\delta /4$ and a sufficiently fine subdivision $K^{\prime }$ admitting a simplicial approximation $F:K^{\prime }\to K$ uniformly within $\delta /4$ of $f$. The subdivision chain map $s:C_{\ast }(K)\to C_{\ast }(K^{\prime })$ sends each simplex to simplices inside it. For a vertex $v$ of such a small simplex, $d(v,F(v))>3\delta /4$, so $F(v)$ cannot lie in the original coarse simplex. Consequently $F_{\#}s$ has zero diagonal coefficient on every coarse simplex. Its trace vanishes in every degree, while it induces $f_{\ast }$ on homology. The trace identity yields $L(f)=0$. Taking the contrapositive provesThis is the Lefschetz fixed-point theorem, established here rather than invoked. For oriented $M$ the same identity is the graph-diagonal formula for the Lefschetz number: $L(f)=⟨(1,f{)}^{\ast }{\epsilon }_{\Delta },[M]⟩$, which vanishes when the graph misses the diagonal. Finally, homotopic maps induce the same cohomology maps, so their Lefschetz numbers agree.

A $d$-dimensional embedded submanifold $F\subseteq M^m$ has the subspace topology and, near each of its points, a manifold chart identifying it with a coordinate slice ${\mathbb{R}}^d\times \{0\}$. These slice charts for an embedded submanifold give its smooth structure, and its inclusion is a smooth embedding.

A regular value $q\in N^n$ of a smooth map $f:M^m\to N^n$ is one for which $d{f}_p:T_{p}M\to T_{q}N$ is surjective at every $p\in f^{-1}(q)$. If the preimage is nonempty this forces $m\ge n$.

The inverse function theorem used here says that if $G:U\subseteq {\mathbb{R}}^m\to {\mathbb{R}}^m$ is smooth and $D{G}_p$ is an invertible matrix, then its restriction to a suitable neighbourhood of $p$ is a diffeomorphism onto an open neighbourhood of $G(p)$, with smooth inverse.

To prove the regular level set theorem, fix $p\in f^{-1}(q)$ and choose coordinates centred at $p,q$. The coordinate representative $\tilde{f}$ has a rank-$n$ Jacobian matrix. After permuting domain coordinates, its first $n$ columns form an invertible matrix. DefineIts Jacobian matrix is block triangular with invertible diagonal blocks, so the inverse function theorem makes $G$ a local coordinate change. In these coordinates, $f^{-1}(q)$ is precisely the slice with first $n$ coordinates zero. Thus it is an embedded submanifold of dimension $m-n$, withAn empty preimage is harmless; the local assertion is then vacuous.

Three equivalent formulations of orientability of a smooth manifold are: an oriented atlas whose transition Jacobian determinants are positive; a smoothly varying orientation of a vector space on each tangent space; and a nowhere-vanishing smooth top-degree differential form. An oriented atlas gives compatible oriented coordinate frames. A partition of unity glues their positive local top forms, and a nonzero top form declares a frame positive precisely when it evaluates positively on that frame. These constructions prove the equivalence.

For the orientability of a regular fibre, write $F=f^{-1}(q)$. The derivative gives a short exact sequence of vector bundlesThis is the reason the normal bundle of a regular fibre is trivial. Choose any fixed positive basis $b_1,\dots ,b_n$ of $T_{q}N$ and an ambient orientation on $M$. Declare a frame $v_1,\dots ,v_{m-n}$ of $T_{p}F$ positive whenis positive in $T_{p}M$, where $d{f}_p({\tilde{b}}_i)=b_i$. Changing a lift adds a tangent vector and leaves the determinant sign unchanged. Local smooth lifts exist because $df$ is surjective, so this rule varies smoothly and agrees on overlaps. It gives a global orientation on $F$. Only an orientation of the single vector space $T_{q}N$ was chosen; $N$ itself need not be orientable.

An orthogonal structure on a real vector bundle $E$ of rank $r$ is a choice of local frames whose transition matrices lie in the orthogonal group $O(r)$. In such frames declareOrthogonal transitions preserve this expression, so it defines a smooth positive-definite fiber metric. Conversely, a fiber metric turns any local frame into a smooth orthonormal frame by the Gram-Schmidt process; the resulting transition matrices are orthogonal. The two constructions are inverse, giving the required equivalence.

An orthogonal local trivialization $E{|}_U\cong U\times {\mathbb{R}}^r$ identifies each fiber isometrically with standard Euclidean space. Equivalently, its coordinate frame satisfies $h(e_i,e_j)={\delta }_{ij}$. This is local and does not assert the existence of a global frame.

Every real vector bundle over a smooth manifold admits a fiber metric. Choose a trivializing open cover and local Euclidean metrics $h_i$. The partition of unity theorem for a Hausdorff second-countable smooth manifold supplies nonnegative smooth functions ${\rho }_i$, with locally finite supports contained in the respective cover members, such that ${\sum }_i{\rho }_i=1$. Setextending each weighted metric by zero off its chart. Local finiteness makes this smooth. For every nonzero fiber vector $v$, at least one positive weight gives $h(v,v)={\sum }_i{\rho }_i{h}_i(v,v)>0$. Thus $h$ is a fiber metric and provides an orthogonal structure.

One definition of an orthogonal connection is that its covariant derivative preserves the fiber metric:for all vector fields $X$ and smooth sections of a vector bundle $s,t$. Another is that its parallel transport along every smooth curve is an isometry between the endpoint fibers.

For the first implication, take parallel sections $s,t$ along a curve. Their covariant derivatives vanish, so the displayed identity makes $h(s,t)$ constant: parallel transport preserves a fibre metric. Conversely, if parallel transport preserves the inner product, choose parallel extensions of arbitrary initial fiber vectors along a curve with prescribed initial tangent $X$. Differentiating their constant inner product gives $({\nabla }_{X}h)(s,t)=0$ at the initial point. Since $X,s,t$ were arbitrary, the connection is metric-compatible.

The equivalent local description is that the connection matrix in an orthonormal frame is skew-symmetric. Writing $\nabla e_j={\sum }_i{A}^i{}_j{e}_i$ in an orthogonal trivialization, differentiate $h(e_i,e_j)={\delta }_{ij}$ to obtainConversely this matrix identity gives the metric-preservation formula by expanding arbitrary sections. It is the orthogonal group connection description of the same condition.

Throughout, the braces denote a multiset, so repeated splitting degrees retain their multiplicities. If the multisets agree after a common shift, permuting summands gives $E_a\cong E_b\otimes \mathcal{O}(c)$. Tensoring all the fibres of a vector bundle by the same line bundle leaves their one-dimensional quotient spaces unchanged. HenceThis proves one implication in the isomorphism classification of abstract rational scrolls.

For the converse first take $n\ge 3$. Every divisor class on $F_a$ has the form $xL+yM$. If its square is zero, restriction to a fibre ${\mathbb{P}}^{n-1}$ giveswhere $h$ is the hyperplane class and $h^2\ne 0$ because $n-1\ge 2$. Thus $y=0$. It follows that $L$ is intrinsically characterized as the primitive nonzero nef class whose square is zero. An isomorphism $u:F_a\to F_b$ therefore carries $L_b$ to $L_a$; the sign is positive because both classes are nef and effective. The complete pencil $|L|$ determines the ruling, sofor an automorphism $\sigma$ of ${\mathbb{P}}^1$.

On a fibre, the isomorphism preserves the positive generator ${\mathcal{O}}_{{\mathbb{P}}^{n-1}}(1)$ of its Picard group. Consequentlyfor some integer $c$. Push forward along the ruling, using ${\pi }_{\ast }\mathcal{O}(M)=E$ and the projection formula for sheaves:Every automorphism of ${\mathbb{P}}^1$ preserves each $\mathcal{O}(d)$. The uniqueness of the splitting degrees in the Birkhoff–Grothendieck theorem now implies equality of the two multisets up to a shift. Replacing this $c$ by $-c$ gives the sign convention in the final formula. One can see that uniqueness numerically: for a split bundle,The first difference counts the degrees with $a_i+t\ge 0$, and the second difference counts the multiplicity of $a_i=-t$. Hence these dimensions determine every splitting degree, with multiplicity.

For $n=2$, subtract the smaller degree and write the normalized bundle as $\mathcal{O}\oplus \mathcal{O}(a)$, $a\ge 0$. Its surface is the Hirzebruch surface ${\mathbb{F}}_a$. Let $B$ be its negative section and $L$ its fibre, so $B^2=-a$, $BL=1$, $L^2=0$. The nef cone is generated by $L$ and $B+aL$: a nef divisor $xL+yB$ must have $y\ge 0$ from intersection with a fibre and $x-ay\ge 0$ from intersection with $B$, and these two inequalities also express it as a nonnegative combination of those generators. If $a>0$, a class $uL+v(B+aL)$ with $u,v\ge 0$ has squarewhich vanishes only when $v=0$. Thus the ruling is again intrinsic, and the same bundle argument applies. If $a=0$, the surface is ${\mathbb{P}}^1\times {\mathbb{P}}^1$ and has two nef square-zero rays, its two rulings. This distinguishes it from every $a>0$ surface. Isomorphisms may exchange its rulings, but both normalized splitting degrees are zero, so the stated classification is unchanged.

For $n=1$, every scroll is ${\mathbb{P}}^1$, and any two single splitting degrees differ by a common shift. Thus all ranks are covered:Reading the braces as ordinary sets would lose information and make the assertion false when degrees repeat. For example, $(0,0,1,2)$ and $(0,1,1,2)$ have the same underlying set but not the same splitting multiset; their minima are both zero, so no nonzero common shift can reconcile them.

Write $f_x$ for the linear isometry specified at $x$. Choose a cofinal sequence $V_i$ from the indexing family $I$, and write the target part of the supplied subordinate flag for a family of linear isometries as $W_i$. After a cofinal enlargement it hasBoth flags exhaust their universe for spectra. Compactness of the parameter space ensures that a finite target stage can contain all images of each fixed source stage.

Let ${\xi }_i$ be the vector bundle on $X$ with fiberHere $\ominus$ is the orthogonal complement in the indicated ambient subspace. Its rank is $\dim W_i-\dim V_i$. Let $\mathrm{Th}({\xi }_i)$ be its Thom space, the disk bundle modulo its sphere bundle, so that every fiber is compactified and all its points at infinity become the single basepoint. For the zero bundle this convention gives $\mathrm{Th}(0_X)=X_{+}$. Define a target indexed prespectrum on the flag byThe smash product here is of based spaces, not yet the derived smash product of spectra.

To specify all structure maps, for $i\le j$ use the continuous fiberwise orthogonal isomorphismBoth sides identify with the complement of $f_x(V_i)$ in $W_j$. In particular this is the isometry obtained by orthogonally projecting onto $f_x(V_j\ominus V_i)$ and its complement; it is not an assertion that the displayed summands coincide separately. Pass to fiberwise one-point compactifications, identify the last summand with $V_j\ominus V_i$ using $f_x^{-1}$, and apply the source structure map. This definesMore explicitly, decompose $z+w\in W_j\ominus f_x(V_i)$, where $z\in W_j\ominus W_i$ and $w\in ({\xi }_i{)}_x$, as $w^{\prime }+f_x(v)$ with $w^{\prime }\in ({\xi }_j{)}_x$ and $v\in V_j\ominus V_i$. Send its pair with $e\in E(V_i)$ to the pair $(w^{\prime },{\sigma }_{V_i,V_j}(v,e))$ over $x$. The notation for finite vectors extends to the compactifications, sending infinities and basepoints to the basepoint. Orthogonal decomposition and associativity of the source structure maps prove the prespectrum identities.

Extend the indexed prespectrum from the cofinal flag to the target indexing family and apply spectrification $L$. The point-set definition isOn a map of topological spectra $u:E\to E^{\prime }$ it is induced at stage $i$ by $1_{\mathrm{Th}({\xi }_i)}\wedge u(V_i)$. Thus the prescribed isometry family, indexing family and flag have all been used. A common cofinal refinement identifies the resulting objects in the stable homotopy category; the point-set model retains the chosen flag. If $f$ is a constant standard inclusion, the complement bundles are trivial and this recovers the usual reindexed $X_{+}\wedge E$.

The associativity of twisted half-smash products follows from an actual isomorphism of the complement bundles. First work over compact parameter spaces, so that a common choice of flags can be made. Choose source, intermediate and target stages $V_i,W_i,Z_i$ satisfyingFor the two twisted half-smash products and their composite the relevant vector bundles have fibersPull the first two bundles back to $Y\times X$. Since $g_y$ is a linear isometry, the decomposition of $Z_i$ into $g_y(W_i)$ and its orthogonal complement gives the fiberwise vector bundle isomorphismIts inverse is given by the two orthogonal projections. The isomorphism varies continuously with $(y,x)$, and the ranks agree:The one-point compactification of an external orthogonal direct sum gives the smash product of its Thom spaces. Using $g_y$ to identify $g_y({\xi }_x)$ with ${\xi }_x$, the isomorphism therefore givesThe left side is a stage of $(Y\times X){⋉}_{h}E$ and the right side is the corresponding two-step prespectrum presentation of $Y{⋉}_g(X{⋉}_{f}E)$.

These stage comparisons commute with structure maps. Indeed, increasing a stage decomposes a suspension vector into its component in $g_y{f}_x(V_j\ominus V_i)$ and the two residual complement components. In either construction the same source component is pulled back to $V_j\ominus V_i$ and used in the same structure map of $E$; the residual components are left in the same final complement. Iterated orthogonal decomposition gives the same result whether the intermediate complement is separated first or last. They also commute with maps of $E$, since the comparison only rearranges the suspension coordinates.

For clarity about the intervening spectrification, maps out of $LP$ into a genuine indexed spectrum $T$ are precisely compatible prespectrum maps $P\to T$. At stage $i$, adjunction from the Thom coordinate identifies such a map with a continuous familywhere the last identification is the genuine indexed spectrum structure of $T$. Compatibility between stages is exactly compatibility of this family with the source structure maps. Thus maps out of the twisted construction are continuous families of maps over the specified linear isometries. A family of such maps over $g$ out of $X{⋉}_{f}E$ is, by applying this universal property a second time, exactly a family of maps out of $E$ over $g_y{f}_x$. This is the same orthogonal-coordinate correspondence just constructed; it respects the identifications imposed by each spectrification. Consequently it descends from the prespectrum presentations to the associated spectra. Common cofinal refinements remove any incompatibility between the independently chosen flags.

The printed hypothesis does not require $Y$ to be compact. For an arbitrary compactly generated parameter space $Y$, use the same construction on compact test spaces mapping into $Y$. Each such family admits the common flags above, and the comparisons agree under restriction because their formula is the canonical orthogonal projection. They therefore assemble into the continuous global comparison in the compactly generated point-set construction. Equivalently, the preceding universal-property correspondence is valid for arbitrary $Y$ and does not impose a uniform finite target stage on all of it. Passing to cellular models for the derived construction gives the requested natural isomorphism

Assume $X$ is a connected compact Riemann surface of geometric genus $g\ge 1$. Work first in ${\mathrm{Pic}}^{g-1}(X)$, translating back to the Jacobian variety by tensoring with ${\mathcal{O}}_X(-(g-1)p_0)$ when desired. PutThis is the image of the Abelian sum map from $X^{(g-1)}$. The Riemann singularity theorem asserts that this image is a reduced Cartier divisor and thatHere multiplicity means the order of a local defining equation in the maximal ideal of the smooth ambient Jacobian variety. Consequently the singular points are precisely the classes with at least two independent sections. We will prove the multiplicity equality, including the nonvanishing of its proposed leading term.

Choose an effective divisor $E$ of degree $n\ge g$, and a local holomorphic family $\mathcal{L}$ of degree-$g-1$ line bundles about $L$ on $X\times U$, where $U$ is a small neighborhood in ${\mathrm{Pic}}^{g-1}(X)$. Such a family is obtained by varying the Čech cocycles of a line bundle. Since $\mathrm{deg}{L}^{\prime }(E)>2g-2$ for all $L^{\prime }\in U$, Serre duality gives $H^1(L^{\prime }(E))=0$; the Riemann-Roch theorem gives $h^0(L^{\prime }(E))=n$. The cohomology and base change for line bundles on a curve then identifiesas vector bundles of rank $n$. Evaluation defines a matrix $M:A\to B$. For each $L^{\prime }$, its exact sequence isThus $f=\det M$ vanishes exactly on $\Theta$.

The function $f$ is not identically zero. To exhibit a degree-$g-1$ line bundle with no section, choose $g$ distinct points $p_1,\dots ,p_g$ giving independent evaluation functionals on $H^0(K_X)$. Such points can be chosen successively because a nonzero holomorphic differential form cannot vanish everywhere. For $F=\sum p_i$, the Riemann-Roch theorem gives $h^0({\mathcal{O}}_X(F))=1$. Choose $q$ outside $F$. The unique section of ${\mathcal{O}}_X(F)$ does not vanish at $q$, so ${\mathcal{O}}_X(F-q)$ has no section. Therefore the evaluation determinant genuinely cuts out a hypersurface. Its support is irreducible, since it is the image of the irreducible symmetric product of a curve $X^{(g-1)}$.

Fix now $L\in \Theta$, and put $r=h^0(L)=h^1(L)$, the equality following from the Riemann-Roch theorem. The matrix $M(L)$ has an invertible $(n-r)$ by $(n-r)$ block. After invertible holomorphic row and column operations, its local form iswhere $N$ is $r$ by $r$. More explicitly, eliminate the invertible block by its Schur complement; the discarded determinant is a holomorphic unit. Hence $f$ is a unit times $\det N$. Every entry of $N$ lies in the maximal ideal, so its order is at least $r$.

The linear term of $N$ is the cup-product mapTo verify this, deform the transition functions as $g_{ij}(1+\epsilon {\xi }_{ij})$ and try to lift a section $s$ to $s+\epsilon s^{\prime }$. Its first-order gluing equation has right side ${\xi }_{ij}{s}_j$, whose Čech cohomology class is exactly the obstruction to the lift. The kernel-to-cokernel derivative of the evaluation matrix gives that same obstruction. Under Serre duality, if $s_1,\dots ,s_r$ and $t_1,\dots ,t_r$ are bases of $H^0(L)$ and $H^0(K_X\otimes L^{-1})$, thenWe must show that this matrix is invertible for some $\xi$; otherwise the lower bound on multiplicity would not be an equality.

For each of these two $r$-dimensional section spaces, there are $r$ distinct points with independent evaluations. Choose them successively, using a section in the remaining kernel to find the next point. The set of such $r$-tuples is a nonempty open subset of $X^r$. The two open sets intersect because $X^r$ is irreducible, and we may avoid their diagonals. Choose $q_1,\dots ,q_r$ in that intersection and trivialize $L$ and $K_X$ there. Both matricesare invertible. Let $\xi \in H^0(K_X{)}^{\ast }=H^1({\mathcal{O}}_X)$ be the sum of the evaluation functionals at these points, with any nonzero weights $c_k$. Thenis invertible. It follows that the degree-$r$ polynomial $\det N_1$ is nonzero, so ${\mathrm{ord}}_{L}f=r$. This proves the needed invertible cup-product direction for a line bundle on a curve by an explicit construction.

We still need to identify this determinant hypersurface with the reduced image, rather than a multiple of it. Choose $g-1$ distinct points imposing independent conditions on $H^0(K_X)$. For their sum $D$, the Riemann-Roch theorem gives $h^0(D)=h^0(K_X-D)=1$. At its class, the calculation above gives order one. The determinant hypersurface has irreducible support, so its sole divisor on an algebraic curve coefficient is one everywhere. A local hypersurface in a smooth space with these divisor on an algebraic curve coefficients is reduced. Thus it is exactly the reduced image $W_{g-1}$, and the multiplicity formula holds for that image itself.