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For a smooth manifold $M$ of dimension $n$, the tangent space $T_{p}M$ can be defined as the space of derivations at $p$: linear maps $v$ on germs of smooth real functions satisfying $v(fg)=f(p)v(g)+g(p)v(f)$. Equivalently, $v$ is the velocity of a smooth curve through $p$, with curves identified when their coordinate velocities agree. A chart $(x^1,\dots ,x^n)$ supplies the basis ${\partial }_i{|}_p$, so $v=v^i{\partial }_i{|}_p$.

The tangent bundle is the disjoint union $TM={\coprod }_{p\in M}{T}_{p}M$, with projection $\pi (v)=p$. Its smooth structure is defined by the local local trivializationsOn overlapping charts, the change of trivialization isThese are smooth changes of coordinates on a $2n$-dimensional smooth manifold, linear and invertible in each fibre. Thus $TM$ is a rank-$n$ vector bundle, and its projection is a submersion. Fibrewise addition, scalar multiplication and the zero section are smooth. A smooth section of a vector bundle $X:M\to TM$, satisfying $\pi \circ X={\mathrm{id}}_M$, is exactly a vector field; locally $X=X^i{\partial }_i$, with components transforming by the displayed Jacobian matrix. Its action on functions is a derivation, and the commutator of two such derivations is their Lie bracket.

The dual bundle of $TM$ is the cotangent bundle $T^{\ast }M$. Its smooth sections of a vector bundle are differential 1-forms, locally $\alpha ={\alpha }_{i}d{x}^i$. The coefficient transformation is the inverse transpose of that for tangent vectors, ensuring that $\alpha (X)$ is a well-defined smooth function. Tensoring these two bundles gives the tensor bundlesTheir sections of a vector bundle are tensor fields; their transition functions of a vector bundle are the corresponding tensor products of the tangent and cotangent transformations. For example, the bundle $\mathrm{End}(TM)=TM\otimes T^{\ast }M$ contains fields of linear endomorphisms, while $S^2{T}^{\ast }M$ contains symmetric bilinear fields. Taking alternating covariant tensors gives ${\Lambda }^k{T}^{\ast }M$, whose sections are differential forms. The wedge product, exterior derivative and pullback of a differential form make these bundles central to integration, Generalized Stokes theorem and de Rham cohomology.

A nowhere-vanishing section of ${\Lambda }^n{T}^{\ast }M$ chooses an orientation of a smooth manifold; such a section exists exactly when $M$ is orientable. Without an orientation of a smooth manifold, one can still integrate sections of the density bundle $|{\Lambda }^n{T}^{\ast }M|$, whose coordinate changes use the absolute Jacobian determinant. A Riemannian metric is a positive-definite section $g$ of $S^2{T}^{\ast }M$. It identifies $TM$ with $T^{\ast }M$ by the musical isomorphisms $X\mapsto g(X,\cdot )$ and gives the Riemannian volume densityOn an oriented smooth manifold, this density corresponds to a volume form.

Another organizing construction is the frame bundle $FM$. A frame over $p$ is a linear isomorphism $u:{\mathbb{R}}^n\to T_{p}M$, and the right action $u\cdot A=u\circ A$ makes $FM$ a principal bundle with general linear group $GL(n,\mathbb{R})$ as its structure group. Given a group representation $\rho :GL(n,\mathbb{R})\to GL(V)$, the associated bundle $FM{\times }_{\rho }V$ is obtained from pairs $(u,v)$ by the relation $(uA,v)\sim (u,\rho (A)v)$. The standard representation reconstructs $TM$; dual, tensor and exterior representations reconstruct the associated bundles described above. A Riemannian metric reduces the frame bundle to the orthonormal frame bundle, with orthogonal group $O(n)$ as structure group; an orientation of a smooth manifold reduces it further to the special orthogonal group $SO(n)$. A global section of $FM$ is a global frame, so it exists exactly when $M$ is a parallelizable manifold. Local triviality does not imply a global trivialization: the Hairy ball theorem prevents even one nowhere-zero vector field on $S^2$.

A connection on a vector bundle differentiates its sections in tangent directions and defines parallel transport. For $TM$, the Levi-Civita connection is distinguished by compatibility with a Riemannian metric and vanishing torsion tensor. Its coefficients are not tensor components, because a coordinate change introduces second derivatives; the difference of two connections is a section of $T^{\ast }M\otimes \mathrm{End}(TM)$. The Riemann curvature tensor is a genuine tensor measuring the failure of covariant derivatives to commute.

Finally, a smooth map $f:M\to N$ gives the pullback tangent bundle $f^{\ast }TN$. Its sections are vector fields along a map, including the velocity of a curve and Jacobi fields along a geodesic. For an embedded submanifold $i:M\hookrightarrow N$, the derivative identifies $TM$ with a subbundle of $i^{\ast }TN$, and the quotient is the normal bundle. With an ambient Riemannian metric, the normal bundle is identified with the orthogonal complement of $TM$ in $i^{\ast }TN$; the second fundamental form records the normal component of the ambient derivative of tangent fields. These constructions connect the bundle description to both intrinsic and extrinsic differential geometry.

For an oriented real vector bundle $\pi :E\to X$ of rank $d$, the Thom isomorphism theorem gives a unique Thom class $u_E\in H^d(D(E),S(E);\mathbb{Z})$ restricting to the chosen generator in each fibre, and isomorphismsEquivalently use the pair $(E,E\setminus X)$, identifying $X$ with the zero section. After forgetting relative supports, pull the Thom class back along the zero section $s$ to define the Euler class:In this expression the relative class is first mapped to absolute cohomology.

A compact embedded submanifold equipped with a coorientation $Y\subset X$ has an oriented normal bundle ${\nu }_Y$ of rank $d$. A tubular neighborhood and excision identify its Thom class with a class $u_Y\in H^d(X,X\setminus Y)$. Its image in $H^d(X)$ is the cohomology class of a cooriented submanifold ${\epsilon }_Y$. In particular it restricts to zero on $X\setminus Y$. If the ambient manifold is oriented, orient $Y$ by the normal-first convention. The class is its Poincare dual, characterized byfor the appropriate complementary degree. This is the stated integration identity in cohomological notation.

First prove the Lefschetz fixed-point theorem for an oriented closed n-manifold $X$. Orient its diagonal by the identification $\Delta \cong X$, and choose the normal orientation so this agrees with the normal-first convention in the product. This produces the integral class ${\epsilon }_{\Delta }$. To compute the part that contributes to the trace, tensor with $\mathbb{Q}$: the trace on a finitely generated integral cohomology group means its trace on the free quotient, equivalently on rational cohomology; torsion contributes nothing.

Choose a basis $e_{p,j}$ of $H^p(X;\mathbb{Q})$ and its Poincare duality pairing dual basis $e_{p,j}^{\vee }\in H^{n-p}(X;\mathbb{Q})$, normalized by $⟨e_{p,i}⌣e_{p,j}^{\vee },[X]⟩={\delta }_{ij}$. The rational image of the cohomology class of the diagonal isIndeed, pairing this expression against $\alpha \times \beta$ with degrees $p,n-p$ picks out the p-summands. The cross-product multiplication contributes $(-1{)}^{p^2}$ and moving $\alpha$ past $e_{p,j}^{\vee }$ contributes $(-1{)}^{p(n-p)}$. Together with $(-1{)}^{np}$ their product is $+1$. The result is $⟨\alpha ⌣\beta ,[X]⟩$, exactly the diagonal's defining identity. The rational Künneth theorem and nondegeneracy of the duality pairing determine the class. An integral torsion component, if present, is not determined by this rational expression and does not affect any evaluation below.

For the graph map ${\Gamma }_f=(1,f):X\to X\times X$, write $f^{\ast }{e}_{p,j}={\sum }_i{A}_{ij}{e}_{p,i}$. Pulling back the expression and evaluating gives the graph-diagonal formula for the Lefschetz number:If $f$ has no fixed point, its graph lies in $(X\times X)\setminus \Delta$. The relative Thom class restricts to zero there, so ${\Gamma }_f^{\ast }{\epsilon }_{\Delta }=0$ and $L(f)=0$. The contrapositive proves the required fixed-point assertion, without requiring smoothness of $f$ or transversality of its graph.

The printed hypothesis does not require $X$ to be orientable. For a nonorientable $X$, the diagonal's normal bundle is $TX$, so an ordinary integral coorientation cannot be assumed. We extend the proved theorem as follows, rather than apply the untwisted diagonal computation incorrectly. We first construct a smooth embedding in Euclidean space using finitely many charts ${\chi }_i:U_i\to {\mathbb{R}}^n$ and smooth functions ${\rho }_i$ supported in those charts, with at least one ${\rho }_i(x)>0$ at every point. The map $\Psi (x)=({\rho }_i(x),{\rho }_i(x){\chi }_i(x){)}_i$, with zero extensions, is injective: a positive first coordinate identifies a chart in which its remaining coordinates recover $x$. Its differential is injective as well, since $d{\rho }_i(v)=0$ and $d({\rho }_i{\chi }_i)(v)=0$ force $d{\chi }_i(v)=0$ in a chart with ${\rho }_i>0$. Compactness makes this injective immersion a homeomorphism onto its image. This finite-chart Euclidean embedding of a compact smooth manifold, enlarged by zero coordinates if necessary, has positive codimension. A compact tubular disc neighbourhood $W$ is an oriented N-manifold with boundary, inheriting its orientation from the ambient Euclidean space. Its double of a manifold $M$ is closed and oriented. Projection of each copy of $W$ to $X$ agrees on their common boundary, giving $r:M\to X$; inclusion into the first copy gives $i:X\to M$ with $ri=1_X$.

Set $F=ifr$. A fixed point of $F$ must lie in $i(X)$, and then is exactly the image of a fixed point of $f$. Moreover $F^{\ast }=r^{\ast }{f}^{\ast }{i}^{\ast }$, and cyclicity of trace for maps between finite-dimensional rational cohomology groups givesin each degree, because $i^{\ast }{r}^{\ast }=1$. Thus $L(F)=L(f)$. The oriented theorem for $M$ now gives a fixed point of $F$, hence of $f$, whenever $L(f)\ne 0$. This oriented manifold retract for the Lefschetz theorem proves the full printed result. The double construction also works componentwise if necessary.

The symplectic neighborhood theorem says that a symplectomorphism between compact symplectic submanifolds, together with a symplectic vector bundle isomorphism of their symplectic normal bundles, extends to a symplectomorphism of neighborhoods. To outline the proof, use tubular neighborhoods to extend the bundle identification smoothly. The two pulled-back symplectic forms agree as bilinear forms along the zero section. The relative Poincaré lemma writes their difference as $d\alpha$, with $\alpha$ vanishing there. Their linear interpolation ${\omega }_t$ remains nondegenerate on a sufficiently small neighborhood. Solvingand integrating $V_t$ gives the relative Moser theorem; the flow fixes the submanifold and carries one form to the other.

For the symplectic fiber sum along a square-zero surface, the self-intersection is the Euler number of the oriented rank-two normal bundle. Thus each normal bundle is trivial. Rescale one ambient symplectic form if necessary to equalize the areas of the two copies of $C$. The Moser theorem for the surface then supplies a base identification preserving their area forms. The symplectic neighborhood theorem identifies each neighborhood with a product carryingRemove small disk neighborhoods and glue collars with the base identification andfor a suitable positive constant $a$. Both signs reverse, so the normal two-forms match; the base forms already match. They define a closed nondegenerate two-form across the neck, agreeing with the ambient forms elsewhere. This constructs the symplectic sum. Its construction uses the area normalization and the chosen normal-bundle gluing.

For a smooth counterexample, take $X=Y={\mathbb{CP}}^2$, and in a small four-ball in each choose an unknotted sphere bounding a three-ball. These spheres are null-homologous and have self-intersection zero. Their standard smooth fiber sum isIndeed, the complement of the standard $S^2\times D^2$ in $S^4$ is $S^1\times D^3$; doubling these complements gives $S^1\times S^3$, and the two ambient projective-plane summands remain as connected sums. This is the smooth fiber sum along unknotted null-homologous spheres.

In the induced orientation, both summands in $Z={\mathbb{CP}}^2\#({\mathbb{CP}}^2\#(S^1\times S^3))$ have positive positive index of the intersection form. The symplectic connected-sum obstruction therefore rules out a symplectic form: connected-sum vanishing of the Seiberg–Witten invariant of a four-manifold contradicts Taubes nonvanishing theorem. In the opposite orientation the intersection form is negative definite, also ruling out a symplectic form, whose class has positive square. Thus the smooth fiber sum need not be symplectic.

Put $L=L_1\cap L_2$, and let $j:L\hookrightarrow M$ be inclusion. The transverse intersection theorem makes $L$ a smooth submanifold without boundary, withIt is compact because it is a closed subset of either compact input submanifold.

Write ${\nu }_r=TM{|}_{L_r}/T{L}_r$. On $L$, consider the bundle mapIts kernel is $TL$, and it is onto because $T{L}_1+T{L}_2=TM$ at every intersection point. Thus the normal bundle of a transverse intersection has the actual bundle isomorphismTransfer the direct-sum complex structure $J_1\oplus J_2$ through this isomorphism. This supplies a complex structure on the normal bundle itself, not merely after stabilization.

Geometrically, multiply the two classes by forming their external product in $M\times M$ and pulling it back along the diagonal $\Delta :M\to M\times M$. The transversality hypothesis is precisely what makes the inverse image of $L_1\times L_2$ under $\Delta$ the submanifold $L$. The normal bundle of this inverse image is the direct sum just calculated. Multiplication of the corresponding Thom classes agrees with the Thom class of this direct sum, so the complex cobordism product represented by transverse intersections isIf the inputs have complex normal ranks $a,b$, the product has degree $2(a+b)$. If they are disjoint, the representative is empty and the product is zero.

For a real or complex vector bundle $E$, choose a fiber metric. Its Thom space is the pointed quotientcollapsing the sphere bundle to one basepoint. When the base is compact, this is the one-point compactification of the total space: fiberwise radial compression identifies the total space with the interior of $D(E)$, and all directions approaching its boundary become the single point at infinity.

Decompose ${\mathbb{C}}^{n+m+2}=V\oplus W$ with $V={\mathbb{C}}^{n+1}$, $W={\mathbb{C}}^{m+1}$, so the subspace being collapsed is $\mathbb{P}(V)={\mathbb{CP}}^n$. Every line outside it has a representative $(v,w)$ with $w\ne 0$. Its projection to $W$ gives the line $\ell =\mathbb{C}w\in \mathbb{P}(W)={\mathbb{CP}}^m$, and there is a unique complex-linear map $A:\ell \to V$ such that $A(w)=v$. The line in $V\oplus W$ is precisely the graph of $A$. This construction is unaffected by replacing $(v,w)$ with $(\lambda v,\lambda w)$.

Let $\gamma ={\eta }_1$ be the tautological line over $\mathbb{P}(W)$. The graph construction and its inverse are continuous, indeed smooth in the usual affine charts, and identifyA Hermitian metric gives a complex-linear bundle isomorphism $\bar{\gamma }\cong {\gamma }^{\ast }$. With the Hermitian form linear in its first argument, it sends $\bar{v}$ to the functional $w\mapsto ⟨w,v⟩$. Linearity in the conjugate vector is exactly what makes this a complex, rather than merely real, isomorphism. Thus the bundle is $({\bar{\eta }}_1{)}^{\oplus (n+1)}$, with the conjugation bars present in the original PDF.

Collapsing the closed subspace ${\mathbb{CP}}^n$ in compact Hausdorff space ${\mathbb{CP}}^{n+m+1}$ gives the one-point compactification of its open complement: neighborhoods of the collapsed point have compact complements, and conversely every compact subset of the complement is closed in the ambient space. The graph homeomorphism therefore extends over the added point. This proves the projective-space quotient as a Thom space of conjugate tautological lines:

The graph coordinates also determine the normal complex structures requested in the second part. The copy $\mathbb{P}(W)$ corresponds to the zero section of the graph bundle, whose normal bundle is that bundle itself. For a coordinate embedding ${\mathbb{CP}}^k\subset {\mathbb{CP}}^n$, use $W={\mathbb{C}}^{k+1}$ and its complementary coordinate space $V={\mathbb{C}}^{n-k}$. Then the normal bundle of a linear complex-projective embedding isThe complex structure is the linear structure in these graph coordinates. Equivalently, applying the preceding homeomorphism with its parameters $n-k-1$ and $k$ identifies the Thom space of this normal bundle with ${\mathbb{CP}}^n/{\mathbb{CP}}^{n-k-1}$, after reordering coordinates. Define ${\nu }_m$ in the identical way. The Hermitian-metric identification with the conjugate line does not change the complex normal orientation.

The original coordinate embeddings may be nested, so first move one into transverse position. Keep ${\mathbb{CP}}^k=\mathbb{P}(A)$ for the first $k+1$ coordinates, and move ${\mathbb{CP}}^m$ to $\mathbb{P}(B)$ for the last $m+1$ coordinates. A unitary coordinate permutation accomplishes this. Every unitary matrix can be joined to the identity by diagonalizing it and varying its eigenvalue phases continuously, so this move is an ambient isotopy. Its trace gives a cobordism carrying the transported normal complex structure, and hence leaves the input cobordism class unchanged.

If $k+m\ge n$, thenTheir projectivizations are transverse and meet in ${\mathbb{CP}}^r$, where $r=k+m-n$. The direct-sum normal structure from question 1 restricts toThe quotient map defining the transverse normal-bundle isomorphism is complex-linear here, so it agrees with the standard normal complex structure and contributes no orientation sign. If $k+m<n$, the first and last coordinate subspaces have no nonzero vector in common, so their projectivizations are disjoint.

Consequently the product of linear projective subspaces in complex cobordism isThe product degree is $2(2n-k-m)$, equal to the real codimension of the stated intersection when it exists. In the equality case $k+m=n$, the representative is one point with its complex-oriented normal space.

Choose a complex projective line in each Complex projective plane summand. Make the connected sum at small disks on these lines, so the lines themselves join through the neck to form a smoothly embedded sphere $S\cong S^2$. Its class is the sum $h_1+h_2$ of the two line classes. The intersection form is diagonal with entries one, henceThe oriented normal bundle of $S$ is a rank-two real bundle with Euler number two, because its Euler number equals the self-intersection number. Oriented plane bundles over $S^2$ are classified by that integer, so this normal bundle is isomorphic to $T{S}^2$. The boundary of a closed tubular neighborhood $N$ is consequently the unit tangent bundle of $S^2$.

A unit tangent pair $(x,v)$ determines the oriented orthonormal frame $(x,v,x\times v)$ in ${\mathbb{R}}^3$. Thus $UT{S}^2\cong SO(3)\cong {\mathbb{RP}}^3$, the latter identification coming from the double covering by unit quaternions. Therefore $\partial N$ is the required smoothly embedded ${\mathbb{RP}}^3$. Its complement is disconnected: the nonempty interior of $N$ and the nonempty exterior $X\setminus N$ are disjoint open subsets of $X\setminus \partial N$ that exhaust it. The neighborhood can be chosen small enough that its exterior is nonempty. This is the construction of separating projective three-space in a positive connected sum.

Write $M{U}^{\ast }={\Omega }_U^{\ast }$ for complex cobordism. A complex orientation of a smooth map is chosen stable complex normal data, not a complex structure on either manifold separately. More concretely, replace the continuous map by a homotopic smooth map $f_0$, choose an embeddingand give its normal bundle ${\nu }_j$ a stable complex structure on a real vector bundle. The auxiliary dimension can be chosen so the normal rank is even; further stabilization makes the normal bundle actually complex. Equivalence means stabilization by trivial complex summands and homotopy of the oriented embedding data. The stable identityidentifies this as orientation of the relative stable normal bundle. For odd relative codimension one uses an odd auxiliary suspension coordinate; the degree formula below is unchanged. This is the geometric factorization model for complex-oriented maps used in Algebraic and Geometric Topology 18, geometric cobordism discussion.

Let $d=\dim Y-\dim X$ and $r={\mathrm{rank}}_{\mathbb{R}}{\nu }_j=d+N$. The chosen normal orientation gives a Thom class $u_{{\nu }_j}$ of degree $r$ and the Thom isomorphism in complex cobordism. For compact manifolds the Pontryagin-Thom collapse of a tubular neighborhood to its Thom space isFor $a\in M{U}^q(X)$, define the Gysin map in complex cobordism by the compositeThus $f_{!}$ raises degree by the real relative codimension $d$. This definition depends on the specified orientation and is invariant under the stated equivalences of its normal data. The plus sign on a space denotes adjoining a disjoint basepoint.

For the embedding $i:L\hookrightarrow M$ with normal rank $2n$, let $c:M_{+}\to \mathrm{Th}(\nu )$ be its tubular collapse. The Thom construction gives $i_{!}(1)=c^{\ast }{u}_{\nu }$. Restricted to $L$, this collapse is exactly the zero section $z:L\to \mathrm{Th}(\nu )$. ThereforeThis is the self-intersection formula in complex cobordism. For an actual complex rank-$n$ normal bundle its top Chern class in complex cobordism is the zero-section pullback of its Thom class, so the right side is $c_n^{MU}(\nu )$. Equivalently, under the splitting principle for complex vector bundles both are the product of the first Chern classes of the line summands. The same argument gives $i^{\ast }{i}_{!}(a)=a{e}_{MU}(\nu )$ for every $a$.

There is a real distinction if the printed word “stable” is interpreted literally. The Euler class of a stably complex real bundle need not equal $c_n^{MU}$ of its stable complex class. For the diagonal embedding $S^4\hookrightarrow S^4\times S^4$, the normal bundle is $T{S}^4$. It has a stable complex structure withso its stable complex rank-two class is trivial and has second Chern class zero. However the ordinary integral image of $i^{\ast }{i}_{!}(1)$ is $e(T{S}^4)$, which evaluates to the Euler characteristic $2$. Thus the always-valid answer is the cobordism Euler class. The requested top-Chern identification is valid for an actual complex normal bundle, or if “top Chern class” here is explicitly being used as the Euler-class terminology for the chosen normal orientation.

Write $A=E\times \{p\}$, $B=\{p\}\times E$ and $D={\Delta }_E$. Their self-intersection numbers vanish: each factor curve has trivial normal bundle, and the self-intersection of the diagonal of a curve is $2-2g(E)=0$. Each pair meets transversely once. Their intersection pairing is thereforeIf a rational linear combination of their cycle classes vanished in $H^2(E\times E,\mathbb{Q})$, intersecting it with $A,B,D$ would multiply its coefficients by this invertible matrix and give zero. All coefficients must vanish, proving the three classes are linearly independent.

The cycle class map detects the Chow Künneth failure for an elliptic self-product. The codimension-one image of the external product of Chow classes consists of sums from ${\mathrm{CH}}^1(E)\otimes {\mathrm{CH}}^0(E)$ and ${\mathrm{CH}}^0(E)\otimes {\mathrm{CH}}^1(E)$. Here ${\mathrm{CH}}^0(E)=\mathbb{Z}[E]$, and every divisor's cohomology class is its degree of a divisor times the class of a point. Thus the image's cohomology is contained in the span of $[A],[B]$. The diagonal's independent class lies outside that span, so its Chow class lies outside the external-product image. Therefore the natural map of Chow groups is not surjective.

An irreducible complex analytic hypersurface is a closed analytic subset of pure complex codimension one which cannot be expressed as the union of two proper closed analytic subsets. Here a local defining function of a complex analytic hypersurface means a reduced local equation: a holomorphic germ generating its vanishing ideal, with each local branch occurring once. The word reduced matters. Merely prescribing the zero set would allow both $f$ and $f^2$, and the requested uniqueness would then be false.

The local facts used are these: in holomorphic coordinates the local ring ${\mathcal{O}}_{X,p}$ is $\mathbb{C}\{z_1,\dots ,z_n\}$, a regular local ring and a unique factorization domain; its units are exactly the germs nonzero at $p$; and the analytic Nullstellensatz identifies the zero-set ideal of a germ $f$ with $\sqrt{(f)}$. A reduced germ is a product of distinct irreducible factors. Unique factorization makes its principal ideal radical: if $f\mid h^r$, every irreducible factor of $f$ divides $h$, so $f\mid h$. Thus two reduced germs with the same hypersurface germ have $(f)=(g)$. Write $g=uf$, $f=vg$. The local ring is a domain, so $uv=1$. Hence $u$ is a holomorphic unit and $u(p)\ne 0$. This proves reduced local defining functions differ by a holomorphic unit. Global irreducibility does not imply local irreducibility at a singular point; the same argument uses the product of all local branches and therefore covers that case too.

A divisor on a complex manifold is a locally finite integer combination of irreducible analytic hypersurfaces. On a compact manifold only finitely many have nonzero coefficients. Choose local meromorphic equations $f_i$ for a divisor $D$, by products of powers of reduced local equations. Their ratios are holomorphic units on overlaps. For the holomorphic line bundle associated to a divisor, choose frames satisfyingThe ratios obey the cocycle identity and so define a holomorphic line bundle $[D]$. This convention makes the local meromorphic sections $f_i{e}_i$ agree, with divisor $D$. Changing the equations by units changes frames and gives an isomorphic bundle.

For hyperplanes $H_1,H_2$ in Complex projective space, choose nonzero linear equations ${\ell }_1,{\ell }_2$. The degree-zero quotient $r={\ell }_1/{\ell }_2$ is a global meromorphic function and has divisor $H_1-H_2$. A principal divisor on a complex manifold gives a trivial associated bundle: using the same global meromorphic equation on every chart makes all frame transitions equal to one. The allowed tensor-product rule now givesIf the hyperplanes coincide the same conclusion is immediate. Equivalently, their common bundle is the hyperplane bundle $\mathcal{O}(1)$.

Locally at the centre of a blowup of a complex manifold at a point, replace a coordinate neighbourhood in ${\mathbb{C}}^n$ byAway from zero the direction is uniquely $[z]$, so this map is an isomorphism there; glue it to $X\setminus \{p\}$. In the chart ${\ell }_j\ne 0$, write $z_j=u$ and $z_i=u{t}_i$ for $i\ne j$. These are smooth holomorphic coordinates on the blowup. The exceptional divisor is $u=0$, globally $E={\sigma }^{-1}(p)\cong \mathbb{P}(T_{p}X)$.

The canonical bundle formula for a point blowup isThe local chart also explains the exponent: pulling back a coordinate volume form gives $\pm u^{n-1}du\wedge {\bigwedge }_{i\ne j}d{t}_i$, so the Jacobian section has divisor $(n-1)E$.

Since $Y$ is smooth at the centre, choose coordinates with $Y=\{z_1=0\}$. Its strict transform is a genuine analytic hypersurface. In a blowup chart $j\ne 1$, it is the closed set $t_1=0$: off $E$, the original equation $u{t}_1=0$ reduces to $t_1=0$, and taking the closure restores the points with $u=0$. In chart $j=1$, there are no such off-exceptional points, and the strict transform is empty there. These descriptions agree on overlaps and show analyticity, with $\tilde{Y}\cap E\cong {\mathbb{CP}}^{n-2}$. The total transform has the extra factor $u$ exactly once, soUnder the given anticanonical assumption, this yieldsTo prove the only-if direction, one must rule out an accidental trivial power of $[E]$. The local blowup is the tautological line bundle over ${\mathbb{CP}}^{n-1}$, with $E$ its zero section. Its normal bundle is therefore ${\mathcal{O}}_E(-1)$. The normal bundle of a smooth analytic hypersurface identifies $[E]{|}_E$ with this bundle. If $n\ge 2$, restrict to a projective line inside $E$; the restriction of $[E{]}^{\otimes (2-n)}$ has degree $n-2$. A trivial line bundle has degree zero, so it is trivial only if $n=2$. Conversely the displayed tensor factor is literally trivial when $n=2$. This is the anticanonical strict transform under a point blowup criterion:For completeness, dimension one cannot satisfy the premise in this compact setting: an irreducible analytic hypersurface is one point, and $[-Y]$ has degree $-1$ on its curve component, whereas $K_X$ has degree $2g-2$. Thus no exceptional dimension-one case invalidates the equivalence.

The geometric argument uses the Mixed Heegaard Floer invariant defined in Question 5(iii). First suppose $g(\Sigma )>0$; the genus-zero qualification is discussed below. Write $r=\Sigma \cdot \Sigma \ge 0$. Blow up $r$ distinct points of the surface. Its proper transform $\tilde{\Sigma }$ in $\tilde{M}=M\#r{\overline{\mathbb{CP}}}^2$ has classChoose the extension of the Spin-c structure withSince $E_j^2=-1$, its exceptional evaluations are $-1$, so the Blowup formula for the mixed Heegaard Floer invariant has no extra power of $U$. Thus ${\Phi }_{\tilde{M},\tilde{\mathfrak{s}}}\ne 0$, andThis calculation is why the self-intersection appears with a plus sign.

It remains to bound the Chern pairing for a square-zero surface $S$. Its oriented normal bundle has Euler number zero and is trivial, so $\partial \nu S=S\times S^1$. Choose an embedded positive-square surface $T$ disjoint from $S$, with positive direction left in its complement, and cut along $N=\partial \nu T$. To see why the choice is possible, impose orthogonality to $[S]$ in the intersection form. A nonzero isotropic class uses at most one positive direction; $b_2^{+}\ge 3$ therefore leaves at least two. Choose an integral positive-square class in that complement, represent it by a surface, and remove algebraically cancelling intersections with $S$ by tubing along $S$. This alters genus but not the class. The neighborhood of $T$ supplies one positive direction and the complement supplies another. Since its normal Euler number is nonzero, $H^1(\nu T)\to H^1(N)$ is onto, proving that the cut is admissible.

Put the outgoing puncture inside $\nu S$, so that the plus Heegaard Floer cobordism map on the side containing $S$ passes through the Floer group of $\partial \nu S$. On the chain level this means splitting the cobordism along that boundary and composing its handle maps; with the fixed determinant class, every intermediate generator has restriction $\mathfrak{t}=\tilde{\mathfrak{s}}{|}_{S\times S^1}$. If the Chern pairing exceeded $2g(S)-2$, the given completed minus group would be zero. As in Question 3, this is the completed convention: for a non-torsion structure, ordinary infinity may retain $(1-U^N)$-torsion, whereas $1-U^N$ is invertible after completion. Such a pairing is positive for $g\ge 1$, so $\mathfrak{t}$ is non-torsion and the completed infinity group ${\mathbf{HF}}^{\infty }(S\times S^1,\mathfrak{t})$ is zero as well. The completed fundamental long exact sequence, whose plus group is the usual plus theory, then makes that plus group zero. The Heegaard Floer cobordism map, and consequently the Mixed Heegaard Floer invariant, would vanish, contradicting nonvanishing. ThereforeSubstituting the preceding blowup calculation provesReversing the orientation of the surface gives the usual absolute-value form of the Adjunction inequality for the mixed Heegaard Floer invariant.

The printed assumptions omit the positive-genus qualification. With the actual Floer groups, the asserted three-dimensional vanishing is false for $S=S^2$: the torsion structure on $S^2\times S^1$ has two nonzero minus towers, although its pairing is $0>-2$. Correspondingly, a small sphere embedded in a ball of a K3 surface has $[S]=0$, $S^2=0$ and Chern pairing zero, while the canonical Mixed Heegaard Floer invariant is nonzero; the desired inequality would read $0\le -2$. Thus the displayed proof establishes the intended positive-genus statement. If the supplied vanishing is instead taken literally as a formal additional axiom for every genus, the same factorization would force the inequality even for a sphere, but that axiom contradicts the actual $S^2\times S^1$ computation.

The canonical bundle is $K_M={\bigwedge }^m(T^{1,0}M{)}^{\ast }$, the holomorphic line bundle of holomorphic top forms on a complex $m$-manifold. For an embedded complex submanifold $V$ of dimension $n$, the holomorphic normal bundle is the quotientTaking determinants and duals gives adjunction for a smooth submanifold:For a closed smooth hypersurface, $[V]={\mathcal{O}}_M(V)$ is its divisor line bundle, equipped with a holomorphic section whose zero divisor is $V$ with multiplicity one. If $f_{\alpha }$ locally defines $V$, local meromorphic generators $1/f_{\alpha }$ transform by the holomorphic units $f_{\alpha }/f_{\beta }$. The stated normal-bundle relation isEquivalently the conormal is ${\mathcal{O}}_M(-V){|}_V$. The closed-divisor interpretation is needed to define $[V]$; an arbitrary nonclosed local submanifold does not define that global divisor bundle.

For a holomorphic subbundle $F\subset E$, define $D_{F}s=\pi D_{E}s$ on sections of $F$, with $\pi$ acting on the bundle factor of a bundle-valued form. It obeys the Leibniz rule. Since $F$ is holomorphic, $D_E^{0,1}s={\bar{\partial }}_{E}s$ lies in $F$, so $D_F^{0,1}={\bar{\partial }}_F$. For $s,t\in F$, their inner products with the normal components of $D_{E}s,D_{E}t$ vanish; hence projecting preserves metric compatibility. The uniqueness proved for the Chern connection identifiesThis is the projected Chern connection.

Choose the adapted smooth unitary frame and retain Question 2's input-index convention. The connection matrix is skew-Hermitian. Its upper-left block is ${\theta }_1$ for $D_F$; writing the lower-left block as $-A$ forces the upper-right block to be ${\overline{A}}^t$:Holomorphicity of $F$ makes the normal part of $D_E^{0,1}s$ vanish, so $B={\overline{A}}^t$ has type $(1,0)$, and $A$ has type $(0,1)$. The matrix $B$ represents the second fundamental form of a holomorphic subbundle in this input-index convention. These types would appear in the opposite blocks with coefficient-column indices; they must not be swapped while retaining the printed minus-sign Cartan equation.

Take the upper-left block of the Cartan curvature equation with input indices. It isTherefore the curvature formula for a holomorphic subbundle isThe sign and the order of the factors both follow from the displayed block multiplication.

Now let $M$ be a complex torus and $V$ a closed smooth hypersurface. The torus has a translation-invariant flat Kähler metric, and $T^{1,0}M$ and $K_M$ are holomorphically trivial. Give $E=T^{1,0}M{|}_V$ its flat metric and apply the preceding formula to $F=T^{1,0}V$. Since ${\Theta }_E=0$,For a $(1,0)$ tangent vector $X$, the latter coefficient is $\mathrm{tr}(B(X)B(X{)}^{†})\ge 0$, so $i{\Theta }_{K_V}$ is a semipositive real $(1,1)$ form. Thus $K_V$ is a semipositive holomorphic line bundle. This proves the semipositive canonical bundle of a submanifold of a complex torus assertion directly from curvature. By adjunction for a smooth submanifold, $K_V\cong [V]{|}_V$.

Restrict a positive metric on $L$ to $V$. The tensor-product metric onhas curvature $i{\Theta }_{L{|}_V}+ai{\Theta }_{K_V}$, which is positive for every nonnegative integer $a$. HenceThe integer condition is implicit in a tensor power. No assertion of strict positivity of $[V]{|}_V$ is required.

For vanishing, write $E_a=L\otimes [V{]}^{\otimes a}$. The allowed Kodaira vanishing theorem says that on a compact Kähler manifold $X$, $H^i(X,K_X\otimes P)=0$ for $i>0$ when $P$ is a positive holomorphic line bundle. Since $K_M$ is trivial, its application on the torus gives $H^i(M,E_0)=H^i(M,L)=0$ for $i>0$.

For every $a\ge 1$, the divisor section gives the exact sequencewhere $i:V\hookrightarrow M$. Crucially,and $P_{a-1}$ is positive by the proved restriction statement. The induced Kähler metric makes $V$ compact Kähler, so Kodaira vanishing theorem on $V$ gives $H^i(V,E_a{|}_V)=0$ for all $i>0$. In the long exact sequence in sheaf cohomology, the segmentthen has zero outside terms by induction. ThereforeThis proves vanishing for divisor twists on a complex torus using restriction positivity and the divisor exact sequence, rather than presuming ambient positivity of every divisor twist.

Euler number and self-intersection. Orient the rank-two normal bundle ${\nu }_S$ by the orientations of $S$ and $M$: the ordered tangent and normal spaces have the ambient orientation. Choose a smooth normal section $s$ transverse to the zero section. Its zeros are isolated, and their signed number is the evaluation $⟨e({\nu }_S),[S]⟩$ of the Euler class on the fundamental class. Scale the section sufficiently small to lie in a tubular neighborhood. Its graph is a push-off $S^{\prime }$ isotopic to $S$.

The intersections of $S$ with $S^{\prime }$ occur exactly at the zeros of $s$. In a positively oriented local splitting $TS\oplus {\nu }_S$, the basis formed from the tangent spaces to $S$ and to the graph has block matrixIts determinant is $\det Ds$, so the local intersection sign is precisely the local zero index of the section. Summing proves the Euler number equals self-intersection identity

The conic. The given map is the degree-two Veronese map $[u:v]\mapsto [u^2:uv:v^2]$ followed by the projective linear transformationThe Veronese map is injective with nonvanishing differential: on the chart $u\ne 0$ its coordinates are $[1:t:t^2]$, and on $v\ne 0$ they are $[s^2:s:1]$. It is therefore a smooth embedding, and $S$ is a smooth plane conic diffeomorphic to ${\mathbb{CP}}^1\cong S^2$. Its equation in the given coordinates is $z_0^2-z_2^2=4z_1^2$.

Let $L$ be a projective line, with its complex orientation. The cohomology ring of complex projective space gives $[L]\cdot [L]=1$ in ${\mathbb{CP}}^2$. The hyperplane $z_1=0$ meets $S$ at $[u:v]=[1:0]$ and $[0:1]$. The zeros of $uv$ are simple in the respective local coordinates, and complex intersections have positive signs. Hence $[S]\cdot [L]=2$, so $[S]=2[L]$. The self-intersection number and the normal Euler class arewhere $u\in H^2(S^2;\mathbb{Z})$ has $⟨u,[S^2]⟩=1$.

The boundary of the tubular neighborhood. Write $N=\nu (S)$ for the closed disk tubular neighborhood and $E=\partial N$ for its boundary. This distinguishes the disk neighborhood $N$ from the vector normal bundle ${\nu }_S$. The space $E$ is the oriented circle bundle of ${\nu }_S$, with Euler class $4u$. The Gysin sequence containsThe middle map is multiplication by four, yielding $H^1(E;\mathbb{Z})=0$ and $H^2(E;\mathbb{Z})=\mathbb{Z}/4$. The same Gysin sequence gives $H^0(E;\mathbb{Z})=H^3(E;\mathbb{Z})=\mathbb{Z}$. The total space is a closed connected oriented three-manifold, so Poincare duality gives