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In the product $(S^{2d}{)}^{2\ell }$, for $d,\ell \ge 1$, vary the factors indexed by an $\ell$-element subset $I$ and fix points in the other factors. Orient the resulting submanifold $S_I$ in increasing factor order. These homology classes form an integral middle-dimensional basis by the Künneth theorem. Complementary subsets give a transverse intersection at one point, with positive sign since all factor dimensions are even. For noncomplementary subsets some factor is fixed in both representatives; moving that point makes them disjoint. In particular every basis class has zero self-intersection number. Complementary pairs give hyperbolic blocks in the intersection form.

For an oriented closed surface smoothly embedded in an oriented four-manifold, a transverse section of the normal bundle gives a homologous small push-off. Its signed zeros are exactly the signed intersections with the original surface. This identifies the evaluation of the Euler class with the self-intersection number.

Successive intersection with Cartier divisors gives a cycle class. On an $n$-dimensional complete scheme, its degree is multilinear and depends only on numerical divisor classes; the top product with one divisor is its self-intersection number.

A line bundle is nef when its degree on every integral complete curve is nonnegative. Equivalently a Cartier divisor $D$ is nef when $D\cdot C\ge 0$ for every such curve. On a projective scheme, adding a positive ample class to a nef class makes it ample; taking limits in the intersection product shows that a nef divisor has nonnegative top self-intersection number.

A negative curve is an irreducible algebraic curve on a smooth projective surface whose self-intersection number is negative. The negativity prevents it from moving in a nontrivial effective linear system and constrains how many curves can represent a fixed numerical class.

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 $A=E\times \{p\}$, $B=\{p\}\times E$ 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.

Choose orientations on the two sphere factors whose product is the specified orientation of $X$. Give the four factors of $Y=(S^2{)}^4$ the corresponding orientations in the order $1,2,3,4$. The Künneth theorem has no Tor functor terms here and givesLet $u_i\in H^2(Y;\mathbb{Z})$ be the pullback of the positively normalized cohomology class from factor $i$. The cohomology ring iswith $⟨u_1{u}_2{u}_3{u}_4,[Y]⟩=1$. Its generators commute by graded commutativity of the cup product.

Fix a point in each sphere. For each pair $i<j$, let $S_{ij}$ be the four-dimensional submanifold in which factors $i,j$ vary and the remaining factors equal their fixed points. Orient $S_{ij}$ by the ordered factors $i,j$. The six homology classesgive a coordinate sphere intersection basis by the homology cross product description in the Künneth theorem. Moving a fixed point in one of the other two factors produces a disjoint homologous representative of each $S_{ij}$. Hence every basis element has zero self-intersection number.

More precisely, the intersection pairing isFor complementary pairs the representatives intersect transversely at one point. The sign is positive because swapping two-dimensional factors contributes $(-1{)}^{2\cdot 2}=1$. For pairs with a common index, a factor fixed in both representatives can be displaced to make them disjoint. In the paired order $(12,34),(13,24),(14,23)$, the intersection form is three copies of $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$.

For the diagonal map $\delta :X\to Y$, write $a,b\in H^2(X;\mathbb{Z})$ for the two sphere generators. ThenThe coefficient of $[S_{ij}]$ in $[\Delta ]$ is $⟨u_i{u}_j,[\Delta ]⟩$. Pullback and the preceding cup product relations show that these coefficients are one for $12,14,23,34$ and zero for $13,24$. ThereforeThere are exactly two complementary pairings among these four terms, each counted twice in the square. ThusThis computes both the explicit homology class and its intersection pairing, including all orientation signs.

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

The exterior. Work first with $W={\mathbb{CP}}^2\setminus \mathrm{int}N$. A collar neighborhood of $\partial W$ shows that its interior, the requested ${\mathbb{CP}}^2\setminus N$, has the same homotopy equivalence type: push the boundary a small positive distance into the collar. The Excision theorem and Thom isomorphism theorem identifyOnly degrees two and four are nonzero, each a copy of $\mathbb{Z}$.

In degree four the map $H_4({\mathbb{CP}}^2;\mathbb{Z})\to H_4(N,E;\mathbb{Z})$ sends the ambient fundamental class to the relative fundamental class of $N$. Under the Thom isomorphism theorem this becomes $[S]$. Thus this map is multiplication by one with compatible orientations. The long exact sequence in homology givesso $H_4(W)=H_3(W)=0$.

In degree two, the Thom isomorphism theorem identifies the map $H_2({\mathbb{CP}}^2)\to H_0(S^2)$ with intersection against $[S]$. A projective line intersects the conic twice, so this map is multiplication by two. The long exact sequence in homology isIt yields $H_2(W)=0$ and $H_1(W)=\mathbb{Z}/2$. In degree zero the relative groups vanish, so $H_0(W)\cong H_0({\mathbb{CP}}^2)\cong \mathbb{Z}$. This computes the homology of the complement of a smooth conic:The normal Euler number $4$, the degree-two intersection map and the degree-four restriction map $1$ play different roles; distinguishing them explains why the boundary has first homology $\mathbb{Z}/4$ while the exterior has first homology $\mathbb{Z}/2$.

The divisor is understood to be on $X$; the printed “on $D$” is a typo. Combining asymptotic Riemann–Roch with cohomology growth for nef twists givesSince a nef divisor has nonnegative top self-intersection number, this limit is positive exactly when $D^n>0$. This is the volume of a nef divisor criterion.

Let $Q\subset {\mathbb{CP}}^2$ be a smooth complex conic. Its homology class is $2H$, so its self-intersection number isRescale the Fubini-Study form so that $Q$ and $C$ have the same symplectic area. Their normal bundles have opposite Euler numbers, $+4$ and $-4$, so the symplectic sum can be formed along $Q$ and $C$.

Concretely, remove tubular neighborhoods $\nu (C)$ and $\nu (Q)$ and glue the boundaries by a fiber-reversing bundle map. The symplectic neighborhood theorem supplies the standard models needed for the gluing, and the symplectic-sum construction supplies a symplectic form onThe second piece is a rational homology ball. Thus this operation replaces the neighborhood of the $-4$ sphere by that rational ball and is the symplectic rational blowdown of a minus-four sphere.

A smooth complex degree-two projective plane curve is a projective linear image of the degree-two Veronese map and is diffeomorphic to a sphere. Its class in ${\mathbb{CP}}^2$ is twice the class of a projective line. Its self-intersection number is four, so its oriented normal bundle has Euler class evaluating to four.