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The Complex projective plane is simply connected. If it were the total space of a covering of degree at least two, it would be the universal cover and have a nonidentity deck transformation. Such a transformation acts by $u\mapsto \pm u$ on $H^{\ast }({\mathbb{CP}}^2)=\mathbb{Z}[u]/(u^3)$, giving Lefschetz number 1 or 3. It therefore has a fixed point, impossible for a nonidentity deck transformation.

Poincare duality gives the displayed formula for connected closed oriented four-manifolds. Every integer occurs. A connected sum of $r$ copies of the Complex projective plane and $s$ copies of $S^1\times S^3$ has Euler characteristic $2+r-2s$, because the summand characteristics are three and zero. Given any integer $h$, choose $s$ large enough that $r=h-2+2s\ge 0$; when both counts are zero use $S^4$. If simple connectivity is additionally required, the possible values instead begin at two, since $b_1=0$; connected sums of projective planes realize every such value.

All cohomology below has rational coefficients. First describe the incidence space geometrically. The image of $g$ is $\mathbb{P}(W)\cong {\mathbb{CP}}^1$, inside $\mathrm{Gr}(1,4)\cong {\mathbb{CP}}^3$. For a line $U\subset W$, choosing a plane $V$ containing $U$ is equivalent to choosing the line $V/U$ in ${\mathbb{C}}^4/U$. If $\mathcal{L}$ is the tautological line bundle on $\mathbb{P}(W)$, thenThis is the projective bundle of a rank-three quotient bundle, with fibre ${\mathbb{CP}}^2$. Local frames of the quotient bundle give local product charts. In particular $F_W$ is an oriented real six-dimensional manifold.

For $V\in S_W\setminus \{W\}$, the intersection $V\cap W$ is exactly a line, so $f^{-1}(V)$ is a single point and the inverse map is $V\mapsto (V\cap W,V)$. The intersection line varies continuously, indeed holomorphically, wherever its dimension is constant: in local Grassmannian charts it is the kernel of a matrix of constant rank, whose kernel bases can be chosen locally. Thus $f$ is an isomorphism over the open stratum. At the remaining point,Set $K=R{f}_{\ast }{\mathbb{Q}}_{F_W}$, a derived direct image of sheaves. The proper base change for sheaves theorem identifies its derived stalks with the cohomology complexes of the fibres. Consequently $K$ is the constant rational sheaf in degree zero on $S_W\setminus \{W\}$, andThe cohomology sheaves are therefore finite-rank local systems on the two strata, with no other nonzero degrees. This proves that $K$ is a constructible complex of sheaves and verifies normalization. The open stratum has real dimension six, so the isolated point has codimension six. Its middle perversity is $\bar{m}(6)=2$, and the intersection-complex support and cosupport axioms require stalk vanishing in degrees above two. The fibre calculation verifies the support axiom, as well as vanishing in negative degrees.

For an oriented real six-dimensional manifold, Verdier duality gives $D_{F_W}{\mathbb{Q}}_{F_W}\cong {\mathbb{Q}}_{F_W}[6]$. For a proper map, Verdier duality commutes with derived direct image. Applying these statements here yieldsThe duality identity $i_W^{!}{D}_{S_W}K\cong D_{\{W\}}{i}_W^{\ast }K$ converts this intoHere $i_W^{!}K$ is the costalk, and dualization of rational complexes reverses degrees. Thus its only nonzero groups are $\mathbb{Q}$ in degrees four and six. In particular they vanish for $a\le 3$, which is precisely the cosupport axiom at this isolated six-dimensional singularity. The uniqueness characterization by constructibility, normalization, support and cosupport now identifies $K$ with the unshifted intersection complex of $S_W$. Taking hypercohomology, and using the composition of derived global sections with $R{f}_{\ast }$, givesFinally apply the fibration result allowed in the question to $g:F_W\to \mathbb{P}(W)$. Its base is the simply connected complex projective line, its fibre is the Complex projective plane, and the projective-bundle charts above verify the fibration hypothesis. Hence, as graded rational vector spaces,The resulting Betti numbers and Poincaré polynomial areThis uses $H^{\ast }({\mathbb{CP}}^r)=\mathbb{Q}[h]/(h^{r+1})$ with $\mathrm{deg}h=2$. No assertion that an arbitrary projective fibration has the product cohomology ring is needed.

An $N$-dimensional stratified pseudomanifold has a closed filtration $\varnothing =X_{-1}\subseteq X_0\subseteq \cdots \subseteq X_N=X$ with each nonempty $X_s\setminus X_{s-1}$ an $s$-dimensional manifold, dense top stratum, and $X_{N-1}=X_{N-2}$. Each point of an $s$-dimensional stratum has a stratum-preserving conical neighbourhood ${\mathbb{R}}^s\times cL$, where $L$ is a compact $(N-s-1)$-dimensional stratified pseudomanifold. The definition is inductive, starting with zero-dimensional spaces. There are no singular codimension-one strata; connectedness of links is not required. One may work in the usual triangulable category used for Witt bordism.

A rational Witt space is such a stratified pseudomanifold for which every stratum of odd codimension $2r+1$ has link $L$ satisfying $I{H}_r^{\bar{m}}(L;\mathbb{Q})=0$. No extra condition is imposed at even codimension. A Witt space with collared boundary has Witt interior and Witt boundary, and a stratum-preserving collar $\partial X\times [0,\epsilon )$ with the product stratification. An orientation means an orientation of the regular stratum, inducing the boundary orientation on its collar.

A three-pronged graph, with its central vertex and open edges as strata, is a stratified space but not a one-dimensional pseudomanifold: the central vertex is a singular codimension-one stratum, and deleting it leaves three local branches instead of the two branches of a one-manifold. For a non-Witt pseudomanifold, take the suspension of $T^2$. Its two vertices have codimension three and link $T^2$, whose middle group is $I{H}_1(T^2;\mathbb{Q})=H_1(T^2;\mathbb{Q})={\mathbb{Q}}^2$. It violates the Witt space condition.

The rational Witt bordism classification in dimensions divisible by four states that, for every $k\ge 1$, the middle intersection form induces an isomorphismHere $W(\mathbb{Q})$ is the Witt group of a field, so a nonsingular symmetric form represents zero precisely when it is metabolic, that is, when it has a half-dimensional totally isotropic subspace. The Complex projective plane gives a nonzero example: its middle group is generated by the degree-two class $h$, with $⟨h^2,[{\mathbb{CP}}^2]⟩=1$. Its form is $⟨1⟩$, of signature one; a metabolic form has signature zero. Thus $[{\mathbb{CP}}^2]\ne 0$ in ${\Omega }_4^{\mathrm{Witt}}$.

For a closed oriented Witt space $X$ of dimension $2k+1$, take the closed coneAway from its vertex it is a stratified product with an interval, so all old Witt space conditions persist. At the new vertex the codimension is $2k+2$, which is even; hence its odd-dimensional link $X$ creates no new Witt condition. The collar $X\times (1-\epsilon ,1]$ is explicit, and the product orientation may be chosen to induce the given orientation on $X$. This is a compact oriented Witt space with collared boundary, proving

Now let $\dim X=2$. Its singular strata are zero-dimensional. Their links are compact one-dimensional pseudomanifolds, which have no singular strata and are therefore nonempty finite disjoint unions of circles. A single circle is allowed, since a chosen stratum need not be intrinsically singular. The Witt condition is automatic in dimension two, because singular strata have even codimension.

Construct the normalization of a two-dimensional pseudomanifold concretely. Remove disjoint small conical neighbourhoods of the finitely many zero-stratum points. The remainder is a compact oriented surface with boundary, with one boundary circle for each local branch. Cap every such circle with a separate disk. The result $\tilde{X}$ is a closed compact oriented topological surface. Map each new disk homeomorphically onto its branch of the original cone, sending all the disk centres belonging to one old singular point to that point. Along the disk boundaries this agrees with the identification with the remainder, so it defines a continuous surjection $\pi :\tilde{X}\to X$. It is a homeomorphism off these finitely many centres, in particular over the prescribed open stratum. Every fibre is a point except over a zero-stratum point, where it is the finite set of its branch centres. Thus every fibre is zero-dimensional, as required.

Each component of $\tilde{X}$ bounds an oriented handlebody: a surface of genus $g$ is the boundary of a three-ball with $g$ one-handles attached. Let $H$ be the disjoint union of these handlebodies, oriented so that $\partial H=\tilde{X}$. Formgluing $x\in \partial H$ to $(\pi (x),0)$. The quotient is compact and Hausdorff, and its only remaining boundary is $X\times \{1\}$, with an explicit product collar. Near regular points of the gluing locus, two half-ball neighbourhoods glue to a three-ball; the induced orientations match after choosing the collar orientation in the usual boundary-gluing convention.

It remains to check the singular links, rather than merely assert that this quotient is a Witt space. If an old point $p$ has $r$ circle branches, the new one-dimensional stratum $\{p\}\times (0,1)$ has link the disjoint union of those $r$ circles. This has codimension two and imposes no Witt condition. At $(p,0)$, each branch supplies a half-three-ball from $H$ and another from the collar. The halves glue to a full three-ball; the different branches share only the positive collar axis. Intersecting with a small link therefore gives $r$ copies of $S^2$ identified at one common point. This is a two-dimensional pseudomanifold with zero middle intersection homology: an allowable one-cycle avoids the common point, so decomposes into cycles in the $r$ punctured spheres, each homeomorphic to ${\mathbb{R}}^2$; each such cycle bounds there. Consequently $I{H}_1({\bigvee }_{a=1}^r{S}^2;\mathbb{Q})=0$. This is exactly the Witt condition at the new codimension-three stratum. The boundary has the required product stratification. Hence $Z$ is a three-dimensional Witt space with collared boundary with $\partial Z\cong X$, proving

For the final product assertion, “product” means a nontrivial product with at least two positive-dimensional factors; a single four-dimensional factor would allow the preceding ${\mathbb{CP}}^2$ counterexample. State the standard rational intersection-homology product fact used here: products of Witt spaces are Witt, their intersection cohomology satisfies the Künneth theorem, and the complementary-degree intersection pairing on elementary tensors is the product of the factor pairings, with the sign $(-1{)}^{\mathrm{deg}\beta \mathrm{deg}{\alpha }^{\prime }}$ for $\alpha \otimes \beta$ and ${\alpha }^{\prime }\otimes {\beta }^{\prime }$. Terms whose degrees in a factor do not add to that factor's dimension pair to zero. Regrouping factors reduces total dimension four to cases $1+3$ or $2+2$.

In the $1+3$ case, write $X=A\times B$. The middle intersection cohomology isEach summand is totally isotropic, because the degrees in $A$ are respectively $0+0$ and $1+1$, neither equal to one. They pair perfectly with one another by factor Poincare duality. Thus either summand has half the total dimension and is a Lagrangian for the middle symmetric form.

For dimensions $2+2$, the middle group is the direct sumThe outside summands are isotropic and pair perfectly with one another. Each factor's degree-one pairing is a nonsingular skew-symmetric form, so choose a Lagrangian subspace $L_A\subset I{H}^1(A)$. Inside the central summand, $L_A\otimes I{H}^1(B)$ is isotropic and has half its dimension. It is orthogonal to both outside summands by their degrees. Thereforeis an isotropic subspace of half the full middle dimension. In both cases the form is metabolic. The Witt bordism classification now proves the desired $[X]=0$ in ${\Omega }_4^{\mathrm{Witt}}$. This also proves the assertion for more factors, using the regrouping just described.

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.

The reverse-oriented Complex projective plane has intersection form $(-1)$, so $b^{+}=0$, while $b_1=0$. The Hodge star splits harmonic two-forms into self-dual and anti-self-dual subspaces, and their dimensions are the positive and negative indices of the intersection form. Therefore every harmonic two-form is an anti-self-dual two-form for every chosen metric.

By part (b), every complex line bundle has a unitary connection with harmonic real vector-bundle curvature, so this connection is an ASD connection. Conversely, an ASD connection on a complex line bundle has $dF=0$ and $d\ast F=-dF=0$, so its vector-bundle curvature is harmonic. The uniqueness in part (b) therefore applies to all ASD connections on that fixed complex line bundle. Each complex line bundle has exactly one gauge orbit of ASD connections.

For a CW complex $X$, let $X^k$ be its $k$-skeleton. The cellular chain complex isThe isomorphism uses excision and $H_k(D^k,S^{k-1})\cong \mathbb{Z}$, after choosing an orientation for each cell. In degree zero, take $X^{-1}=\varnothing$ and the free group on the vertices. The differential is the compositeThe long exact sequences of skeleton pairs imply $d_{k-1}{d}_k=0$. The cellular boundary formula describes its coefficient at a $(k-1)$-cell as the mapping degree of the attaching sphere after all other $(k-1)$-cells are collapsed. In degree one this is the signed difference of the two endpoints. The cellular homology theorem identifies the homology of this complex with singular homology: relative homology of successive skeletons is concentrated in their cell dimension, and the exact skeleton sequences leave precisely these kernels modulo images. This is the construction, rather than merely a count of cells.

The filtration ${\mathbb{CP}}^0\subset {\mathbb{CP}}^1\subset {\mathbb{CP}}^2$ gives one cell in each of dimensions zero, two and four, since ${\mathbb{CP}}^k\setminus {\mathbb{CP}}^{k-1}\cong {\mathbb{C}}^k$. There are no odd-dimensional cells, so every cellular differential vanishes. ThusThe complex projective line has one zero-cell and one two-cell. Its product with itself has one zero-cell, two two-cells and one four-cell. Its cellular chain complex again has zero differentials, givingA homeomorphism induces isomorphisms on homology groups. The second groups have different ranks, and therefore

Now orient the Complex projective plane by its complex coordinates. The inclusion of its two-skeleton ${\mathbb{CP}}^1$ identifies the fundamental class of a projective line with a generator $h$ of $H_2({\mathbb{CP}}^2;\mathbb{Z})\cong \mathbb{Z}$: the cellular generator has no incoming three-cell boundary and no outgoing boundary.

Take the projective lines $L_0=\{z_2=0\}$ and $L_1=\{z_1=0\}$. A path of unitary coordinate transformations takes one to the other, so their oriented homology classes are both $h$. They meet in the single point $[1:0:0]$. In its affine chart, let $w=z_1/z_0$ and $v=z_2/z_0$. Their tangent spaces are the complex $w$-axis and complex $v$-axis, so the intersection is a transverse intersection. The ordered real basesconcatenate to the complex orientation of the ambient four-manifold. Thus the local sign is $+1$, andThis computes the intersection form using distinct representatives of the same class, avoiding an ill-defined attempt to count a line intersecting itself as a set. Bilinearity now givesEquivalently, the map $H_2\to \mathrm{Hom}(H_2,\mathbb{Z})$ sending $a$ to $\lambda (a,-)$ is an isomorphism. This is exactly a unimodular intersection pairing, and proves projective lines generate a unimodular intersection form. Reversing the ambient orientation would change the matrix to $(-1)$ and would still be unimodular.

Use the standard complex orientations and the product orientation. Let $a,b\in H^2(S^2\times S^2;\mathbb{Z})$ satisfy $a^2=b^2=0$ and $⟨ab,[S^2\times S^2]⟩=1$, and let $x\in H^2({\mathbb{CP}}^2;\mathbb{Z})$ satisfy $⟨x^2,[{\mathbb{CP}}^2]⟩=1$. These are the cohomology rings of the product of two spheres and the Complex projective plane.

For a map from the Complex projective plane, write $f^{\ast }a=rx$ and $f^{\ast }b=sx$. Naturality of the cup product gives $r^2{x}^2=f^{\ast }(a^2)=0$ and $s^2{x}^2=f^{\ast }(b^2)=0$. Since $x^2$ has infinite order, $r=s=0$. It follows that $f^{\ast }(ab)=0$, so every such map has degree $0$.

In the reverse direction, write $g^{\ast }x=ra+sb$. ThenEvery even mapping degree really occurs. To construct it, identify $S^2$ with ${\mathbb{CP}}^1$ and use the Segre embeddingThe generator $x_3\in H^2({\mathbb{CP}}^3;\mathbb{Z})$ pulls back to $a+b$: restricting to either factor gives a projective line and hence coefficient $1$. By the cellular approximation theorem, $\sigma$ is homotopic to a cellular map. Its four-dimensional domain then maps into the four-skeleton ${\mathbb{CP}}^2$ of ${\mathbb{CP}}^3$. Denote that map by $g_0$; restriction of $x_3$ to ${\mathbb{CP}}^2$ is $x$, so $g_0^{\ast }x=a+b$ and $\mathrm{deg}{g}_0=2$.

For any integer $m$, choose a map $d_m:S^2\to S^2$ of mapping degree $m$. For $m>0$ one may use $z\mapsto z^m$ on the Riemann sphere, for $m<0$ use $z\mapsto {\overline{z}}^{|m|}$, and for $m=0$ use a constant map. Then $g_m=g_0\circ (d_m\times \mathrm{id})$ satisfies $g_m^{\ast }x=ma+b$, so the possible degrees are exactly all even integers, with $g_m$ realizing $2m$. The cellular approximation here deforms this particular map into the skeleton; it does not require a retraction of ${\mathbb{CP}}^3$ onto ${\mathbb{CP}}^2$.

Finally, take the connected sum of oriented manifolds with both summands carrying their standard complex orientations. Its degree-two cohomology has generators $u,v$ withThis follows by choosing the generators supported away from the two balls used to form the connected sum, so mixed cup products vanish while each square gives the common orientation class. WriteIf $\mathrm{deg}h=n$, the three cup product relations sayTaking determinants gives $-(\det P{)}^2=n^2$, which forces $n=0$. Therefore the only possible degree to the connected sum is $0$, realized by a constant map. This is a degree constraint from intersection forms: the indefinite intersection form of the sphere product cannot pull back a definite form with a nonzero degree multiplier.

On $N={\mu }^{-1}(-1/4)$, the diagonal fundamental vector field $X$ is nonzero, since the fixed-point set lies at moment map values $0$ and $-1/2$. The nondegenerate bilinear form ${\omega }_p$ and $d\mu ={\iota }_X\omega$ imply $d\mu \ne 0$, so $-1/4$ is a regular value. By the regular level set theorem, $N$ is a three-dimensional embedded submanifold of the four-dimensional Complex projective plane.

At $p\in N$, set $W=T_{p}N=\ker d{\mu }_p$. Then $\omega (X,v)=0$ for every $v\in W$, so $X\in W^{\omega }$, the symplectic orthogonal complement. This complement is one-dimensional and $d\mu (X)=\omega (X,X)=0$, givingThus the level set is a coisotropic submanifold, neither an isotropic submanifold nor a Lagrangian submanifold. Indeed an isotropic subspace of a symplectic vector space in dimension four has dimension at most two, whereas $\dim N=3$. Its characteristic line field is generated by the diagonal circle action.

For standard unknotted spheres contained in four-balls in $X$ and $Y$, the standard fiber-reversing gluing gives the displayed connected sum. The complement of an unknotted $S^2\times D^2$ in $S^4$ is $S^1\times D^3$, and its double is $S^1\times S^3$. The portions of $X$ and $Y$ outside the four-balls remain as connected summands. Taking both ambient manifolds to be the Complex projective plane gives a smooth fiber sum with no symplectic form, by the symplectic connected-sum obstruction.

An integral intersection form on a finite-rank free abelian group $L$ is unimodular if its adjoint $x\mapsto \lambda (x,-)$ is an isomorphism to the integral dual. In an integral basis, this is equivalent to the pairing matrix having determinant $1$ or $-1$, not merely nonzero determinant. For example, projective lines generate a unimodular intersection form on the Complex projective plane, with matrix $(1)$ in its complex orientation.