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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 a map $f:M\to N$ of connected closed oriented four-manifolds, naturality of the cup product givesWhen $\mathrm{deg}f\ne 0$, nondegeneracy makes $f^{\ast }:H^2(N;\mathbb{Q})\to H^2(M;\mathbb{Q})$ injective. Consequently the target intersection form, multiplied by the mapping degree, must embed into the source form. For equal second Betti numbers, a matrix $P$ for pullback satisfies $P^{\mathsf{T}}{Q}_{M}P=(\mathrm{deg}f)Q_N$; taking determinants often rules out nonzero degrees. In particular, a two-dimensional indefinite source form cannot realize a nonzero multiple of a two-dimensional definite target form.

On a connected compact Kähler manifold of complex dimension two, the intersection form restricted to real $(1,1)$ cohomology has signature $(1,h^{1,1}-1)$. The Kähler class spans a positive line, and its orthogonal complement consists of primitive classes, which are negative definite. Thus a totally isotropic subspace of real $(1,1)$ cohomology has dimension at most one.

The intersection form of a 4-manifold is the middle-dimensional signed-intersection pairing of an oriented compact four-manifold. On a closed manifold it can equivalently be defined on second cohomology by evaluating the cup product on the fundamental class. The general intersection form also applies to higher dimensions.

A Kirby diagram depicts a handle decomposition of a four-dimensional manifold using a framed link for the two-handles and dotted circles for the one-handles. The framing specifies the attaching normal directions; it controls the resulting intersection form and the boundary Dehn surgery.

A nontrivial product of positive-dimensional closed oriented Witt spaces of total dimension four has metabolic rational middle intersection form. For dimensions $1+3$, the two Künneth summands in degree two pair only with one another, so either is a half-dimensional isotropic subspace. For dimensions $2+2$, the outside degree-two summands pair with one another, and the $I{H}^1\otimes I{H}^1$ summand has the tensor pairing of two symplectic forms. A Lagrangian in the first symplectic factor tensored with the entire second factor is a half-dimensional isotropic subspace of that summand. Adding one outside summand gives a Lagrangian in the complete middle form. The Witt bordism classification then makes the product null-bordant.

A one-sided section loop and a two-sided fiber loop in the reflection mapping torus have mod-two self-intersections one and zero and meet once. Their intersection form is $Q$. In the evaluation-dual cohomology basis, the Poincare duality pairing has matrix $Q^{-1}$, giving $t^2=0$ and $x^2=tx\ne 0$.

When compact oriented four-manifolds are glued along an entire closed boundary three-manifold, the signature of the result is the sum of their intersection form signatures. It applies to $X_0{\cup }_{\partial X_0}(-X_1)$ with the difference of the two signatures. Gluing along only part of a boundary can require a correction; that is a different situation.

Let $F$ be an oriented Seifert surface of genus $g$ for $K$, and choose a basis $a_1,\dots ,a_{2g}$ of $H_1(F;\mathbb{Z})$. Push a representative curve slightly in the positive normal direction to obtain $a_i^{+}$. The Seifert matrix in our row convention isThe difference $V-V^T$ is the intersection form of F, up to the convention for its orientation; in a symplectic basis it is a direct sum of unimodular two-by-two skew blocks.

Write $X=S^3\setminus \mathrm{int}N(K)$ for the knot exterior. Linking number with K gives the epimorphism ${\pi }_1(X)\to \mathbb{Z}$ sending a positive meridian to one. Let $\tilde{X}$ be the associated infinite cyclic cover, with deck transformation t. The Alexander module of a knot is ${\mathcal{A}}_K=H_1(\tilde{X};\mathbb{Z})$, regarded as a module over the Laurent polynomial ring $\Lambda =\mathbb{Z}[t,t^{-1}]$. Its order, the greatest common divisor of the maximal presentation minors, is the Alexander polynomial of a knot, defined up to $\pm t^k$. Equivalently these minors generate its zeroth Fitting ideal; we will obtain a square presentation, so its determinant gives the order. No assertion that $\Lambda$ is a principal ideal domain is needed.

Here is the Seifert-matrix presentation of the Alexander module with its topological justification. Cut X along F, obtaining a connected manifold Y with two boundary copies $F^{+}$ and $F^{-}$. Up to collars this is the complement of a thickening of F in $S^3$. Alexander duality gives $H_1(Y;\mathbb{Z})\cong {\mathbb{Z}}^{2g}$ and the perfect linking pairing with $H_1(F;\mathbb{Z})$. Choose generators $b_j$ satisfying $\mathrm{lk}(b_j,a_i)={\delta }_{ij}$. This homology statement does not require Y to be a handlebody. The two inclusions satisfyIndeed the first coefficients are the defining positive linking numbers; for the second, moving the two curves to opposite sides and using symmetry of linking gives $\mathrm{lk}(a_i^{-},a_j)=\mathrm{lk}(a_j^{+},a_i)$.

Stack copies of Y, identifying the positive copy of F in one layer with the negative copy in the next. This is $\tilde{X}$. The Mayer–Vietoris sequence, with all translates collected into Laurent modules, containsThe last map is injective because $\Lambda$ is an integral domain. Thus the entire Alexander module is presented by the row relations $tV-V^T$ (transpose the matrix if using column vectors). At $t=1$ its determinant is $\det (V-V^T)=1$, so the determinant is nonzero and the module is a torsion module. Consequentlywhere $\doteq$ denotes equality up to a Laurent unit. The second conclusion follows because a determinant of size $2g$ with entries linear in t has breadth at most $2g$; minimize over Seifert surfaces.

For the displayed diagram, start on the outer lower arc near the lower left undercrossing and follow the arc through the lower outer loop toward the upper right undercrossing. Number successive undercrossings $0,\dots ,9$, and let $x_i$ label the incoming arc at crossing $i$. Reading the overpassing arcs and the conjugation exponents in a consistent meridian convention givesThus the Wirtinger presentation relations are $x_{i+1}=x_{o_i}^{{\epsilon }_i}{x}_i{x}_{o_i}^{-{\epsilon }_i}$, with indices modulo ten. Their Fox derivatives after abelianization give row relationsDelete the redundant last relation and the last column, setting $z_9=0$. This is a nine-generator presentation of the Alexander module. The cofactor determinant is $t^{-6}(t^4+t^3-3t^2+t+1)$; here is a smaller elimination check. Retain $z_1,z_4$. Seven relations solve the other generators asThe two remaining rows, at undercrossings two and seven, are the matrixIts determinant is $t^{-3}(t^4+t^3-3t^2+t+1)$, differing from the original cofactor only by a Laurent unit. ConsequentlyThe factor has value one at $t=1$ and is reciprocal, as required. Its Conway-normalized Alexander polynomial is $t^2+t-3+t^{-1}+t^{-2}$, and its knot determinant is three.

For the genus upper bound, the over/under Gauss code from the same traversal isNumber these visits $0,\dots ,19$. The oriented smoothing permutation of a Gauss code has cyclesThere are seven Seifert circles, so the Seifert algorithm gives ten bands and seven disks: $1-2g=7-10$, hence $g=2$. The Alexander breadth bound on Seifert genus supplies the matching lower bound from breadth four. Therefore $\boxed{g_s(K)=2}$.

Modulo two, the primitive Alexander polynomial becomes $t^4+t^3+t^2+t+1$. It has neither zero nor one as a root, and it is not $(t^2+t+1{)}^2=t^4+t^2+1$, the only possible product of irreducible monic quadratics over ${\mathbb{F}}_2$. It is therefore irreducible over ${\mathbb{F}}_2$, hence over $\mathbb{Q}$ and $\mathbb{Z}$ by Gauss lemma for polynomials. The full-degree irreducible Alexander polynomial implies a prime knot criterion now applies, since its breadth is exactly twice the genus. Thus the knot is prime. The full-degree condition is needed to rule out a nontrivial summand with unit Alexander polynomial.

To test reflection, calculate the Jones polynomial from the same crossing data. The diagram has nine positive crossings and one negative crossing, so its writhe of a link diagram is eight. Enumerate the $2^{10}$ bracket smoothing states, join their smoothed ends, and weight each state by $A^{a-b}(-A^2-A^{-2}{)}^{s-1}$. Correct by $(-A^3{)}^{-8}$ and put $t=A^{-4}$. This givesAs checks, $V_K(1)=1$ and $V_K(-1)=-3={\Delta }_K(-1)$. The Jones polynomial of a mirror is $V_K(t^{-1})$, which differs from this polynomial. Therefore the knot is not equivalent to its reflection. Reversing all crossing conventions reciprocates the displayed Jones polynomial and leaves the conclusion unchanged.

A simply connected manifold is connected and orientable: its orientation character is a group homomorphism ${\pi }_1(X)\to \{\pm 1\}$, and this group is trivial. Choose an orientation. The Hurewicz theorem in degree one gives $H_1(X;\mathbb{Z})=0$. The universal coefficient theorem for cohomology then gives $H^1(X;\mathbb{Z})=0$ andbecause the possible $\mathrm{Ext}(H_1(X),\mathbb{Z})$ term is zero.

The homology groups are finitely generated for a closed manifold. The displayed dual group is therefore free abelian. Poincare duality identifies $H_2(X;\mathbb{Z})$ with $H^2(X;\mathbb{Z})$, so $H_2$ itself has no torsion. Since the second Betti number is one, both groups are isomorphic to $\mathbb{Z}$. This torsion argument is essential: rank one by itself would not rule out finite summands.

Again by Poincare duality, $H^3(X;\mathbb{Z})\cong H_1(X;\mathbb{Z})=0$, and $H^4(X;\mathbb{Z})\cong H_0(X;\mathbb{Z})=\mathbb{Z}$. Thus the only nonzero cohomology groups are $\mathbb{Z}$ in degrees $0,2,4$.

Let $u$ generate $H^2$. The integral intersection formis unimodular. Indeed, cap product with $[X]$ is the Poincare duality isomorphism to $H_2$, and the preceding universal-coefficient identification makes evaluation the full integral dual. Composing these isomorphisms identifies the pairing with an isomorphism $H^2\to \mathrm{Hom}(H^2,\mathbb{Z})$. In rank one its matrix must therefore be $[1]$ or $[-1]$. Consequently $⟨u^2,[X]⟩=\pm 1$, so $u^2$ generates $H^4$.

All products in degrees above four vanish. Hence the assignment of the degree-two generator giveswhich is the cohomology ring of complex projective space for ${\mathbb{CP}}^2$. If the intersection form is negative, this ring isomorphism reverses the chosen top orientation generator; preserving that additional orientation is not part of the requested ring isomorphism.

An index-$\lambda$ handle in dimension $d$ is $D^{\lambda }\times D^{d-\lambda }$. A handle attachment glues its attaching region $S^{\lambda -1}\times D^{d-\lambda }$ to a corresponding neighborhood in the boundary of the preceding manifold with boundary. Besides the embedding of the attaching sphere, one specifies a framing of an embedded sphere, namely the normal-disk directions used in this gluing. Corners are then smoothed.

In dimension four, a two-handle is $D^2\times D^2$ and attaches along $S^1\times D^2$. Its attaching circle is a knot in the boundary of the zero-handle $D^4$. Identify that boundary with $S^3$ and remove a point away from the circles to draw them in ${\mathbb{R}}^3$. Disjoint attaching circles give a framed link. Each integer records the twist of its normal framing relative to the zero Seifert framing; the integer is also the linking number of the circle and its framed push-off. The resulting boundary change is Dehn surgery on a framed link. This is the two-handle part of a Kirby diagram.

For $S^2\times S^2$, use the following explicit handle decomposition. Write each factor as two disks, $S^2=D_{-}^2\cup D_{+}^2$, glued along their boundary circles. The four product pieces giveThe first is a four-dimensional zero-handle after rounding its corners. Its boundary is the union of the solid tori $S^1\times D^2$ and $D^2\times S^1$. The two middle product pieces attach along these tori. Their attaching circles $S^1\times \{0\}$ and $\{0\}\times S^1$ form the standard Hopf link, with linking number one after choosing orientations.

Both framings are zero. For example, the normal disk of $S^1\times \{0\}$ is the second disk factor. Its constant normal push-off has linking number zero with the original circle: the circle bounds $D^2\times \{0\}$ in the four-ball, and a parallel push-off bounds a disjoint translated disk. Thus the framed self-intersection is zero; the same argument applies with the factors interchanged. These product framings are the zero Seifert framings.

After these two two-handles, the remaining product piece $D_{+}^2\times D_{+}^2$ is a four-ball. Its boundary is exactly the boundary of the assembled first three pieces, so it attaches as a four-handle. Equivalently, zero surgery on the Hopf link gives $S^3$, which this handle caps. We have therefore constructed the actual product, provingThere are no one- or three-handles. As a check, the two factor spheres have self-intersection zero and meet once, giving the intersection form $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. The explicit product construction, rather than the intersection form alone, identifies the smooth four-manifold. This is the zero-framed Hopf-link handle decomposition of the sphere product.

Let F be an oriented genus-g Seifert surface for K, and choose a basis $a_1,\dots ,a_{2g}$ of $H_1(F;\mathbb{Z})$. Push a representative curve slightly in the positive normal direction to obtain $a_i^{+}$. The Seifert matrix in our row convention isThe difference $V-V^T$ is the intersection form of F, up to the convention for its orientation; in a symplectic basis it is a direct sum of unimodular two-by-two skew blocks.

Write $X=S^3\setminus \mathrm{int}N(K)$ for the knot exterior. Linking number with K gives the epimorphism ${\pi }_1(X)\to \mathbb{Z}$ sending a positive meridian to one. Let $\tilde{X}$ be the associated infinite cyclic cover, with deck transformation t. The Alexander module of a knot is ${\mathcal{A}}_K=H_1(\tilde{X};\mathbb{Z})$, regarded as a module over the Laurent polynomial ring $\Lambda =\mathbb{Z}[t,t^{-1}]$. Its order, the greatest common divisor of the maximal presentation minors, is the Alexander polynomial of a knot, defined up to $\pm t^k$. Equivalently these minors generate its zeroth Fitting ideal; we will obtain a square presentation, so its determinant gives the order. No assertion that $\Lambda$ is a principal ideal domain is needed.

Here is the Seifert-matrix presentation of the Alexander module with its topological justification. Cut X along F, obtaining a connected manifold Y with two boundary copies $F^{+}$ and $F^{-}$. Up to collars this is the complement of a thickening of F in $S^3$. Alexander duality gives $H_1(Y;\mathbb{Z})\cong {\mathbb{Z}}^{2g}$ and the perfect linking pairing with $H_1(F;\mathbb{Z})$. Choose generators $b_j$ satisfying $\mathrm{lk}(b_j,a_i)={\delta }_{ij}$. This homology statement does not require Y to be a handlebody. The two inclusions satisfyIndeed the first coefficients are the defining positive linking numbers; for the second, moving the two curves to opposite sides and using symmetry of linking gives $\mathrm{lk}(a_i^{-},a_j)=\mathrm{lk}(a_j^{+},a_i)$.

Stack copies of Y, identifying the positive copy of F in one layer with the negative copy in the next. This is $\tilde{X}$. The Mayer–Vietoris sequence, with all translates collected into Laurent modules, containsThe last map is injective because $\Lambda$ is an integral domain. Thus the entire Alexander module is presented by the row relations $tV-V^T$ (transpose the matrix if using column vectors). At $t=1$ its determinant is $\det (V-V^T)=1$, so the determinant is nonzero and the module is torsion. Consequentlywhere $\doteq$ denotes equality up to a Laurent unit. The second conclusion follows because a determinant of size $2g$ with entries linear in t has breadth at most $2g$; minimize over Seifert surfaces.

For the displayed knot we give a reproducible diagram calculation. Label its ten crossings A through J as follows: A and B are the two upper crossings from left to right; C,D,E are the next three crossings in descending order; F is the crossing below E on the almost vertical strand; G is the left crossing of the bottom loop; H,I,J are the remaining three bottom crossings from left to right. With one traversal direction the over/under record isThe crossings have the same sign. Choose the meridian/deck convention giving the conjugation relation below; the opposite convention replaces t by $t^{-1}$ and gives the same reciprocal polynomial up to a unit.

Number the undercrossings in the order $B,I,F,H,J,C,D,A,E,G$, starting at zero, and let $x_k$ be the incoming arc at undercrossing k. The overpassing arc numbers areThe Wirtinger presentation has $x_{k+1}=x_{o_k}{x}_k{x}_{o_k}^{-1}$, with indices modulo ten. Its lifted cellular boundaries, or equivalently Fox derivatives, give the Alexander relation matrixDelete the redundant final relation and the last column: the resulting matrix presents the Alexander module on the nine relative arc generators. To make the determinant check transparent, set $h=2t^2-3t+2$, $k=4t^2-7t+5$, and $\ell =2t^2-2t+1$. Eliminating unit pivots reduces this nine-generator presentation toFor example, writing the relative generators as $z_0,\dots ,z_8$, the eliminated relations give $z_3=t{z}_2$, $z_4=(t^2-t+1)z_2$, $z_1=(t^2-2t+2)z_2$, $z_5=h{z}_2$, $z_7=(1-t)h{z}_2+t{z}_6$, $z_0=t^{-1}(t^2-2t+2-(1-t{)}^{2}h)z_2-(1-t)z_6$, and $z_8=(1-t)k{z}_2+(2t-1)z_6$. The two remaining relations are precisely the rows of N. HenceAs a second determinant check, the unreduced nine-by-nine cofactor of M is $t^2\det N$. The polynomial has value one at $t=1$ and is reciprocal, as required.

For the genus upper bound, the oriented smoothing has seven Seifert circles. This can also be checked from the traversal record: number its twenty entries starting at one; smoothing sends an incoming visit to the entry following its other visit to the same crossing. The cycles are $(1,17,13,19,7)$, $(2,16)$, $(3,11,15)$, $(4,10)$, $(5,9)$, $(6,20,8)$, and $(12,18,14)$. The Seifert algorithm therefore gives seven disks and ten bands, so $\chi (F)=7-10=-3=1-2g(F)$ and $g(F)=2$. The Alexander breadth is four, so the Alexander breadth bound on Seifert genus gives the matching lower bound. Thus $\boxed{g_s(K)=2}$.

Finally, modulo two the primitive polynomial becomes $t^4+t^3+t^2+t+1$. It has neither zero nor one as a root. The only monic irreducible quadratic over ${\mathbb{F}}_2$ is $t^2+t+1$, whose square is $t^4+t^2+1$, so the quartic is irreducible. Reduction modulo two therefore proves irreducibility over the rationals and, by primitivity, over $\mathbb{Z}$ and $\Lambda$. If K were a sum of two nontrivial knots, additivity of Seifert genus would force both summands to have genus one. Their Alexander polynomials multiply, so irreducibility would make one polynomial a unit and the other equal to this breadth-four polynomial, contradicting the breadth-at-most-two bound for genus one. This uses the full-degree irreducible Alexander polynomial implies a prime knot criterion, rather than irreducibility alone. Therefore K is prime.

Use rational coefficients and finite singular chains. For an $N$-dimensional stratified pseudomanifold with filtration $X_j$, the lower middle perversity isA singular simplex $\sigma :{\Delta }^i\to X$ is an allowable singular simplex when, for every $k\ge 2$, ${\sigma }^{-1}(X_{N-k})$ lies in the $(i-k+\bar{m}(k))$-skeleton of ${\Delta }^i$; negative-dimensional skeletons are empty. Let $A_i$ be the span of these simplices in $S_i(X;\mathbb{Q})$. The intersection chains areThe condition on $\partial c$ is imposed after combining equal simplices and cancelling coefficients: the faces of an allowable simplex need not themselves be allowable. If $c\in I{S}_i$, then $\partial c\in A_{i-1}$ and $\partial (\partial c)=0$, so $\partial c\in I{S}_{i-1}$. Thus $I{S}_{\ast }$ is a chain subcomplex of $S_{\ast }$, and $I{H}_i(X)=H_i(I{S}_{\ast })$ defines intersection homology.

Write $v$ for the vertex of $cY$. Its codimension is $2n$, so the vertex condition for an allowable $i$-simplex isIn particular all simplices of degree at most $n$ avoid $v$. For $i<n$, both the degree-$i$ chains and the degree-$(i+1)$ chains that can bound them lie in $cY\setminus \{v\}=Y\times (0,1)$. The two chain complexes therefore have the same cycles and boundaries in degree $i$, proving the low-degree isomorphism.

For $i\ge n$, let $z$ be an allowable $i$-cycle. Cone every simplex of $z$ radially to $v$, using the cone on ${\Delta }^i$ as ${\Delta }^{i+1}$, and write the resulting chain as $cz$. At the vertex, coning an old inverse image in the $(i-n-1)$-skeleton gives a subset of the $(i-n)$-skeleton, and the new cone vertex also lies in this skeleton because $i-n\ge 0$. For every other singular stratum, coning the inverse image increases its permitted skeleton dimension by one, exactly as increasing the simplex dimension does. Thus $cz$ is allowable. The usual cone identity is $\partial (cz)=z-c(\partial z)=z$; since $z$ is allowable, the boundary condition for intersection chains holds as well. Every cycle is consequently a boundary. This proves the intersection homology cone formula here:The assumption $\dim Y=2n-1$ entails $n\ge 1$, so no augmented degree-zero exception arises in this coning argument.

For the suspension of a topological space, the two endpoints are collapsed separately, giving two distinct vertices. Use the following standard computational facts for intersection homology: a product with an interval has the same intersection homology as the original stratified space, via the projection and its stratum-preserving homotopy inverse; and an open cover has a Mayer–Vietoris sequence whose overlap map is the signed pair of inclusion maps. Cover the suspension by its two open cones. Their overlap is $Y\times (-\epsilon ,\epsilon )$. Below degree $n$, the overlap map is, under the interval identifications, $a\mapsto (a,-a)$ in $I{H}_i(Y)\oplus I{H}_i(Y)$, hence is injective with cokernel $I{H}_i(Y)$. In degree $n$ both cone groups vanish and the next overlap map, in degree $n-1$, is still injective. Above degree $n$ both adjacent cone groups vanish. Exactness therefore givesThis differs from ordinary reduced suspension homology: below the cutoff the cycles cannot use either vertex, and in the middle degree coning kills them. If $Y$ is a compact oriented Witt space, its Poincare duality pairs $I{H}_i(Y)$ with $I{H}_{2n-1-i}(Y)$. The displayed groups of the suspension are correspondingly paired in complementary degrees $i$ and $2n-i$, while its middle group is zero. The new vertices have even codimension $2n$ and thus impose no additional Witt space condition; the suspension is again Witt. Its middle intersection form has zero rank and, when its dimension is divisible by four, zero signature.

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.

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.

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.

In the cohomology basis $h_1,\dots ,h_n$ supplied by the projective lines in the summands, the intersection form isFor each permutation $\sigma \in S_n$, the linear map $h_i\mapsto h_{\sigma (i)}$ preserves this pairing. Distinct permutations induce distinct maps, and their compositions agree with permutation composition. Hence the permutation matrices give a subgroup $G\cong S_n$ of the isometry group of $Q_X$.

To realize these isometries smoothly, describe $X$ as a punctured $S^4$ with $n$ punctured copies of ${\mathbb{CP}}^2$ attached to its boundary spheres. An orientation-preserving ambient isotopy of $S^4$ can interchange any two chosen disjoint attachment balls while carrying their parametrizations along with them: move their centers along disjoint arcs and extend the motion to small balls. There is room to make these arcs disjoint in dimension four. Extend the endpoint diffeomorphism across the two exchanged punctured copies using the chosen identifications of those identical oriented summands; use the identity on the remaining copies. Collar coordinates make the maps fit smoothly along the gluing spheres. This exchanges the two line classes without changing their signs. Since transpositions generate $S_n$, composing these diffeomorphisms realizes each element of $G$. Pullback permutes the cohomology basis by the inverse permutation, which still realizes the same subgroup and all its elements. These are permutation diffeomorphisms of identical connected summands.

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.

False. Take $M={\mathbb{CP}}^2\#{\mathbb{CP}}^2$ and $N=S^2\times S^2$, with their standard product and connected-sum orientations. They are closed simply connected four-manifolds. For the connected sum, removing a four-ball leaves the fundamental group unchanged, and the Seifert-van Kampen theorem across the separating three-sphere preserves simple connectivity. The Mayer–Vietoris sequence computes integral cohomology $\mathbb{Z},{\mathbb{Z}}^2,\mathbb{Z}$ in degrees zero, two and four, and zero in other degrees. The Künneth theorem gives the same groups for $S^2\times S^2$. Thus their Betti numbers agree and all their integral cohomology is torsion-free.

Their cohomology rings nevertheless differ. In $M$, let $u,v$ come from the two projective summands and let $z$ generate degree four. The cohomology ring of the connected sum of two complex projective planes hasThese relations follow by using cycles in the separate summands and the degree-one pinch projections onto them. In $N$, the sphere-factor generators $x,y$ satisfyConsequently every degree-two class of $N$ has squarean even multiple of the top generator, whereas $u^2$ in $M$ is a primitive generator. An integral graded-ring isomorphism cannot change this property, since it sends a generator of $H^4$ to its positive or negative. A homotopy equivalence induces a graded-ring isomorphism, so none exists. Equivalently, their intersection forms are $\mathrm{diag}(1,1)$ and $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, an odd form and an even form respectively.