Branched-cover correction terms and sliceness

By [F1], $B$ has the homotopy type of a finite simplicial complex. Its integral chain groups are finite free abelian groups. Finite integer row and column operations, using the Euclidean algorithm, give Smith forms for their boundary maps, so each integral homology group is finitely generated. Tensoring these finite matrices with $\mathbb{Q}$ kills exactly their finite cyclic factors and preserves their free ranks. The assumed vanishing of positive rational homology therefore makes both $H_1(B;\mathbb{Z})$ and $H_2(B;\mathbb{Z})$ finite torsion groups. No assertion about a choice of infinitely many generators is needed.

Every $x\in H_2(B;\mathbb{Z})$ is torsion by step 1.1. Apply [F2]: its embedded oriented surface has integer normal Euler number zero. Hence $⟨w_2(TB),{\rho }_{2}x⟩=0$. Thus the evaluation homomorphism of [F3] with $G={\mathbb{F}}_2$ sends $w_2(TB)$ to zero on the entire integral group $H_2(B;\mathbb{Z})$. By exactness, $w_2(TB)$ is the image of a class $v\in {\mathrm{Ext}}^1(H_1(B;\mathbb{Z}),{\mathbb{F}}_2)$. This also covers zero $H_2$, without requiring mod-two $H_2$ to vanish.

The coefficient lift is onto, including two-torsion. Write the finite group $H_1(B;\mathbb{Z})$ as a finite direct sum of cyclic groups $\mathbb{Z}\mathrm{/}n$. From the free resolution $0\to \mathbb{Z}\stackrel{n}{\to }\mathbb{Z}\to \mathbb{Z}\mathrm{/}n\to 0$, applying $\mathrm{Hom}(-,G)$ gives ${\mathrm{Ext}}^1(\mathbb{Z}\mathrm{/}n,G)=G\mathrm{/}nG$. The coefficient reduction therefore induces the explicit map $\mathbb{Z}\mathrm{/}n\to {\mathbb{F}}_2\mathrm{/}n{\mathbb{F}}_2$. If $n$ is odd its target is zero; if $n$ is even its target is ${\mathbb{F}}_2$ and the reduction is onto. Finite direct sums give a surjection ${\mathrm{Ext}}^1(H_1(B),\mathbb{Z})\to {\mathrm{Ext}}^1(H_1(B),{\mathbb{F}}_2)$. When $H_1=0$, both groups are zero and the same conclusion holds. Choose a preimage $\tilde{v}$ of the one class $v$ in step 2.1.

Let $c\in H^2(B;\mathbb{Z})$ be the image of $\tilde{v}$ under the integral UCT injection. Naturality of [F3] under $\mathbb{Z}\to {\mathbb{F}}_2$ identifies its reduction with the mod-two UCT image of $v$, which is exactly $w_2(TB)$ by step 2.1. Thus $c$ is an actual integral lift. In fact it lies in the finite Ext subgroup; no assertion that distinct such lifts label distinct Spin${}^c$ structures is made.

Choose a smooth Riemannian metric by a finite trivializing atlas and a smooth partition, as in the construction of [F1]. Apply [F4] to $B$ with this metric and the integral lift $c$. Its transition-cocycle construction produces the required Spin${}^c(4)$ principal bundle and its equivariant map to the oriented tangent frame bundle. This proves the original existence claim. All possible finite cyclic two-primary factors were retained in step 3.1; neither a vanishing mod-two homology assumption nor a determinant-class bijection was used.

For any second lift $P_1$, define $L(P_0,P_1{)}_x$ to be the set of Spin${}^c$-equivariant maps from $(P_0{)}_x$ to $(P_1{)}_x$ covering the identity map of the common oriented frame fiber. Since both are lifts of that same frame fiber, two such maps differ by a unique element of the central kernel $U(1)$ from [F1]. In local lift trivializations, the transition functions of $P_0$ and $P_1$ project to the same $\mathrm{S}\mathrm{O}(n)$ functions, so their pointwise quotient lies in central $U(1)$; the cocycle identity for each lift makes these quotients a $U(1)$-valued cocycle. Thus the difference fibers form a principal $U(1)$-bundle $L(P_0,P_1)$ over $X$.

Associate to the principal difference bundle in step 1.1 the standard complex line. By [F2], its unit circle bundle is that principal bundle. The diagonal central action defines the twist $P_0\otimes L=(P_0{\times }_{X}L)\mathrm{/}U(1)$; evaluation of each difference map gives a Spin${}^c$-equivariant isomorphism $P_0\otimes L(P_0,P_1)\cong P_1$. If $P_0\otimes L\cong P_0$, its difference bundle is trivial, so the action is free; the construction for arbitrary $P_1$ proves transitivity. Tensor product of line bundles gives the group law and pullback commutes with the fiberwise construction.

By [F3], the isomorphism class of the associated line $L$ is uniquely identified with $c_1(L)\in H^2(X;\mathbb{Z})$, compatibly with tensor product and pullback. Thus the free transitive line-bundle action is precisely a free transitive $H^2(X;\mathbb{Z})$ action and is natural under $f^{\ast }$. If $X$ is a point, $H^2(X;\mathbb{Z})=0$ and the given lift is the only class in its torsor; no existence of an initial lift is asserted. AC is used only through [A1] and [F3].

Since $B$ has finite CW type by [F1] and has zero positive rational homology, each $H_i(B;\mathbb{Z})$ for $i>0$ is finite. Put $a=\mathrm{\mid }{H}_1(B;\mathbb{Z})\mathrm{\mid }$ and $b=\mathrm{\mid }{H}_2(B;\mathbb{Z})\mathrm{\mid }$. The UCT [F3] gives $\mathrm{\mid }{H}^2(B;\mathbb{Z})\mathrm{\mid }=a$ and $\mathrm{\mid }{H}^3(B;\mathbb{Z})\mathrm{\mid }=b$, because the Hom terms from finite groups vanish and ${\mathrm{Ext}}^1(\mathbb{Z}\mathrm{/}m,\mathbb{Z})\cong \mathbb{Z}\mathrm{/}m$. Poincare-Lefschetz duality [F2] gives $\mathrm{\mid }{H}^2(B,Y;\mathbb{Z})\mathrm{\mid }=b$ and $\mathrm{\mid }{H}^3(B,Y;\mathbb{Z})\mathrm{\mid }=a$.

In the pair sequence, $H^1(Y;\mathbb{Z})=0$ because $Y$ is a rational homology sphere and its first integral cohomology is $\mathrm{Hom}(H_1(Y),\mathbb{Z})$. Thus $H^2(B,Y)\to H^2(B)$ is injective. Also $H^3(Y;\mathbb{Z})\cong \mathbb{Z}$, so the map from the finite group $H^3(B;\mathbb{Z})$ to $H^3(Y;\mathbb{Z})$ is zero; exactness makes $H^3(B,Y)\to H^3(B)$ surjective. It follows that the restriction image $M$ has order $a\mathrm{/}b$, and that $H^2(Y)\mathrm{/}M$ has order $a\mathrm{/}b$. Hence $\mathrm{\mid }{H}^2(Y)\mathrm{\mid }=(a\mathrm{/}b{)}^2$ and $\mathrm{\mid }M{\mathrm{\mid }}^2=\mathrm{\mid }{H}^2(Y)\mathrm{\mid }=\mathrm{\mid }{H}_1(Y)\mathrm{\mid }$, the last equality following from UCT and Poincare duality for the rational homology sphere.

For any $m\in M$, choose $u\in H^2(B;\mathbb{Z})$ restricting to $m$, and let $c$ be an integral cocycle representing $u$. Since $H^2(B;\mathbb{Q})=0$, the rational cocycle $c$ is $\delta q$ for a rational $1$-cochain $q$; reducing $q$ modulo $\mathbb{Z}$ gives a cocycle $v\in H^1(B;\mathbb{Q}\mathrm{/}\mathbb{Z})$ with $\beta (v)=u$. It is unique because $H^1(B;\mathbb{Q})=0$. Restriction of this cochain lift gives $\beta (v{\mathrm{\mid }}_Y)=m$. For $Y$, both $H^1(Y;\mathbb{Q})$ and $H^2(Y;\mathbb{Q})$ vanish, so the same coefficient sequence proves that $\beta :H^1(Y;\mathbb{Q}\mathrm{/}\mathbb{Z})\to H^2(Y;\mathbb{Z})$ is an isomorphism. By UCT, $H^1(Y;\mathbb{Q}\mathrm{/}\mathbb{Z})\cong \mathrm{Hom}(H_1(Y),\mathbb{Q}\mathrm{/}\mathbb{Z})$; by Poincare duality, $H^2(Y;\mathbb{Z})\cong H_1(Y;\mathbb{Z})$. Cup evaluation is therefore the perfect character-evaluation pairing under these identifications. If $X,D$ are torsion $1$-cycles with $mX=\mathrm{\partial }F$, the cochain represented by the rational bounding chain $F\mathrm{/}m$, reduced modulo integral cochains, is ${\beta }^{-1}({\mathrm{P}\mathrm{D}}^{-1}[X])$; its evaluation against ${\mathrm{P}\mathrm{D}}^{-1}[D]$ is $F\cdot D\mathrm{/}m$ modulo $\mathbb{Z}$. Thus this cohomological formula agrees with the chain linking form up to the global orientation sign, which does not change annihilators.

If $x\in M$ is the restriction of $w\in H^2(B;\mathbb{Z})$, naturality of cup product [F5] gives ${\lambda }_Y(m,x)=⟨v{\mathrm{\mid }}_Y⌣w{\mathrm{\mid }}_Y,[Y]⟩=⟨v⌣w,i_{\ast }[Y]⟩=0$, since $i_{\ast }[Y]=0$ in $H_3(B;\mathbb{Z})$ by the homology pair sequence: $[Y]$ is the boundary of the relative fundamental class $[B,Y]$. Therefore $M\subseteq M^{\perp }$. For a nonsingular finite pairing on a group $G$, restriction of characters $G\to \mathrm{Hom}(G,\mathbb{Q}\mathrm{/}\mathbb{Z})$ to $\mathrm{Hom}(M,\mathbb{Q}\mathrm{/}\mathbb{Z})$ is onto. To see this, extend a character from a subgroup $H$ one generator $g$ at a time: if the least $n>0$ with $ng\in H$ is fixed, choose $z\in \mathbb{Q}\mathrm{/}\mathbb{Z}$ with $nz=f(ng)$ (possible because multiplication by $n$ is onto), and set $f^{\mathrm{\prime }}(h+kg)=f(h)+kz$; the relation $ng\in H$ makes $f^{\mathrm{\prime }}$ well defined. Since $G$ is finite this process terminates. Hence $\mathrm{\mid }{M}^{\perp }\mathrm{\mid }=\mathrm{\mid }G\mathrm{\mid }\mathrm{/}\mathrm{\mid }M\mathrm{\mid }$. Apply this with $G=H^2(Y;\mathbb{Z})$ and use step 2.1 to obtain $\mathrm{\mid }{M}^{\perp }\mathrm{\mid }=\mathrm{\mid }M\mathrm{\mid }$. Thus $M=M^{\perp }$.

By [F7], choose a Spin${}^c$ structure on $B$. Then [F6] makes the Spin${}^c$ structures on $B$ an affine $H^2(B;\mathbb{Z})$-space. Restriction is equivariant, so the image of this nonempty affine space is one affine coset of the image subgroup $M$. This proves the extension-set assertion. If $M=0$, the same sequence gives $\mathrm{\mid }{H}^2(Y)\mathrm{\mid }=1$, so the trivial subgroup is its own annihilator; the empty-boundary case is excluded by the connected-boundary hypothesis.

The minus chain complex is finite free over ${\mathbb{F}}_2[U]$, with generators $[\mathbf{x},-1]$ for the finitely many intersection points in $\mathfrak{t}$. Its homology is finitely generated and its localization is the infinity homology, since localization is exact and the localized differentials are identical. Thus the graded image $L$ of $H{F}^{-}\to R\rho$ is nonzero. Its degrees are bounded above by the largest degree of a minus-chain generator. The set of integers $j$ with $U^j\rho \in L$ is consequently nonempty, bounded below, and closed under addition of nonnegative integers. If $a$ is its minimum, that set is exactly $\{a,a+1,\dots \}$. Every homogeneous element of the Laurent tower is a monomial multiple of $\rho$, and the image is graded; hence $L=U^a{\mathbb{F}}_2[U]\rho$.

The long exact sequence of the actual short exact sequence $C{F}^{-}\to C{F}^{\mathrm{\infty }}\to C{F}^{+}$ identifies the image in plus homology, with its grading, as $R\rho \mathrm{/}{U}^a{\mathbb{F}}_2[U]\rho$. Its nonzero homogeneous elements have exponents $j<a$, so their least degree is $\mathrm{deg}\rho -2(a-1)$, attained at $U^{a-1}\rho$. Sending the standard tower element $U^k$ to $U^{k+a-1}\rho$ identifies the image with $R\mathrm{/}U{\mathbb{F}}_2[U]$ shifted by this least degree. Multiplying any individual element sufficiently many times by $U$ kills it, while decreasing exponents supplies a preimage under any power of $U$. This proves all the tower assertions.

The pointed-move maps preserve absolute grading and the minus/infinity/plus exact sequence, so they carry $L$ onto the corresponding image lattice and induce grading-preserving tower isomorphisms. Therefore the least degree is independent of the diagram and auxiliary data. Choosing $U^b\rho$ instead changes $a$ to $a-b$ and $\mathrm{deg}\rho$ to $\mathrm{deg}\rho -2b$, leaving that degree unchanged. In the genus-zero diagram of $S^3$ there is one generator $[{\mathbf{x}}_0,0]$ of degree zero, no nonconstant disks, and minus lattice $U{\mathbb{F}}_2[U][{\mathbf{x}}_0,0]$. Here $a=1$, so the least grading is zero.

The two boundary inclusions are rational homology isomorphisms and $H^2(W;\mathbb{Q})=0$. In particular the square of the Chern class is zero, and $\chi (W)=\sigma (W)=0$. The same assertions hold for the reverse cobordism from $Y$ to $S^3$. For each of the fixed induced Spin${}^c$ structures, the constructed infinity map is an isomorphism with degree $(0-2\cdot 0-3\cdot 0)\mathrm{/}4=0$, and the corresponding plus map commutes with the infinity-to-plus projection. This uses the individual fixed-structure conclusion, including torsion extensions.

Apply the supplied commuting-square comparison first to $W:S^3\to Y$ and then to its reverse $Y\to S^3$. The degree-zero isomorphisms give respectively $d(Y,\mathfrak{t})\ge d(S^3)=0$ and $d(S^3)=0\ge d(Y,\mathfrak{t})$. These are exactly the two stated inequalities in both orientations.

The alpha-gamma diagram, and the diagram with those two attaching curves reversed, each have one intersection in each Spin${}^c$ structure. Their infinity differentials are zero: the Maslov index of a self-disk in a torsion Spin${}^c$ structure is $2n_z$, so cannot be one. The minus lattice in each summand is therefore exactly $U{\mathbb{F}}_2[U][x_i,0]$, and its plus image has bottom $[x_i,0]$. Its correction term is the absolute grading of that generator, with no unknown exponent shift.

For the negative-trace map from $S^3$ to $-L(p,1)$, the fixed triangle has index and basepoint multiplicity zero and count one. It sends the sphere generator at exponent zero to $[x_i,0]$. The four-dimensional data are $c_1^2=-k^2\mathrm{/}p$, $\chi =1$ and $\sigma =-1$. Hence its degree is $D_i=(1-k^2\mathrm{/}p)\mathrm{/}4$, and $d(-L(p,1),i)=D_i$. This computes the negative orientation directly.

The cyclically reordered triangle gives the actual fixed-structure map from $L(p,1)$ to $S^3$ and sends $[x_i,0]$ to the sphere generator at exponent zero. It is the same oriented triple four-manifold with the other sphere input filled, so it has the same degree $D_i$. Consequently $0=\mathrm{gr}[x_i,0]+D_i$, and $d(L(p,1),i)=-D_i=((2i-p{)}^2-p)\mathrm{/}(4p)$. This argument uses an actual turned-around triangle map rather than a general grading-duality assertion.

The characteristic labels under orientation reversal or conjugation are paired by $k\mapsto -k$, equivalently $i\mapsto p-i$ modulo $p$. Their squares agree, so the two computations give exactly the asserted negation for the opposite orientation with the conjugate label. For $p=1$ the same computation has $k=-1$ and both correction terms zero.

Choose the filtration ${\mathbb{C}\mathbb{P}}^0\subset P\subset {\mathbb{C}\mathbb{P}}^2$ with $P=\{y=0\}\cong {\mathbb{C}\mathbb{P}}^1$ and ${\mathbb{C}\mathbb{P}}^0=[1:0:0]$. Its successive complements are $\mathbb{C}$ and ${\mathbb{C}}^2$, so the cellular boundary maps vanish for degree reasons and $H_2({\mathbb{C}\mathbb{P}}^2;\mathbb{Z})=\mathbb{Z}[P]$, $H_4({\mathbb{C}\mathbb{P}}^2;\mathbb{Z})=\mathbb{Z}[{\mathbb{C}\mathbb{P}}^2]$; UCT gives free cohomology, and put $h=\mathrm{P}\mathrm{D}[P]$. The projective maps $T_t([x:y:z])=[x:y+tx:z]$ ($0\le t\le 1$) give a homotopy from the inclusion of $P$ to that of $P^{\mathrm{\prime }}=\{y=x\}$. The two lines meet transversely at $[0:0:1]$ with positive sign, since a complex-linear isomorphism between their tangent complex lines and the ambient complex plane has positive real determinant. By [F3], $[P{]}^2=1$. On the charts $s\mathrm{\ne }0$ and $t\mathrm{\ne }0$, the conic map $[s:t]\mapsto [s^2:st:t^2]$ has target coordinates $(u,u^2)$ with $u=t\mathrm{/}s$ and $(v^2,v)$ with $v=s\mathrm{/}t$, respectively; these recover the source coordinate and have injective derivatives. Its image is exactly $xz-y^2=0$: if $x\mathrm{\ne }0$ the point is $[1:y\mathrm{/}x:(y\mathrm{/}x{)}^2]$, and if $x=0$ then $y=0$ and the point is $[0:0:1]$. The same chart descriptions show the zero set is a smooth complex-oriented sphere and the map is an embedding. The line $P$ meets it at the two simple roots $st=0$, both transversely and positively in the complex orientations. Thus [F3] gives $[Q]\cdot [P]=2$; because $[P]$ generates $H_2$, this implies $[Q]=2[P]$, hence $[Q]\cdot [Q]=4$.

The normal Euler number of $Q$ is therefore $4$ by [F4]. Its unit normal circle bundle is the boundary of its tubular disk bundle. Compute its genus-one gluing directly from the conic charts, without assuming an Euler-to-surgery identification. In the chart $x=1$ write the surface as $(y,z)=(u,u^2)$ and use normal coordinate $n=z-y^2$. In the chart $z=1$ use normal coordinate $n^{\mathrm{\prime }}=x-y^2$. Their change is $n^{\mathrm{\prime }}=(z-y^2)\mathrm{/}{z}^2$; differentiating in the normal direction on the conic gives $n^{\mathrm{\prime }}=u^{-4}n$ in the normal line bundle. On the equator $u=e^{i\phi }$ its unit normal circle transition is therefore ${\theta }_N={\theta }_S-4\phi$. Each closed hemisphere has its actual displayed complex normal frame, so its unit circle bundle is $D^2\times S^1$, a solid torus. On their common torus let $m$ be the equator loop at fixed ${\theta }_S$, and $l$ the fiber loop. The southern solid-torus meridian is $m$, and the northern meridian is $m+4l$, since holding ${\theta }_N$ fixed gives that exact slope. Identify the southern solid torus with the standard unknot exterior by $l={\mu }_U$, $m={\lambda }_U$: its fiber is the core circle, while its base disk is the exterior meridian disk. The second filling slope is therefore $4{\mu }_U+{\lambda }_U$, with the global boundary orientation determined by the product identifications. If those identifications reverse the desired global orientation, reverse the orientation of the total space; the statement permits exactly that choice. Thus [F5] identifies the total circle bundle with $L(4,1)$ up to orientation, by an actual meridian calculation. The complement has the opposite boundary orientation to its tubular neighborhood in the complex-oriented projective plane; choose the orientation of $B_2$ as allowed in the Statement so its boundary agrees with the declared lens-space convention.

By [F2], the relative cohomology groups of $({\mathbb{C}\mathbb{P}}^2,B_2)$ are $\mathbb{Z}$ in degrees $2$ and $4$ and zero otherwise, since $Q\cong S^2$. The relative-to-absolute map in degree two sends the Thom generator to $\mathrm{P}\mathrm{D}[Q]$: the normal-Thom sign is $(-1{)}^{2\cdot 2}=+1$ by [F2]. By step 1.1, $[Q]=2[P]$, so this map is multiplication by $2$ under $h=\mathrm{P}\mathrm{D}[P]$. For degree four, let $a$ be the positive generator of $H^2(Q;\mathbb{Z})$, with $⟨a,[Q]⟩=1$. The Thom image of $a$ evaluates on $[{\mathbb{C}\mathbb{P}}^2]$ as $⟨a,[Q]⟩$: excision reduces the evaluation to the normal disk fibers, and the normalized Thom class evaluates $+1$ on each such fiber by [F2]. Thus the degree-four relative-to-absolute map sends a generator to a generator and is an isomorphism.

By [F6], the cohomology long exact sequence of the pair gives $H^1(B_2;\mathbb{Z})=0$, $H^2(B_2;\mathbb{Z})\cong \mathbb{Z}\mathrm{/}2$, and $H^j(B_2;\mathbb{Z})=0$ for every $j\ge 3$: in degree one the map $\mathbb{Z}\to \mathbb{Z}$ is injective multiplication by $2$, in degree two its cokernel is $\mathbb{Z}\mathrm{/}2$ and the next relative group is zero, and in degree four the relative-to-absolute map is an isomorphism. In degree zero, the relative group vanishes, so $H^0({\mathbb{C}\mathbb{P}}^2;\mathbb{Z})\to H^0(B_2;\mathbb{Z})$ is an isomorphism; hence $B_2$ is connected. Repeat the relative calculation over $\mathbb{Q}$: the degree-two map is multiplication by $2$ and the degree-four map is the identity, both isomorphisms, so the pair sequence gives $H^j(B_2;\mathbb{Q})=0$ for $j>0$ and $H^0(B_2;\mathbb{Q})=\mathbb{Q}$. The UCT over $\mathbb{Q}$ then gives the rational homology of a point. Together with step 2.1 this proves that $B_2$ is a rational homology ball bounded by $L(4,1)$.

The forward punctured-cobordism map gives $d(Y,\mathfrak{t})\ge 0$ by [F1].