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A four-handle is an index-four handle. In a four-dimensional handle decomposition it is a four-ball attached along its whole boundary, a three-sphere. It caps a spherical boundary component left by earlier handle attachments.

Every finitely presented group is the fundamental group of a closed connected smooth oriented four-manifold. Build a compact four-dimensional handle decomposition with one zero-handle, a one-handle for each generator of a group and a two-handle for each relator. Turning the handle decomposition upside down shows that its boundary surjects on the fundamental group. Its double of a manifold has the same fundamental group by the Seifert-van Kampen theorem.

A smooth oriented cobordism $W:Y_1\to Y_2$ with a Spin-c structure induces $U$-equivariant maps $F_{W,\mathfrak{s}}^{\circ }$ on the corresponding Heegaard Floer homology groups. A handle decomposition defines the maps using generator selection for one- and three-handles and holomorphic triangle counts for two-handles. Composition sums over extensions of the prescribed Spin-c structures. When the boundary determinant classes are torsion, the absolute grading shift isThe square is defined using the rational relative lift. If $b_2^{+}(W)>0$, the infinity map vanishes. Between rational homology spheres, when $b_1(W)=b_2^{+}(W)=0$, the infinity map is an isomorphism of Laurent towers.

For integral Dehn surgery on a framed link in $S^3$, Kirby calculus uses two moves: a handle slide and addition or removal of a split unknot with framing $+1$ or $-1$. Two surgery diagrams give orientation-preservingly diffeomorphic closed three-manifolds exactly when these moves and link isotopy relate them.^[ref]

A handle slide replaces one attaching component by its band sum with a framed push-off of another; on the surgery linking matrix, it performs the matching integral row and column operation. It preserves the four-dimensional surgery trace. Adding a split $\pm 1$-framed unknot preserves the boundary three-manifold but changes the trace by connected sum with ${\mathbb{CP}}^2$ or its orientation reverse. This differs from creation or cancellation of a canceling handle pair in a four-dimensional handle decomposition, which preserves the four-dimensional manifold itself.

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.

For a closed connected orientable surface of genus $g$, the Künneth theorem gives the Betti numbers of $S^1\times {\Sigma }_g$ as $(1,2g+1,2g+1,1)$. The weak Morse inequalities give the displayed lower bound. A surface Morse function with one minimum, $2g$ saddles and one maximum exists from its standard handle decomposition. Adding a positive multiple of the cosine Morse function on the circle makes a product of Morse functions, whose critical counts are precisely these four Betti numbers. The lower bound is attained.

Reverse a handle decomposition of a compact $n$-dimensional manifold. Absolute $k$-handles become relative $(n-k)$-handles based on the boundary. The absolute boundary matrices count attaching sphere-belt sphere intersections; the corresponding relative matrices are their transposes over ${\mathbb{F}}_2$. The absolute chain complex is thus the complementary-degree dual of the relative chain complex. The universal coefficient theorem for cohomology proves Poincare-Lefschetz duality without an orientation hypothesis.

The two-dimensional Schoenflies theorem says that a simple closed piecewise linear curve in $S^2$ separates it into two components whose closures are piecewise linear balls of dimension two. In particular, a homeomorphism of the curve to the equator extends to a piecewise linear homeomorphism of the whole sphere.

A piecewise linear ball is a space piecewise linearly homeomorphic to a standard simplex. Write $D=B_1\cap B_2$. The complementary disk to $D$ in each $\partial B_j$ is a disk by the two-dimensional Schoenflies theorem. Thus each pair $(\partial B_j,D)$ is piecewise linearly homeomorphic to a sphere with a chosen hemisphere. Choose these boundary maps to agree on $D$. Extend them to the interiors by coning from an interior point, after compatible subdivisions. The resulting maps identify $B_1$ and $B_2$ with the two halves of a standard ball, glued along its equatorial disk. ConsequentlyThe same boundary parametrization proves a useful companion fact: if a ball is indented by removing a smaller ball attached along a boundary disk, the closure of what remains is again a ball. In the standard boundary-disk model, move the indented disk through a collar neighbourhood to a flat disk.

A handle decomposition is a filtration obtained by attaching handles $D^r\times D^{n-r}$ along $\partial D^r\times D^{n-r}$ in the boundary of the preceding stage. The number $r$ is the handle index; index-zero handles are disjoint balls, and index-$n$ handles close the remaining spherical boundary components. Starting with a triangulation of $M$, thicken its vertices, then its edges, and successively its higher-dimensional simplexes. In a barycentric subdivision these thickenings have precisely the indicated handle and attaching-region product structures. Adding collars between successive attachments gives a collared handle decomposition.

For a critical level handle embedding, assign successive distinct heights $t_j$ to the handles. At height $t_j$ the horizontal slice contains the $j$th handle and its attaching region. Between handle heights, the embedded boundary of the previously attached handles travels along its product collar; there is no topological change there. The lower boundary collar meets the attaching region of the next handle, and the upper collar leaves its new boundary. The stated ambient isotopy therefore separates all topology changes into horizontal handles and all intervening motion into collars. For an embedded surface, the local changes are the appearance of a disk, attachment of a band, and disappearance of a disk.

Here is the three-dimensional conclusion, with the distinction between the bounded and unbounded regions made explicit: a piecewise linear embedded $S^2$ in ${\mathbb{R}}^3$ bounds a piecewise linear three-ball; in the one-point compactification $S^3$, both complementary closures are three-balls. The unbounded closure in ${\mathbb{R}}^3$ itself is not a compact ball.

Put the embedded sphere in critical level handle embedding position. At a regular horizontal level its intersection with the plane is a finite disjoint union of simple closed curves. An innermost such curve bounds a planar disk whose interior misses the sphere. Cut the sphere along this curve and insert two slightly displaced copies of the disk. This replaces one sphere by two spheres. Repeat at regular levels between the critical heights, using innermost curves first. The number of curves at these chosen levels decreases at each operation, so the process stops. Each resulting sphere then lies in a slab containing at most one horizontal handle.

These terminal spheres bound balls directly. With no handle the surface is a disk transported through a collar and capped at its ends, the boundary of $D^2\times I$. A minimum or maximum merely supplies one of those caps. At a band level there are two planar boundary circles on one side and one on the other. Cap the two circles, straighten each planar disk using the two-dimensional Schoenflies theorem, and straighten the joining band in its planar collar. If the circles are not nested, this gives two balls joined along a boundary disk; if they are nested, it gives a ball with one boundary-attached ball removed. The gluing three-balls along a boundary disk argument and its indentation version show in both cases that the terminal sphere bounds a ball. This also explains why the planar position of the band matters: no assertion about an arbitrary knotted band is being made.

Reverse the disk surgeries. Suppose the two resulting spheres bound balls $A$ and $B$. Their disjoint boundaries imply that the balls are disjoint or one contains the other. Shrink the small displacement of the two surgery caps back to the original planar disk. In the disjoint case, the pre-surgery sphere bounds the union of two balls along that disk. In the nested case, it bounds the larger ball with the smaller boundary-attached ball removed. Both are balls by the preceding elementary facts. Induction through the finitely many surgeries proves the Three-dimensional Schoenflies theorem. To treat the other region in $S^3$, remove a point inside the ball just constructed and apply the same argument in the resulting copy of ${\mathbb{R}}^3$.

A 3-manifold is an irreducible three-manifold when every piecewise linear embedded two-sphere bounds a ball. Suppose $p:{\mathbb{R}}^3\to M$ is a universal cover, and lift an embedded sphere $S$ to $\tilde{S}$. Since $S$ is simply connected, its lift maps homeomorphically to $S$. By the Three-dimensional Schoenflies theorem, $\tilde{S}$ bounds a compact ball. The inverse image of $S$ is locally finite, so only finitely many other lifted spheres lie inside this ball. Choose an innermost one, bounding a ball $B$ whose interior meets no lift of $S$.

The interior of $B$ maps to one component $U$ of $M\setminus S$. This restriction is a covering map onto $U$: a path in $U$ starting in the image lifts into $\mathrm{int}B$, since it cannot cross a lifted boundary sphere. The covering is proper, because a compact subset of $U$ has preimage closed in the compact ball and cannot meet $\partial B$. Near $S$, the boundary map $\partial B\to S$ is one-to-one, so this connected covering has one sheet. The map on $B$ is therefore a homeomorphism onto $U\cup S$. Thus every embedded sphere in $M$ bounds a ball, and $M$ is an irreducible three-manifold.

Work over $\mathbb{Q}$. A closed connected orientable surface of genus $g$ has Betti numbers $(1,2g,1)$; the circle has Betti numbers $(1,1)$. The Künneth theorem therefore givesWe use the following precise consequence of the Morse handle-attachment theorem: a closed manifold with a Morse function has the homotopy type of a finite CW complex with one index-$j$ cell for each index-$j$ critical point. Its degree-$j$ rational cellular chain group has dimension $c_j$, so its homology has dimension at most $c_j$. Consequently $c_j\ge b_j$, the weak Morse inequalities. Summing yieldsThis counts every critical point, regardless of whether some critical values coincide.

To attain the bound, construct a surface Morse function $h:{\Sigma }_g\to \mathbb{R}$ from the standard orientable handle decomposition: start with a disk, attach $2g$ orientable bands in the usual paired pattern forming the genus-$g$ surface with one boundary component, and cap that boundary with a disk. Give the first disk a minimum, each band one saddle with local form $-x^2+y^2$, and the final cap a maximum. Regular collar coordinates join successive levels without additional critical points. This constructs $h$ with counts $(1,2g,1)$.

On the product takeIts critical points are exactly the pairs of a critical point of $h$ with $\theta =0,\pi$. The Hessian matrix is block diagonal, with surface block $\mathrm{Hess}h$ and circle entry $-\delta \cos \theta$. It is nonsingular and its Morse index is the sum of the two factor indices. Thus this product of Morse functions has countsand exactly $4g+4$ critical points. If distinct critical values are desired, choose $\delta$ outside the finite set producing collisions after first choosing distinct critical values for $h$. The minimal Morse critical count on a circle times an orientable surface is therefore attained for every $g\ge 0$.

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 $x,y$ be the generators associated with the two pairs of one-handle feet. Following the attaching circle across the identifications marked $a,b$ and $c,d$ gives the word $xy{x}^{-1}{y}^{-1}$, up to reversal and changing generator orientations. Thus the van Kampen theorem givesThe handle decomposition has cellular groups $C_0=\mathbb{Z}$, $C_1={\mathbb{Z}}^2$, $C_2=\mathbb{Z}$, with no higher groups. The boundary of the two-cell is the abelianized attaching word, which is zero. HenceThe zero framing is the product framing of the commutator handlebody of the torus, so the represented manifold is $T^2\times D^2$ and its intersection form is $(0)$. In particular its boundary is $T^2\times S^1=T^3$.

The boundary homology can also be obtained without using the product identification. By Poincare-Lefschetz duality, $H_3(W,\partial W)=H^1(W)={\mathbb{Z}}^2$ and $H_2(W,\partial W)=H^2(W)=\mathbb{Z}$. The map $H_2(W)\to H_2(W,\partial W)$ is its zero intersection form. The long exact sequence of a pair then gives split sequencesThe boundary is connected and oriented, so its zeroth and third homology are both $\mathbb{Z}$. Therefore

Take a finite presentation $T=⟨x_1,\dots ,x_r\mid R_1,\dots ,R_s⟩$. For a closed-manifold realization of a finitely presented fundamental group, begin with a five-dimensional ball and attach $r$ one-handles, giving a compact oriented five-manifold $W_1$ with fundamental group the free group on the $x_i$. Represent the relators by disjoint embedded circles in its four-dimensional boundary. General position allows them to be embedded and disjoint, and their oriented normal bundles admit framings. Attach a two-handle along each such circle. The resulting compact five-manifold $W$ hasby the Seifert-van Kampen theorem. Now set $B=\partial W$. It is a compact connected smooth four-manifold without boundary. Turning this handle decomposition upside down gives a relative handle decomposition of $(W,B)$ with handles of indices three, four and five only. Attaching such handles does not change the fundamental group: there are no new one-handles or two-handle relations. Thus ${\pi }_1(B)\to {\pi }_1(W)$ is an isomorphism andThese are the differential-topological handle facts used in the construction; the dimension-five thickening ensures the boundary retains the prescribed fundamental group.

Now assume $T$ finite. Choose any smooth Riemannian metric on $B$ and lift it to its universal cover $\tilde{B}$. This cover is compact, since its degree is $|T|$, and simply connected. Its deck transformations give a free isometric $T$-action. For the two Gassmann equivalent subgroups putEach is a closed Riemannian manifold, and $\tilde{B}$ is its simply connected universal cover, with deck transformation group $U_i$. Consequently ${\pi }_1(M_i)\cong U_i$.

Here is the complete spectral argument behind the Sunada theorem in this situation. Each finite-dimensional upstairs eigenspace $E_{\lambda }$ is a representation of $T$, because the action is isometric and commutes with the Laplace-Beltrami operator. Functions downstairs correspond exactly to invariant functions upstairs, so their multiplicities are $\dim E_{\lambda }^{U_i}$. Averaging the representation matrices over a subgroup is a projection onto its invariant subspace. Taking its trace givesThe character ${\chi }_{\lambda }$ is constant on each conjugacy class of $T$. Gassmann equivalence means $|U_1\cap C|=|U_2\cap C|$ for every such class $C$; summing these equalities also gives $|U_1|=|U_2|$. The two averages are therefore equal for every $\lambda$. The quotients are isospectral, with ${\pi }_1(M_i)\cong U_i$.

If $U_2=t{U}_1{t}^{-1}$, the map $[x{]}_{U_1}\mapsto [tx{]}_{U_2}$ is well defined, invertible and isometric because $t$ acts by an isometry upstairs. Conjugate subgroups give isometric quotients.

For an explicit example with different fundamental groups, use the order-sixteen Gassmann pairEmbed each in $T=S_{16}$ by left multiplication on its own sixteen elements, identified with the same sixteen-point set. These regular representations are faithful. Both groups have one identity element, three elements of order two, and twelve of order four: in $C_4^2$ the elements of order dividing two are precisely its four even-coordinate elements; in $Q_8\times C_2$ they are the four elements in $\{1,-1\}\times C_2$. Every remaining element has order four.

In a regular action, an order-$k$ element is a product of $16/k$ disjoint $k$-cycles. Thus each subgroup intersects the identity conjugacy class of $S_{16}$ in one element, the class of type $2^8$ in three, the class of type $4^4$ in twelve, and every other class in zero. They are Gassmann equivalent. However $H_1$ is abelian and $H_2$ is nonabelian, so they are not isomorphic. Apply the preceding construction with fundamental group $S_{16}$ and these subgroups. It producesIn particular they cannot even be homeomorphic, and therefore cannot be isometric.

All homology and cohomology in this solution use ${\mathbb{F}}_2$ coefficients, so no orientation hypothesis is necessary. The Poincare-Lefschetz duality degrees are complementary:The starred notation in the question must be understood with this degree reversal.

Turn the handle decomposition upside down. An absolute $k$-handle becomes a relative $(n-k)$-handle based on $\partial W$. The absolute cellular boundary coefficient counts intersections of an attaching sphere with a belt sphere. In the reversed handle decomposition, the same intersections are counted in the opposite order. Over ${\mathbb{F}}_2$ the two incidence matrices are transposes, with no sign ambiguity. Thus the absolute chain complex is the complementary-degree dual of the relative chain complex. Taking homology and using the universal coefficient theorem for cohomology over a field proves the displayed isomorphism. This is mod-two handle duality.

To obtain a perfect pairing from the assumed closed case, form the double of a manifold $D(W)$ from two copies of $W$. The fold map $D(W)\to W$ is a retraction, so the inclusion of either copy induces an injection on homology. If $a\ne 0$ in $H_k(W)$, its image in $H_k(D(W))$ is nonzero. Closed-manifold mod-two Poincare duality supplies a class $z\in H_{n-k}(D(W))$ with intersection $1$ against that image. Cut a representative of $z$ along the boundary and retain its part in the first copy; it defines $b\in H_{n-k}(W,\partial W)$. Equivalently, apply the quotient map $D(W)\to D(W)/W_{-}\cong W/\partial W$ and excision. Intersections with a representative of $a$ in the interior are unchanged, so $a\cdot b=1$. This proves nondegeneracy in the first variable. Mod-two handle duality gives equal finite dimensions for the two spaces, hence nondegeneracy in the second variable as well.

For the three-dimensional conclusion, setThe long exact sequence in relative homology says $L=\mathrm{im}\partial$, where $\partial :H_2(W,\partial W)\to V$. The closed-surface intersection pairing on $V$ is nondegenerate. A collar calculation gives the adjoint identityIndeed, push $x$ slightly into the collar: intersections with the relative surface representing $b$ correspond to its boundary intersections with $x$. By the absolute-relative perfect pairing already proved, $x$ is orthogonal to every $\partial b$ precisely when $i_{\ast }x=0$. Therefore $L^{\perp }=L$.

For any finite-dimensional space with a perfect pairing, $\dim L+\dim L^{\perp }=\dim V$. Hence the half-lives-half-dies theorem givesBut $H_1({\mathbb{RP}}^2;{\mathbb{F}}_2)\cong {\mathbb{F}}_2$ has dimension one. It cannot be the entire boundary of a compact three-manifold, because the displayed dimension would be $1/2$. This excludes nonorientable three-manifolds as well as orientable ones.

A closed connected three-manifold has a handle decomposition with one zero-handle and one three-handle. To arrange this, cancel the zero-handles along a spanning tree of connecting one-handles; perform the corresponding construction in the dual handle decomposition to consolidate the three-handles. This does not require the three-manifold to be orientable.

If there are $r$ one-handles and $s$ two-handles, the Euler characteristic isMod-two Poincare duality pairs the Betti numbers of a closed odd-dimensional manifold, giving $\chi (Y)=0$. Thus $r=s$. The one-handles give $r$ generators of a group of the fundamental group, the attaching circles of the two-handles give $r$ relators, and the three-handle does not change the fundamental group. Therefore ${\pi }_1(Y)$ admits a balanced presentation.

Now letbe any finite group presentation. Start with a four-dimensional zero-handle and $r$ one-handles. Its fundamental group is the free group on the $x_i$. Represent the finitely many relators by disjoint embedded circles in its three-dimensional boundary; a small perturbation makes the circles disjoint. Choose framings of an embedded sphere and attach $s$ two-handles. The resulting compact connected oriented four-manifold $W$ has ${\pi }_1(W)\cong G$ by the Seifert-van Kampen theorem.

The map ${\pi }_1(\partial W)\to {\pi }_1(W)$ is surjective. One way to see this is to turn the handle decomposition upside down: the relative handles based on $\partial W$ have indices $2,3,4$, and so add no fundamental group generators. Also $\partial W$ is connected, since surgery on embedded circles in a connected three-manifold preserves connectedness.

Take the double of a manifoldIt is a closed connected smooth oriented four-manifold. The Seifert-van Kampen theorem givesBoth maps from ${\pi }_1(\partial W)$ are the same surjection, under the natural identification of the two copies. The pushout identifies the two copies of every element of $G$ and imposes no new relations: the fold homomorphism is inverse to either inclusion. ThusThis is four-manifold realization of finitely presented groups.

An index-$h$ handle is $D^h\times D^{n-h}$, attached along $S^{h-1}\times D^{n-h}$. Its attaching sphere is $S^{h-1}\times \{0\}$ with its attaching framing of an embedded sphere, and its belt sphere in the new boundary is $\{0\}\times S^{n-h-1}$. A handle decomposition is an ordered sequence of such handle attachments.

A handle slide of one $h$-handle over another replaces its framed attaching sphere by a band sum with a parallel framed copy of the second attaching sphere, along a band in the boundary of the previously attached handles. The second handle stays fixed. The band transports the framing of an embedded sphere. The union of the two handles and their attaching collars can be reidentified by a diffeomorphism, so this changes the decomposition, rather than the underlying manifold. A sequence of these moves, allowing the usual isotopies of attaching data, gives handle decompositions related by slides. Reading the decomposition backwards interchanges attaching spheres and belt spheres; a slide of dual handles is a slide of the corresponding original handles.

For the standard handle-calculus assertion, take $1\le k\le n-2$, so both the relevant attaching and belt spheres are connected. We need the local geometric move behind handle-slide isolation of a cancelling pair. Write $p=A_1\cap B_1$. If $q$ is an intersection of some $A_j$, $j>1$, with $B_1$, choose an arc on $B_1$ from $q$ to $p$, and take a narrow parallel band to the unique sheet of $A_1$ at $p$. Slide $A_j$ over an appropriate parallel copy of $A_1$. Choose the copy's orientation so that the new sheet at $p$ has the opposite local smooth intersection number to the sheet at $q$.

Within a thin neighbourhood of the arc, the old sheet at $q$ and the added sheet at $p$ are the two ends of one band. An isotopy pushes that band through the neighbourhood of $B_1$ and removes both ends. Outside this neighbourhood the original $A_j$ is unchanged, apart from the added copy of the parts of $A_1$ away from $B_1$. Since $A_1$ meets $B_1$ only at $p$, this copy contributes no further intersections with $B_1$. Thus one slide followed by an isotopy removes one original geometric intersection with $B_1$. This is a local band move, not an appeal to handle cancellation or to an algebraic cancellation of signs.

The arcs and bands may be chosen one at a time; in the one-dimensional sphere case, start with an intersection adjacent to $p$, so that the arc interior contains no other intersection. Perturb to make all intersections transverse. The relevant spheres are compact, so there are only finitely many intersections. Repeating the move for every $j>1$ leaves $A_1$ and $B_1$ fixed and givesThe bands can avoid the remaining attaching data; any consequent change to intersections with other belt spheres is allowed at this stage.

For the second stage, view the same interface from the dual decomposition. Now $B_i^{\prime }$ are attaching spheres and $A_j^{\prime }$ are belt spheres. Apply the same local move to each intersection of $B_i^{\prime }$, $i>1$, with $A_1^{\prime }$, sliding that dual handle over the one with attaching sphere $B_1^{\prime }$. This changes the original $k$-handles. It keeps $A_1^{\prime }$ and $B_1^{\prime }$ fixed, while removing the unwanted intersections with $A_1^{\prime }$. The parallel copies and bands may be taken disjoint from all $A_j^{\prime }$, $j>1$, because those spheres are already disjoint from $B_1^{\prime }$ and from $A_1^{\prime }$. Hence their intersections with $B_1^{\prime }$ remain empty. We obtainNo handle has been removed.

There is an implicit index-range qualification in this geometric assertion. If arbitrary zero-handles are included, the second conclusion need not hold: two zero-handles joined by one one-handle have an attaching zero-sphere with one endpoint on each zero-handle belt sphere. Slides preserving this handle inventory cannot make its second endpoint avoid both belts. The dual obstruction occurs for top-index handles. Thus the argument uses the customary intermediate-index setting of the handle-slide lemma; the unqualified assertion at the extreme indices would require additional hypotheses or additional moves.

For the numerical formula, fix coherent orientations for the integer smooth intersection numbers, and letKeep the orientation at the surviving point, so its number remains $\delta$. A slide of a $(k+1)$-handle adds or subtracts the first row of the intersection matrix. Removing all geometric intersections in the first column has net row operationEven if opposite-sign intersections require extra slides, their net signed coefficient is the displayed one. In particular,The dual slides then perform the column operationsSince $C_1^{\prime }=(\delta ,0,\dots ,0{)}^T$, these operations leave the lower-right block unchanged. Thus the final answer, including both stages of slides, isThe remaining block is the Schur complement of the unit pivot $\delta$. If the original orientation is chosen so that $\delta =1$, it is simply $m_{lk}-m_{l1}{m}_{1k}$. For unoriented handles use intersection numbers modulo $2$, with the same formula interpreted in ${\mathbb{F}}_2$.

A Stein surface is a Stein manifold of complex dimension two, hence real dimension four. Its global holomorphic functions separate points and provide local coordinates, and the holomorphic hull of every compact subset is compact. The word “surface” here refers to complex dimension, rather than to a real two-dimensional surface.

The affine complex plane ${\mathbb{C}}^2$ is a Stein surface, as is the smooth closed hypersurface $xy-z^2=1$ in ${\mathbb{C}}^3$: its gradient $(y,x,-2z)$ never vanishes on the hypersurface. A compact connected positive-dimensional complex manifold cannot be Stein, because its holomorphic functions are constant. In real dimension four, the topology of Stein surfaces has especially rich links with handle decompositions and contact structures on three-dimensional boundaries.^[ref]

A three-handle on an $n$-dimensional manifold with boundary, $n\ge 3$, is $D^3\times D^{n-3}$ attached along $S^2\times D^{n-3}$. Its core is $D^3\times \{0\}$ and its attaching sphere is two-dimensional. In dimension three it caps a boundary sphere with a three-ball; in dimension four its attaching region is $S^2\times D^1$. Viewed from the opposite end of a handle decomposition, it becomes an index-$(n-3)$ handle. This dual viewpoint explains why handles of different indices appear in complementary pairs.

A zero-handle in an $n$-dimensional handle decomposition is $D^0\times D^n=D^n$. Its attaching region is empty, so it starts a connected component of the manifold. In dimension four it is a four-ball; adding one-handles and two-handles gives a four-dimensional handlebody with a two-dimensional CW spine.