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For the three-dimensional lens space $L(p)$, the integral coefficient sequence shows that the integral connecting map $H^1(L(p);{\mathbb{F}}_p)\to H^2(L(p);\mathbb{Z})$ is an isomorphism between groups of order $p$. Reduction modulo $p$ is also an isomorphism in degree two, so their composite Bockstein is an isomorphism.

For a connected complete Riemannian manifold of dimension $n\ge 2$ and constant sectional curvature $k>0$, its complete simply connected universal cover is the round sphere of radius $k^{-1/2}$. This is the constant-curvature case of Cartan-Ambrose-Hicks theorem. Thus the manifold is a quotient by a finite group of isometries acting freely. In even dimension, the determinant embeds this group in $\{1,-1\}$: every determinant-one orthogonal transformation in odd ambient dimension has eigenvalue one and therefore a fixed point. The only possible nonidentity element is the antipodal involution, giving Real projective space. In odd dimensions, free cyclic scalar actions on complex unit spheres instead produce lens spaces.

The quotient model of a lens space splits into two solid tori. In the boundary angle lattice ${\mathbb{Z}}^2+\mathbb{Z}(1/p,q/p)$, choose ${\mu }_1=(0,1)$ and ${\lambda }_1=(1/p,q/p)$. The other meridian is $(1,0)=-q{\mu }_1+p{\lambda }_1$. A second column completing this primitive slope to a determinant-one matrix specifies a gluing map after reversing one boundary-basis convention. Different completions differ by a meridional Dehn twist extending across the second solid torus, so they give the same three-manifold.

Identify $S^{2b-1}$ with the unit sphere in ${\mathbb{C}}^b$, and put $\zeta =e^{2\pi i/a}$. The generator of the cyclic group $\mathbb{Z}/a$ acts byIt preserves the sphere because $|\zeta |=1$. If its $k$th power fixes a point, then $({\zeta }^k-1)z_j=0$ for every $j$. Some coordinate is nonzero, so ${\zeta }^k=1$, meaning $k\equiv 0(\mathrm{mod}a)$. Thus the scalar root-of-unity action is free. The case $a=1$ is the trivial group and also satisfies freeness. The quotient is a standard lens space.

By part (b), $M=S^n/\Gamma$, where the group of deck transformations acts freely by round isometries. Every round isometry is the restriction of an orthogonal transformation of ${\mathbb{R}}^{n+1}$: it preserves the inner product $⟨x,y⟩=\cos d(x,y)$, so its images of an orthonormal basis determine its linear extension.

Suppose $n$ is even. Every $A\in SO(n+1)$ has eigenvalue one. Indeed, the nonreal eigenvalues come in conjugate pairs with product one, while the remaining real eigenvalues are $\pm 1$. Their number is odd and their product is one, forcing at least one $+1$. Such a transformation fixes a unit vector. Freeness of the deck transformation action consequently implies that the determinant map $\Gamma \to \{\pm 1\}$ has trivial kernel. Thus $|\Gamma |\le 2$.

If $\Gamma$ is nontrivial with generator $A$, then $A^2=I$. Its eigenvalues are $\pm 1$, and freeness excludes the eigenvalue $+1$. Hence $A=-I$, the antipodal involution. We obtainThis proves the even-dimensional part of the classification of complete positive constant-curvature manifolds.

The conclusion fails in odd dimensions. For $n=2m-1\ge 3$, identify $S^{2m-1}$ with the unit sphere in ${\mathbb{C}}^m$. Scalar multiplication by $\zeta =e^{2\pi i/r}$, with $r\ge 3$, generates a freely acting cyclic group of isometries: ${\zeta }^{j}z=z$ for nonzero $z$ forces ${\zeta }^j=1$. The quotient is complete with constant sectional curvature one and has fundamental group $C_r$, so it is neither the simply connected sphere nor Real projective space with fundamental group $C_2$. In particular $S^3/C_3$ is a lens space counterexample.

Put $P=|p|$ and $\zeta =e^{2\pi i/p}$. The lens space is the quotientThe cyclic group has order $P$. Its action is free: if $g^j$ fixes a point with $z_1\ne 0$, then $p\mid j$; if $z_1=0$, then $z_2\ne 0$ and $p\mid qj$, which again implies $p\mid j$ by coprimality. A finite free smooth action gives a closed orientable three-manifold, and the quotient projection is a covering space.

Since $S^3$ is simply connected, this is the universal covering and its deck group is the cyclic group of order $P$. HenceFor $P=1$ the quotient is $S^3$ and the group is trivial. With this unoriented quotient convention, replacing $p$ by $-p$ reverses the chosen generator and leaves the quotient unchanged; oriented surgery conventions may record the signs differently.

Use the quotient lens spaces and set $\zeta =e^{2\pi i/7}$. Let $g_j(z_1,z_2)=(\zeta z_1,{\zeta }^j{z}_2)$. The mapis a homeomorphism and satisfiesbecause $3\cdot 2\equiv -1(\mathrm{mod}7)$. Since $2$ is invertible modulo $7$, $g_3^2$ generates the entire target deck group. Hence $F$ maps cyclic orbits bijectively to cyclic orbits, as does its inverse, and descends to a homeomorphism:This is an explicit map, so no classification theorem for lens spaces is required. It reverses orientation: complex conjugation in one coordinate reverses the ambient real orientation, while swapping the two complex coordinates preserves it.

Use the boundary-orientation convention in which $-p/q$ surgery on the unknot gives $L(p,q)$. The negative continued fraction isAttach three two-handles to a zero-handle along a chain of unknots, with adjacent linking numbers $+1$ and framings $-3,-2,-2$. Equivalently, take the plumbing of oriented disk bundles over three spheres with Euler numbers $-3,-2,-2$. Successive slam-dunk moves change the boundary surgery coefficient to $-3-1/(-2-1/(-2))=-7/3$, proving that its oriented boundary is $L(7,3)$ in this convention.

The zero sections are a basis of second homology, and the intersection form isIts successive leading principal minors are $-3,5,-7$. Equivalently those of $-Q$ are $3,5,7>0$, so Sylvester's criterion proves that $Q$ is negative definite. This explicitly constructs the required filling.

If lens spaces are instead labeled by positive surgery with its boundary orientation, use the chain with framings $-2,-4$, whose coefficient is $-7/4$; its boundary is orientation-preservingly the positive-surgery $L(7,3)$, since orientation reversal changes $q=4$ to $q=3$. Its negative-definite form is $\left(\begin{array}{cc}-2& 1\\ 1& -4\end{array}\right)$. This is only a convention change in the named boundary, and specifies a valid filling in either labeling system.

For the unknot the three-manifolds are $S^3$, $U_0=S^1\times S^2$, and $U_n=L_n$, where $L_n$ denotes positive $n$-surgery with its surgery orientation. This fixes the orientation independently of competing conventions for the name $L(n,1)$.

Use the grading convention $\mathrm{gr}{\Theta }_{S^3}^{+}=0$ and $\mathrm{gr}{\Theta }_{S^3}^{-}=-2$. Let $R=\mathbb{Z}[U]$ and $\hat{R}=\mathbb{Z}[[U]]$. The ordinary minus groups arewhere $0\le k<n$ andThe lens-space absolute grading can be calculated from the standard torus surgery triple. For $0\le k<n$, the reverse trace structure with evaluation $2k-n$ has a small embedded holomorphic triangle with basepoint multiplicity zero and signed count $\pm 1$, taking $y_k$ to ${\Theta }^{-}$. Its degree shift is ${\Delta }_k=(1-(2k-n{)}^2/n)/4$. Thus $(d_k-2)+{\Delta }_k=-2$, which gives the boxed formula for $d_k$. Additional triangles wind around the surgery region and give the higher powers of $U$ calculated below.

Their infinity groups are obtained by inverting $U$: one Laurent tower for $S^3$ and each lens-space structure, two for the torsion structure on $S^1\times S^2$, and $\mathbb{Z}[U,U^{-1}]/(1-U^{|i|})$ in its non-torsion structures. Their plus groups are respectively ${\mathcal{T}}_0^{+}$, ${\mathcal{T}}_{1/2}^{+}\oplus {\mathcal{T}}_{-1/2}^{+}$, and ${\mathcal{T}}_{d_k}^{+}$. The hat groups are $\mathbb{Z}$ in degree zero, $\mathbb{Z}$ in each degree $\pm 1/2$, and $\mathbb{Z}$ in degree $d_k$. These follow directly by straightening the standard genus-one Heegaard diagrams: the lens space has one generator in each structure and no chain differential, while the two bigons in the torsion $S^1\times S^2$ diagram cancel with coherent signs. The sphere adjunction condition eliminates its non-torsion plus and hat groups, but not its ordinary minus group. The Non-torsion Heegaard Floer homology of the sphere-circle product calculation comes from a two-generator chain complex with chain differential $1-U^{|i|}$; its cokernel is the displayed quotient. For the triangle, complete the free minus towers by replacing $R$ with $\hat{R}$. The non-torsion summands disappear after completion, because $1-U^{|i|}$ has an inverse $1+U^{|i|}+U^{2|i|}+\cdots$ in $\hat{R}$.

For $k=0$, choose signs so that the exact triangle becomesThe zero-surgery trace has $\chi =1$, $\sigma =0$, and $c_1^2=0$ in the torsion structure; hence its absolute degree shift is $-1/2$, taking ${\Theta }^{-}$ to $x_{-}$. The auxiliary-generator map takes $x_{+}$ to $y_0$, so its effective grading shift isFor the underlying cobordism $Z_n$ in its torsion structure, $c_1^2=0$, $\chi =1$, $\sigma =0$, and its actual cobordism shift is $-1/2$. Tensoring with the canonical lens generator adds $d_0=(n-1)/4$ under the connected-sum grading convention, explaining the effective shift. For $n=1$ the auxiliary lens space is $S^3$, and this is simply the other degree-$-1/2$ surgery trace map.

The reverse trace components are explicit in every $k$. For $j=k+n\ell$, put $m_j=2j-n$. Its intersection form is $(-n)$, soIndeed $c_1^2=-m_j^2/n$, $\chi =1$, $\sigma =-1$, and matching source and target degrees gives precisely the displayed exponent. The exponent is nonnegative, and the coefficient is a unit: the infinity trace map is an isomorphism, and restriction to the free minus towers leaves the unique grading-compatible monomial.

With a compatible choice of Coherent orientations in Heegaard Floer homology, the sum $C_k$ is multiplication byFor $k=0$, pair $\ell$ with $1-\ell$: their exponents agree and their signs are opposite, so $P_0=0$, in agreement with the first triangle. For $0<k<n$ the term $\ell =0$ is the unique constant term, and every other exponent is positive. Thus $P_k=1+U(\cdots )$ is a unit in $\hat{R}$. The zero-surgery summand for such a $k$ vanishes, and the exact triangle isIt is isomorphic, as an ungraded completed-module triangle, to one with the final map the identity. The sum generally has no single absolute Heegaard Floer grading shift; each individual trace component has shift ${\Delta }_j$. This distinction and the power series explain why replacing the completed triangle by a polynomial-module triangle is not innocuous.

Give the three Hopf fibration components coherent orientations, so their pairwise linking numbers are one. In the link exterior, their longitudes satisfy $[{\lambda }_i]={\sum }_{j\ne i}{\mu }_j$. Filling along $p_i{\mu }_i+q_i{\lambda }_i$ imposes that relation on first homology. For the three rational coefficients the presentation matrix isIts determinant is $30$, and the greatest common divisor of its two-by-two minors is one (for example, $-5$ and $-3$ occur). Its Smith normal form is $\mathrm{diag}(1,1,30)$, soA nonzero rational coefficient with numerator one does not by itself make a multi-component surgery an integral homology sphere: the nonzero linking numbers must be included.

For the integer filling, use the Seifert fibered space structure $P\times S^1$. Its central regular fiber is $h=abc$, and each preferred longitude is $h{\mu }_i^{-1}$. The three filling relations are consequentlySubstituting $b=h$ into $abc=h$ gives $c=a^{-1}$. The remaining equations say $a^2=a^4$, hence $a^2=1$ and $h=1$. Thus its fundamental group is cyclic of order two.

Geometrically, a filling coefficient $r$ attaches a Seifert fibered space solid torus with multiplicity $|r-1|$, the distance of $r\mu +\lambda$ from the regular fiber $\mu +\lambda$. The multiplicities are $2,1,4$; the middle filling creates no exceptional fiber. The result is a Seifert fibered space over the sphere with at most two exceptional fibers, hence a union of two solid tori, or a lens space. A lens space with fundamental group of order two is $L(2,1)$. ThereforeAs a separate arithmetic check, the integer surgery linking matrix has determinant $-2$, in agreement with its first homology of order two.

Use the meridian of a knot $\mu$ and Seifert longitude $\lambda$ of the unknot on its boundary torus. The unknot's knot exterior is a solid torus whose disk-bounding curve is $\lambda$. A genus-one Heegaard diagram is thereforeHere the horizontal coordinate is $\mu$ and the vertical coordinate is $\lambda$. The drawing uses a small translate of $\alpha$ to avoid intersections on the edge of the square; opposite edges are identified. The algebraic intersection number of curves on an oriented surface has absolute value five.

Starting with $\Sigma \times [0,1]$, attach a three-dimensional two-handle along $\alpha \times \{0\}$, using its surface framing, and cap the resulting sphere with a three-handle. This gives the solid torus on that side. Attach a two-handle along $\beta \times \{1\}$, again using the surface framing, and cap its sphere with a three-handle. The second solid torus has its disk-bounding curve identified with $5\mu +3\lambda$, so the resulting oriented lens space is precisely the specified $5/3$ Dehn filling.

The elliptization theorem says that a closed oriented three-manifold with finite fundamental group has spherical geometry. When the fundamental group is a cyclic group, it is a lens space. Consequently the candidates are $L(5,q)$ with $q\in \{1,2,3,4\}$, using the same surgery orientation convention throughout.

The oriented classification of lens spaces says that $L(p,q)$ and $L(p,q^{\prime })$ are related by an orientation-preserving homeomorphism exactly when $q^{\prime }\equiv q^{\pm 1}(\mathrm{mod}p)$. Since $1^{-1}=1$, $4^{-1}=4$, and $2^{-1}=3$ modulo five, the complete list isThese are three distinct oriented homeomorphism classes. Reversing orientation interchanges $L(5,1)$ and $L(5,4)$; the class containing $L(5,2)$ and $L(5,3)$ admits an orientation-reversing homeomorphism. The elliptization theorem and the stated oriented classification of lens spaces are the general classification results used here.

With zero framings, the surgery linking matrix isIts eigenvalues are $n-1$ on the span of $(1,\dots ,1)$ and $-1$ on the complementary subspace, so $\det Q_2=(-1{)}^{n-1}(n-1)$. For $n>1$, its Smith normal form is $\mathrm{diag}(1,\dots ,1,n-1)$, and therefore

Handle slides reduce this surgery diagram to the standard surgery diagram of the lens space $L(n-1,1)$; equivalently, the generator of the cokernel has linking pairing $1/(n-1)$. Thusup to the orientation convention for surgery. When $n=1$, the matrix is $(0)$ and the exceptional answer is $S^1\times S^2$, with $H_1\cong H_2\cong \mathbb{Z}$.

Every connected closed orientable three-manifold with finite fundamental group admits spherical geometry. With cyclic fundamental group, it is a lens space.