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If $L$ is the complexification of a real vector bundle of the tautological bundle and $\mu =[L]-1$, thenRealizing Real projective space as the circle bundle of the square of the tautological bundle on Complex projective space gives this through the K-theory Gysin sequence of a sphere bundle.

If $E\to B$ is an oriented real plane bundle, then $\mathcal{L}Gr(E)=\mathbb{P}(E)$ is an oriented circle bundle. Its transition rotations have twice the angle of those of $E$, so

At a regular free circle group level of a moment map, the inverse image of a Lagrangian submanifold of the reduced space is a Lagrangian submanifold of the original space. Its restricted symplectic form is the pullback of the reduced form and hence vanishes. Its dimension is one more than that of the reduced Lagrangian, exactly half the ambient dimension. It is a circle bundle over the original Lagrangian.

Fix the sign convention ${\iota }_{{\xi }_M}\omega =-d{\mu }^{\xi }$, where ${\xi }_M$ is the fundamental vector field and ${\mu }^{\xi }=⟨\mu ,\xi ⟩$. A Hamiltonian group action is a symplectic Lie group action together with an equivariant moment map $\mu :M\to {\mathfrak{g}}^{\ast }$ satisfying this identity for every $\xi \in \mathfrak{g}$. Equivariance means $\mu (gx)={\mathrm{Ad}}_{g^{-1}}^{\ast }\mu (x)$. Merely requiring each fundamental field to have some Hamiltonian is the weaker condition of a weakly Hamiltonian action.

For a circle group action, identify its Lie algebra with $\mathbb{R}$, and let $Y$ generate the action. Put $Z={\mu }^{-1}(t)$ and let $i:Z\hookrightarrow M$. Since $t$ is a regular value,The action is free, so $Y_x\ne 0$. The symplectic orthogonal complement of this hyperplane is exactly $\mathbb{R}{Y}_x$, which is contained in $T_{x}Z$. ThusThe free action of the compact circle group is proper, and the quotient $Q=Z/S^1$ is a smooth manifold with projection $\pi :Z\to Q$ a principal bundle. The form $i^{\ast }\omega$ is invariant and kills the orbit directions, hence is basic and descends to a unique two-form ${\omega }_Q$ satisfying ${\pi }^{\ast }{\omega }_Q=i^{\ast }\omega$. It is closed because pullback by a submersion detects differential forms. If a quotient tangent vector annihilates ${\omega }_Q$, a lift lies in the displayed kernel and hence projects to zero. Therefore ${\omega }_Q$ is nondegenerate. This proves symplectic reduction directly.

If $L\subset Q$ is a Lagrangian submanifold, its inverse image $\tilde{L}={\pi }^{-1}(L)$ is a smooth circle bundle over $L$. The restricted symplectic form is ${\pi }^{\ast }({\omega }_Q{|}_L)=0$. If $\dim M=2N$, then $\dim Q=2N-2$, $\dim L=N-1$, and $\dim \tilde{L}=N$, proving that $\tilde{L}$ is Lagrangian. This is the Lagrangian lift through circle reduction.

For scalar multiplication on ${\mathbb{C}}^{n+1}$, the generator is $Y={\sum }_j(-y_j{\partial }_{x_j}+x_j{\partial }_{y_j})$. Direct contraction givesChoose $t>0$. Its level is the sphere of radius $R=\sqrt{2t}$ and its quotient is ${\mathbb{CP}}^n$. Complex conjugation on the sphere reverses the ambient symplectic form and descends to an anti-symplectic involution of the quotient. Its fixed set is ${\mathbb{RP}}^n$: a projective point fixed by conjugation has a real representative after a phase change. For two tangent vectors to this fixed set, anti-symplecticity makes the reduced form equal to its negative, so it vanishes. The real dimension is $n$, half the quotient dimension, proving the Lagrangian fixed locus of an anti-symplectic involution property.

The lifted Lagrangian submanifold can be written explicitly asThe only redundancy in this parametrization is the displayed free involution: two nonzero real representatives on the same complex ray differ by a real sign. The quotient embeds smoothly in the sphere, and projection to $[x]\in {\mathbb{RP}}^n$ exhibits its circle bundle structure. Its Lagrangian property follows from the reduction argument, including the case $n=0$.

Euler number and self-intersection. Orient the rank-two normal bundle ${\nu }_S$ by the orientations of $S$ and $M$: the ordered tangent and normal spaces have the ambient orientation. Choose a smooth normal section $s$ transverse to the zero section. Its zeros are isolated, and their signed number is the evaluation $⟨e({\nu }_S),[S]⟩$ of the Euler class on the fundamental class. Scale the section sufficiently small to lie in a tubular neighborhood. Its graph is a push-off $S^{\prime }$ isotopic to $S$.

The intersections of $S$ with $S^{\prime }$ occur exactly at the zeros of $s$. In a positively oriented local splitting $TS\oplus {\nu }_S$, the basis formed from the tangent spaces to $S$ and to the graph has block matrixIts determinant is $\det Ds$, so the local intersection sign is precisely the local zero index of the section. Summing proves the Euler number equals self-intersection identity

The conic. The given map is the degree-two Veronese map $[u:v]\mapsto [u^2:uv:v^2]$ followed by the projective linear transformationThe Veronese map is injective with nonvanishing differential: on the chart $u\ne 0$ its coordinates are $[1:t:t^2]$, and on $v\ne 0$ they are $[s^2:s:1]$. It is therefore a smooth embedding, and $S$ is a smooth plane conic diffeomorphic to ${\mathbb{CP}}^1\cong S^2$. Its equation in the given coordinates is $z_0^2-z_2^2=4z_1^2$.

Let $L$ be a projective line, with its complex orientation. The cohomology ring of complex projective space gives $[L]\cdot [L]=1$ in ${\mathbb{CP}}^2$. The hyperplane $z_1=0$ meets $S$ at $[u:v]=[1:0]$ and $[0:1]$. The zeros of $uv$ are simple in the respective local coordinates, and complex intersections have positive signs. Hence $[S]\cdot [L]=2$, so $[S]=2[L]$. The self-intersection number and the normal Euler class arewhere $u\in H^2(S^2;\mathbb{Z})$ has $⟨u,[S^2]⟩=1$.

The boundary of the tubular neighborhood. Write $N=\nu (S)$ for the closed disk tubular neighborhood and $E=\partial N$ for its boundary. This distinguishes the disk neighborhood $N$ from the vector normal bundle ${\nu }_S$. The space $E$ is the oriented circle bundle of ${\nu }_S$, with Euler class $4u$. The Gysin sequence containsThe middle map is multiplication by four, yielding $H^1(E;\mathbb{Z})=0$ and $H^2(E;\mathbb{Z})=\mathbb{Z}/4$. The same Gysin sequence gives $H^0(E;\mathbb{Z})=H^3(E;\mathbb{Z})=\mathbb{Z}$. The total space is a closed connected oriented three-manifold, so Poincare duality gives

The exterior. Work first with $W={\mathbb{CP}}^2\setminus \mathrm{int}N$. A collar neighborhood of $\partial W$ shows that its interior, the requested ${\mathbb{CP}}^2\setminus N$, has the same homotopy equivalence type: push the boundary a small positive distance into the collar. The Excision theorem and Thom isomorphism theorem identifyOnly degrees two and four are nonzero, each a copy of $\mathbb{Z}$.

In degree four the map $H_4({\mathbb{CP}}^2;\mathbb{Z})\to H_4(N,E;\mathbb{Z})$ sends the ambient fundamental class to the relative fundamental class of $N$. Under the Thom isomorphism theorem this becomes $[S]$. Thus this map is multiplication by one with compatible orientations. The long exact sequence in homology givesso $H_4(W)=H_3(W)=0$.

In degree two, the Thom isomorphism theorem identifies the map $H_2({\mathbb{CP}}^2)\to H_0(S^2)$ with intersection against $[S]$. A projective line intersects the conic twice, so this map is multiplication by two. The long exact sequence in homology isIt yields $H_2(W)=0$ and $H_1(W)=\mathbb{Z}/2$. In degree zero the relative groups vanish, so $H_0(W)\cong H_0({\mathbb{CP}}^2)\cong \mathbb{Z}$. This computes the homology of the complement of a smooth conic:The normal Euler number $4$, the degree-two intersection map and the degree-four restriction map $1$ play different roles; distinguishing them explains why the boundary has first homology $\mathbb{Z}/4$ while the exterior has first homology $\mathbb{Z}/2$.

Choose the Hopf map $\eta$ and orientations so its Hopf invariant is one. The cellular cohomology of $Y_d$ is $\mathbb{Z}$ in degrees $0,2,4$ and zero otherwise. Let $x,z$ be generators in degrees two and four. Precomposing $\eta$ with a degree-$d$ self-map of $S^3$ represents $d\eta$. The induced map of mapping cones is the identity on the two-cell and has degree $d$ on the four-cell. Since the Hopf mapping cone has cup square equal to its top generator, naturality of the cup product gives $x^2=dz$. Equivalently, precomposition scales the Hopf invariant by degree. HenceThe coefficient is $d$, not $d^2$: multiplication in ${\pi }_3(S^2)$ is being used, rather than postcomposition by a degree-$d$ map of the target sphere.

For $X=X_{d,f}$, the only nonzero cellular boundary is $C_4\to C_3$, multiplication by $f$. Thus its cellular cohomology givesLet $x$ again generate $H^2$, and let $z$ be the class of the four-cell cochain. Collapsing the three-sphere gives a map $X\to Y_d$ which pulls the degree-two generator back to $x$ and the degree-four generator back to $z$. Hence $x^2=dz$, now interpreted modulo $f$. Every other product of positive-degree classes vanishes by dimension. This is the cohomology ring of a Hopf attachment with a sphere summand. More explicitly, if $f\ne 0$,If $f=0$, there is additionally a generator $y$ of degree three, with $xy=y^2=yz=0$. The displayed description of the groups uses $\mathbb{Z}/0\mathbb{Z}=\mathbb{Z}$, so the zero case is included. In particular, a nonzero $f$ kills the degree-three cohomology but can leave torsion in degree four.

There are no one-cells, and attaching cells of dimension at least three does not change ${\pi }_1$. Thus $X$ is simply connected. The Hurewicz theorem in degree two gives

To compute ${\pi }_3$, choose a map $g:X\to {\mathbb{CP}}^{\infty }=K(\mathbb{Z},2)$ representing $x$. This Eilenberg–MacLane space can be realized by the classifying space of $S^1$. On $S^2$ the map is the standard inclusion; on $S^3$ it is constant, and it extends over the four-cell since ${\pi }_3(K(\mathbb{Z},2))=0$. Its homotopy fibre $T$ is the pullback of the universal circle bundle $S^{\infty }\to {\mathbb{CP}}^{\infty }$. The map on ${\pi }_2$ is an isomorphism, so the long exact sequence of homotopy groups of a fibration gives ${\pi }_1(T)={\pi }_2(T)=0$ and ${\pi }_3(T)\cong {\pi }_3(X)$. Therefore the Hurewicz theorem identifies the latter with $H_3(T;\mathbb{Z})$.

Over $W=S^2\vee S^3$, the circle-bundle total space isThe first summand is the Hopf fibration total space over $S^2$, and the second is the trivial bundle over $S^3$. Their intersection is the fibre over the wedge point. The Mayer–Vietoris sequence gives $H_3(P)=\mathbb{Z}A\oplus \mathbb{Z}B$, with generators from the two three-spheres. These are the lifts of ${\eta }_1$ and $i_2$. The circle bundle over $W$ also has a 2-connected total space, so lifting and the Hurewicz theorem identify $d{\eta }_1+f{i}_2$ with $dA+fB$.

Over the attached four-disc the bundle is trivial. The resulting relative pair has the homology of $(S^1\times D^4,S^1\times S^3)$, so $H_4(T,P)=\mathbb{Z}$ and $H_3(T,P)=0$. Its boundary map sends a generator to the lift of the attaching map, namely $dA+fB$. The long exact sequence in relative homology now givesApplying Smith normal form to this one relation proves the third homotopy group of a Hopf attachment with a sphere summand:Since $d>0$, the greatest common divisor is positive even when $f=0$. This derivation uses the required map to an Eilenberg–MacLane space and determines the extension, rather than merely the orders of its pieces.

Use the standard complex K-theory Thom class convention. For a rank-$r$ complex vector bundle, its K-theory Euler class is the pullback of the K-theory Thom class along the zero section, namelyHere ${\Lambda }^j$ denotes an exterior power, and a Hermitian metric identifies the dual bundle with the conjugate vector bundle. Choosing ${\lambda }_{-1}(E)$ instead changes the K-theory Euler class by the unit $(-1{)}^r[\det E]$, so its kernel and cokernel computations are unchanged.

For the disk bundle $D(E)$ and sphere bundle $S(E)$, the K-theory Thom class identifies $K^i(D(E),S(E))$ with $K^i(X)$. The exact sequence of this pair becomes the K-theory Gysin sequence of a sphere bundle:The groups are two-periodic by Bott periodicity.

View Real projective space as the circle bundle of ${\gamma }^{\otimes 2}$ over Complex projective space, where $\gamma \to {\mathbb{CP}}^n$ is the tautological bundle. Indeed, $S^{2n+1}/\{\pm 1\}={\mathbb{RP}}^{2n+1}$ maps to $S({\gamma }^{\otimes 2})$ by $[v]\mapsto v\otimes v$; in each fiber squaring identifies exactly $v$ and $-v$.

Use Complex K-theory of complex projective space in the formThe K-theory Euler class of ${\gamma }^{\otimes 2}$ is $1-[\gamma {]}^{-2}$. Multiplying it by the unit $-[\gamma {]}^2$ gives $1-[\gamma {]}^2=2t-t^2$. The K-theory Gysin sequence of a sphere bundle therefore givesUnder the projection $q:{\mathbb{RP}}^{2n+1}\to {\mathbb{CP}}^n$, the pullback vector bundle $q^{\ast }\gamma$ is the complexification of a real vector bundle $L$ of the tautological bundle of Real projective space: the fiber map sends $[v,z]$ to $zv$. Thus $t$ pulls back to $1-[L]$. Put $\mu =[L]-1$. The relation $L^{\otimes 2}\cong 1$ gives ${\mu }^2=-2\mu$, and $t^{n+1}=0$ gives $2^n\mu =0$. HenceFor completeness, these relations give precisely the abelian group $\mathbb{Z}\oplus \mathbb{Z}/2^n$. In the basis $t,\dots ,t^n$, the image of multiplication by $2t-t^2$ is generated by $2t-t^2,2t^2-t^3,\dots ,2t^n$; its relation matrix is triangular with determinant $2^n$, and the relations express every $t^j$ as $2^{j-1}t$. Thus there is exactly one cyclic torsion subgroup of order $2^n$.

If $n\ge 1$ and $a=a_0+a_{1}t+\cdots +a_n{t}^n$ is in the kernel, successive coefficients of $(2t-t^2)a$ give $2a_0=0$, $2a_1-a_0=0$, and so on, up to $2a_{n-1}-a_{n-2}=0$. Hence $a_0=\cdots =a_{n-1}=0$, with $a_n$ arbitrary. For $n=0$ the multiplier is zero on $R=\mathbb{Z}$. In all cases,

Finally let $E=L^{\oplus k}$ with $k\ge 1$, over $B={\mathbb{RP}}^{2n+1}$. Since $L^{\ast }\cong L$, multiplicativity of the K-theory Euler class givesOn $K^0(B)=\mathbb{Z}\oplus (\mathbb{Z}/2^n)\mu$, the image of multiplication by this K-theory Euler class is the subgroup generated by $2^{k-1}\mu$: multiplication by an integer produces every such multiple, while multiplication by $\mu$ only adds multiples of $2^k\mu$. ThereforeMultiplication by $\mu$ is zero on $K^{-1}(B)\cong \mathbb{Z}$, since $2^n\mu =0$ and this abelian group has no nonzero torsion elements. Thus the K-theory Gysin sequence of a sphere bundle gives a short exact sequenceIt splits as an exact sequence of abelian groups because its quotient is a free abelian group. ConsequentlyThe convention $\mathbb{Z}/1=0$ covers $n=0$ and $k=1$. For example, when $k=1$ the sphere bundle is $(S^{2n+1}\times S^1)/\{(v,z)\sim (-v,-z)\}$, a mapping torus of the antipodal map; its Topological K-theory has no nonzero torsion elements, agreeing with the formula. If $k=0$ is allowed, the rank-zero vector bundle has an empty unit sphere bundle, so $K^0(S(E))=0$ instead.

Any symplectic form on $S^2$ orients its tangent bundle. Choose a compatible complex structure and inner product; this reduces the structure group to $SO(2)$. Since every line in an oriented symplectic plane is Lagrangian,is an oriented circle bundle.

If a transition function of $T{S}^2$ rotates vectors through an angle $\alpha$, its action on unoriented lines rotates the coordinate $e^{2i\theta }$ through $2\alpha$. The Euler class of the Lagrangian-line bundle of an oriented plane bundle therefore givesBy the Poincaré-Hopf theorem,for the chosen orientation, up to changing both signs. Hence the Euler number of $\mathcal{L}Gr(T{S}^2)$ is $\pm 4$, which is nonzero. A smoothly trivial oriented circle bundle has zero Euler class, so $\mathcal{L}Gr(T{S}^2)\to S^2$ cannot be smoothly trivial for any choice of symplectic form.

Map $X_{d,f}$ to $K(\mathbb{Z},2)={\mathbb{CP}}^{\infty }$ using its degree-two generator. The homotopy fibre is the total space of the corresponding circle bundle, is 2-connected, and has ${\pi }_3$ equal to that of $X_{d,f}$. Over $S^2\vee S^3$ its total space is $P=S^3{\cup }_{S^1}(S^1\times S^3)$, with $H_3(P)=\mathbb{Z}A\oplus \mathbb{Z}B$. The four-cell bundle is trivial; its relative degree-four generator has boundary $dA+fB$, the lifted attaching map. The long exact sequence in relative homology and Hurewicz theorem therefore give ${\pi }_3(X_{d,f})={\mathbb{Z}}^2/⟨(d,f)⟩$. Smith normal form gives the displayed expression when $d>0$, including $f=0$.

The diagonal quotient $L(p)=S^3/(\mathbb{Z}/p)$ is the unit circle bundle of a complex line bundle over $S^2$ whose Euler class is $p$ times a generator. Its Gysin sequence giveswhile $H^j(L(p);{\mathbb{F}}_p)\cong {\mathbb{F}}_p$ for $0\le j\le 3$.