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Use ${\mathbb{F}}_2$ coefficients throughout. Write $h=c_1(\mathcal{O}(1))\mathrm{mod}2$ on Complex projective space and $a=w_1({\gamma }_d)$ on Real projective space. We use the standard cohomology ring of complex projective space ${\mathbb{F}}_2[h]/(h^{d+1})$, with $|h|=2$, and the real projective calculation from the preceding solution. The complex projective calculation follows, for example, from its one cell in each even dimension together with the fact that transverse projective hyperplanes represent the powers of $h$; $d$ such hyperplanes have one intersection point.

For the standard inclusion $i:{\mathbb{RP}}^d\hookrightarrow {\mathbb{CP}}^d$, the restricted complex tautological line is ${\gamma }_d{\otimes }_{\mathbb{R}}\mathbb{C}$. As a real vector bundle it is ${\gamma }_d\oplus {\gamma }_d$, whose total Stiefel–Whitney class is $(1+a{)}^2=1+a^2$. The Stiefel–Whitney class of the underlying real bundle of a complex line identifies its degree-two class with its First Chern class reduced modulo two. Passing to the dual line defining $h$ changes only a sign, which disappears modulo two. Thus mod-two restriction from complex to real projective space gives $i^{\ast }h=a^2$. In particular the answer for dimension two is the ring mapIt sends $1$ to $1$, is an isomorphism in degree two, and sends $h^2$ to zero.

Now assume $k\ge 1$ and put $M={\mathbb{CP}}^{2k}$, $N={\mathbb{RP}}^{2k}$, $U=M\setminus N$, with inclusions $i:N\hookrightarrow M$ and $j:U\hookrightarrow M$. The real dimensions are $4k$ and $2k$, respectively. A tubular neighborhood and excision identify the cohomology of $(M,U)$ with that of the normal disk bundle relative to its sphere bundle. Every real normal bundle is oriented over ${\mathbb{F}}_2$, so the Thom isomorphism theorem givesUnder this identification the relative-to-absolute map is the cohomological Gysin map of an embedding $i_{!}$. The exact sequence of the pair is thereforeHere $\delta$ denotes the ordinary connecting homomorphism followed by the inverse of the isomorphism supplied by the Thom isomorphism theorem.

To compute $i_{!}$, use mod-two Poincare duality and its evaluation formulaFor $0\le r\le k$, take $b=a^{2r}$ and $z=h^{k-r}$. The right side is $⟨a^{2k},[N{]}_2⟩=1$. Since the target group is generated by $h^{k+r}$ and $⟨h^{2k},[M{]}_2⟩=1$, this provesOdd-degree classes have zero image because $M$ has no odd-degree cohomology. Exactness now givesFor precision, below degree $2k-1$ the restriction map is an isomorphism. In degree $2k-1$ the next map is $i_{!}:H^0(N)\to H^{2k}(M)$, an isomorphism, so the complement group is zero. In even degrees $2k+2r$, $0\le r<k$, the preceding $i_{!}$ is an isomorphism and the following odd-degree $i_{!}$ is zero; consequently $\delta :H^{2k+2r}(U)\to H^{2r+1}(N)$ is an isomorphism. The intervening odd groups and the top group $H^{4k}(U)$ vanish.

It remains to establish multiplication. Put $x=j^{\ast }h$. Since $h^k=i_{!}1$, exactness gives $x^k=0$. Choose $y\in H^{2k}(U)$ with $\delta y=a$, which is possible by the just-computed isomorphism. Naturality of the relative cup product, and the module property of the Thom isomorphism theorem, giveThus $\delta (x^{r}y)=a^{2r+1}$ for $0\le r<k$. Each class $x^{r}y$ is nonzero and generates its upper-half group, while $1,x,\dots ,x^{k-1}$ generate the lower-half groups. Also $y^2=0$ because $H^{4k}(U)=0$. These classes account for every group, so there are no further relations:By the field-coefficient Künneth theorem this is precisely the cohomology ring of ${\mathbb{CP}}^{k-1}\times S^{2k}$. For $k=1$ the relation $x=0$ leaves the sphere's ring. The calculation identifies rings and does not claim a homotopy equivalence of the spaces; $k=0$ is outside the expression involving ${\mathbb{CP}}^{k-1}$.

The tautological bundle $\mathcal{O}(-1)$ over Complex projective space has fibre over $[\ell ]$ equal to the line $\ell \subset {\mathbb{C}}^{r+1}$. Define $\mathcal{O}(1)=\mathcal{O}(-1{)}^{\ast }$ and the twisting sheaf on projective space, viewed analytically as a holomorphic line bundle, by

For ${\mathbb{CP}}^1$, use coordinates $z=Z_1/Z_0$ and $w=Z_0/Z_1=1/z$. The tautological frames $(1,z)$ and $(w,1)$ differ by $z^{-1}$, so frames $e_0,e_1$ for $\mathcal{O}(d)$ satisfy $e_1=z^d{e}_0$. A global holomorphic section is therefore described by entire functions $s_0(z),s_1(w)$ withWrite $s_0(z)={\sum }_{k\ge 0}{a}_k{z}^k$. Holomorphicity of $s_1$ at zero permits only the powers $w^{d-k}$ with $d-k\ge 0$. For $d\ge 0$, this says $s_0$ is a polynomial of degree at most $d$; for $d<0$, every coefficient must vanish. Equivalently, when $d\ge 0$ the sections are homogeneous polynomials of degree $d$ in $Z_0,Z_1$. Thus

As usual for Complex projective space and Real projective space, homogeneous coordinates are represented by nonzero vectors. Componentwise complex conjugation descends to the diffeomorphismIt is well defined because $\overline{\lambda z}=\overline{\lambda }\overline{z}$, and ${\tau }^2=I$. Every standard manifold chart is invariant under $\tau$, with ${\phi }_i\circ \tau =\sigma \circ {\phi }_i$. Part (d) and the gluing of the Fubini-Study form therefore imply ${\tau }^{\ast }{\omega }_{FS}=-{\omega }_{FS}$.

A point $[z]$ in the fixed-point set satisfies $\overline{z}=\lambda z$. Applying complex conjugation again gives $|\lambda |=1$. Choose $a\ne 0$ with $a/\overline{a}=\lambda$; then $\overline{az}=az$, so $[z]$ has a real representative. Conversely every real representative determines a point in the fixed-point set. Hence $\mathrm{Fix}(\tau )={\mathbb{RP}}^n$, and part (c) gives

For the First Chern class identity, use finite-dimensional classifying maps. A complex line bundle on a compact Hausdorff space embeds in a finite-dimensional trivial vector bundle, so it is a pullback vector bundle of the tautological bundle on some Complex projective space. For the two complex line bundles, the universal tensor product of vector bundles is classified by the mapIts restriction to either factor pulls the tautological bundle back to the tautological bundle on that factor. The degree-two cohomology of the product is the direct sum of the degree-two cohomology of its factors, so these two restrictions determine its First Chern class. Pulling back to $X$ provesThis proves the integral identity, including possible torsion elements in the cohomology of $X$; an argument using only curvature forms would not detect those torsion elements.

The splitting principle for complex vector bundles says that for a vector bundle $E\to X$ there is an iterated projective bundle $p:F(E)\to X$ for which $p^{\ast }$ is injective on integral cohomology, and $p^{\ast }E$ is a direct sum of complex line bundles. The same assertion holds with rational coefficients, and a common such space can split any finite collection of vector bundles. The injectivity follows at each stage from the projective bundle formula for complex vector bundles; a Hermitian metric splits the resulting filtration into a direct sum.

Write $x_i=c_1(L_i)$ for the Chern roots of the pulled-back vector bundle. Define the degree-$2j$ component of its Chern character byThe Fundamental theorem of symmetric polynomials expresses each power-sum symmetric polynomial ${\sum }_i{x}_i^j$ uniquely as an integral polynomial in the elementary symmetric polynomials, which are the Chern classes of $E$. Thus the definition makes sense on $X$ itself, without choosing actual Chern roots there. On a compact Hausdorff space, each vector bundle is classified by a finite-dimensional Grassmannian; its high-degree Chern character components therefore vanish. Hence this defines a class in the direct sum $H^{2\ast }(X;\mathbb{Q})$ even if $X$ has no finite dimension. On more general spaces an infinite Chern character is instead interpreted in the product of the even-degree cohomology groups.

On a common splitting space, the Chern roots of a direct sum are the combined root lists, whereas the First Chern class identity makes the Chern roots of a tensor product of vector bundles the pairwise sums $x_i+y_j$. ConsequentlyThe injectivity in the splitting principle for complex vector bundles descends both identities to $X$. Additivity extends the Chern character to the Grothendieck group by $\mathrm{ch}([E]-[F])=\mathrm{ch}(E)-\mathrm{ch}(F)$. Multiplicativity and $\mathrm{ch}(1)=1$ then show that the Chern character is a unital ring homomorphism.

For Complex K-theory of complex projective space, let $\gamma$ be the tautological bundle, put $t=1-[\gamma ]$, and set $h=-c_1(\gamma )$. The cohomology ring of complex projective space is $\mathbb{Z}[h]/(h^{n+1})$. The CW filtration gives a cofibrationThe K-theory six-term exact sequence and Complex K-theory of a sphere give, inductively,Here $b_n$ is the image of a Bott element from $S^{2n}$, normalized so that $\mathrm{ch}(b_n)=h^n$. This normalization follows from the multiplicative form of Bott periodicity: the Chern character of the degree-two Bott element is an integral generator, and multiplication yields the top-degree generator on $S^{2n}$. In particular, the abelian group $K^0({\mathbb{CP}}^n)$ is a free abelian group of rank $n+1$.

Assume inductively that $1,t,\dots ,t^{n-1}$ form an integral basis on ${\mathbb{CP}}^{n-1}$. Their lifts, together with $b_n$, form an integral basis on ${\mathbb{CP}}^n$. Sincethe Chern characters of this basis are linearly independent over $\mathbb{Q}$: their first nonzero degrees are $0,2,\dots ,2n$. Thus the Chern character is injective on $K^0({\mathbb{CP}}^n)$. By induction $t^n$ restricts to zero on ${\mathbb{CP}}^{n-1}$, so $t^n=a{b}_n$ for an integer $a$. Comparing Chern characters gives $a=1$. Also $\mathrm{ch}(t^{n+1})=0$, so $t^{n+1}=0$. This proves the integral answer, rather than merely its rationalization:The induction begins with ${\mathbb{CP}}^0$ a point and $t=0$.

Let $E$ have rank $r\ge 1$, let $p:\mathbb{P}(E)\to X$ be its projectivization of a real vector bundle, and let $\lambda \subset p^{\ast }E$ be its tautological bundle. Put $u=w_1(\lambda )\in H^1(\mathbb{P}(E);{\mathbb{F}}_2)$. The Projective bundle definition of Stiefel–Whitney classes starts from the isomorphismIn particular, $p^{\ast }$ is injective. Expand $u^r$ uniquely in this basis. The coefficients define the Stiefel–Whitney classes through the relationSet $w_0(E)=1$ and $w_i(E)=0$ for $i>r$; for a rank-zero vector bundle set $w(E)=1$. Uniqueness also proves naturality of the Stiefel–Whitney classes under a pullback vector bundle.

To prove the Whitney product formula for Stiefel–Whitney classes, apply the splitting principle for real vector bundles simultaneously to $E$ and $E^{\prime }$. Its injective pullback on mod-two cohomology follows by iterating the Projective bundle definition of Stiefel–Whitney classes. It suffices to compute for a split vector bundle $E={⨁}_{i=1}^r{L}_i$. Put $x_i=w_1(L_i)$.

On $\mathbb{P}(E)$, the coordinate projections of the tautological inclusion $\lambda \hookrightarrow p^{\ast }E$ give sections of the real line bundles ${\lambda }^{\ast }\otimes p^{\ast }{L}_i$ with no common zero. The mod-two Euler class of a real line bundle is its first Stiefel–Whitney class, and the First Stiefel–Whitney class of a tensor product of real line bundles gives $u+p^{\ast }{x}_i$. Multiplication of Euler classes under a direct sum, and vanishing of the Euler class of a vector bundle with a nowhere-zero section, therefore giveComparing with the unique monic relation defining the Stiefel–Whitney classes shows that $w_j(E)$ is the $j$th elementary symmetric polynomial in the $x_i$. Thus $w(E)={\prod }_i(1+x_i)$. A direct sum joins the two lists, so injectivity of the splitting pullback gives

For the immersion obstruction, use the notation $g=(f,f)$: the printed $f\times f$ here means the map from one copy of Real projective space into the product. Write $x=w_1({\gamma }_{\mathbb{R}})$ and let $z$ be the mod-two reduction of $c_1({\gamma }_{\mathbb{C}}^{\ast })$. The mod-two cohomology ring of real projective space and the cohomology ring of complex projective space giveNontriviality in degree two forces $f^{\ast }z=x^2$. Necessarily $k\ge 1$, since the degree-two cohomology of ${\mathbb{CP}}^0$ vanishes.

The standard tangent bundle identities for Real projective space and Complex projective space areFor the underlying real vector bundle of a complex line bundle, its total Stiefel–Whitney class is $1+c_1\mathrm{mod}2$. HenceIf $g$ is homotopic to an immersion $j$, its normal bundle $\nu$ has rank $4k-(2k+1)=2k-1$, andThe Whitney product formula for Stiefel–Whitney classes and homotopy invariance of cohomology yieldThe division is legitimate in the truncated cohomology ring, since $1+x$ is a unit. The degree-$2k$ component isIt must vanish because $2k>\mathrm{rank}\nu$, while $x^{2k}\ne 0$. Therefore $k$ must be odd.

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.

A holomorphic line bundle is ample when some positive tensor power is very ample, so its sections define an embedding into Complex projective space. It is a positive holomorphic line bundle when it has a Hermitian metric with positive Chern curvature $i{F}_{\nabla }$. The Kodaira embedding theorem says that a compact complex manifold carrying a positive holomorphic line bundle is projective; sufficiently high tensor powers give a holomorphic embedding into projective space.

For a compact space $X$, $K^0(X)$ is the Grothendieck group of isomorphism classes of finite-rank complex vector bundles under direct sum. Tensor product descends to this group and makes it a ring, with $[{\mathbb{C}}_X]$ as its unit.

The hypothesis $E_0\oplus {\mathbb{C}}_X\cong E_1\oplus {\mathbb{C}}_X$ says that $E_0$ and $E_1$ define the same stable class. Rank-$d$ bundles are classified by maps to $BU(d)$, and the stabilization $BU(d)\to BU$ is $2d$-connected. Since the finite CW complex $X$ has dimension $k\le 2d$, stabilization is injective on $[X,BU(d)]$. Thus stable cancellation for complex vector bundles gives

Now let $E_0,E_1$ have rank $d$ over Complex projective space ${\mathbb{CP}}^d$ and have the same Chern classes. Their Chern characters agree because each component of $\mathrm{ch}(E)$ is a universal rational polynomial in the Chern classes. The ringis torsion-free, while the Chern character becomes an isomorphism after tensoring with $\mathbb{Q}$; it is therefore injective. Hence $[E_0]=[E_1]$ in K-theory, so after adding trivial bundles they are isomorphic. Since ${\mathbb{CP}}^d$ has real dimension $2d$, cancellation applies once more:

The tautological bundle over Complex projective space isOn $U_i=\{z_i\ne 0\}$, the vector $e_i([z])=z/z_i$ is a holomorphic frame. On $U_i\cap U_j$,so the transition functions are holomorphic. Define $\mathcal{O}(1)=\mathcal{O}(-1{)}^{\ast }$ and, for $k\in \mathbb{Z}$, define $\mathcal{O}(k)=\mathcal{O}(1{)}^{\otimes k}$ when $k\ge 0$ and $\mathcal{O}(k)=\mathcal{O}(-1{)}^{\otimes (-k)}$ when $k<0$. Its transition functions are the corresponding $k$th powers of those of $\mathcal{O}(1)$, equivalently the $(-k)$th powers of those of $\mathcal{O}(-1)$.

A nonzero linear functional $\ell \in ({\mathbb{C}}^{n+1}{)}^{\ast }$ restricts on each projective line to define a global holomorphic section $s_{\ell }$ of $\mathcal{O}(1)$. Its zero divisor is the projective hyperplane $H_{\ell }=\{[z]:\ell (z)=0\}$, so every hyperplane is an effective divisor of $\mathcal{O}(1)$.

Conversely, let $D$ be the divisor of a nonzero section of $\mathcal{O}(1)$. In the affine chart $z_0=1$, the section is represented by an entire function $f$ on ${\mathbb{C}}^n$. Compatibility with the other projective charts gives the linear growth bound in the question along every complex affine line. The supplied Liouville-type result makes $f$ affine-linear, and homogenizing it gives a linear functional $\ell$ on ${\mathbb{C}}^{n+1}$. Thus $D=H_{\ell }$.

In local coordinates centred at $x$, the blowup of a complex manifold at a point is modeled byOn the chart $v_i\ne 0$, write $u=z_i$ and $t_j=v_j/v_i$; then $z_j=u{t}_j$, giving holomorphic coordinates $(u,(t_j{)}_{j\ne i})$. These charts show that the blowup is a complex manifold, that $\sigma$ is holomorphic, and that its exceptional fiber is ${\mathbb{CP}}^{n-1}$. A biholomorphic coordinate change at the centre lifts by sending a punctured point together with its limiting tangent direction to its image; the chart formulas extend across the exceptional divisor. Hence the construction is independent of coordinates up to biholomorphism.

For $a(z)=-z$ on ${\mathbb{C}}^2$, defineThis is holomorphic and satisfies $\sigma \circ \tilde{a}=a\circ \sigma$. The blowup $\widetilde{mathbb{C}^2}$ is the total space of $\mathcal{O}(-1)\to {\mathbb{CP}}^1$, and $\tilde{a}$ acts as multiplication by $-1$ in every fiber. The invariant fiber coordinate is $w=u^2$. If $u_j=g_{ji}{u}_i$ are the transition laws for $\mathcal{O}(-1)$, then $w_j=g_{ji}^2{w}_i$, which are the transition laws for $\mathcal{O}(-2)$. Thus the local quotient maps $(t,u)\mapsto (t,u^2)$ glue, give the quotient a complex-manifold atlas even along the fixed zero section, and yield

A Hermitian metric on a holomorphic vector bundle $E\to X$ is a smoothly varying family of positive-definite Hermitian forms $h_x$ on its fibers. A Chern connection is a connection on a vector bundle $\nabla$ that is compatible with $h$ and whose $(0,1)$ part is the bundle's Dolbeault partial connection, ${\nabla }^{0,1}={\bar{\partial }}_E$.

Choose a local holomorphic frame $e$ and write $h=h(e,e)>0$. The connection form in this frame isIf $e^{\prime }=ge$ for a nowhere-zero holomorphic function $g$, then $h^{\prime }=|g{|}^{2}h$ andwhich is exactly the connection-form transformation law. The local formulas therefore define a global connection.

Identify $\mathcal{O}(-1)$ with the tautological bundle over Complex projective space. The standard Hermitian inner product of ${\mathbb{C}}^{n+1}$ restricts to each tautological line. On the affine chart $U_i=\{Z_i\ne 0\}$, put $w_j=Z_j/Z_i$ for $j\ne i$ and use the holomorphic frameThenThe curvature form of a connection isConsequently Chern-Weil theory gives the closed representativewhere ${\omega }_{FS}$ is the Fubini-Study form.

Write a point of Real projective space as a nonzero vector $(w_1,w_2)\in {\mathbb{C}}^2$ modulo real nonzero scaling. The map to Complex projective space sends its class to the complex line through that vector. A free right action of the circle group is $[w]\cdot u=[\lambda w]$ where ${\lambda }^2=u$; the two possible roots differ by a real sign and therefore give the same projective point. Local smooth square-root branches prove smoothness. Over $[z:1]$, the fibre coordinate is $(w_2/|w_2|{)}^2$; over $[1:\zeta ]$, it is $(w_1/|w_1|{)}^2$. On the overlap $\zeta =1/z$, the transition function of a principal bundle is $u\mapsto (z/\overline{z})u$. Its phase winds twice. Ordinary scalar multiplication by $u$ instead has a kernel of order two and is not a free action of a group.

The divisor on a complex manifold of a nonzero global meromorphic function records its zero and pole orders along irreducible analytic hypersurfaces. It is called principal. Its holomorphic line bundle associated to a divisor is trivial: choosing the same global meromorphic function as the local equation on every open set gives identity transition functions. In particular, two hyperplanes in Complex projective space have isomorphic associated bundles, since the quotient of their homogeneous linear equations is a global meromorphic function with their difference as divisor.

A complex-linear endomorphism $A$ of ${\mathbb{C}}^{n+1}$ induces a holomorphic vector field on Complex projective space by differentiating the projective transformations $[z]\mapsto [e^{tA}z]$. Its zeroes are precisely the projective eigenlines of $A$; a diagonalizable $A$ with distinct eigenvalues therefore gives exactly $n+1$ zeroes.

For $H={\sum }_j(q_j^2+p_j^2)/2$ and $E>0$, the oscillator flow is a free circle action on the sphere of radius $\sqrt{2E}$. Its quotient is Complex projective space. For two modes, the reduced form in $w=z_2/z_1$ is $iEdw\wedge d\bar{w}/(1+|w{|}^2{)}^2$, of total area $2\pi E$. Reduction at zero energy is singular instead of a smooth quotient of the asserted dimension.