Complex projective space (${\mathbb{CP}}^n$)

Complex projective space is the space of complex lines through the origin in ${\mathbb{C}}^{n+1}$.

In the chart $x_j=X_j/X_0$ of Complex projective space, differentiate the quotient coordinates to obtain this formula for $\pi :{\mathbb{C}}^{n+1}\setminus \{0\}\to {\mathbf{P}}^n$. The kernel is the radial line $\mathbb{C}X$. A degree-one homogeneous vector field on the punctured cone descends to a holomorphic vector field, since its scaling cancels the inverse scaling of $d\pi$. This gives the fiberwise construction of the Euler sequence on complex projective space.

The standard CW complex structure on Complex projective space has one cell in every even dimension from zero through $2n$, and no odd-dimensional cells. Its cellular chain complex therefore has zero differential. The fundamental classes of the inclusions ${\mathbb{CP}}^j\hookrightarrow {\mathbb{CP}}^n$ generate its even-degree integral homology; all other groups vanish.

Every continuous self-map of ${\mathbb{CP}}^{2r}$ has a fixed point. If its pullback sends the integral degree-two generator to $d$ times itself, multiplicativity gives trace $d^j$ in degree $2j$. Thus its Lefschetz number is $1+d+\cdots +d^{2r}$, which is positive: for $d\ne 1$ it is $(1-d^{2r+1})/(1-d)>0$, and at $d=1$ it is $2r+1$. The Lefschetz fixed-point theorem applies. In particular no nontrivial finite group acts freely on this simply connected space.

The Complex projective plane is simply connected. If it were the total space of a covering of degree at least two, it would be the universal cover and have a nonidentity deck transformation. Such a transformation acts by $u\mapsto \pm u$ on $H^{\ast }({\mathbb{CP}}^2)=\mathbb{Z}[u]/(u^3)$, giving Lefschetz number 1 or 3. It therefore has a fixed point, impossible for a nonidentity deck transformation.

Identify ${\mathbb{CP}}^m$ with nonzero homogeneous binary forms of degree $m$, modulo nonzero scalar multiplication. Multiplication of $m$ projective linear forms gives a well-defined holomorphic map: no product is zero, and scaling one form scales the entire coefficient vector. A form with $m$ distinct roots has exactly $m!$ ordered factorizations. In affine root coordinates the derivative is a Vandermonde determinant, nonzero at distinct roots. Each such preimage contributes $+1$ with complex orientations, so the mapping degree is $m!$. This also realizes the least positive degree allowed by the factorial degree obstruction for products of two-spheres.

The complex projective plane is the two-dimensional Complex projective space. Its second homology is generated by the class $H$ of a projective line, with $H^2=1$.

The fundamental class of a complex projective line generates $H_2({\mathbb{CP}}^2;\mathbb{Z})$. Two distinct projective lines are homologous and meet at one transverse point. In complex affine coordinates their two complex tangent lines concatenate to the complex orientation of the plane, so the signed intersection is $+1$. The integral intersection form therefore has matrix $(1)$, giving a unimodular intersection pairing. This computes a self-intersection through distinct representatives of the same class.

A projective hyperplane in ${\mathbb{CP}}^n$ is the zero set of a nonzero complex-linear functional on ${\mathbb{C}}^{n+1}$. It is isomorphic to ${\mathbb{CP}}^{n-1}$ and determines a hyperplane divisor.

The Euler sequence is the short exact sequence of holomorphic vector bundlesTaking top exterior powers gives $\det T{\mathbb{CP}}^n\cong \mathcal{O}(n+1)$.

Dualizing the determinant identity in the Euler sequence on complex projective space giveswhere $H$ is a hyperplane divisor.

For a smooth degree-$d$ hypersurface $X_d\subset {\mathbb{CP}}^n$, the adjunction formula and $[X_d]\cong \mathcal{O}(d)$ give

For a smooth degree-$d$ projective plane curve $X$, the adjunction formula gives $K_X\cong {\mathcal{O}}_X(d-3)$. A hyperplane section has degree $d$, so

The complex projective line is the space of complex lines in ${\mathbb{C}}^2$. It is diffeomorphic to the two-sphere and is the quotient $S^3/S^1$ in the Hopf fibration.

The complex Hopf fibration is the principal circle bundleIt is the unit-sphere bundle of the tautological complex line bundle.

Let $a,b\ge 2$ with $a+b=2n+1$. Collapse the lower cells of $S^a\times S^b$ to obtain $q:S^a\times S^b\to S^{2n+1}$ and compose with the Hopf fibration $p:S^{2n+1}\to {\mathbb{CP}}^n$. Both factor inclusions are killed by q, so every induced homotopy group map is zero; the induced reduced homology maps also vanish. If the composite were nullhomotopic, homotopy lifting property would deform q into the circle fibre. Every map from the simply connected product to that circle is nullhomotopic, contradicting the degree-one top homology map of q.

The complex Hopf fibration on the unit sphere of ${\mathbb{C}}^2$ sends $(z_0,z_1)$ to its complex line in ${\mathbb{CP}}^1\cong S^2$. Its fibres are circles. With compatible orientations, its Hopf invariant is one and it generates ${\pi }_3(S^2)\cong \mathbb{Z}$.

If $x=c_1(\mathcal{O}(1))$ has degree two, thenReducing coefficients modulo two gives

Give Complex projective space its complex orientation. In the homology cross product basis of $H_{2n}({\mathbb{CP}}^n\times {\mathbb{CP}}^n;\mathbb{Z})$, the diagonal has the displayed class. If $a,b$ are the two normalized degree-two hyperplane classes, its Poincare dual is ${\sum }_{j=0}^n{a}^j{b}^{n-j}$. Indeed every degree-$2n$ monomial $a^r{b}^{n-r}$ restricts on the diagonal to the top power of the hyperplane class, whose evaluation on the fundamental class is one. These evaluations determine all coefficients.

Give $(S^2{)}^m$ its product orientation and ${\mathbb{CP}}^m$ its complex orientation. If $u_i$ are the degree-two sphere classes and $h$ the normalized projective class, write $f^{\ast }h={\sum }_i{a}_i{u}_i$. In the source cohomology ring, $u_i^2=0$ and the generators commute. Hence $f^{\ast }{h}^m=m!({\prod }_i{a}_i)u_1\cdots u_m$. Evaluation on the fundamental class gives the displayed mapping degree. Every positive degree is therefore at least $m!$, and the polynomial multiplication map to complex projective space attains that bound.

Let $i:{\mathbb{RP}}^d\hookrightarrow {\mathbb{CP}}^d$ be the standard inclusion. If $h=c_1(\mathcal{O}(1))\mathrm{mod}2$ and $a=w_1({\gamma }_d)$, thenThe restricted complex tautological bundle is the complexification of a real vector bundle ${\gamma }_d{\otimes }_{\mathbb{R}}\mathbb{C}$. Its underlying real bundle is ${\gamma }_d\oplus {\gamma }_d$. The Whitney product formula for Stiefel–Whitney classes gives total class $(1+a{)}^2=1+a^2$, while the Stiefel–Whitney class of the underlying real bundle of a complex line identifies its second class with the reduced First Chern class. Dualizing changes the integral sign but has no effect modulo two. Thus the full induced map on cohomology sends $h^j$ to $a^{2j}$.

For $k\ge 1$, put $U={\mathbb{CP}}^{2k}\setminus {\mathbb{RP}}^{2k}$. Its cohomology ring isHere $x$ is the restricted hyperplane class. The cohomological Gysin map of an embedding sends $a^{2j}$ to $h^{k+j}$ and every odd-degree class to zero, by Poincare duality and mod-two restriction from complex to real projective space. Its exact sequence gives one-dimensional groups in even degrees $0,2,\dots ,4k-2$, and zero otherwise. Choose $y$ with $\delta y=a$. The connecting-map module identity gives $\delta (x^{j}y)=a^{2j+1}$ for $0\le j<k$, so these are the upper-half generators. The lower-half generators are $x^j$. The relations $x^k=0$ and $y^2=0$ follow respectively from $i_{!}1=h^k$ and $H^{4k}(U;{\mathbb{F}}_2)=0$. The result is a ring isomorphism with Complex projective space ${\mathbb{CP}}^{k-1}$ times $S^{2k}$; it does not by itself assert a homotopy equivalence of the spaces.

For $n\ge 1$, every graded-ring automorphism ofacts on $H^2$ by a signed permutation matrix. All eight such matrices are realized by independently applying complex conjugation to the factors and by interchanging the factors.

For $0<k<n$, let $X={\mathbb{CP}}^n/{\mathbb{CP}}^k$ and let $x_i\in H^{2i}(X;\mathbb{Z})$ pull back to $x^i\in H^{2i}({\mathbb{CP}}^n;\mathbb{Z})$. Additively,and its positive-degree multiplication is