The smooth quintic threefold as a compactification geometry

A projective hyperplane is the transverse zero locus of a section of $\mathcal{O}(1)$, so [F3] and [F4] identify its Poincaré dual with $H_{{\mathbb{P}}^4}=c_1(\mathcal{O}(1))$. Take the four coordinate hyperplanes $z_1=0,z_2=0,z_3=0,z_4=0$ in homogeneous coordinates $[z_0:z_1:z_2:z_3:z_4]$. Their common intersection is the single point $[1:0:0:0:0]$; on the chart $z_0\mathrm{\ne }0$, their local equations are the independent complex coordinates $z_i\mathrm{/}{z}_0$, $1\le i\le 4$, so the intersection is transverse and has positive sign for the complex orientations. Repeated application of [F5] gives $⟨H_{{\mathbb{P}}^4}^4,[{\mathbb{P}}^4]⟩=1$.

Choose a homotopy equivalence $r:K\to X$ from a path-connected CW complex, available by [F10]. The tangent bundle $TX$ is numerable by [F10], and $r^{\ast }TX$ is numerable because its numerating cover and subordinate partition pull back as in [F11]. By [F13], the projective-bundle quotient identifies $P(r^{\ast }TX)$ with $K{\times }_{X}P(TX)$ and pulls back the tautological line; Euler naturality [F14] pulls back its class. Pulling back the unique projective-bundle relation for $TX$ from [F15] therefore gives a monic relation for $r^{\ast }TX$ with coefficients $r^{\ast }{c}_i(TX)$; uniqueness in [F15] yields $r^{\ast }{c}_3(TX)=c_3(r^{\ast }TX)$. The complex orientation pulls back by [F14], so the top-Chern identity [F6] on $K$ gives $r^{\ast }{c}_3(TX)=r^{\ast }e(T{X}_{\mathbb{R}})$. By [F12], $r^{\ast }:H^6(X;\mathbb{Z})\to H^6(K;\mathbb{Z})$ is an isomorphism, hence $c_3(TX)=e(T{X}_{\mathbb{R}})$ on $X$.

By [A1] and [F3], $c_1(\mathcal{O}(5))=5c_1(\mathcal{O}(1))$. The zero locus $X$ is a transverse complex divisor, and [F4] with rank $2$ and codimension $2$ gives $[X]=5H_{{\mathbb{P}}^4}\cap [{\mathbb{P}}^4]$ with positive sign. Therefore [F5] gives ${\int }_X{H}^3=⟨H_{{\mathbb{P}}^4}^3⌣5H_{{\mathbb{P}}^4},[{\mathbb{P}}^4]⟩=5$ by step 1.1. This verifies the degree-five normalization used in [F2].

The unit sphere $S^9\subset {\mathbb{C}}^5$ is compact by [F9]. The map $q:S^9\to {\mathbb{P}}_{\mathbb{C}}^4$, $z\mapsto \mathbb{C}z$, is surjective because every nonzero vector has a unit representative. It is continuous: over the projective chart $z_i\mathrm{\ne }0$, its coordinate functions are the continuous ratios $z_j\mathrm{/}{z}_i$ on that open subset of the sphere. Given an open cover of ${\mathbb{P}}^4$, the inverse images under $q$ form an open cover of $S^9$; a finite subcover exists by [F9], and surjectivity makes the corresponding original opens cover ${\mathbb{P}}^4$. Thus projective space is compact by the open-cover definition. In each projective chart the equation of $X$ is the zero set of the continuous polynomial $f(z\mathrm{/}{z}_i)$; its complement is open in that chart, and hence ${\mathbb{P}}^4\setminus X$ is open. For any relative open cover $\{W_{\lambda }\}$ of $X$, use the stated AC to choose ambient opens $V_{\lambda }$ with $W_{\lambda }=X\cap V_{\lambda }$. The family $\{V_{\lambda }\}\cup \{{\mathbb{P}}^4\setminus X\}$ covers projective space and therefore has a finite subcover. Removing the complement leaves a finite subcover of $X$ by the $W_{\lambda }$. This proves compactness of $X$ directly. Connectedness remains the given hypothesis. Smoothness gives its complex-threefold charts. On an affine projective chart the Fubini–Study form is $i\mathrm{\partial }\bar{\mathrm{\partial }}\log (1+\mathrm{\mid }z{\mathrm{\mid }}^2)$, whose Hermitian matrix is $((1+\mathrm{\mid }z{\mathrm{\mid }}^2){\delta }_{ab}-{\bar{z}}_a{z}_b)\mathrm{/}(1+\mathrm{\mid }z{\mathrm{\mid }}^2{)}^2$; Cauchy–Schwarz makes it positive definite. Its local potentials differ on overlaps by the logarithm of the squared modulus of a nonzero holomorphic transition function, whose mixed derivative is zero. Thus these positive closed forms glue, and their restriction to the complex tangent bundle of $X$ is Kähler. This verifies the geometric input of [F1] directly. Applying [F2] with the verified intersection from step 2.1 yields $K_X\cong {\mathcal{O}}_X$, $c_3(TX)=-40H^3$, and hence ${\int }_X{c}_3(TX)=-40\cdot 5=-200$. The Chern computation and the degree normalization are conditional exactly as stated; in particular, no integral-homology vanishing above the manifold dimension is used.

By [F7] the evaluation $⟨e(T{X}_{\mathbb{R}}),[X]⟩$ equals $\chi (X)$; step 1.2 identifies this Euler class with $c_3(TX)$, whose integral is $-200$ by step 3.1. Thus $\chi (X)=-200$. The broad canonical-bundle conclusion does not imply exact holonomy or Hodge numbers, as [F1] states.