Free CHSW action and Euler characteristic of its quintic quotient

The point $[1:-1:0:0:0]$ lies on $Y_0$, so it is nonempty. The partial derivatives of $f=\sum z_i^5$ are $5z_i^4$; they cannot all vanish at a projective point, so $Y_0$ is smooth. Both transformations preserve $f$. The diagonal entries of $AB{A}^{-1}$ are $(1,\zeta ,{\zeta }^2,{\zeta }^3,{\zeta }^4)={\zeta }^4(\zeta ,{\zeta }^2,{\zeta }^3,{\zeta }^4,1)$, hence $AB{A}^{-1}={\zeta }^{4}B$ as matrices and $A,B$ commute in $\mathrm{PGL}(5,\mathbb{C})$. If $A^k{B}^{\mathrm{\ell }}$ is projectively scalar and $k\equiv ̸0(\mathrm{m}\mathrm{o}\mathrm{d}5)$, its coordinate permutation is a nontrivial 5-cycle, a contradiction; if $k=0$ and $\mathrm{\ell }\mathrm{\ne }0$, the entries ${\zeta }^{\mathrm{\ell }},{\zeta }^{2\mathrm{\ell }},{\zeta }^{3\mathrm{\ell }},{\zeta }^{4\mathrm{\ell }},1$ are pairwise distinct and $B^{\mathrm{\ell }}$ is not scalar. Thus the 25 elements $A^k{B}^{\mathrm{\ell }}$ are distinct and form $G\cong {\mathbb{Z}}_5^2$.

Let $H$ be the topological first Chern class of ${\mathcal{O}}_{{\mathbb{P}}^4}(1){\mathrm{\mid }}_{Y_0}$. Three coordinate hyperplanes $z_1=z_2=z_3=0$ cut $Y_0$ in the five points of $z_4^5+z_5^5=0$ on ${\mathbb{P}}^1$. In the chart $z_5=1$, these are the five distinct roots of $t^5+1$; its derivative $5t^4$ is nonzero at every root, so the intersections are transverse. Each point has positive local intersection sign because all varieties carry their complex orientations. Therefore ${\int }_{Y_0}{H}^3=5$. Now [F1] applies and gives ${\int }_{Y_0}{c}_3(T{Y}_0)=-40\cdot 5=-200$.

A projective fixed point of $A^k{B}^{\mathrm{\ell }}$ is an eigenline. For $k=0$ and $\mathrm{\ell }\mathrm{\ne }0$, the five diagonal eigenvalues of $B^{\mathrm{\ell }}$ are distinct, so its eigenlines are the coordinate axes, none of which satisfies $\sum z_i^5=0$. For $k\mathrm{\ne }0$, $A^k{B}^{\mathrm{\ell }}$ is a monomial matrix whose permutation is a 5-cycle and whose five nonzero coefficients have product ${\zeta }^{10\mathrm{\ell }}=1$; hence $(A^k{B}^{\mathrm{\ell }}{)}^5=I$. An eigenvector $z$ has all coordinates nonzero, and the eigenvector equations raised to the fifth power show that the five values $z_i^5$ are equal. Their sum is then five times a nonzero complex number, not zero. No nonidentity element fixes a point of $Y_0$, so the action is free.

Choose a very ample line bundle $L$ on $Y_0$ and form $L^{\mathrm{\prime }}={⨂}_{g\in G}{g}^{\ast }L$. Each factor is very ample, so the product of the corresponding embeddings followed by the Segre embedding makes $L^{\mathrm{\prime }}$ very ample; permuting the factors gives a $G$-linearization. Embed $Y_0$ by the complete linear system of $L^{\mathrm{\prime }}$ and let $S$ be its homogeneous coordinate ring, generated in degree one. Put $R=S^G$. The orbit polynomial ${\prod }_{g\in G}(T-gs)$ is monic over $R$ for every $s\in S$, so [F7] makes $S$ a finite $R$-module. Choose module generators $m_j$, including $1$, and express each $m_i{m}_j$ and each degree-one algebra generator of $S$ as an $R$-linear combination of them. Let $A_0$ be the complex algebra generated by the finitely many coefficients. The $A_0$-span of the $m_j$ is a subalgebra containing every generator of $S$, hence equals $S$. By [F6], $A_0$ is Noetherian and its submodule $R\subset S$ is finite. Thus $R$ is a finitely generated complex algebra. It is graded with $R_0=\mathbb{C}$ and finite-dimensional homogeneous pieces; taking the homogeneous parts of any finite generating list gives homogeneous generators. By [F8], $Y:=\mathrm{Proj}R$ is projective.

For any finite orbit choose $s\in S_1$ nonzero at every orbit point: each forbidden vanishing is a proper linear hyperplane and finitely many such hyperplanes do not cover a complex vector space. Its norm $N(s)={\prod }_{g\in G}gs$ is homogeneous invariant and nonzero on that orbit. Hence the positive-degree invariant principal opens cover $\mathrm{Proj}S=Y_0$, and the inclusion $R\hookrightarrow S$ defines a morphism to $Y$. For each homogeneous $f\in R$ of positive degree, the inverse image of $V_f=D_{+}(f)\subset Y$ is $U_f=D_{+}(f)\subset Y_0$. These are affine with rings $B_f=(R[f^{-1}]{)}_0$ and $A_f=(S[f^{-1}]{)}_0$. Averaging a numerator over $G$ shows $B_f=A_f^G$. The algebra $A_f$ is finite type over $\mathbb{C}$: if $e=\mathrm{deg}f$, the fractions of monomials of degree $e$ divided by $f$ generate it, by grouping any degree-$ke$ monomial into $k$ degree-$e$ monomials. Each element of $A_f$ is integral over $B_f$ by its orbit polynomial, so [F7] makes $A_f$ finite over $B_f$ and makes $\mathrm{Spec}{A}_f\to \mathrm{Spec}{B}_f$ surjective. Since the $V_f$ cover $Y$, the morphism is finite and surjective. In each source chart distinct finite closed orbits are separated by invariant functions: prescribe values $0$ and $1$ on the two orbits by [F8], and average over $G$. Thus each closed-point fiber is exactly one orbit. The finite morphism is proper analytically, so this continuous map from compact $Y_0$ onto Hausdorff projective $Y$ realizes the ordinary topological orbit quotient. Around any free orbit choose mutually disjoint analytic neighborhoods permuted by $G$; on one neighborhood the invariant local holomorphic functions on their union identify with all holomorphic functions there. This gives local biholomorphic quotient charts. Consequently $Y$ is smooth and $Y_0\to Y$ is a 25-sheeted covering; compactness and connectedness descend from $Y_0$.

Let $\vartheta ={\iota }_{\mathcal{E}}(d{z}_1\wedge \cdots \wedge d{z}_5)$ on ${\mathbb{C}}^5\setminus \{0\}$, where $\mathcal{E}={\sum }_i{z}_i{\mathrm{\partial }}_{z_i}$ is the Euler vector field. It is horizontal for scalar multiplication and homogeneous of weight $5$, so in the affine slices $z_j=1$ it defines compatible nowhere-zero local representatives of a section of $K_{{\mathbb{P}}^4}\otimes \mathcal{O}(5)$: changing the slice scales the horizontal form by the fifth power of the scalar, precisely the $\mathcal{O}(5)$ transition factor. This explicitly trivializes that line bundle and proves $K_{{\mathbb{P}}^4}\cong \mathcal{O}(-5)$. The polynomial $f={\sum }_i{z}_i^5$ is a section of $\mathcal{O}(5)$. Since $Y_0$ is smooth, near every point choose holomorphic coordinates $(w,u_1,u_2,u_3)$ with $Y_0=\{w=0\}$. The meromorphic form $\vartheta \mathrm{/}f$ has a simple pole there and can be written $a(w,u)(dw\mathrm{/}w)\wedge d{u}_1\wedge d{u}_2\wedge d{u}_3+\beta$, with $a$ holomorphic and nowhere zero and $\beta$ holomorphic. Its residue is $a(0,u)d{u}_1\wedge d{u}_2\wedge d{u}_3$. Under a change of such coordinates $w^{\mathrm{\prime }}=b(w,u)w$ with $b$ nowhere zero, $d{w}^{\mathrm{\prime }}\mathrm{/}{w}^{\mathrm{\prime }}=dw\mathrm{/}w+db\mathrm{/}b$; the additional holomorphic term has zero residue, while the tangential form changes by the canonical Jacobian. Thus the local residues glue to a nowhere-zero holomorphic three-form ${\mathrm{\Omega }}_0$ on $Y_0$, the adjunction/residue construction. The matrices defining $A$ and $B$ both have determinant $1$ ($A$ is a 5-cycle and $\det B={\zeta }^{10}=1$) and both preserve $f$. Since $g^{\ast }\vartheta =\det (g)\vartheta$ for a linear transformation commuting with $\mathcal{E}$, they preserve $\vartheta \mathrm{/}f$ and hence ${\mathrm{\Omega }}_0$. Invariance lets ${\mathrm{\Omega }}_0$ descend uniquely through the quotient charts of step 3.1 to a nowhere-zero holomorphic three-form on $Y$, so $K_Y$ is trivial. The finite-group invariant-ring construction in step 2.2 also makes $Y$ projective; an ample class on $Y$ supplies a Kähler class.

By [F5], equip $Y_0$ itself with the finite CW structure of a finite triangulation. It is path-connected because it is a connected manifold. Its complex rank-three tangent bundle is numerable: a finite trivializing cover has a subordinate partition of unity on this compact smooth base. With its complex orientation, [F4] therefore identifies its top Chern class with its Euler class; [F3] gives $\chi (Y_0)={\int }_{Y_0}{c}_3(T{Y}_0)=-200$. Step 3.1 gives a 25-sheeted cover $Y_0\to Y$; applying [F2] to this cover yields $\chi (Y_0)=25\chi (Y)$, so $\chi (Y)=-200\mathrm{/}25=-8$.