# Two finite torus quotients and their line bundles {#cg-s6-01} CG-S6 · Lesson 1 Two of the local pieces in the S6 construction are controlled by integral matrices of orders three and four. Their reduced central surfaces can be built from products of one-dimensional complex tori. We will construct the maps, identify every point in their kernels, and prove that the resulting finite group actions are free. The two covering degrees are nine and eight. A line bundle carried by each action will then give an explicit example of a smooth trivialization that cannot be holomorphic. This lesson supplies the finite central-fibre calculation. The varying family, its cusp, and the global topology are subsequent parts of the course. The matrices below act on the original marked lattice; no change of period, sign, or orientation is implicit. ## 1. The period map as an actual real isomorphism {#period-map} Use the ordered lattice \[ \Lambda=\mathbb Z\langle\widehat\gamma,\widehat u, \widehat w,\widehat\delta\rangle\subset\mathbb R^4. \tag{1.1} \] Its coordinates are columns \(x=(a,b,c,d)^{\mathsf t}\); the covectors \(\gamma,u,w,\delta\) extract these four entries. If \(\tau,\mu,\beta\in\mathbb C\), set \[ \Pi=\begin{pmatrix}6\mu&\tau&1&0\\\beta&\mu&0&1\end{pmatrix}, \qquad D=\operatorname{Im}\beta- \frac{6(\operatorname{Im}\mu)^2}{\operatorname{Im}\tau}. \tag{1.2} \] We work with \(\operatorname{Im}\tau>0\) and \(D<0\). To see exactly what these inequalities do, suppose \(\Pi x=0\). Imaginary parts of its two components give \[ \begin{pmatrix}6\operatorname{Im}\mu&\operatorname{Im}\tau\\ \operatorname{Im}\beta&\operatorname{Im}\mu\end{pmatrix} \binom ab=0. \tag{1.3} \] The determinant is \(6(\operatorname{Im}\mu)^2-\operatorname{Im}\tau\operatorname{Im}\beta =-\operatorname{Im}\tau D>0\). Thus \(a=b=0\), and the real parts then give \(c=d=0\). Consequently \(\Pi:\mathbb R^4\to\mathbb C^2\) is a bijective real-linear map. It identifies \(\Pi\Lambda\) with a discrete lattice and defines a compact complex torus \[ T=\mathbb C^2/\Pi\Lambda. \tag{1.4} \] Here a complex torus means the quotient of a complex vector space by a full lattice. For completeness, discreteness follows by applying the continuous inverse \(\Pi^{-1}\) to a convergent sequence of lattice points. Only finitely many lattice points can lie in a bounded set. A sufficiently small ball therefore has disjoint translates by nonzero lattice vectors, and its projection is a holomorphic coordinate chart. Distinct orbits can be separated by smaller balls, so the quotient is Hausdorff. The image of \(\Pi[0,1]^4\) covers the quotient, proving compactness. Since \(\mathbb C^2\) is connected, so is its quotient. The two period matrices needed here are obtained from \[ \begin{array}{c|c|c|c} j&\tau_j&\mu_j&\kappa_j\\\hline 1&\rho=e^{\pi i/3}&(2-\rho)/3&\beta_1+2\mu_1\\ 2&i&(1-i)/2&\beta_2+3\mu_2 \end{array} \tag{1.5} \] with \(D_j<0\). Write \(\Pi_j\) for (1.2) at these values and \(T_j=\mathbb C^2/\Pi_j\Lambda\). The parameters \(\beta_1,\beta_2\) retain their full real and imaginary parts. In particular, \[ \operatorname{Im}\kappa_1=\operatorname{Im}\beta_1- \tfrac23\operatorname{Im}\rho=D_1<0, \qquad \operatorname{Im}\kappa_2=\operatorname{Im}\beta_2-\tfrac32=D_2<0. \tag{1.6} \] The ordered periods \((1,\kappa_j)\) have negative imaginary orientation. This causes no defect in the quotient construction, and we keep that orientation. ## 2. Fixed vectors do not necessarily split the integral lattice {#integral-lattices} The two monodromy matrices, with the declared column convention, are \[ A_1=\begin{pmatrix}1&0&0&0\\6&0&1&0\\-6&-1&-1&0\\-2&1&0&1\end{pmatrix}, \qquad A_2=\begin{pmatrix}1&0&0&0\\0&0&-1&0\\-6&1&0&0\\3&0&1&1\end{pmatrix}. \tag{2.1} \] Put \[ \begin{gathered} \varepsilon=(1,2,-4,0)^{\mathsf t},\quad \varepsilon'=(1,3,-3,0)^{\mathsf t},\quad \delta_*=(0,0,0,1)^{\mathsf t},\\ e_1=(0,1,-2,0)^{\mathsf t},\quad f_1=(0,0,-3,1)^{\mathsf t},\\ e_2=(0,1,-1,0)^{\mathsf t},\quad f_2=(0,0,-2,1)^{\mathsf t}. \end{gathered} \tag{2.2} \] **Proposition 2.1.** The full fixed and image lattices are \[ \begin{array}{c|c|c} &\ker(A_j-I)\cap\Lambda&(A_j-I)\Lambda\\\hline j=1&\mathbb Z\varepsilon\oplus\mathbb Z\delta_*&\mathbb Ze_1\oplus\mathbb Zf_1\\ j=2&\mathbb Z\varepsilon'\oplus\mathbb Z\delta_*&\mathbb Ze_2\oplus\mathbb Zf_2. \end{array} \tag{2.3} \] Their sums have indices three and two in \(\Lambda\), respectively. Each image lattice itself is saturated: an integer vector whose nonzero integer multiple lies in that image already lies in the image. **Proof.** The equation \((A_1-I)x=0\) reads \(6a-b+c=0\), \(-6a-b-2c=0\), and \(-2a+b=0\). It gives \(b=2a,c=-4a\), with \(a,d\) arbitrary integers. For \(A_2\) the equations are \(-b-c=0\), \(-6a+b-c=0\), and \(3a+c=0\); they give \(b=3a,c=-3a\). This proves the fixed-lattice claims. In the original basis the columns of the image matrices are \[ \begin{array}{c|rrrr} A_1-I&6e_1-2f_1&-e_1+f_1&e_1&0\\ A_2-I&3f_2&-e_2&-e_2+f_2&0. \end{array} \tag{2.4} \] Every column is in the asserted span; conversely these columns recover each \(e_j,f_j\). The first span consists exactly of the integer solutions of \(a=0,\ 2b+c+3d=0\), because such a solution is \(be_1+df_1\). The second consists exactly of \(a=0,\ b+c+2d=0\), with expression \(be_2+df_2\). Multiplication by a nonzero integer cannot turn a nonzero value of these integer linear expressions into zero, proving saturation. One can compute the indices without presuming an integral direct-sum decomposition. Let \(B_1=[\varepsilon,\delta_*,e_1,f_1]\) and \(B_2=[\varepsilon',\delta_*,e_2,f_2]\). For \(x=(a,b,c,d)^{\mathsf t}\), solving \(x=B_j(A,B,C,E)^{\mathsf t}\) gives \[ \begin{aligned} B_1^{-1}x&=\left(a,\ d+\frac{c+2b}{3},\ b-2a,\ -\frac{c+2b}{3}\right)^{\mathsf t},\\ B_2^{-1}x&=\left(a,\ d+\frac{c+b}{2},\ b-3a,\ -\frac{c+b}{2}\right)^{\mathsf t}. \end{aligned} \tag{2.5} \] Thus the first sum is the kernel of the surjective map \(x\mapsto c+2b\pmod3\), and the second is the kernel of \(x\mapsto c+b\pmod2\). Their quotients are \(\mathbb Z/3\) and \(\mathbb Z/2\). Expanding the same matrices gives \(\det B_1=-3\) and \(\det B_2=-2\), including the original orientations. ∎ The images under the periods display the complex geometry behind these integer calculations: \[ \begin{array}{c|cc} &j=1&j=2\\\hline \text{first fixed vector}&\Pi_1\varepsilon=(0,\kappa_1)&\Pi_2\varepsilon'=(0,\kappa_2)\\ \delta_*&(0,1)&(0,1)\\ e_j&(\rho-2,-(\rho-2)/3)&(i-1,-(i-1)/2)\\ f_j&(-3,1)&(-2,1). \end{array} \tag{2.6} \] These are direct matrix products. For instance \(6(2-\rho)/3+2\rho-4=0\). The fixed real plane becomes the complex line \((0,\xi)\). The moving plane becomes \((\eta,-\eta/3)\) or \((\eta,-\eta/2)\). They are complementary complex lines. Their sum is all of \(\mathbb C^2\), even though the integral sum in Proposition 2.1 has a nontrivial finite quotient. ## 3. The product covers and every kernel point {#product-covers} Define \[ \begin{gathered} E_j=\mathbb C/(\mathbb Z+\kappa_j\mathbb Z),\\ F_1=\mathbb C/\bigl(\mathbb Z(\rho-2)+\mathbb Z(-3)\bigr),\qquad F_2=\mathbb C/\bigl(\mathbb Z(i-1)+\mathbb Z(-2)\bigr). \end{gathered} \tag{3.1} \] Each pair of periods is independent over \(\mathbb R\). The quotient-chart and compactness argument of Section 1 applies in real dimension two, so these are compact one-dimensional complex tori. **Theorem 3.1.** The linear maps \[ L_1(\xi,\eta)=(\eta,\xi-\eta/3),\qquad L_2(\xi,\eta)=(\eta,\xi-\eta/2) \tag{3.2} \] induce holomorphic group homomorphisms \(\ell_j:E_j\times F_j\to T_j\). They are surjective covering maps, with \[ \ker\ell_1=\{0,k_1,2k_1\},\quad k_1=(2/3,\rho), \qquad \ker\ell_2=\{0,k_2\},\quad k_2=(1/2,i). \tag{3.3} \] Coordinates of \(k_j\) in (3.3) mean classes in the indicated product torus. **Proof.** The images of the four product periods under \(L_j\), listed in (2.6), generate exactly \(\Pi_j B_j\mathbb Z^4\). Therefore the maps descend. Each \(L_j\) has complex determinant \(-1\), hence is invertible; its induced map is locally biholomorphic and surjective. Its kernel is the quotient of the target lattice by the image lattice, which has the respective cardinality three or two by Proposition 2.1. This kernel identification follows explicitly by sending a target lattice vector \(\lambda\) to the class of \(L_j^{-1}\Pi_j\lambda\). For \(\widehat u=(0,1,0,0)^{\mathsf t}\) the exact decompositions are \[ \widehat u=\tfrac23\delta_*+e_1-\tfrac23f_1 =\tfrac12\delta_*+e_2-\tfrac12f_2. \tag{3.4} \] In the first expression the moving period is \((\rho-2)-(2/3)(-3)=\rho\); in the second it is \((i-1)-(1/2)(-2)=i\). Thus \(L_j(k_j)=\Pi_j\widehat u\), so both indicated points are in the kernel. The first coordinates have exact orders three and two: since \(\operatorname{Im}\kappa_j\ne0\), a real number lies in \(\mathbb Z+\kappa_j\mathbb Z\) precisely when it is an integer. The moving coordinates obey \(3\rho=3(\rho-2)-2(-3)\) and \(2i=2(i-1)-(-2)\). The points therefore have the asserted exact orders and exhaust the kernels. Translations by a finite subgroup act freely. Small disjoint neighbourhoods of its orbit points give covering charts, proving the covering assertion without an additional compactness theorem. ∎ ## 4. Adding the affine rotations {#affine-quotients} Set \(m_1=3,m_2=4\), \(v_1=\varepsilon\), and \(v_2=-\varepsilon'\). On \(\mathbb R^4/\Lambda\) consider \[ g_j[x]=[A_jx+v_j/m_j]. \tag{4.1} \] This is well defined because the matrices are integral. We now check its holomorphic nature in the stated periods rather than assume that an integral matrix is automatically complex-linear. The two complex matrices are \[ R_1=\begin{pmatrix}\rho-1&0\\(2-\rho)/3&1\end{pmatrix}, \qquad R_2=\begin{pmatrix}-i&0\\(1+i)/2&1\end{pmatrix}. \tag{4.2} \] Multiplying the four columns gives \(R_j\Pi_j=\Pi_j A_j\). Alternatively, (2.6) identifies a real basis on which both sides agree: the fixed line has eigenvalue one and the moving line has eigenvalue \(\rho-1\) or \(-i\). Thus the action on \(T_j\) is exactly \([z]\mapsto[R_jz+\Pi_jv_j/m_j]\). The moving eigenvalues have orders three and four, respectively, so \(A_j^{m_j}=I\). Directly \(A_jv_j=v_j\), whence \(g_j^k[x]=[A_j^kx+kv_j/m_j]\) and \(g_j^{m_j}=1\). On the product torus define \[ h_1(\xi,\eta)=(\xi+\kappa_1/3,(\rho-1)\eta),\qquad h_2(\xi,\eta)=(\xi-\kappa_2/4,-i\eta). \tag{4.3} \] The moving rotations preserve the exact lattices in (3.1), as the following images of their two generators show: \[ \begin{aligned} (\rho-1)(\rho-2)&=-2(\rho-2)+(-3),& (\rho-1)(-3)&=-3(\rho-2)+(-3),\\ (-i)(i-1)&=(i-1)-(-2),& (-i)(-2)&=2(i-1)-(-2). \end{aligned} \tag{4.4} \] The corresponding integer matrices have determinant one. Each rotation therefore maps its period lattice onto itself. Substitution into (3.2) gives the equality on \(\mathbb C^2\) \[ L_jh_j=R_jL_j+\Pi_jv_j/m_j, \qquad \ell_jh_j=g_j\ell_j. \tag{4.5} \] The minus sign in the second translation is required by \(v_2=-\varepsilon'\). **Theorem 4.1.** Let \(k_j\) also denote translation by the kernel point (3.3). The groups generated on the product tori are \[ \langle k_1,h_1\rangle\cong C_3\times C_3, \qquad \langle k_2,h_2\rangle\cong C_2\times C_4. \tag{4.6} \] Both actions are free. Their quotients are compact smooth complex surfaces, and the diagrams of holomorphic covering maps are \[ E_1\times F_1\xrightarrow[\deg=3]{\ell_1}T_1 \xrightarrow[\deg=3]{}S_1=T_1/\langle g_1\rangle, \qquad E_2\times F_2\xrightarrow[\deg=2]{\ell_2}T_2 \xrightarrow[\deg=4]{}S_2=T_2/\langle g_2\rangle. \tag{4.7} \] The composite degrees are exactly nine and eight. **Proof.** First check commutativity on the quotient tori. On the moving coordinates, \[ (\rho-1)\rho-\rho=-(\rho-2)+(-3), \qquad (-i)i-i=-(i-1). \tag{4.8} \] These differences are periods, so \(h_j\) commutes with \(k_j\). The maps \(h_j\) have orders \(m_j\): their \(m_j\)-th translations are \(\kappa_1\) and \(-\kappa_2\), while a smaller positive power has a nonidentity moving linear part. To establish both the group order and freeness, write \((d_1,d_2)=(3,2)\), \((c_1,c_2)=(2/3,1/2)\), and \((\epsilon_1,\epsilon_2)=(1,-1)\). A fixed point of \(k_j^a h_j^b\), where \(0\le a