On the compactness of the set of invariant Einstein metrics

Abstract

Let $M = G/H$ be a connected simply connected homogeneous manifold of a compact, not necessarily connected Lie group $G$. We will assume that the isotropy $H$-module $\mathfrak{g/h }$ has a simple spectrum, i.e. irreducible submodules are mutually non-equivalent. There exists a convex Newton polytope $N=N(G,H)$, which was used for the estimation of the number of isolated complex solutions of the algebraic Einstein equation for invariant metrics on $G/H$ (up to scaling). Using the moment map, we identify the space $\mathcal{M }_1$ of invariant Riemannian metrics of volume 1 on $G/H$ with the interior of this polytope $N$. We associate with a point ${x \in \partial N}$ of the boundary a homogeneous Riemannian space (in general, only local) and we extend the Einstein equation to $\partial N$. As an application of the Alekseevsksky–Kimel’fel’d theorem, we prove that all solutions of the Einstein equation associated with points of the boundary are locally Euclidean. We describe explicitly the set $T\subset \partial N$ of solutions at the boundary together with its natural triangulation. Investigating the compactification ${\overline{\mathcal{M }}}_{1} = N$ of $\mathcal{M }_1$, we get an algebraic proof of the deep result by Böhm, Wang and Ziller about the compactness of the set $\mathcal{E }_1 \subset \mathcal{M }_1$ of Einstein metrics. The original proof by Böhm, Wang and Ziller was based on a different approach and did not use the simplicity of the spectrum. In Appendix, we consider the non-symmetric flag manifolds $G/H$ with the second Betti number $b_2=1$. We calculate the normalized volumes $2,6,20,82,344$ of the corresponding Newton polytopes and discuss the number of complex solutions of the algebraic Einstein equation and the finiteness problem.

Appendix: Case of Kähler homogeneous space with \(b_2=1\)

Consider now a Kähler homogeneous space $G/H$ with the second Betti number $b_2=1$. Assume that the isotropy $H$-module $\mathfrak{g }/\mathfrak{h }$ is split into $d>1$ irreducible submodules.

Lemma 8

Given a Kähler homogeneous space $G/H$ with $d>1$ of a simple Lie group $G$, then $2^{-b_2(G/H)}\nu (G/H) \in \mathbb{Z }$.

Idea of proof

$2^{b_2(G/H)} = [\mathbb{Z }^d:L]$, where $L \subset \mathbb{Z }^d$ is the subgroup generated by vertices of the polytope $\Delta$. (Remark that $\Delta = \Delta _\mathrm{min}=\Delta _{\max }$). $\square$

Let, moreover, $b_2(G/H)=1$. Then $2\le d \le 6$. For $d=2$ the polytope $\Delta$ is the segment with ends $e_2$ and $2e_1-e_2$. For $3\le d \le 6$ the vertices of the polytope $\Delta$ are the points $e_i+e_j-e_k \in \mathbb{R }^d$ with $1\le i,j,k \le d$, $i\ne k$, $j\ne k$, $i\pm j\pm k =0$. Here $e_1=(1,0,\ldots ,0),\ldots ,e_d=(0,\ldots ,0,1)$. Using MAPLE, one can triangulate these polytopes and find their normalized volumes $\nu =\nu (G/H)$. Thus, we obtain:

Claim

$2^{-1}\nu \in \{1,3,10,41,172\}.$

The following table gives some information about polytopes $\Delta$ corresponding to Kähler homogeneous spaces $G/H$ with the second Betti number $b_2=1$ and $d>1$.

For completeness, we find the volume of a similar $(d-1)$-dimensional polytope with $d=7$. Here $f$ is the number of facets of $\Delta$, and $m$ the number of all faces $\gamma$ of $\Delta$ with $0<\dim (\gamma )<d-1$ (which we call marked) that NOT satisfy conditions of Test 1 or Test 2 of [9, Section 7.1]. A marked face $\gamma$ is not a vertex. ⁵ Moreover, one can prove that $\gamma$ is not an edge, so $1<\dim (\gamma )<\dim (\Delta )$.

$$\begin{aligned} \begin{array}{c|llllll|llll} d &{} &{}\quad 2 &{}\quad 3 &{}\quad 4 &{}\quad 5 &{}\quad 6 &{} 7 \\ f &{} &{}\quad 2 &{}\quad 4 &{}\quad 7 &{}\quad 16 &{}\quad 36 &{} 100 \\ \nu &{} &{}\quad 2 &{}\quad 6 &{}\quad 20 &{}\quad 82 &{}\quad 344 &{}1598 \\ \hline \varepsilon &{} &{}\quad \nu &{}\quad \nu &{}\quad \nu &{}\quad 81 &{}\quad ? &{} - \\ \delta &{} &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 1 &{}\quad ? &{}- \\ m &{} &{}\quad 0 &{}\quad 0 &{}\quad 3 &{}\quad 13 &{}\quad 40 &{} \end{array} \end{aligned}$$

We write also the known numbers $\varepsilon$ and $\delta = \nu -\varepsilon$. By [9], if $m=0$, then $\nu =\varepsilon$. The first non-trivial case is $d=4$.

For $d\le 5$ all the positive solutions of the algebraic Einstein equations are known.

In the case $d=5$ they calculated by Chrysikos and Sakane [6].

They also prove that all the complex solution are isolated ($d=5$).

Remark

(the case $d=4,\;\nu =20$) (Fig. 4). Here $\Delta$ is a three-dimensional polytope with three marked faces $\gamma ,$ namely, a trapezoid $\gamma _1$, a parallelogram $\gamma _2$, and a pentagon $\gamma _3$. To prove that $\nu =\varepsilon$ one can associate with each marked face $\gamma$ a complex hypersurface $s_{\gamma }(x_1,\ldots ,x_4)=0$ in $(\mathbb{C }\setminus 0)^4$ and check that it is non-singular. Here $s(x)$ is the above Laurent polynomial (scalar curvature), and $s_{\gamma }(x)$ is the sum of all monomials of $s(x)$ whose vector exponents belong to $\gamma$. See [8, 9, Section 1.7.2]. This is essentially a two-dimensional problem ($2=\dim (\gamma _i)$), i.e., we must check that a plane curve is non-singular. It is easy to prove this for $\gamma =\gamma _1$, but for $\gamma _k$, $k=2,3$ the problem reduces to $D_k[s]\ne 0$, where $D_k[s]$ is a homogeneous polynomial (a $k$-monomial) in coefficients of $s(x)$ with $\deg (D_k) = k$. The coefficients of $s(x)$ depend on $G/H$. There are four Kähler homogeneous spaces $G/H$ with $b_2=1$ and $d=4$, namely, the spaces ${E_8/T^1\cdot A_1\cdot A_6},\quad {E_8/T^1\cdot A_2\cdot D_5},\quad {E_7/T^1\cdot A_1\cdot A_2\cdot A_3},\quad {F_4/T^1\cdot \widetilde{A}_1\cdot A_2}$ (we use the Dynkin’s notation $\widetilde{A}_1$ for the three-dimensional subgroup associated with a short simple root). The corresponding scalar curvature polynomials $s(x)$ are computed by A.Arvanitoyeorgos and I.Chrysikos (arXiv:0904.1690). For each of them one can check that $D_2[s]D_3[s]\ne 0$. This proves that $\varepsilon =\nu$ for $d=4$.

Fig. 4

The $3$-dimensional polytope $\Delta$ with $7$ facets corresponding to four Kähler homogeneous spaces with $b_2=1$ and $d=4$

The case $d=5$, $\nu =82$. There is a unique Kähler homogeneous space $G/H$ with $b_2(G/H)=1$ and $d=5$, namely, the space ${G/H=E_8/T^1\cdot A_3\cdot A_4}.$ By [6, Section 3, the text after eq.(25)] it implies that the algebraic Einstein equation has, up to scale, $81$ complex solutions, corresponding to roots of some polynomial $(x_5-5)h_1(x_5)$ of degree $81$ in one variable $x_5$. There exist six positive solutions [6, Theorem A]; in particular, the root $x_5=5$ corresponds to a unique, up to scale, invariant Kähler metric on $G/H$. Using MAPLE, one can check that the polynomial $h_1(x_5)$ has $80$ simple roots, and $81$ solutions of Einstein equation are distinct (moreover, it has $30$ real roots, and Einstein equation has $31$ real solutions). Thus

$$\begin{aligned} \nu - \varepsilon = 82-81 = 1. \end{aligned}$$

We prove independently that $\nu > \varepsilon$. Let $s(x_1, \dots ,x_5)$ be the scalar curvature of a invariant metric $g_x$, as above. It is a Laurent polynomial in $x_i^{-1}$. We claim that there exists a limit

$$\begin{aligned} s_{\infty }(x_1, \dots ,x_5) = \lim _{t\,\rightarrow \,+ \infty } s(t^{2}\,{x_{1}}, t^{4}\,{x_{2}}, t^{3}\,{x_{3}}, t\,{x_{4}}, t\,{x_{5}}), \end{aligned}$$

and the homogeneous function $s_{\infty }$ depends essentially on $3$ variables. Indeed, it follows from Proposition 4 and the above description of the Newton polytope $\Delta$ that

$$\begin{aligned} s_{\infty }&= - \frac{\scriptstyle [1,4,5]}{2}{\frac{{x_{1}}}{{x_{4}}\,{x_{5}}}} - \frac{\scriptstyle [1,3.4]}{2}{\frac{{x_{3}}}{{x_{1}}\,{x_{4}}}}\\&\quad - \frac{\scriptstyle [2,3,5]}{2}{\frac{{x_{2}}}{{x_{3}}\,{x_{5}}}} - \frac{\scriptstyle [1,1,2]}{4}{\frac{{x_{2}}}{{x_{1}}^{2}}}. \end{aligned}$$

Then there is a two-dimensional face of the polytope $\Delta$, the parallelogram $P$ with vertices $e_4+e_5-e_1$, $e_1+e_4-e_3$, $2e_1-e_4$, and $e_3+e_5-e_2$. The face $P$ is orthogonal to the vector

$$\begin{aligned} \mathbf{f} = (2,4,3,1,1). \end{aligned}$$

According to [6, Proposition 7] we have $[1, 1, 2]=12, [1, 2, 3]=8, [1, 3, 4]=4, [1, 4, 5]=4/3, [2, 2, 4]=4, [2, 3, 5]=2$. Then the product of monomials, corresponding to each pair of opposite vertices of $P$, coincides with $2 {\frac{{x_{2}}}{{x_{1}}\,{x_{4}}\,{x_{5}}}}$, and $s_{\infty }$ can be represented as

$$\begin{aligned} s_{\infty } = z_0(1+z_1+z_2+z_1z_2) = z_0(z_1+1)(z_2+1), \end{aligned}$$

where $z_0=-\frac{{x_{2}}}{{x_{3}}\,{x_{5}}}$. Since for $z_1=z_2=-1$ we have

$$\begin{aligned} s_{\infty } = ds_{\infty } =0, \end{aligned}$$

the complex hypersurface $s_{\infty }(x)=0$ has a singular point $x$ with $\prod x_i\ne 0$.

By [8, 9, Section 1.7.2] this implies that $\nu - \varepsilon >0$.

Note that $\Delta$ has $m-1=12$ marked faces, other than $P$; namely, $6$ three-dimensional faces with normal vectors

$$\begin{aligned} \mathbf{f} = (1, 2, 3, 4, 5), (1, 2, 1, 2, 1), (2, 1, 1, 2, 0), (1, 0, 1, 0, 1),(1, 2, 2, 1, 0), (1, 2, 1, 0, 1), \end{aligned}$$

and $6$ parallelograms defined by the following normal vectors (such as $\mathbf{f}=(2, 4, 3, 1, 1)$):

$$\begin{aligned} \mathbf{f} = (1, 1, 2, 2, 3), (1, 2, 3, 4, 4), (1, 2, 2, 3, 4), (2, 4, 5, 3, 1), (5, 3, 2, 6, 1), (3, 1, 2, 2, 1) \end{aligned}$$

The corresponding $12$ complex hypersurfaces are non-singular.

Additional remarks (the case $d=5$). Consider $(z_0,z_1,z_2)=(1,-1,-1)$ as a point $p$ in the four-dimensional toric variety $\Delta ^{\mathbb{C }}$. Let $O\subset \Delta ^{\mathbb{C }}$ be the orbit of the group $(\mathbb{C }\setminus 0)^5/\mathbb{C }^\times$ through $p$. The closure of $O$ is the two-dimensional toric subvariety $P^{\mathbb{C }}$. The point $p\in O$ is a solution at infinity (in the sence of Sect. 10) of the algebraic Einstein equation.

Our example is excellent as the following lemma shows.

Lemma 9

We claim now that $\Delta ^{\mathbb{C }}$ is smooth at each point $q\in O$. Moreover, assuming $\varphi :\Delta ^{\mathbb{C }}\rightarrow \mathbb{P }^{N-1}(\mathbb C )$ be the natural map into the complex projective space $\mathbb{P }^{N-1}(\mathbb C )$, $N=\#(\mathbb{Z }^5\cap \Delta )$, then $\varphi ^{-1}(\varphi (q))=\{q\}$, and $\varphi (\Delta ^{\mathbb{C }})$ is smooth at $\varphi (q)$.

We will apply the localization along $O$ to prove that the point $p$ is an isolated solution (at infinity) with the multiplicity $1$ of the algebraic Einstein equation.

Proof

Let $v_0,v_1,v_{12},v_2$ are vertices of the parallelogram $P$, so $v_0+v_{12}=v_1+v_2$, and

$$\begin{aligned} u_1&= -e_1+e_4+e_5 = v_1, \\ u_2&= -e_2 + 2e_4+2e_5 = 2v_1+v_2,\\ u_3&= -e_3+2e_4+e_5= v_1+v_{12}, \quad \, u_4=e_4,\quad \, u_5=e_5. \end{aligned}$$

The set of vectors $\{u_i : i=1,\ldots ,5\}$, and, hence, $\{v_0,v_1,v_2,u_4,u_5\}$ are biases in $\mathbb{Z }^5=\bigoplus \mathbb{Z }e_i$. Let $\pi (a):=(a_4,a_5)$ for each $a=\sum a_i u_i \in \mathbb{Z }^5$. We prove, that $\pi (\mathbb{Z }^5\cap \Delta )$ generates the semigroup $\mathbb{Z }_+^2$. The face $P$ of $\Delta$ is the intersection of two facets with normal vectors $\mathbf{f}_i$, $i=1,2$, so that

$$\begin{aligned} \mathbf{f} = (2,4,3,1,1) = (1, 2, 2, 1, 0) + (1, 2, 1, 0, 1) =\mathbf{f}_1 + \mathbf{f}_2. \end{aligned}$$

For any $a\in \mathbb{Z }^5\cap \Delta$ we have $a_4 = \langle \mathbf{f}_1,a\rangle \ge 0$, and $a_5 = \langle \mathbf{f}_2,a\rangle \ge 0$. Then $\pi (a)\in \mathbb{Z }_+^2$. This prove the assertion, since $\pi (e_4) = (1,0),\;\pi (e_5)=(0,1)$, $e_4,e_5\in \Delta$. The lemma follows $\square$

Now let $(z_0,z_1,z_2,y_1,y_2)$ be coordinates on $(\mathbb{C }\setminus 0)^3\times \mathbb{C }^2$. Assume that for $y_1y_2\ne 0$

$$\begin{aligned} - {\displaystyle \frac{{x_{2}}}{{x_{3}}\,{x_{5}}}} ={z_{0}}, \quad - {\displaystyle \frac{2}{3}} \,{\displaystyle \frac{{x_{1}} }{{x_{4}}\,{x_{5}}}} ={z_{0}}\,{z_{1}}, \quad - 3\,{\displaystyle \frac{{x_{2}}}{{x_{1}}^{2}}} ={z_{0}}\,{z_{2}}, \quad \\ - 2\, {\displaystyle \frac{{x_{3}}}{{x_{1}}\,{x_{4}}}} ={z_{0}}\,{z_{1 }}\,{z_{2}}, \quad {\displaystyle \frac{1}{{x_{4}}}} ={y_{1}}, \quad {\displaystyle \frac{1}{{x_{5}}}} ={y_{2}}, \end{aligned}$$

so

$$\begin{aligned} \left\{ \! {x_{3}}={\displaystyle \frac{3}{4}} \, {\displaystyle \frac{{z_{0}}^{2}\,{z_{1}}^{2}\,{z_{2}}}{{y_{1}} ^{2}\,{y_{2}}}} , \,{x_{2}}= - {\displaystyle \frac{3}{4}} \, {\displaystyle \frac{{z_{0}}^{3}\,{z_{1}}^{2}\,{z_{2}}}{{y_{1}} ^{2}\,{y_{2}}^{2}}} , \,{x_{4}}={\displaystyle \frac{1}{{y_{1}}} } , \,{x_{1}}= - {\displaystyle \frac{3}{2}} \,{\displaystyle \frac{{z_{0}}\,{z_{1}}}{{y_{1}}\,{y_{2}}}} , \,{x_{5}}= {\displaystyle \frac{1}{{y_{2}}}} \! \right\} . \end{aligned}$$

Then

$$\begin{aligned} s&= -{\displaystyle \frac{16}{9}} \,{\displaystyle \frac{{y_{1}} ^{3}\,{y_{2}}^{4}}{{z_{0}}^{6}\,{z_{1}}^{4}\,{z_{2}}^{2}}} + {\displaystyle \frac{{\displaystyle \frac{16}{9}} \,{y_{1}}^{4} \,{y_{2}}^{2}}{{z_{0}}^{5}\,{z_{1}}^{4}\,{z_{2}}^{2}}} - {\displaystyle \frac{32}{3}} \,{\displaystyle \frac{{y_{1}}^{3} \,{y_{2}}^{2}}{{z_{0}}^{4}\,{z_{1}}^{3}\,{z_{2}}^{2}}}\\&\quad + {\displaystyle \frac{{\displaystyle \frac{16}{9}} \,{y_{1}}^{2} \,{y_{2}}^{2}}{{z_{0}}^{3}\,{z_{1}}^{3}\,{z_{2}}}} - {\displaystyle \frac{32\,{y_{1}}^{2}\,{y_{2}}^{2}}{{z_{0}}^{3}\, {z_{1}}^{2}\,{z_{2}}}} + {\displaystyle \frac{{\displaystyle \frac{80}{3}} \,{y_{1}}^{2}\,{y_{2}}}{{z_{0}}^{2}\,{z_{1}}^{2}\, {z_{2}}}} - {\displaystyle \frac{8}{3}} \,{\displaystyle \frac{{y_{1}}\,{y_{2}}^{2}}{{z_{0}}^{2}\,{z_{1}}}} \\&\quad + {\displaystyle \frac{4\,{y_{1}}^{2}}{{z_{0}}\,{z_{1}} \,{z_{2}}}} + {\displaystyle \frac{{\displaystyle \frac{4}{9} } \,{y_{1}}^{2}}{{z_{0}}\,{z_{1}}}} - {\displaystyle \frac{80}{ 3}} \,{\displaystyle \frac{{y_{1}}\,{y_{2}}}{{z_{0}}\,{z_{1}}}} + {\displaystyle \frac{{y_{2}}^{2}}{{z_{0}}}} + {\displaystyle \frac{{\displaystyle \frac{4}{9}} \,{y_{2}}^{2} }{{z_{0}}\,{z_{1}}}} + 8\,{y_{1}} - {\displaystyle \frac{8}{3} } \,{\displaystyle \frac{{y_{1}}}{{z_{1}}}} + 4\,{y_{2}} + {z_{ 0}}\,{z_{1}} \\&\quad + {z_{0}}\,{z_{1}}\,{z_{2}} + {z_{0}} + {z_{0}}\,{z_{2}}, \end{aligned}$$

$s$ is a polynomial in $y_1$, $y_2$, and

$$\begin{aligned} s= z_0 + z_0z_1+z_0z_2 +z_0z_1z_2 + 8y_1 - \frac{8}{3}\frac{y_1}{z_1} +4y_2 + [2], \end{aligned}$$

where $[2]$ denotes the terms with degree $\ge 2$. Similarly, for $s_i = x_i\partial s/\partial x_i$, $i=1,\ldots ,5$ we have

$$\begin{aligned} s_1&= z_0z_1 -2z_0z_2 -z_0z_1z_2 + \frac{8}{3}\frac{y_1}{z_1} + [2], s_2= + z_0 + z_0z_2 - \frac{8}{3}\frac{y_1}{z_1} + [2], \\s_3&= -z_0 + z_0z_1z_2 + \frac{8}{3}\frac{y_1}{z_1} + [2], s_4= -z_0z_1 - z_0z_1z_2 - 8y_4 + [2], \\s_5&= -z_0 - z_0z_1 - 4y_5 + [2], \end{aligned}$$

where $[2]$ denotes $(y_1^2, y_1y_2, y_2^2)$. Computing the matrix $J= \frac{\partial (s_1,s_2,s_3,s_4,s_5)}{\partial (z_1,z_2,y_1,y_2)}$, setting $z_0=1,z_1=z_2=-1$, $y_1=y_2=0$, adding the row $(d_1,\ldots ,d_5)$ of dimensions $d_i=\dim (\mathfrak{m }_i)$, and finding the determinant, we obtain

$$\begin{aligned} \left| \begin{array}{r@{\quad }r@{\quad }r@{\quad }r@{\quad }r} d_1&{}d_2&{}d_3&{}d_4&{}d_5\\ 2&{}0&{}-1&{}0&{}-1\\ -1&{}1&{}-1&{}1&{}0 \\ -8/3&{}8/3&{}-8/3&{}-8&{}0 \\ 0&{}0&{}0&{}0&{} -4 \end{array}\right| =\frac{128}{3} (d_1+3d_2+2d_3) >0. \end{aligned}$$

Then the solution $p\in \Delta ^{\mathbb{C }}$ of the algebraic Einstein equation with local coordinates $z_0=1,z_1=z_2=-1, y_1=y_2=0$ is isolated, and non-degenerate.

The case $d=6$, $\nu = 344$. There is a unique Kähler homogeneous space $G/H$ with $b_2(G/H)=1$ and $d=6:$

$$\begin{aligned} G/H = E_8/T^1\cdot A_4\cdot A_2\cdot A_1. \end{aligned}$$

The corresponding $5$-dimensional polytope $\Delta$ in $\mathbb{R }^6$ has $36$ facets, i.e., $4$-dimensional faces. Each of them can be defined by the orthogonal vector $\mathbf{f}=(y_1, \dots ,y_6)$ such that $\gcd (y_1, \dots ,y_6)=1$, and $y_i\ge 0$; then $\langle \mathbf{f},x \rangle \,\ge 0$ for any $x \in \Delta$.

For example, the vector $\mathbf{f}=(1, 2, 3, 4, 5, 6)$ is orthogonal to the facet with $9$ vertices

$$\begin{aligned} \begin{array}{llllllll} \mathrm{e}^{2}_{1 1},&{}\mathrm{e}^{3}_{1 2},&{}\mathrm{e}^{4}_{1 3},&{}\mathrm{e}^{5}_{1 4},&{}\mathrm{e}^{6}_{1 5},\\ &{}\mathrm{e}^{4}_{2 2},&{}\mathrm{e}^{5}_{2 3},&{}\mathrm{e}^{6}_{2 4},&{} \\ &{} &{}\mathrm{e}^{6}_{3 3},&{} &{} \end{array} \end{aligned}$$

where $\mathrm{e}^k_{ij}=e_i+e_j-e_k$. We write all the facets:

(1) $16$ four-dimensional simplices with normal vectors

(2) $8$ four-dimensional pyramids with normal vectors

(3) $5$ other facets with normal vectors with positive entries:

(4) $7$ facets with normal vectors with non-negative entries:

Facets (1) and (2) are not marked faces. E.g., simplices (1) and its sub-faces satisfy [9, Test 7.1].

Facets (3) and (4) are marked faces. There are $13$ three-dimensional and $15$ two-dimensional marked faces.

We get $13$ vectors, orthogonal to three-dimensional marked faces (each of them is proportional to the sum of two distinct vectors (2)–(4)):

In the two-dimensional case we obtain:

(a) $6$ parallelograms with normal vectors

(b) $9$ parallelograms with normal vectors

Each of parallelograms listed in (a) and (b) [with the exception of two last entries in (b)] belongs exactly to $3$ facets.

We claim now, that $6$ marked parallelograms (a) correspond to singular complex hypersurfaces as above (consequently $\varepsilon <\nu$), and $9$ marked parallelograms (b) correspond to non-singular hypersurfaces. For the proof, one can calculate $6+9$ determinants $\left| \begin{array}{ll}a&{}b\\ b^{\prime }&{}a^{\prime }\end{array}\right|$, where $a,a^{\prime },b,b^{\prime }$ are some coefficients of $s(x_1,\ldots ,x_6)$, using equalities ([6, Prop.13]): $[1, 1, 2] = 8, [1, 2, 3] = 6, [1, 3, 4] = 4, [1, 4, 5] = 2, [1, 5, 6] = 1, [2, 2, 4] = 6, [2, 3, 5] = 2, [2, 4, 6] = 2, [3, 3, 6] = 2$.

Thus we may unmark $9$ of $40$ marked faces.

Corollary

The hypothesis that all complex solutions of the algebraic Einstein equation on $G/H = E_8/T^1\cdot A_4\cdot A_2\cdot A_1$ are isolated reduces to examination of $31 = 40-9$ cases, corresponding to $12$ four-dimensional, $13$ three-dimensional, and $6$ two-dimensional faces of the polytope $\Delta$.

In each case we may unmark the $k$-dimensional face, if the corresponding complex hypersurface is non-singular (this is the $k$-dimensional problem, $k<5$); otherwise we must examine Einstein equation in a neighborood $U \subset \Delta ^{\mathbb{C }}$ of the ‘solution at infinity’ defined by each singular point (cf. Sect. 10).