Weyl groups and the automorphisms of the configuration of 27 lines

Each adjacent transposition is an involution. Two adjacent transpositions with disjoint supports commute, so their product has order two, while \[ ((i,i+1)(i+1,i+2))^3=1. \] Thus the defining Coxeter relations are satisfied. The adjacent transpositions generate the full symmetric group, so the homomorphism is surjective.

Put \[ r_j=s_{n-1}s_{n-2}\cdots s_j \qquad(1\le j\le n-1), \] and put \(r_n=1\). We show that the union \[ \mathcal U=\bigcup_{j=1}^nW_{n-1}r_j \] is stable under right multiplication by every generator \(s_i\).

If \(i<j-1\), then \(s_i\) commutes with every factor of \(r_j\), so \[ r_js_i=s_ir_j \] and the coset is unchanged. If \(i=j-1\), then \[ r_js_{j-1}=r_{j-1}. \] If \(i=j\), the last factor cancels and \[ r_js_j=r_{j+1}. \]

Finally suppose \(i>j\). Repeatedly using the braid relation \[ s_is_{i-1}s_i=s_{i-1}s_is_{i-1} \] and commuting nonadjacent generators gives \[ r_js_i=s_{i-1}r_j. \] Since \(i-1\le n-2\), the left factor lies in \(W_{n-1}\). Thus every \(W_{n-1}r_j\) is carried into one of the displayed cosets by right multiplication by a generator. Because \(1\in\mathcal U\), every word in the generators lies in \(\mathcal U\).

Step 2 gives at most \(n\) right cosets of \(W_{n-1}\). Hence \[ |W_n|\le n|W_{n-1}|. \] Starting with \(|W_2|\le2\) and iterating gives \[ |W_n|\le n!. \]

Step 1 gives a surjection onto the group of order \(n!\), while step 3 gives at most \(n!\) elements in the source. Thus the source has exactly \(n!\) elements and the surjection is injective. This proves part (a).

For each \(i\), \[ \alpha_i^2=e_i^2+e_{i+1}^2=-2, \] and \[ K_S\cdot\alpha_i=0. \] Also \[ \beta^2=1-1-1-1=-2 \] and \[ K_S\cdot\beta=-3+1+1+1=0. \]

For \(\alpha_i=e_i-e_{i+1}\), \[ e_i\cdot\alpha_i=-1, \qquad e_{i+1}\cdot\alpha_i=1, \] so \[ s_{\alpha_i}(e_i)=e_{i+1}, \qquad s_{\alpha_i}(e_{i+1})=e_i. \] Every other basis vector is orthogonal to \(\alpha_i\).

For \(\beta\), one finds \[ s_\beta(h)=2h-e_1-e_2-e_3, \] and \[ \begin{aligned} s_\beta(e_1)&=h-e_2-e_3,\\ s_\beta(e_2)&=h-e_1-e_3,\\ s_\beta(e_3)&=h-e_1-e_2, \end{aligned} \] while \(e_4,e_5,e_6\) are fixed. These are exactly the pullback formulas for the standard quadratic transformation centered at the first three points, as in P-AGH542QUADTRANSFORM.

The only nonzero pairings between distinct simple roots are \[ \alpha_i\cdot\alpha_{i+1}=1 \qquad(1\le i\le4) \] and \[ \beta\cdot\alpha_3=1. \] All other distinct pairings are zero.

If two square-\(-2\) roots are orthogonal, their reflections commute, so the product has order two. If \(\gamma^2=\delta^2=-2\) and \[ \gamma\cdot\delta=1, \] then on the span of \(\gamma,\delta\) the two reflections are represented in that ordered basis by \[ \begin{pmatrix}-1&1\\0&1\end{pmatrix}, \qquad \begin{pmatrix}1&0\\1&-1\end{pmatrix}, \] whose product has order three. Thus the defining \(E_6\) Coxeter relations hold.

Every reflection in step 6 preserves the intersection form and fixes \(K_S\). It therefore preserves the set of classes satisfying \[ L^2=-1, \qquad K_S\cdot L=-1, \] which is exactly the set of \(27\) line classes FE-SRFCUBIC. Since it also preserves pairwise intersection numbers, it preserves the incidence graph of the \(27\) lines. Thus the reflections define elements of \(G\), and step 7 shows that the Coxeter presentation gives the asserted homomorphism.

Substitute the divisor classes \[ E_i=e_i, \quad F_{ij}=h-e_i-e_j, \quad G_i=2h-\sum_{r\ne i}e_r \] into the intersection form \[ h^2=1, \qquad e_i^2=-1, \qquad h\cdot e_i=e_i\cdot e_j=0\quad(i\ne j). \] Each displayed formula follows immediately.

We classify a six by the number of \(E\)-lines it contains, using step 9.

If it contains all six \(E_i\), we obtain the standard six \[ \{E_1,\ldots,E_6\}. \] If it contains at least two but fewer than six \(E\)-lines, no \(G\)-line can occur, because a \(G_j\) is skew to only the single \(E_j\). Every accompanying \(F_{ab}\) must have both indices outside the chosen \(E\)-indices. Distinct \(F\)-lines are mutually skew exactly when their two-element index sets meet.

For three chosen \(E\)-lines, the complementary index set has size three and its three pairs form a mutually intersecting family; this gives one six for each three-element subset, hence \[ \binom63=20. \] For two, four, or five chosen \(E\)-lines, there are not enough pairwise intersecting two-element subsets to complete a six. Indeed, on four vertices a pairwise-intersecting family of edges has size at most three: if three edges form a triangle, no fourth edge meets all three, while otherwise all edges share one vertex. On two or one vertices the bound is immediate.

Now suppose there is exactly one \(E_i\). The only possible \(G\)-line is \(G_i\). Without \(G_i\), one would need five pairwise-intersecting edges of the complete graph on the remaining five indices, but the largest such family has size four. With \(G_i\), one needs four such edges. A family of four pairwise-intersecting edges on five vertices is necessarily a star: after two edges \(12,13\), any edge avoiding vertex \(1\) must be \(23\), and then no fourth distinct edge can meet all three \(12,13,23\). Thus a four-edge family has a common vertex. There are five choices for its center. Hence this case contributes \[ 6\cdot5=30 \] sixes.

The same argument with \(E\) and \(G\) interchanged gives \(20\) sixes containing three \(G\)-lines and one six containing all six \(G_i\). No other number of \(G\)-lines can occur. Therefore the total is \[ 1+20+30+20+1=72. \]

The five cases in step 10 are distinguished by the numbers of \(E\)- and \(G\)-lines: \[ (6,0), \quad (3,0), \quad (1,1), \quad (0,3), \quad (0,6). \] The symmetric group on the six indices is transitive on each combinatorial choice occurring in the classification: on three-element subsets, and on ordered pairs consisting of the index \(i\) of the unique \(E_i\) and the center of the four-edge star among the other five indices. Thus the displayed five sets are orbit representatives.

The subgroup generated by \(x_1,\ldots,x_5\) is the full permutation group on the six indices by part (a), so every \(s_{ijk}\) is a conjugate of \(s_\beta\) and belongs to the image of \(\Phi\).

For \[ \beta_{ijk}=h-e_i-e_j-e_k, \] the reflection formula gives the following useful rules. It sends an \(E\)-line indexed by one of \(i,j,k\) to the \(F\)-line on the other two indices; it fixes an \(F_{ab}\) having exactly one index in \(\{i,j,k\}\); and if neither \(a\) nor \(b\) lies in that triple, it sends \(F_{ab}\) to the \(G\)-line whose index is the unique remaining sixth index. Applying these rules to the four displayed representatives gives exactly the stated identities.

Every permutation of the \(E_i\) is realized by the elements \(x_1,\ldots,x_5\), so \(\Sigma_6\) is contained in the stabilizer.

Conversely, an automorphism of the incidence configuration preserving \(\mathcal S_0\) induces a permutation of its six lines. Once that permutation is known, all other \(21\) lines are determined by incidence with \(\mathcal S_0\): the line \(F_{ij}\) is the unique line meeting exactly \(E_i,E_j\) among the six, while \(G_i\) is the unique line meeting exactly the five \(E_j\) with \(j\ne i\). Thus an element fixing every \(E_i\) fixes all \(27\) lines. The stabilizer therefore embeds in \(\Sigma_6\), and equality follows.

By step 13, the subgroup \(H=\operatorname{im}(\Phi)\) is transitive on the \(72\) sixes. It contains the full stabilizer \(\Sigma_6\) of \(\mathcal S_0\). Hence orbit–stabilizer gives \[ |H|\ge72\cdot6!=51840. \]

On the other hand \(G\) also acts on the \(72\) sixes, and step 14 says that the stabilizer of \(\mathcal S_0\) in \(G\) has order \(6!\). Therefore \[ |G|\le72\cdot6!=51840. \] Since \[ H\subseteq G, \] both inequalities are equalities and \[ H=G. \] This proves part (b).

Write a root as \[ r=ah-\sum_{i=1}^6b_ie_i. \] The conditions \[ K_S\cdot r=0, \qquad r^2=-2 \] are \[ \sum_i b_i=3a, \qquad \sum_i b_i^2=a^2+2. \] Cauchy–Schwarz gives \[ 9a^2 = \left(\sum_i b_i\right)^2 \le 6\sum_i b_i^2 = 6a^2+12, \] so \[ |a|\le2. \]

If \(a=0\), the equations become \[ \sum b_i=0, \qquad \sum b_i^2=2, \] so exactly one \(b_i\) is \(1\), one is \(-1\), and the rest are zero. This gives \(30\) roots \(e_i-e_j\).

If \(a=1\), then \[ \sum_i b_i=\sum_i b_i^2=3. \] Since \(m(m-1)\ge0\) for every integer \(m\), equality of these sums forces each \(b_i\) to be \(0\) or \(1\); exactly three are \(1\). This gives \[ \binom63=20 \] roots. The case \(a=-1\) gives their negatives, for another \(20\).

If \(a=2\), the same argument gives \[ \sum b_i=\sum b_i^2=6, \] so every \(b_i=1\), giving one root; \(a=-2\) gives its negative. Thus the total number is \[ 30+20+20+1+1=72. \]

The subgroup generated by \(s_{\alpha_1},\ldots,s_{\alpha_5}\) permutes the indices, so it is transitive on the roots \(e_i-e_j\) up to the evident change of ordered pair.

Now \[ s_\beta(e_3-e_4) = h-e_1-e_2-e_4, \] which is a root of the second type. Permuting indices reaches every positive root of that type. If \(\eta=w(\alpha_3)\) is any root already reached from a simple root, then \[ s_\eta=w s_{\alpha_3}w^{-1} \] belongs to the Weyl group and sends \(\eta\) to \(-\eta\). Hence all roots of the second type, with both signs, lie in the same orbit.

Finally, for \[ \gamma=h-e_4-e_5-e_6 \] one has \[ \gamma\cdot\beta=1, \] and therefore \[ s_\beta(\gamma) = 2h-\sum_{i=1}^6e_i. \] Thus the final pair of roots also belongs to the same orbit.

Directly, \[ \rho\cdot(e_i-e_j)=0. \] For a root \[ h-e_i-e_j-e_k, \] the intersection with \(\rho\) is \[ 2-3=-1, \] and \(\rho^2=-2\), so neither of the other two root types is orthogonal to \(\rho\). Hence the orthogonal root subsystem consists exactly of the \(30\) differences \(e_i-e_j\), with simple roots \[ e_1-e_2, e_2-e_3, \ldots, e_5-e_6. \] This is type \(A_5\), whose Weyl group is \(\Sigma_6\) by part (a).

The geometric representation theorem for Coxeter groups identifies the abstract group given by the Coxeter presentation with the reflection group generated by the six simple-root reflections of step 7; in particular this representation is faithful.

For a finite Weyl group, the stabilizer of a root is the Weyl group of the root subsystem orthogonal to that root. By step 18, the stabilizer of \(\rho\) is therefore the type-\(A_5\) Weyl group, of order \(720\). Step 17 shows that the orbit of \(\rho\) consists of all \(72\) roots. Orbit–stabilizer now gives \[ |\mathbf E_6| = 72\cdot720 = 51840. \]

Step 15 gives a surjective homomorphism \[ \mathbf E_6\twoheadrightarrow G \] and shows \[ |G|=51840. \] Step 19 gives \[ |\mathbf E_6|=51840. \] A surjection between finite groups of equal order is an isomorphism. This proves part (c).

Steps <1>1–<1>4 prove part (a), steps <1>5–<1>15 prove part (b), and steps <1>16–<1>20 prove part (c).