E₆/F₄ · Proposition III family

Verdict

Condition (1) holds: $Z_{G^\theta}(S)=\mathrm{Spin}_8$ is connected. Coverage fails exactly at the labels $3k\omega_1^\vee$ and $3k\omega_6^\vee$ with $k$ odd, the smallest being $3\omega_1^\vee$.

By Proposition III, step 7, H1–H5 fail for $E_6/F_4$.

Smallest failure: $3\omega_1^\vee$

$E_6/F_4$ has no parameter. Its labels fill the chamber of a restricted root system of type $A_2$, and the failures sit on the walls:

$$\Lambda_S^+=\{a\omega_1^\vee+b\omega_6^\vee:a,b\ge0,\ a\equiv b\bmod3\}$$ $$\Lambda_S^+\setminus\varphi(\Lambda_T^+)=\{3k\omega_1^\vee,\,3k\omega_6^\vee:k\ \text{odd}\}.$$

The 60-degree chamber spanned by omega 1 check and omega 6 check, with the labels a omega 1 check plus b omega 6 check, a congruent to b mod 3, drawn as points of a triangular lattice. All are green except red rings at 3 omega 1 check, 9 omega 1 check on the wall b = 0 and 3 omega 6 check, 9 omega 6 check on the wall a = 0. An inset shows the Satake diagram E IV: the E6 diagram with white end nodes 1 and 6 and black nodes 2, 3, 4, 5. — **Figure 1.** $E_6/F_4$: the labels without a dominant lift are the odd multiples of $3\omega_1^\vee$ and $3\omega_6^\vee$ on the walls; inset, the Satake diagram E IV.

Intuition on the smallest failureworked here

What $\varphi$ does. It kills the black coroots; for $i=2,\dots,5$, $\varphi(\omega_i^\vee)$ has both coefficients $\ge1$, so only $\omega_1^\vee$ lands on the wall $b=0$ and only $\omega_6^\vee$ on $a=0$: $$\varphi(\omega_1^\vee)=2\omega_1^\vee,\qquad\varphi(\omega_6^\vee)=2\omega_6^\vee.$$
Why a wall label can fail. A dominant $\lambda=\sum_ic_i\omega_i^\vee$ with $\varphi(\lambda)$ on the wall $b=0$ is a multiple of $\omega_1^\vee$, so its image is even; the lattice condition makes the label a multiple of $3$: $$\varphi(\lambda)=a\omega_1^\vee\ \Longrightarrow\ \lambda=c_1\omega_1^\vee,\ \ a=2c_1$$ $$a\equiv b=0\bmod3.$$
Same picture as $\mathrm{SL}_6/\mathrm{Sp}_6$. Sending the pair values $a$ of that family to $(a_1-a_2)\omega_1^\vee+(a_2-a_3)\omega_6^\vee$ matches the two label sets and their failures (its Figure 1): $$(2,2,-1,-1,-1,-1)\leftrightarrow3\omega_1^\vee$$ $$(1,1,1,1,-2,-2)\leftrightarrow3\omega_6^\vee.$$
For H5. $Z_{G^\theta}(S)=\mathrm{Spin}_8$ is connected, so by Proposition III, step 6 H5 fails exactly at these labels.

1 · Model

Group. $G=E_6$ is simply connected. Its Bourbaki labels form the chain 1–3–4–5–6 with node 2 attached to node 4, and its Cartan matrix has entries $\langle\alpha_i,\alpha_j^\vee\rangle$:

$$A=\begin{pmatrix}2&0&-1&0&0&0\\0&2&0&-1&0&0\\-1&0&2&-1&0&0\\0&-1&-1&2&-1&0\\0&0&0&-1&2&-1\\0&0&0&0&-1&2\end{pmatrix}.$$

Involution. $\theta$ is the involution of type E IV, and its fixed group $G^\theta=F_4$ is connected because $G$ is simply connected. Its Satake diagram has white nodes $1,6$, black nodes $2,3,4,5$ forming a $D_4$, and no arrows.

Torus. $S$ is a maximal θ-split torus and $T\supseteq S$ a θ-stable maximal torus, written in coroot coordinates. The Satake diagram makes $\theta$ act on $T$ by $-w_L$, where $w_L$ is the longest element of the Weyl group of the black $D_4$:

$$T=\{\alpha_1^\vee(t_1)\,\alpha_2^\vee(t_2)\cdots\alpha_6^\vee(t_6)\}\cong(\mathbb C^\times)^6,\qquad\theta=-w_L.$$

$w_L$ is $-1$ on the span of the black coroots and fixes its orthogonal complement, the span of $\omega_1^\vee,\omega_6^\vee$. So $\theta$ fixes $\alpha_2^\vee,\dots,\alpha_5^\vee$ and negates $\omega_1^\vee,\omega_6^\vee$.

Split torus. $\Lambda_S$ has the basis $\varphi\alpha_1^\vee,\varphi\alpha_6^\vee$ (section 2), so $S$ has two coordinates $u,v\in\mathbb C^\times$:

$$S=\{s(u,v)\},\qquad s(u,v)=\alpha_1^\vee(u^2)\,\alpha_2^\vee(uv)\,\alpha_3^\vee(u^2v)\,\alpha_4^\vee(u^2v^2)\,\alpha_5^\vee(uv^2)\,\alpha_6^\vee(v^2).$$

$\theta(s)=s^{-1}$ on $S$, and $\dim S=2$ is the rank of $E_6/F_4$, the number of white nodes.

Fixed torus. $(T^\theta)^\circ$ is the torus of $\alpha_2^\vee,\dots,\alpha_5^\vee$. Every $t\in T$ is $s(u,v)\,t'$ with $u^2=t_1$, $v^2=t_6$ and $t'\in(T^\theta)^\circ$, so $T=(T^\theta)^\circ\cdot S$.

2 · Lattices

Cocharacters. $G$ is simply connected, so $\Lambda_T=\mathbb Z\alpha_1^\vee\oplus\dots\oplus\mathbb Z\alpha_6^\vee$. In this basis $\theta$ is the integer matrix whose column $j$ is $\theta\alpha_j^\vee$:

$$\theta=\begin{pmatrix}-1&0&0&0&0&0\\-1&1&0&0&0&-1\\-2&0&1&0&0&-1\\-2&0&0&1&0&-2\\-1&0&0&0&1&-2\\0&0&0&0&0&-1\end{pmatrix},\qquad\theta^2=1.$$

The first column comes from splitting $\alpha_1^\vee=p+q$, with $p$ in the black span and $q$ orthogonal to it. Pairing with $\alpha_2,\dots,\alpha_5$ determines $p$, and $\theta\alpha_1^\vee=p-q=2p-\alpha_1^\vee$:

$$p=-\tfrac12\alpha_2^\vee-\alpha_3^\vee-\alpha_4^\vee-\tfrac12\alpha_5^\vee,\qquad\theta\alpha_1^\vee=-\alpha_1^\vee-\alpha_2^\vee-2\alpha_3^\vee-2\alpha_4^\vee-\alpha_5^\vee.$$

The last column follows by the symmetry $1\leftrightarrow6$, $3\leftrightarrow5$. Since $A\theta=\theta^TA$, the same matrix gives $\theta$ on the roots, and $-\theta\alpha_1=\alpha_1+\alpha_2+2\alpha_3+2\alpha_4+\alpha_5$ is a root: the highest root of the $D_5$ on nodes $1,\dots,5$.

Split lattice. $\Lambda_S$ is the $(-1)$-eigenlattice, $\Lambda_T\cap(\mathbb Q\omega_1^\vee\oplus\mathbb Q\omega_6^\vee)$. The inverse Cartan matrix gives:

$$3\omega_1^\vee=4\alpha_1^\vee+3\alpha_2^\vee+5\alpha_3^\vee+6\alpha_4^\vee+4\alpha_5^\vee+2\alpha_6^\vee,$$ $$3\omega_6^\vee=2\alpha_1^\vee+3\alpha_2^\vee+4\alpha_3^\vee+6\alpha_4^\vee+5\alpha_5^\vee+4\alpha_6^\vee.$$

So $a\omega_1^\vee+b\omega_6^\vee$ has integer coroot coordinates exactly when $a\equiv b\bmod3$. Equivalently, $\omega_1^\vee$ and $\omega_6^\vee$ have classes $1$ and $2$ in the coweight lattice modulo $\Lambda_T$, a cyclic group of order $3$:

$$\Lambda_S=\{a\omega_1^\vee+b\omega_6^\vee:a,b\in\mathbb Z,\ a\equiv b\bmod3\}.$$

Leading-term map. $\varphi=1-\theta$ annihilates the black coroots:

$$\varphi=\begin{pmatrix}2&0&0&0&0&0\\1&0&0&0&0&1\\2&0&0&0&0&1\\2&0&0&0&0&2\\1&0&0&0&0&2\\0&0&0&0&0&2\end{pmatrix},$$ $$\varphi\alpha_1^\vee=2\omega_1^\vee-\omega_6^\vee,\qquad\varphi\alpha_6^\vee=-\omega_1^\vee+2\omega_6^\vee,\qquad\varphi\alpha_j^\vee=0\ \ (j=2,\dots,5).$$

The two images satisfy $a\equiv b\bmod3$ and span a sublattice of index $\det\left(\begin{smallmatrix}2&-1\\-1&2\end{smallmatrix}\right)=3$ in $\mathbb Z\omega_1^\vee\oplus\mathbb Z\omega_6^\vee$. So $\varphi(\Lambda_T)=\Lambda_S$.

3 · Positive roots

Borel. $B$ is the Borel of the simple roots $\alpha_1,\dots,\alpha_6$, so $\Phi^+$ is the $36$ roots with nonnegative coefficients. A root is trivial on $S$ exactly when its $\alpha_1$- and $\alpha_6$-coefficients vanish, so $L=Z_G(S)$ is the Levi of the black nodes:

$$\Phi_L=\Phi\cap(\mathbb Z\alpha_2+\mathbb Z\alpha_3+\mathbb Z\alpha_4+\mathbb Z\alpha_5),\qquad|\Phi_L^+|=12.$$

θ on roots. $\theta$ fixes $\Phi_L$, since it fixes the black span. For $\alpha\in\Phi^+\setminus\Phi_L$, the root $-\theta\alpha=w_L\alpha$ keeps the $\alpha_1$- and $\alpha_6$-coefficients of $\alpha$, one of which is positive, because the black Weyl group changes only the coefficients of $\alpha_2,\dots,\alpha_5$:

$$-\theta\alpha\in\Phi^+\qquad(\alpha\in\Phi^+\setminus\Phi_L).$$

4 · Centralizer

The Levi. The derived group of $L$ has maximal torus $(T^\theta)^\circ$, whose cocharacter lattice $\Lambda_T\cap(\mathbb Q\alpha_2^\vee\oplus\dots\oplus\mathbb Q\alpha_5^\vee)$ is the coroot lattice $\mathbb Z\alpha_2^\vee\oplus\dots\oplus\mathbb Z\alpha_5^\vee$ of the black $D_4$. So the derived group is the simply connected $\mathrm{Spin}_8$, and $L=S\cdot\mathrm{Spin}_8$ because $S$ is the identity component of the centre of $L$.

θ on the Levi. $\theta$ acts trivially on $L/S$ (Proposition III, step 4), so $g\mapsto\theta(g)g^{-1}$ maps the connected group $\mathrm{Spin}_8$ into the finite group $S\cap\mathrm{Spin}_8$. Hence $\theta$ fixes $\mathrm{Spin}_8$ pointwise and inverts $S$:

$$\theta(s\,m)=s^{-1}m,\qquad L^\theta=S[2]\cdot\mathrm{Spin}_8,\qquad S[2]=\{s\in S:s^2=1\}.$$

The 2-torsion of $S$. In the coordinates of section 1, $S[2]=\{s(u,v):u,v=\pm1\}$. Its generators lie in $(T^\theta)^\circ\subseteq\mathrm{Spin}_8$:

$$s(-1,1)=\alpha_2^\vee(-1)\,\alpha_5^\vee(-1),\qquad s(1,-1)=\alpha_2^\vee(-1)\,\alpha_3^\vee(-1).$$

So the centralizer is connected. It is the complexification of the group $M=\mathrm{Spin}(8)$ of the real form $E_{6(-26)}$:

$$Z_{G^\theta}(S)=L^\theta=\mathrm{Spin}_8.$$

Chen–Yi's Example 88 asserts that the $L^+G$-stabilizers on $LX$ are connected for $E_6/F_4$, without naming the centralizer. By step 4 that is the same condition.

5 · Coverage

Dominant cones. Write $\lambda=\sum_ic_i\omega_i^\vee$; it is dominant when every $c_i\ge0$. Modulo $\Lambda_T$, the coweights $\omega_1^\vee,\omega_3^\vee,\omega_5^\vee,\omega_6^\vee$ have classes $1,2,1,2$ and $\omega_2^\vee,\omega_4^\vee$ lie in $\Lambda_T$, so:

$$\Lambda_T^+=\{\textstyle\sum_ic_i\omega_i^\vee:\ c_i\in\mathbb Z_{\ge0},\ c_1-c_3+c_5-c_6\equiv0\bmod3\},$$ $$\Lambda_S^+=\{a\omega_1^\vee+b\omega_6^\vee:\ a,b\in\mathbb Z_{\ge0},\ a\equiv b\bmod3\}.$$

The map on coweights. $\varphi$ annihilates $\alpha_2^\vee,\dots,\alpha_5^\vee$, so $\varphi(\omega_i^\vee)$ depends only on the $\alpha_1^\vee$- and $\alpha_6^\vee$-coefficients $y_i,z_i$ of $\omega_i^\vee$. The inverse Cartan matrix supplies them:

$$\varphi(\omega_i^\vee)=y_i\,\varphi\alpha_1^\vee+z_i\,\varphi\alpha_6^\vee=(2y_i-z_i)\,\omega_1^\vee+(2z_i-y_i)\,\omega_6^\vee.$$

Table 1. $i$ and $1$ and $2$
$i$	$1$	$2$	$3$	$4$	$5$	$6$
$(y_i,z_i)$	$(\frac43,\frac23)$	$(1,1)$	$(\frac53,\frac43)$	$(2,2)$	$(\frac43,\frac53)$	$(\frac23,\frac43)$
$\varphi(\omega_i^\vee)$	$2\omega_1^\vee$	$\omega_1^\vee+\omega_6^\vee$	$2\omega_1^\vee+\omega_6^\vee$	$2\omega_1^\vee+2\omega_6^\vee$	$\omega_1^\vee+2\omega_6^\vee$	$2\omega_6^\vee$

$$\varphi\Bigl(\sum_ic_i\omega_i^\vee\Bigr)=(2c_1+c_2+2c_3+2c_4+c_5)\,\omega_1^\vee+(c_2+c_3+2c_4+2c_5+2c_6)\,\omega_6^\vee.$$

Lemma. A label $a\omega_1^\vee+b\omega_6^\vee\in\Lambda_S^+$ lies in $\varphi(\Lambda_T^+)$ unless $b=0$ with $a$ odd, or $a=0$ with $b$ odd. Since $a\equiv b\bmod3$, the labels without a lift are

$$3k\omega_1^\vee,\quad3k\omega_6^\vee\qquad(k\ \text{odd}).$$

Proof. If $b=0$, the $\omega_6^\vee$-coefficient forces $c_2=\dots=c_6=0$, and then $a=2c_1$ is even; the case $a=0$ is symmetric. Otherwise the parities of $(a,b)$ pick one of four dominant lifts:

$$\text{even, even}:\ \frac a2\omega_1^\vee+\frac b2\omega_6^\vee,\qquad\text{odd, odd}:\ \omega_2^\vee+\frac{a-1}2\omega_1^\vee+\frac{b-1}2\omega_6^\vee,$$ $$\text{odd, even}:\ \omega_5^\vee+\frac{a-1}2\omega_1^\vee+\frac{b-2}2\omega_6^\vee,\qquad\text{even, odd}:\ \omega_3^\vee+\frac{a-2}2\omega_1^\vee+\frac{b-1}2\omega_6^\vee.$$

The last two need $b\ge2$, resp. $a\ge2$, which hold outside the exceptions. Each lift lies in $\Lambda_T$, because the image formula gives $a-b\equiv-(c_1-c_3+c_5-c_6)\bmod3$.

The smallest failure. $3\omega_1^\vee$ lifts in $\Lambda_T$, as Chen–Yi, Lemma 90 asserts, but not dominantly:

$$\varphi(2\alpha_1^\vee+\alpha_6^\vee)=3\omega_1^\vee,\qquad\varphi(\lambda)=3\omega_1^\vee\ \Longrightarrow\ c_2=\dots=c_6=0,\ \ 2c_1=3.$$

A search over all labels with $a,b\le15$ finds exactly six without a dominant lift: $3\omega_1^\vee$, $9\omega_1^\vee$, $15\omega_1^\vee$ and their images under $1\leftrightarrow6$.

\(i\)	\(1\)	\(2\)	\(3\)	\(4\)	\(5\)	\(6\)
\((y_i,z_i)\)	\((\frac43,\frac23)\)	\((1,1)\)	\((\frac53,\frac43)\)	\((2,2)\)	\((\frac43,\frac53)\)	\((\frac23,\frac43)\)
\(\varphi(\omega_i^\vee)\)	\(2\omega_1^\vee\)	\(\omega_1^\vee+\omega_6^\vee\)	\(2\omega_1^\vee+\omega_6^\vee\)	\(2\omega_1^\vee+2\omega_6^\vee\)	\(\omega_1^\vee+2\omega_6^\vee\)	\(2\omega_6^\vee\)