Proof. Note that if $G=D_\ell $ , by our convention, the values of k are $1,\dots ,\ell -1,\ell ',\ell "$ . In all cases, in the statement of the proposition, the group G is transitive on $\mathcal {S}_k(V)$ . It is well known that the maximal parabolic subgroups of a classical group are stabilizers of totally singular subspaces; therefore, these must be the generic stabilizers. Unless we are in type D and $k=\ell -1$ , the stabilizer of $y\in \mathcal {S}_k(V)$ is a conjugate of $P_k$ . If $G=D_\ell $ and $k=\ell -1$ , the stabilizer of $y\in \mathcal {S}_k(V)$ is a conjugate of $P_{\ell -1,\ell }$ .

Proof. In each of these cases, $k=1$ and the module V is symplectic. Therefore, $\mathcal {S}_k(V)=\mathcal {G}_k(V)$ , and the result follows directly from [Reference Guralnick and Lawther10]. In the last column of Table 5, we give a reference for each individual case.

Proof. Let $q=p^i$ and let $\sigma =\sigma _q$ be the standard Frobenius morphism acting on K as $t\mapsto t^\sigma =t^q$ and on G as $x_{\pm \alpha _1}(t)\mapsto x_{\pm \alpha _1}(t^q)$ . Let $G=SL_2(K)$ . We can view V as the space $M_{2\times 2}(K)$ of $2\times 2$ matrices on which G acts by $g.v=gv(g^\sigma )^T$ for $v\in M_{2\times 2}(K)$ and $g\in G$ . Since G preserves the determinant of v for all $v\in M_{2\times 2}(K)$ , we can take the quadratic form $Q:V\rightarrow K$ as $Q(v)=\det v$ . The singular $1$ -spaces of V are therefore the $1$ -spaces spanned by matrices with determinant $0$ . Let y be the singular $1$ -space spanned by $\left (\begin {matrix} 0 & 1\\ 0 & 0 \\ \end {matrix}\right )$ . Then $\left (\begin {matrix} a & b\\ c & d \\ \end {matrix}\right )\in G_y$ if and only if $b=c=0$ and $d=a^{-1}$ . Since $\dim G - \dim \mathcal {S}_1(V) = 1 =\dim G_y$ , the element y is in a dense orbit for the G-action on $\mathcal {S}_1(V)$ . Therefore, $C_{\mathcal {S}_1(V)} = T_1$ .

Proof. By [Reference Rizzoli27, Lemma 5.5], there is $y\in \mathcal {S}_1(V)$ with $G_y=U_2T_1$ . Since $\dim G - \dim \mathcal {S}_1(V)=8-5 =\dim G_y$ , the element y is in a dense orbit for the G-action on $\mathcal {S}_1(V)$ . Therefore, $C_{\mathcal {S}_1(V)} = U_2T_1$ .

Proof. By [Reference Revoy24, 2.3.1(II)], there is $y\in \mathcal {S}_1(V)$ with $G_y=U_8A_2T_1$ . Since $\dim G - \dim \mathcal {S}_1(V)=35-18 =\dim G_y$ , the element y is in a dense orbit for the G-action on $\mathcal {S}_1(V)$ . Therefore, $C_{\mathcal {S}_1(V)} = U_8A_2T_1$ .

Proof. By [Reference Liebeck, Saxl and Seitz17, Thm B], the group G is transitive on $\mathcal {S}_1(V)$ . The generic stabilizer is the $P_3$ -parabolic (i.e. $C_{\mathcal {S}_1(V)} = U_6A_2T_1$ ).

Proof. By [Reference Liebeck, Saxl and Seitz17, Thm B], the group G is transitive on $\mathcal {S}_4'(V)$ . Let $W\in \mathcal {S}_4'(V)$ . The group G is the group of fixed points of a triality automorphism of $D_4 = Cl(V)$ . Therefore, $G_W$ is isomorphic to the generic stabilizer for the action on $\mathcal {S}_1(V)$ , i.e. $U_6A_2T_1$ . Then it is easy to see ([Reference Guralnick, Liebeck, Macpherson and Seitz11, Lemma 3.5]) that $G_W$ acts on $\mathcal {G}_2(W)$ with two orbits, one with stabilizer $U_5A_1A_1T_1$ and one with stabilizer $U_7A_1A_1T_1$ . Since every totally singular $2$ -space is contained in an element of $\mathcal {S}_4'(V)$ , we conclude that there are at most two G-orbits on $\mathcal {S}_2(V)$ . Since $\dim G-\dim \mathcal {S}_2(V)=12$ , there must be a $12$ -dimensional stabilizer for the G-action on $\mathcal {S}_2(V)$ . The only possibility is therefore $C_{\mathcal {S}_2(V)} = U_5A_1A_1T_1$ .

Similarly, $G_W$ acts on $\mathcal {G}_3(W)$ with two orbits, one with stabilizer $U_3A_2T_1$ and one with stabilizer $U_8A_1T_2$ . Since $\dim G-\dim U_3A_2T_1 = \dim \mathcal {S}_3(V)$ , we conclude that $C_{\mathcal {S}_3(V)} = U_3A_2T_1$ .

Proof. By [Reference Igusa12, Prop. 5, Prop. 6] (when $p\neq 2$ ) and [Reference Guralnick, Liebeck, Macpherson and Seitz11, Lemma 2.11] (when $p=2$ ), there is only one orbit on $\mathcal {G}_1(V)$ with a $22$ -dimensional stabilizer when $\ell =4$ , with structure $U_7G_2T_1$ , and only one orbit on $\mathcal {G}_1(V)$ with $25$ -dimensional stabilizer when $\ell =5$ , with structure $U_{14}B_2T_1$ . Since $\dim G-\dim \mathcal {S}_1(V) = 22$ when $\ell =4$ and $25$ when $\ell =4$ , we conclude that $C_{\mathcal {S}_1(V)} = U_7G_2T_1$ when $\ell =4$ and $U_{14}B_2T_1$ when $\ell =5$ .

Proof. By [Reference Rizzoli27, Lemma 5.15], there is $y\in \mathcal {S}_1(V)$ with $G_y=U_6A_1T_1$ . Since $\dim G - \dim \mathcal {S}_1(V)=21-11 =\dim G_y$ , the element y is in a dense orbit for the G-action on $\mathcal {S}_1(V)$ . Therefore, $C_{\mathcal {S}_1(V)} = U_6A_1T_1$ .

Proof. By the proof of [Reference Guralnick, Liebeck, Macpherson and Seitz11, Lemma 2.11], there is only one orbit on $\mathcal {G}_1(V)$ with a $36$ -dimensional stabilizer, with structure $U_{14}B_3T_1$ . Since $\dim G - \dim \mathcal {S}_1(V)=66-30 =\dim G_y$ , the element y is in a dense orbit for the G-action on $\mathcal {S}_1(V)$ . Therefore, $C_{\mathcal {S}_1(V)} = U_{14}B_3T_1$ .

Proof. By [Reference Liebeck, Saxl and Seitz17, Thm. A], G is transitive on singular $1$ -spaces of V. Therefore, a representative can be taken to be the $1$ -space spanned by the highest weight vector, with stabilizer $P_1=U_{5}A_1T_1$ .

Proof. Let $\mu _1,\mu _2,\mu _3$ be the positive weights $2\alpha _1+\alpha _2,\alpha _1+\alpha _2,\alpha _1$ , respectively. Given weight vectors $v_{\mu _1},v_{-\mu _2} $ , let $y:=\langle v_{\mu _1},v_{-\mu _2}\rangle $ , an element of $\mathcal {S}_2(V)$ . Since $n_2n_1n_2n_1n_2.\langle v_{\mu _1} \rangle = \langle v_{-\mu _2}\rangle $ , we can easily determine that $G_{\langle v_{\mu _1}\rangle } \cap G_{\langle v_{-\mu _2}\rangle } = P_1\cap P_1^{n_2n_1n_2n_1n_2} = U_3T_2$ , by checking which root subgroups are in common between $P_1$ and $P_1^{n_2n_1n_2n_1n_2}$ . Therefore, $G_y\leq U_3A_1T_1$ . Since $\dim \mathcal {S}_2(V) = 7$ (in both cases $p\neq 2$ and $p=2$ ), the minimum dimension of the stabilizer of any totally singular $2$ -space is $7$ . Therefore, $G_y = U_3A_1T_1 = C_{\mathcal {S}_2(V)}$ .

Proof. By the proof of [Reference Guralnick, Liebeck, Macpherson and Seitz11, Lemma 3.4], there is an $A_2$ -subgroup which is the stabilizer of $y\in \mathcal {S}_3(V)$ . Since $\dim G - \dim \mathcal {S}_3(V)=14-6 =\dim G_y$ , the element y is in a dense orbit for the G-action on $\mathcal {S}_3(V)$ . Therefore, $C_{\mathcal {S}_3(V)} = A_2$ .

Proof. By [Reference Brundan6, Lemma 4.13], there is $y\in \mathcal {S}_1(V)$ such that $G_y=U_{14}G_2T_1$ . Since $\dim G - \dim \mathcal {S}_1(V)=29 =\dim G_y$ , the element y is in a dense orbit for the G-action on $\mathcal {S}_1(V)$ . Therefore, $C_{\mathcal {S}_1(V)} = U_{14}G_2T_1$ .

Proof. By [Reference Liebeck and Saxl16, Lemma 4.3], there is only one orbit on $\mathcal {G}_1(V)$ with a $79$ -dimensional stabilizer, with structure $U_{26}F_4T_1$ . Since $\dim G - \dim \mathcal {S}_1(V)=133-54 =\dim U_{26}F_4T_1$ , we must have $C_{\mathcal {S}_1(V)} = U_{26}F_4T_1$ .

Proof. The composition factors of $\mathrm {Lie}(G)$ are listed in [Reference Liebeck and Seitz19, Prop. 1.10]. By assumption on V, we have $V=\mathrm {Lie}(G)/Z$ , where Z is the centre of $\mathrm {Lie}(G)$ . If $p=2$ , and $G=A_\ell $ with $\ell \equiv 1\ \mod 4$ , or $G=D_\ell $ with $\ell \equiv 2\ \mod 4$ , or $G=E_7$ , then the module V is symplectic and $\mathcal {S}_1(V) = \mathcal {G}_1(V)$ . In all other cases, the module V is orthogonal. The proof closely mimics [Reference Guralnick and Lawther10, Lemma 4.5(ii)], but in the interest of clarity, it is fully reproduced with the appropriate changes. Note that our setup corresponds to the specific case $\theta = 1$ in [Reference Guralnick and Lawther10, Lemma 4.5(ii)], which in particular means $G=H$ in the proof of [Reference Guralnick and Lawther10, Lemma 4.5(ii)]. Let

$$\begin{align*}W^\ddagger = \{ w\in W:\exists \xi\in K^*,\forall v\in V_0, w.v=\xi v\},\end{align*}$$

and let $N^\ddagger $ be the pre-image of $W^\ddagger $ under the quotient map $N\rightarrow W$ . Let Y be $\mathcal {S}_1(V_0)$ and let $\hat {Y}_1$ be the subset of Y consisting of $1$ -spaces of $V_0$ spanned by images of regular semisimple elements in $\mathrm {Lie}(T)$ . Since Z does not contain regular semisimple elements, the set $\hat {Y}_1$ is nonempty, and thus dense in Y. Since $\dim V_0\geq 3$ , the span of Y is the full $V_0$ . Therefore, any element in N which fixes all $y\in Y$ must be in $N^\ddagger $ . Thus, given $w\in W\setminus W^\ddagger $ , take $n\in N$ with $nT=w$ ; the set of elements of Y fixed by n is a proper closed subvariety of Y. Let $\hat {Y}_2$ be the complement of the union of these subvarieties as w runs over $W\setminus W^\ddagger $ . Set $\hat {Y} = \hat {Y}_1\cap \hat {Y}_2$ , a dense open subset of Y.

Let $y\in \hat {Y}$ . By [Reference Guralnick, Liebeck, Macpherson and Seitz11, Lemma 2.1], two elements of $\mathcal {G}_1(V_0)$ are in the same G-orbit if and only if they are in the same W-orbit. Therefore, $G.y\cap Y$ is finite and $\dim (\overline {G.y\cap Y})=0$ . Also, since y is spanned by a regular semisimple element, we have $G_y^0 = T$ , and therefore, $\dim (\overline {G.y})=\dim G - \dim T $ . Finally, note that since $V = \mathrm {Lie(G)}/Z$ , we have $(C_G(V_0))^0 = T$ , and therefore, $\dim V-\dim V_0=\dim G-\dim T$ (see the proof of [Reference Guralnick, Liebeck, Macpherson and Seitz11, Lemma 2.4]). Thus,

$$ \begin{align*} \dim \mathcal{S}_1(V)-\dim (\overline{G.y}) &= \dim V- 2 -\dim (\overline{G.y}) =\\ &= \dim G-\dim T+\dim V_0 -2 -\dim (\overline{G.y}) =\\ &= \dim Y - \dim(\overline{G.y\cap Y}) = \dim Y + \operatorname{\mathrm{codim}} \mathrm{Tran}_G(y,Y)-\dim (\overline{G.y}),\end{align*} $$

where the last step uses Lemma 2.13. By definition, this proves that y is Y-exact. The conditions of Lemma 2.15 hold and $C_{\mathcal {S}_1(V)} = T.W^\ddagger = C_{\mathcal {G}_1(V)}$ .

Proof. If $p=2$ and $\ell \equiv 2\ \mod 4$ , then the module V is symplectic and $\mathcal {S}_1(V)=\mathcal {G}_1(V)$ . Therefore, assume that when $p=2$ , we have $\ell \not \equiv 2\ \mod 4$ , which implies that the module V is orthogonal. We use the setup of [Reference Guralnick and Lawther10, Prop. 5.13] combined with the approach of [Reference Guralnick and Lawther10, Lemma 4.5(ii)], which we saw in action in Proposition 4.1. Inside $\bigwedge ^2 V_{nat}$ , we have submodules $X_1 = \{ \sum _{i<j}\rho _{ij}e_i\wedge e_j+\sum _{i<j}\sigma _{ij}f_i\wedge f_j+\sum _{i,j}\tau _{ij}e_i\wedge f_j:\sum _i\tau _{ii}=0\}$ and $X_2=\langle \sum _i e_i\wedge f_i \rangle $ . If $p\nmid \ell $ then $V=X_1$ , otherwise $X_2<X_1$ and $V=X_1/X_2$ . In all cases, $V=X_1/(X_1\cap X_2)$ . Let $x_i = e_i\wedge f_i$ . The $0$ -weight space is $V_0 = \{ \sum a_ix_i+(X_1\cap X_2):\sum a_i = 0\}$ . Then G fixes a nondegenerate quadratic form on V such that $Q( a_ix_i+(X_1\cap X_2)) = \sum a_i^2 +\sum _{i<j}a_ia_j$ (see [Reference Garibaldi and Nakano9, p. 8.1.2] when $p=2$ ). Let $Y = \mathcal {S}_1(V_0)$ and let

$$\begin{align*}\hat{Y}_1 = \left\{ \langle v \rangle\in Y , v=\sum a_ix_i+(X_1\cap X_2): a_i\neq a_j\text{ if }i\neq j\right\},\end{align*}$$

a dense subset of Y. Then $y\in \hat {Y}_1$ is fixed by $C=A_1^\ell = \bigcap _{i} G_{\langle e_i,f_i \rangle }$ . Any minimal connected overgroup of C in G is isomorphic to $C_2 A_1^{\ell -2}$ , which does not fix $y\in \hat {Y}_1$ because of the condition on the coefficients. Therefore, for any $y\in \hat {Y}_1$ , we have $(G_y)^0 = C$ . Let $N = N_G(C)/C$ , a group isomorphic to $Sym(\ell )$ . Let

$$\begin{align*}N^\ddagger = \{ n\in N:\exists \xi\in K^*,\forall v\in V_0, n.v=\xi v\}.\end{align*}$$

Now assume that $n\in N$ fixes all $y\in Y$ . Since $\dim V_0\geq 3$ , the span of Y is the full $V_0$ . Therefore, any element in N which fixes all $y\in Y$ must be in $N^\ddagger $ . Thus, the set of elements fixed by $n\in N\setminus N^\ddagger $ is a proper closed subvariety of Y. Let $\hat {Y}_2$ be the complement of the union of these subvarieties as n runs over $N\setminus N^\ddagger $ . Set $\hat {Y} = \hat {Y}_1\cap \hat {Y}_2$ , a dense open subset of Y. Finally, note that the proof of [Reference Guralnick and Lawther10, Prop. 5.13] shows that $N^\ddagger = 1$ . Let $y\in \hat {Y}$ . Two elements of $\mathcal {G}_1(V_0)$ are in the same G-orbit if and only if they are in the same N-orbit, and therefore, $\dim (\overline {G.y\cap Y})=0$ . Also, $\dim V-\dim V_0=\dim G-\dim C$ , by the proof of [Reference Guralnick, Liebeck, Macpherson and Seitz11, Lemma 2.4]. Therefore, as in Proposition 4.1, we get

$$ \begin{align*} \dim \mathcal{S}_1(V)-\dim (\overline{G.y})&= \dim V- 2 -\dim (\overline{G.y}) =\\ &= \dim G-\dim C+\dim V_0 -2 -\dim (\overline{G.y}) = \\ &=\dim Y - \dim(\overline{G.y\cap Y}),\end{align*} $$

proving that y is Y-exact. The conditions of Lemma 2.15 hold and $C_{\mathcal {S}_1(V)} = C.N^\ddagger = C = A_1^\ell $ .

Proof. In all of these cases, the module V is a composition factor of $\mathrm {Lie}(G)$ and the $0$ -weight space is $2$ -dimensional, and we will show that the generic stabilizer is the stabilizer of one of the two singular $1$ -spaces of $V_0$ .

Let $G=B_2$ and $\lambda = 2\lambda _2$ with $p\neq 2$ . Then $V=\mathrm {Lie}(G)$ , and we can take $V_0 = \{\mathrm {diag}(a,b,0,-b,-a) :a,b\in K\}$ . Since $p\neq 2$ , the group G fixes the nondegenerate quadratic form Q induced by the Killing form. Let $v=\mathrm {diag}(a,b,0,-b,-a)$ be a singular element of $V_0$ . Then since $Q(v)=0$ , we know that $a^2+b^2=0$ . Since v is regular semisimple, we must have $(G_{\langle v\rangle })^0 = T_2$ . We then find that $W_{\langle v\rangle }=\langle w\rangle $ , where w is an element of order $4$ sending $\mathrm {diag}(a,b,0,-b,-a)\mapsto \mathrm {diag}(b,-a,0,a,-b) $ . Since $\dim G-\dim \mathcal {S}_1(V) = 2 =\dim G_{\langle v\rangle }$ , we conclude that $C_{\mathcal {S}_1(V)} = T_2.\mathbb {Z}_4$ .

Let $G=A_2$ and $\lambda =\lambda _1+\lambda _2$ with $p\neq 3$ . Then $V=\mathrm {Lie}(G)$ , and we can take $V_0 = \{\mathrm {diag}(a,b,-a-b):a,b\in K\}$ . If $p\neq 2$ , a nondegenerate symmetric bilinear form preserved by G is given by the Killing form. If $p=2$ , we find an explicit description of a nondegenerate quadratic form preserved by G in [Reference Babic and Chernousov3, §5.1]. Let $v=\mathrm {diag}(a,b,-a-b)$ be a singular element of $V_0$ , which implies that $a^2+b^2+ab = 0$ . Since v is regular semisimple, we must have $(G_{\langle v\rangle })^0 = T_2$ . As $a^2+b^2+ab = 0$ , we then find that $W_{\langle v\rangle }=\langle w\rangle $ , where w is a $3$ -cycle in W. Since $\dim G-\dim \mathcal {S}_1(V) = 2 =\dim G_{\langle v\rangle }$ , we conclude that $C_{\mathcal {S}_1(V)} = T_2.\mathbb {Z}_3$ .

Let $G=A_3$ and $\lambda =\lambda _1+\lambda _3$ with $p=2$ . Then $V=\mathfrak {sl}_4/\langle I\rangle $ , where I is the identity $4\times 4$ matrix, and we can take $V_0 = \{\mathrm {diag}(a,b,a+b,0)+\langle I\rangle :a,b\in K\}$ . Let $v=\mathrm {diag}(a,b,a+b,0)+\langle I \rangle $ be a singular element of $V_0$ , which implies that $a^2+b^2+ab = 0$ (again see [Reference Babic and Chernousov3, §5.1]). Since $\mathrm {diag}(a,b,a+b,0)$ is regular semisimple, we must have $(G_{\langle v\rangle })^0 = T_3$ . As $a^2+b^2+ab = 0$ , we then find that $W_{\langle v\rangle }=\langle \tau _1,\tau _2,w\rangle $ , where w is a $3$ -cycle in W sending $\mathrm {diag}(a,b,a+b,0)\mapsto \mathrm {diag}(a+b,a,b,0)$ , $\tau _1$ is an element of order $2$ sending $\mathrm {diag}(a,b,a+b,0)\mapsto \mathrm {diag}(0,a+b,b,a)$ and $\tau _2$ is an element of order $2$ sending $\mathrm {diag}(a,b,a+b,0)\mapsto \mathrm {diag}(a+b,0,a,b)$ . Since $\dim G-\dim \mathcal {S}_1(V) = 3 =\dim G_{\langle v\rangle }$ , we conclude that $C_{\mathcal {S}_1(V)} = T_3.Alt(4)$ .

Let $G=G_2$ and $\lambda = \lambda _2$ with $p\neq 3$ . Then $V=\mathrm {Lie}(G)$ . We view G as a subgroup of $B_3$ , so that we can take $V_0 =\{ \mathrm {diag}(a,b,-a-b,0,-a,-b,a+b):a,b\in K\}$ . Let $v=\mathrm {diag}(a,b,-a-b,0,-a,-b,a+b)$ be a singular element of $V_0$ , which implies that $a^2+b^2+ab = 0$ (see [Reference Babic and Chernousov3, §5.1]). Since v is regular semisimple, we must have $(G_{\langle v\rangle })^0 = T_2$ . As $a^2+b^2+ab = 0$ , we then find that $W_{\langle v\rangle }=\langle \tau , w\rangle $ , where $\tau $ is an element of order $2$ sending $v\mapsto -v$ and w is an element of order $3$ sending $\mathrm {diag}(a,b,-a-b,0,a+b,-b,-a)\mapsto \mathrm {diag}(-a-b,a,b,0,-b,-a,a+b)$ . Since $\dim G-\dim \mathcal {S}_1(V) = 2 =\dim G_{\langle v\rangle }$ , we conclude that $C_{\mathcal {S}_1(V)} = T_2.\mathbb {Z}_6$ .

Proof. In all of these cases, the module V is a composition factor of $\bigwedge ^2 V_{nat}$ , the $0$ -weight space is $2$ -dimensional, and the generic stabilizer is the stabilizer of one of the two singular $1$ -spaces of $V_0$ . Let $X_1,X_2$ be as in the proof of Proposition 4.2, so that $V = X_1/X_2$ .

Let $G=C_3$ and $\lambda = \lambda _2$ with $p\neq 3$ . Then $X_2 =0$ , and we can take

$$\begin{align*}V_0 = \{a e_1\wedge f_1+b e_2\wedge f_2 -(a+b)e_3\wedge f_3 :a,b,\in K\}.\end{align*}$$

Let $v=a e_1\wedge f_1+b e_2\wedge f_2 -(a+b)e_3\wedge f_3$ be a singular element of $V_0$ . Then since $Q(v)=0$ , we know that $a^2+b^2+ab=0$ . Since $a\neq b$ , we must have $(G_{\langle v\rangle })^0 = A_1^3$ as in the proof of Proposition 4.2. We then find that $W_{\langle v\rangle }=\langle w\rangle $ , where w is a $3$ -cycle in W. Since $\dim G-\dim \mathcal {S}_1(V) = 9 =\dim G_{\langle v\rangle }$ , we conclude that $C_{\mathcal {S}_1(V)} = A_1^3.\mathbb {Z}_3$ .

The remaining two cases are entirely similar, with the result for $D_4$ being derived from $C_4$ , since $V_{D_4}(\lambda _2) = V_{C_4}(\lambda _2)\downarrow D_4$ when $p=2$ .

Proof. In each of these cases, except for $(B_6,\lambda _6,p ,1)$ with $p\neq 2$ , it was proven in [Reference Rizzoli26] or [Reference Rizzoli27] that G has a dense orbit on $\mathcal {S}_k(V)$ . The structure of the generic stabilizer is as given in [Reference Rizzoli26][Reference Rizzoli27], with the appropriate reference listed in Table 6. Note that for the case $(B_6,\lambda _6,2,1)$ , Proposition 5.2 clarifies the lackluster explanation found in [Reference Rizzoli26].

Finally, in the case $(B_6,\lambda _6,p,1)$ with $p\neq 2$ , the module V is symplectic, and we have $\mathcal {S}_k(V) = \mathcal {G}_k(V)$ , with the result following from [Reference Guralnick and Lawther10].

Proof. This result is already listed in [Reference Rizzoli26, Thm. 1], albeit without a full explanation. The quadruple $(D_5,\lambda _5,2,1)$ has generic stabilizer $(G_2G_2).\mathbb {Z}_2$ by [Reference Guralnick and Lawther10]. By [Reference Rizzoli26, Lemma 5.17], we deduce that $C_{\mathcal {S}_1(V)}$ is isomorphic to the generic stabilizer of the action of $(G_2G_2).\mathbb {Z}_2$ on $V_{G_2}(\lambda _1)\oplus V_{G_2}(\lambda _1)$ . Since $G_2$ is transitive on nonzero vectors of $V_{G_2}(\lambda _1)$ , the generic stabilizer for this action is easily seen to be $(U_5A_1)^2.\mathbb {Z}_2$ .

Proof. As in Proposition 3.13, we know that there is $y\in \mathcal {S}_3(V)$ with $G_y = A_2$ . Since $\dim G - \dim A_2 = 6 = \dim \mathcal {S}_3(V)$ , we conclude that $C_{\mathcal {S}_3(V)} = A_2$ .

Proof. This is proved in [Reference Rizzoli27, Lemma 5.69].

Proof. We can recover this result by a slight change of the proof of [Reference Guralnick and Lawther10, Prop. 6.7]. Let $G=SL_2(K)^3$ with basis $e,f$ for $V_{nat}$ , so that $V=V_{nat}\otimes V_{nat}\otimes V_{nat}$ . Then, like in the proof of [Reference Guralnick and Lawther10, Prop. 6.7], given $\mathbf {a}=(a_1,a_2,a_3,a_4)$ , let

$$\begin{align*}v^{(1)}&=a_1e\otimes e\otimes e+ a_2e\otimes f\otimes f+a_3f\otimes e\otimes f+a_4f\otimes f\otimes e,\\v^{(2)}&=a_1f\otimes f\otimes f+ a_2f\otimes e\otimes e+a_3e\otimes f\otimes e+a_4e\otimes e\otimes f. \end{align*}$$

Then $y_{\mathbf {a}}:=\langle v^{(1)},v^{(2)} \rangle $ is totally singular if and only if $a_1^2+a_2^2+a_3^2+a_4^2 = 0$ . In the proof of [Reference Guralnick and Lawther10, Prop. 6.7], they then proceed to define $\hat {Y}$ as a dense subset of $Y:=\{ y_{\mathbf {a}}:\mathbf {a}\neq (0,0,0,0)\}$ by requiring certain polynomials on the coefficients to be nonzero. One of these conditions is that $a_1^2+a_2^2+a_3^2+a_4^2 \neq 0$ . However, this is later only used for the case $p=2$ . Therefore, we can modify the definition of $\hat {Y}$ in the proof of [Reference Guralnick and Lawther10, Prop. 6.7] and still end up with an open dense subset of $\mathcal {G}_2(V)$ where all stabilizers are conjugate to $\mathbb {Z}_2.\mathbb {Z}_2$ . The difference is that now this set will also contain totally singular $2$ -spaces, and therefore, $C_{\mathcal {S}_2(V)}=C_{\mathcal {G}_2(V)}=\mathbb {Z}_2.\mathbb {Z}_2$ .

Proof. We shall describe how to use the proof of [Reference Guralnick and Lawther10, Proposition 6.28] to reach the conclusion. In [Reference Guralnick and Lawther10, Proposition 6.28], the authors determine the generic stabilizer for the G-action on all $2$ -spaces. They do so in the following manner. They define a certain $8$ -space $V_{[0]}$ spanned by pairs of opposite weight vectors. This $8$ -space is the fixed point space of a subgroup A of G, where $A=D_4,A_1^3,T_2$ according to whether $G=E_7,D_6,A_5$ , respectively. They define a dense subset $\hat {Y}_1$ of $Y:=\mathcal {G}_2(V_{[0]})$ with the property that for any $y\in Y$ , we have $\mathrm {Tran}_G(y,Y) = A A_1^3.Sym(3)$ , where $A_1^3$ acts on $V_{[0]}$ as $\lambda _1\otimes \lambda _1\otimes \lambda _1$ . The set $\hat {Y}_1$ is defined by requiring certain expressions in terms of the coefficients of the given $V_{[0]}$ basis to be nonzero. Here, the key observation is that these conditions do not exclude all totally singular $2$ -spaces of $V_{[0]}$ , and therefore, $\hat {Y}^{\mathcal {S}}_1:=\hat {Y}_1\cap \mathcal {S}_2(V_{[0]})$ is a dense subset of $Y^{\mathcal {S}}:=\mathcal {S}_2(V_{[0]})$ . Given $y\in \hat {Y}^{\mathcal {S}}_1 $ , since $\mathrm {Tran}_G(y,Y) = A A_1^3.Sym(3)$ , we also have $\mathrm {Tran}_G(y,Y^{\mathcal {S}}) = A A_1^3.Sym(3)$ . Furthermore, they show that $G_y = A(A_1^3)_y$ for all $y\in \hat {Y}_1$ .

Assume $p\neq 2$ . Then by Lemma 5.5, there exists a dense open subset $\hat {Y}^{\mathcal {S}}_2$ of $Y^{\mathcal {S}}$ such that every stabilizer is $A_1^3$ -conjugate to $\mathbb {Z}_2.\mathbb {Z}_2$ . Taking the intersection with $\hat {Y}^{\mathcal {S}}_1$ , we get an open dense subset $\hat {Y}^{\mathcal {S}}$ of $Y^{\mathcal {S}}$ . For all $y\in \hat {Y}^{\mathcal {S}}$ , we know that $G_y$ is conjugate to $A.\mathbb {Z}_2.\mathbb {Z}_2$ . In each case, the codimension of the transporter of $y\in \hat {Y}^{\mathcal {S}}$ into $Y^{\mathcal {S}}$ is equal to the codimension of $Y^{\mathcal {S}}$ in $\mathcal {S}_2(V)$ . Therefore, every $y\in \hat {Y}^{\mathcal {S}}$ is $Y^{\mathcal {S}}$ -exact. By Lemma 2.15, we conclude that $C_{\mathcal {S}_2(V)} $ is $D_4.\mathbb {Z}_2.\mathbb {Z}_2,A_1^3.\mathbb {Z}_2.\mathbb {Z}_2,T_2.\mathbb {Z}_2.\mathbb {Z}_2$ according to whether $G=E_7,D_6,A_5$ , respectively. These generic stabilizers are the same as for the action on all $2$ -spaces.

Now assume that $p=2$ . By Lemma 5.4, there exists a dense open subset $\hat {Y}^{\mathcal {S}}_2$ of $Y^{\mathcal {S}}$ such that every stabilizer has a $1$ -dimensional connected component and stabilizers are pairwise non-conjugate. Taking the intersection with $\hat {Y}^{\mathcal {S}}_1$ , we get a dense open subset $\hat {Y}^{\mathcal {S}}_3$ of $Y^{\mathcal {S}}$ . Let $y_1,y_2\in \hat {Y}^{\mathcal {S}}_3$ and assume that $x.y_1 = y_2$ . Then $x\in N_G(A)$ , and since $V_{{0}}$ is the fixed space of A, we must have $x\in \mathrm {Tran}_G(y_1,Y^{\mathcal {S}}) = A A_1^3.Sym(3)$ . Therefore, $y_1$ and $y_2$ must be in the same $A_1^3.Sym(3)$ -orbit. However by construction, $y_1$ and $y_2$ are not in the same $A_1^3$ -orbit, and therefore, there exists a dense open subset $\hat {Y}^{\mathcal {S}}$ of $Y^{\mathcal {S}}$ , contained in $\hat {Y}^{\mathcal {S}}_3$ , such that any two distinct elements have non-conjugate stabilizers. In each case, the codimension of the transporter of $y\in \hat {Y}^{\mathcal {S}}$ into $Y^{\mathcal {S}}$ is equal to the codimension of $Y^{\mathcal {S}}$ in $\mathcal {S}_2(V)$ . Therefore, every $y\in \hat {Y}^{\mathcal {S}}$ is $Y^{\mathcal {S}}$ -exact. By Lemma 2.16, there is no generic stabilizer in the action of G on $\mathcal {S}_2(V)$ . In fact, Lemma 2.14 shows that we have semi-generic stabilizers $D_4.U_1.\mathbb {Z}_2,A_1^3.U_1.\mathbb {Z}_2,T_2.U_1.\mathbb {Z}_2$ according to whether $G=E_7,D_6,A_5$ , respectively.

Proof. Take $G = SL_3(K)$ acting on $V = \mathfrak {sl}_3(K)$ by conjugation. Here, $Z(G) = \langle \mathrm {diag} (\omega ,\omega ,\omega ) \rangle $ , where $\omega $ is a nontrivial third-root of unity. Note that $Z(G)$ acts trivially on V. Let $\tau $ be the graph automorphism acting on G as $g\mapsto g^{-T}$ and on V as $v\mapsto -v^T$ . We have that G fixes a nondegenerate quadratic form on V given by

$$\begin{align*}Q\left( (m_{ij})_{ij} \right) = m_{11}^2+m_{22}^2+m_{11}m_{22}+\sum_{i<j}m_{ij}m_{ji}.\end{align*}$$

For $1\leq i,j\leq 3$ , let $e_{ij}$ denote a $3\times 3$ matrix with a $1$ in position $(i,j)$ and zeroes everywhere else. There are three G-orbits on $\mathcal {S}_1(V)$ , which we label as $\Delta _1$ , $\Delta _2$ and $\Delta _3$ , respectively with representatives $\langle e_{13}\rangle $ , $\langle e_{12}+e_{23}\rangle $ , $\langle e_{11}+\omega e_{22}+\omega ^2 e_{33}\rangle $ . This follows directly from considering the Jordan Canonical Form of elements in V. The stabilizers are respectively $B = U_3T_2$ , $Z(G).U_2T_1$ and $T_2.\mathbb {Z}_3$ .

Let

$$\begin{align*}u_{bc} = \left(\begin{matrix} & 1 \\ & & b \\c & \\ \end{matrix}\right),\quad v_{ad} = \left(\begin{matrix} & & 1 \\ a & \\ & d \\ \end{matrix}\right); \end{align*}$$

$$\begin{align*}Y = \left\lbrace \langle u_{bc}, v_{ad}\rangle : a+c+bd = 0 \right\rbrace.\end{align*}$$

The set Y is a $3$ -dimensional subvariety of $\mathcal {S}_2(V)$ . Let

$$\begin{align*}\hat{Y} = \left\lbrace \langle u_{bc}, v_{ad}\rangle\in Y : abcd \neq 0,\, \frac{(bd-c)^2}{bcd}\neq 0,-\frac{3}{2},-3 \right\rbrace, \end{align*}$$

where we disregard the expression $\frac {(bd-c)^2}{bcd}\neq -\frac {3}{2}$ if $p = 2$ . Then $\hat {Y}$ is a dense subset of Y. Let $y = \langle u_{bc}, v_{ad}\rangle \in \hat {Y}$ . Then $\langle u_{bc} \rangle $ and $\langle v_{ad}\rangle $ are in $\Delta _3$ since they are spanned by rank- $3$ matrices. Now consider $v = u_{bc}+\lambda v_{ad}$ . We have $\det v = bc+ad \lambda ^3$ , which implies that there are precisely three $1$ -spaces of y not belonging to $\Delta _3$ . It is clear that none of these have rank $1$ , and therefore, all three of these $1$ -spaces belong to $\Delta _2$ . Let $\lambda _1,\lambda _2,\lambda _3$ be the three distinct roots of $q(x) = bc+ad x^ 3 $ , so that $\langle u_{bc}+\lambda _i v_{ad}\rangle \in \Delta _2$ . Then

$$\begin{align*}G_y\leq (G_{\langle u_{bc}+\lambda_1 v_{ad}\rangle }\cap G_{\langle u_{bc}+\lambda_2 v_{ad}\rangle }).Sym(3). \end{align*}$$

Let $g^* = \operatorname {\mathrm {diag}}(1,\omega ,\omega ^2)\in G$ . Then $g^*. u_{bc} = \omega ^2 u_{bc}$ and $g^*. v_{ad} = \omega v_{ad}$ , implying $g^*\in G_y$ and $Z(G)\langle g^* \rangle \simeq Z(G).\mathbb {Z}_3\leq G_y$ . Take $\mu ,\,\nu \in K$ with $\mu ^3 = \frac {1}{ac}$ and $\nu ^3 = \frac {ab}{d}$ , and let

$$\begin{align*}g^\ddagger =\operatorname{\mathrm{diag}} (\mu,\nu,{(\mu\nu)^{-1}})\tau ,\end{align*}$$

an element of $N_{G\langle \tau \rangle }(\langle g^* \rangle )$ fixing y. All elements of the form $h\tau $ with $h\in T$ are conjugate under T. Therefore, for any $y_1,y_2\in \hat {Y}$ , we know that $(G\langle \tau \rangle )_{y_1}$ and $(G\langle \tau \rangle )_{y_2}$ contain a subgroup conjugate to $Z(G).\langle g^*\rangle \langle g^\ddagger \rangle \simeq Z(G).Sym(3)$ .

We now proceed in the following way. We show that the stabilizer in $G_{\langle u_{bc}+\lambda _1 v_{ad}\rangle }$ of y is $Z(G)$ . This in turn implies that $G_y = Z(G).\mathbb {Z}_3$ . Since $\langle u_{bc}+\lambda _1 v_{ad} \rangle \in \Delta _2$ , we are able to find an element of G sending $u_{bc}+\lambda _1 v_{ad} \mapsto e_{12}+e_{23}$ . This is achieved by a scalar multiple of

$$\begin{align*}R = \left(\begin{matrix} 0 & 0 & 1 \\ c &\lambda_1 d & 0 \\ -\frac{bc}{\lambda_1} & c & -\lambda_1 a\\ \end{matrix}\right).\end{align*}$$

We now have

$$\begin{align*}R.y = \left\langle \left(\begin{matrix} a_1 & 0 & 1 \\ a_2 & a_1 & 0 \\ a_3 & -a_2 & -2 a_1\\ \end{matrix}\right) , e_{12}+e_{23}\right\rangle,\end{align*}$$

where $a_1=\lambda _1a,\,a_2 = \frac {a}{d}(bd-c) = \frac {c^2}{d}-b^2d$ , and $a_3 = \frac {3abc}{\lambda _1 d} = -3a_1^2$ . By assumption, $a_1a_2a_3\neq 0$ . Now let $g\in G_{\langle e_{12}+e_{23} \rangle }$ . Multiplying by an element of $Z(G)$ , we can assume that $g=ns$ , where $s = \mathrm {diag}(\frac {1}{t},1,t)$ and $n = \left (\begin {matrix} 1 & n_1 & n_2 \\ 0 & 1 & n_1 \\0 & 0 & 1\\ \end {matrix}\right ).$ Then

$$ \begin{align*} (m_{ij})_{ij}&:=g. \left(\begin{matrix} a_1 & 0 & 1 \\ a_2 & a_1 & 0 \\-3a_1^2 & -a_2 & -2 a_1\\ \end{matrix}\right) =\\ &= \left(\begin{matrix} a_1 + a_2 t n_1 - 3 a_1^2 t^2n_2 & m_{12} & m_{13} \\ t (a_2 - 3 a_1^2 t n_1) & a_1 - 2 a_2 t n_1 + 3 a_1^2 t^2 n_1^2 & m_{23} \\ -3 a_1^2 t^2 & -t (a_2 - 3 a_1^2 t n_1) & -2 a_1 + a_2 t n_1 - 3 a_1^2t^2 (n_1^2- n_2)\\ \end{matrix}\right),\end{align*} $$

where

$$ \begin{align*} m_{12}&=t (a_2 (-n_1^2 - n_2) + 3 a_1^2 t n_1 n_2),\\ m_{13}&= \frac{1}{t^2} + a_2 t n_1^3 - 3 a_1n_2 - 3 a_1^2 t^2 n_2 (n_1^2 - n_2),\\ m_{23}&= -3 a_1 n_1 - 3 a_1^2 t^2 (n_1^2-n_2) + a_2 t (n_1^2 - n_2).\end{align*} $$

Assume that g fixes $R.y$ . Since $m_{31}=-3a_1^2t^2$ , we must have $(m_{ij})_{ij} = t^2 R.y + \alpha (e_{12}+e_{23})$ for some $\alpha \in K$ . Therefore, $m_{21} = t^2 a_2$ which implies $n_1 = \frac {a_2(1-t)}{3 a_1^2 t}.$ Similarly, we must have $m_{11}=m_{22}$ which implies $n_2 = \frac {a_2 n_1}{a_1^2 t }-n_1^2$ . Then $m_{11} = t^2 a_1$ implies that either $t=\pm 1$ or $a_2^2= 3a_1^3$ . Assume that $a_2^2= 3a_1^3$ . Since $a_2=\frac {a}{d}(bd-c)$ and $a_1=\lambda _1 a$ , the equation $a_2^2= 3a_1^3$ implies $(bd-c)^2 = -3bcd$ , contrary to the definition of $\hat {Y}$ . Therefore, $t=\pm 1$ . If $t=1$ , we immediately get $n_1=n_2=0$ , concluding. Assume therefore that $t=-1$ and $p\neq 2$ . Then $m_{12}=m_{23}$ forces $2a_2^2 = 3 a_1^3$ , which implies $ 2(bd-c)^2 = -3bcd$ , which is impossible by assumption on $\hat {Y}$ .

This concludes the proof that $G\langle \tau \rangle _y = Z(G)\langle g^*\rangle \langle g^\ddagger \rangle $ . Now for any $y\in \hat {Y}$ , since $G_y= Z(G)\langle g^*\rangle $ , any element in $\operatorname {\mathrm {Tran}} _G(y,Y)$ must be in $N( \langle g^*\rangle )$ . We know that $N( \langle g^*\rangle ) = T_2.\mathbb {Z}_3$ , and it is easy to check that $T_2\in \operatorname {\mathrm {Tran}}_G(y,Y)$ . Therefore, $\dim \operatorname {\mathrm {Tran}}_G(y,Y) =2 $ , and then since $\dim \mathcal {S}_2(V)-\dim G = 1$ , the set $\hat {Y}$ is Y-exact. By Lemma 2.15, we conclude that the quadruple $(G\langle \tau \rangle , \lambda ,p,2)$ has generic $ts$ -stabilizer $Sym(3)$ .

Proof. This is entirely similar to the proof of Proposition 5.6 and relies on the construction used in [Reference Guralnick and Lawther10, Prop. 6.26]. All we have to observe is that the set $\hat {Y}_1$ defined in the proof of [Reference Guralnick and Lawther10, Prop. 6.26] does indeed contain totally singular $2$ -spaces. This follows from the observation in the proof of [Reference Guralnick and Lawther10, Prop. 6.25], where the authors need to show that the set $\hat {Y}_1$ is nonempty. They do so by saying that $v^{(1)}=a_{33}e_{\gamma _{33}}+a_{12}e_{\gamma _{12}}+a_{21}e_{\gamma _{21}}$ , $v^{(2)}=b_{11}e_{\gamma _{11}}+b_{23}e_{\gamma _{23}}+b_{32}e_{\gamma _{32}}$ , and $v^{(3)}=e_{\gamma _{22}}+e_{\gamma _{31}}+e_{\gamma _{13}}$ span a $3$ -space in $\hat {Y}_1$ if $(a_{12}b_{23}-a_{33}b_{11})(a_{21}b_{32}-a_{12}b_{23})(a_{33}b_{11}-a_{21}b_{32})\neq 0$ . Clearly, there are totally singular $2$ -spaces $\langle v^{(1)},v^{(2)}\rangle $ with coefficients satisfying this condition, and therefore, the set $\hat {Y}_1$ defined in the proof of [Reference Guralnick and Lawther10, Prop. 6.26] does contain totally singular $2$ -spaces.

Once we understand this, the generic stabilizer is, respectively, $A_2.X$ , $T_1.X$ , where X is the generic stabilizer for the action of $A_2.\mathbb {Z}_2$ on $\mathcal {S}_2(\lambda _1+\lambda _2)$ . By Lemma 5.7, we conclude.