3·2·1 $\mathfrak{g}$ non-abelian

Suppose that $\mathfrak{g}$ is non-abelian. Then, for dimensional reasons, $\mathfrak{g}$ must be the affine two-dimensional Lie algebra. Choose a basis A, B of $\mathfrak{g}$ with $[A,B]=A$ . Then A is linearly equivalent to some $a_i$ , $1\leq i\leq 6$ . In the first step we need to find all $B\in\mathfrak{g}$ linearly independent from a given $A=a_i$ , $1\leq i\leq 6$ , in (3·7), so that $[a_i,B]=a_i$ . We can immediately exclude all $a_i$ for $2\leq i\leq 5$ . This follows from the fact that in these cases $(a_i)_{33}\ne 0$ , but for all $B=(B_{ij})$

\begin{equation*} {([a_i,B])}_{33}=0.\end{equation*}

Hence, we only need to consider $a_1$ and $a_6$ . We start with $a_1$ and find that $[a_1,B]=a_1$ holds if and only if B is of the form

\begin{equation*} B=\left(\begin{array}{c@{\quad}c@{\quad}c} B_{11} & B_{12} & B_{13}\\ 0 & B_{11}+1 & B_{12}\\ 0 & 0 & B_{11}+2 \end{array}\right).\end{equation*}

We can without loss of generality assume that $B_{12}=0$ by replacing $B\to B-B_{12}a_1$ . Then we can write B as

(3·8) \begin{equation} B=\left(\begin{array}{c@{\quad}c@{\quad}c} b-1 & 0 & c\\ 0 & b & 0\\ 0 & 0 & b+1 \end{array}\right),\quad b,c\in\mathbb{R},\end{equation}

and we see that B has three distinct real eigenvalues $b-1,b,b+1$ , independent of $b,c\in\mathbb{R}$ . Thus, Lemma 2·12 restricts b so that at least one of these eigenvalues is positive and one is negative and, hence, $b\in({-}1,1)$ is a necessary condition. In the next step we will determine all homogeneous polynomials h of degree at least three that are invariant under $a_1$ and B, and check which ones are hyperbolic.

A homogeneous polynomial

(3·9) \begin{equation} h=\sum\limits_{k=0}^\tau\sum\limits_{\ell=0}^k P_{k \ell} z^{\tau-k}x^{k-\ell}y^\ell,\end{equation}

$P_{k \ell}\in\mathbb{R}$ for all $k\in\{0,\ldots,\tau\}$ , $\ell\in\{0,\ldots,k\}$ , is invariant under $a_1$ if and only if

\begin{align*} \sum\limits_{k=1}^\tau\sum\limits_{\ell=0}^k (k-\ell)P_{k \ell}z^{\tau-k}x^{k-\ell-1}y^{\ell+1} + \sum\limits_{k=1}^\tau\sum\limits_{\ell=1}^k\ell P_{k \ell} z^{\tau-k+1}x^{k-\ell}y^{\ell-1}=0.\end{align*}

By shifting indices and combining sums, we find that the above equation is equivalent to

(3·10) \begin{align} &\sum\limits_{k=1}^{\tau-1}\sum\limits_{\ell=1}^k\bigg(\underbrace{(k-\ell+1)}_{\ne 0} P_{k (\ell-1)} + \underbrace{(\ell+1)}_{\ne 0} P_{(k+1) (\ell+1)}\bigg) z^{\tau-k}x^{k-\ell}y^\ell\\ &+\sum\limits_{\ell=1}^\tau \underbrace{(\tau-\ell+1)}_{\ne 0} P_{\tau (\ell-1)} x^{\tau-\ell}y^{\ell} + \sum\limits_{k=1}^{\tau-1} P_{(k+1) 1}z^{\tau-k}x^k + P_{1 1} z^\tau =0.\notag\end{align}

Hence, $P_{k 1}=0$ for all $k\in\{1,\ldots,\tau\}$ and $P_{\tau\ell}=0$ for all $\ell\in\{0,\tau-1\}$ , see Figures 3, 4 for a visualisation of which indices must vanish for an odd and an even example of $\tau$ .

Figure 3. Table of the indices $(k,\ell)$ for $\tau=7$ in (3·9) that are marked with a solid black dot represent $P_{k \ell}=0$ .

Figure 4. Compared with Figure 3, note that $P_{\tau\tau}$ need not necessarily vanish at this point for $\tau$ even, in this example $\tau=8$ .

From

(3·11) \begin{equation} \underbrace{(k-\ell+1)}_{\ne 0} P_{k (\ell-1)} + \underbrace{(\ell+1)}_{\ne 0} P_{(k+1) (\ell+1)}=0\end{equation}

for all $k\in\{1,\ldots,\tau-1\}$ , $\ell\in\{1,\ldots,k\}$ we inductively obtain $P_{k \ell}=0$ for all odd $\ell$ with $\ell\leq\tau$ , and all $k\in\{\ell,\ldots,\tau\}$ , see Figures 5, 6.

Figure 5. From (3·11) we obtain that more values $P_{k\ell}$ must vanish in comparison with Figure 3.

Figure 6. As for $\tau$ odd in Figure 5, $P_{k\ell}$ vanishes for more indices $(k,\ell)$ for $\tau$ even.

Figure 7. Repeating the argument for Figure 5 from (3·11), we obtain more indices $(k,\ell)$ for with $P_{k\ell}=0$ must hold.

Similarly, we find that $P_{k\ell}=0$ for $\ell$ even, $\ell<\tau$ , and all $k\in\left\{\tau-\lfloor({\tau-\ell-1}/{2})\rfloor,\ldots,\tau\right\}$ , see Figures 7 and 8.

Figure 8. Analogous to Figure 7 and $\tau$ odd, we obtain vanishing conditions for $P_{k\ell}$ for $\tau$ even.

The remaining sets of coefficients for k even, $0\leq k\leq \tau$ ,

\begin{equation*} \left\{P_{(k-i)\,(k-2i)}\ \left|\ 0\leq i\leq \frac{k}{2}\right.\right\},\end{equation*}

are each determined by the choice of the value $P_{kk}\in\mathbb{R}$ , respectively. Hence, the number of free variables in h for it to be invariant under $a_1$ is $1+\lfloor{\tau}/{2}\rfloor$ . Now, instead of studying when h is invariant under B of the form (3·8), observe that we can diagonalise B via $C\,:\!=\,\left(\begin{array}{c@{\quad}c@{\quad}c} 1 & 0 & \frac{c}{2}\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right)$ and obtain, independent of $c\in\mathbb{R}$ ,

\begin{equation*} C^{-1} B C=\left(\begin{array}{c@{\quad}c@{\quad}c} b-1 & 0 & 0\\ 0 & b & 0\\ 0 & 0 & b+1 \end{array}\right),\quad C^{-1} a_1 C =a_1.\end{equation*}

Thus we can change basis so that $c=0$ in B. Note that while $a_1$ does not preserve the eigenspaces of B spanned by the column vectors of C, we find that $a_1$ instead either maps them to zero, or shifts eigenspaces of B to the previous one when numbering by column of C. Now for h to be invariant under B for $c=0$ , we note that for any monomial M, ${\,\mathrm{d}}M\left(B\cdot\left(\begin{smallmatrix}x\\y\\z\end{smallmatrix}\right)\right)\in\mathbb{R}\cdot M$ , implying that for $M=z^{\tau-k}x^{k-\ell}y^\ell$ , $0\leq k\leq \tau$ , $0\leq k\leq \ell$ , M is invariant under B if and only if

\begin{equation*} b=b(k,\ell)= \frac{-\tau+2k-\ell}{\tau}.\end{equation*}

Hence, h is invariant under B if and only if all of its monomials are invariant under B. The condition $b\in({-}1,1)$ thus implies $(k,\ell)\ne (0,0)$ and $(k,\ell)\ne (\tau,0)$ . We further find that $b(k,\ell)=b(K,L)$ if and only if (K, L) can be written as

\begin{equation*} (K,L)=(k+m,\ell+2m)\end{equation*}

for some $m\in\mathbb{Z}$ . Hence, h is invariant under B if and only if it is of the form

(3·12) \begin{equation} h=\sum\limits_{m=0}^{\mathrm{min}\{k,\tau-k,\lfloor\frac{\tau}{2}\rfloor\}} P_{(k+m)\,2m}z^{\tau-k-m}x^{k-m}y^{2m}\end{equation}

with $k\in\{1,\ldots,\tau-1\}$ fixed and $P_{(k+m)\,2m}\in\mathbb{R}$ for all $0\leq m\leq \mathrm{min}\{k,\tau-k,\lfloor {\tau}/{2}\rfloor\}$ . See Figures 9, 10. for a visual representation of the potentially non-zero coefficients $P_{k\ell}$ for two examples of h.

Figure 9. The potentially non-vanishing coefficients $P_{k\ell}$ in h (3·12) for $\tau=7$ , $k=3$ , marked with white dots.

Figure 10. The analogue picture of Figure 9 for $\tau=8$ , $k=4$ .

By combining our results, we find that h is invariant under $a_1$ and B if and only if it can be written as

\begin{equation*} h=\sum\limits_{m=0}^{k} P_{(k+m)\,2m}z^{\tau-k-m}x^{k-m}y^{2m}\end{equation*}

for some fixed $k\in\left\{1,\ldots,\lfloor {\tau}/{2}\rfloor\right\}$ with $b=({-\tau+2k})/{\tau}$ , and $P_{(k+m)\,2m}$ fulfilling

(3·13) \begin{equation} P_{(k+m)\,2m} = \left({-}\frac{1}{2}\right)^{m} \binom{m}{k} P_{k\, 0}\end{equation}

for all $1\leq m\leq k$ with $P_{k\,0}\in\mathbb{R}$ fixed. Since $1\leq k\leq \lfloor {\tau}/{2}\rfloor$ , we might further rewrite h as

(3·14) \begin{equation} h= P_{k\, 0} z^{\tau-2k}\left(zx-\tfrac{1}{2}y^2\right)^k.\end{equation}

Now we need to check which of the above polynomials are hyperbolic and thereafter which are pairwise inequivalent, and we need to describe the connected components of $\mathrm{hyp}_1(h)$ . By rescaling we might assume without loss of generality that $P_{k\,0}=\pm 1$ , and changing coordinates $(x,y,z)\to \left(x-z,x+z,\sqrt{2}y\right)$ transforms h to

\begin{equation*} h= \pm (x+z)^{\tau-2k}\big(x^2-y^2-z^2\big)^k.\end{equation*}

Observe that in any case

\begin{equation*} \{h=0\}=\{x+z=0\}\cup\{x^2-y^2-z^2=0\},\end{equation*}

meaning that the complement of $\{h=0\}$ in the ambient space $\mathbb{R}^3$ has four connected components. Only two of those are cones, namely the ones containing the points $(1,0,0)^\mathrm{T}$ and $({-}1,0,0)^\mathrm{T}$ . Thus, in order to check which of the above polynomials are hyperbolic, it suffices to check whether at least one of the aforementioned cones contain a hyperbolic point.

We start with the case of $\tau$ being odd. Then $k< {\tau}/{2}$ is equivalent to $k\leq \lfloor {\tau}/{2}\rfloor$ , and an overall sign change $(x,y,z)\to({-}x,-y,-z)$ maps h to $-h$ , independent of k. Hence we might assume $P_{k\,0}=1$ , so that

(3·15) \begin{equation} h= (x+z)^{\tau-2k}\big(x^2-y^2-z^2\big)^k.\end{equation}

This further shows that if h is hyperbolic, $\mathrm{hyp}_1(h)$ has one connected component. For $p=(1,0,0)^\mathrm{T}$ we then check that $h(p)=1$ and

(3·16) \begin{equation} -\partial^2h_p = \left(\begin{array}{c@{\quad}c@{\quad}c} -\tau(\tau-1) & 0 & (\tau-1)({-}\tau+2k)\\ 0 & 2k & 0\\ (\tau-1)({-}\tau+2k) & 0 & -\tau(\tau-1)+4k(\tau-k) \end{array}\right).\end{equation}

Using $\tau\geq 3$ , $1\leq k< {\tau}/{2}$ , and $\mathrm{det}\left({-}\partial^2h_p\right)=-8k^2(\tau-1)(\tau-k)<0$ , we obtain that $-\partial^2h_p$ has Minkowski signature, so p is in fact a hyperbolic point of h. The single connected component of $\mathrm{hyp}_1(h)$ is precisely given by the connected component of $\{h=1\}$ that contains p. This follows from the fact connected homogeneous special surfaces span a cone, and the only other potential candidate for a connected component of $\mathrm{hyp}(h)$ would have to contain $-p$ . But since $\tau$ is odd, $h({-}p)=-1$ . The full automorphism group $G^h$ of h is thus generated by the stabiliser of p, $\mathrm{stab}^h_p$ , and the connected component of the identity transformation, $G^h_0$ . The latter is generated by $a_1$ and B by construction. Any element of $\mathrm{stab}^h_p$ is in particular an automorphism of the set $\{h=0\}$ , and more precisely must preserve $\{x+z=0\}$ and $\{x^2-y^2-z^2=0\}$ , and their intersection $\{x+z=0,y=0\}=\mathrm{span}\left\{\left(\begin{smallmatrix}1\\0\\-1\end{smallmatrix}\right)\right\}$ . A matrix $A\in\mathrm{GL}(3)$ maps p to itself and leaves $\{x+z=0\}$ and $\{x+z=0,y=0\}=\mathrm{span}\left\{\left(\begin{smallmatrix}1\\0\\-1\end{smallmatrix}\right)\right\}$ invariant if and only if it is of the form

\begin{equation*} A=\left(\begin{array}{c@{\quad}c@{\quad}c} 1 & u & v\\ 0 & w & 0\\ 0 & -u & 1-v \end{array}\right)\end{equation*}

for some $u,v,w\in\mathbb{R}^3$ , such that A is invertible. We verify that A preserves $\{x^2-y^2-z^2=0\}$ if and only if $u=v=0$ and $w=\pm 1$ . Hence, $\mathrm{stab}^h_p$ contains only the identity and the transformation $y\to -y$ . They commute with all elements of $G^h_0$ that are generated by

\begin{equation*} N^{-1}BN=\left(\begin{array}{c@{\quad}c@{\quad}c} b & 0 & 1\\ 0 & b & 0\\ 1 & 0 & b \end{array}\right),\end{equation*}

but do not commute with elements generated by

\begin{equation*} N^{-1}a_1N=\frac{1}{\sqrt{2}}\left(\begin{array}{c@{\quad}c@{\quad}c} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & -1 & 0 \end{array}\right),\end{equation*}

where N describes the coordinate transformation $(x,y,z)\to(x-z,\sqrt{2}y,x+z)$ . Hence, $G^h=G^h_0\ltimes \mathbb{Z}_2$ .

Next let $\tau$ be even. In this case we can exclude $k=\lfloor {\tau}/{2}\rfloor={\tau}/{2}$ since that will lead to the one-dimensional stabiliser case studied in Section 3·1. So from here on we will assume that $1\leq k< {\tau}/{2}$ . If k is additionally even, for $\{h>0\}$ to not be empty $P_{k\,0}=1$ is a necessary requirement, meaning that h is again of the form (3·15). In general, if $P_{k\,0}=1$ , we check that $p=(1,0,0)^\mathrm{T}$ is a hyperbolic point of h just as we did for $\tau$ odd. In this case, $\mathrm{hyp}_1(h)$ has two isometric connected components that are related by $(x,y,z)\to ({-}x,-y,-z)$ . The negative identity transformation commutes with any element in $G^h$ , and we find similar to the case $\tau$ odd that $G^h\cong \left(G^h_0\ltimes \mathbb{Z}_2\right)\times\mathbb{Z}_2$ . If $\tau$ is even and k is odd, we obtain the same result as for $\tau$ even and k even if $P_{k\,0}=1$ . The missing case we need to study is thus $\tau$ even, k odd, and $P_{k\,0}=-1$ . The two connected components of the complement of $\{h=0\}$ that are cones contain the points p and $-p$ , respectively. In both cases however, $h(p)=h({-}p)=-1$ . Hence, neither cone contains hyperbolic points of h, and by the supposed homogeneity and consequent property that connected components of $\mathrm{hyp}_1(h)$ are closed in the ambient space [ Reference Cortés, Nardmann and Suhr7 , Proposition 1·8], we obtain that $\{h=1\}$ does not contain a special homogeneous surface.

It remains to show that the polynomials of the form $h=(x+z)^{\tau-2k}\big(x^2-y^2-z^2\big)^k$ , $1\leq k< {\tau}/{2}$ , are pairwise inequivalent. To see this, recall that a supposed linear transformation mapping h corresponding to k to $\overline{h}$ corresponding to $\overline{k}\ne k$ must restrict to an automorphism of the vector subspace $\{x+z=0\}\subset\mathbb{R}^3$ that furthermore restricts to an automorphism of $\mathrm{span}\left\{\left(\begin{smallmatrix}1\\ 0\\ -1\end{smallmatrix}\right)\right\}$ . Observe that h has a zero of degree $\tau-2k$ in every point $\{x+z=0\}\cap\{y\ne0\}$ , but $\overline{h}$ has a zero of degree $\tau-2\overline{k}\ne \tau-2k$ in every such point. This excludes h and $\overline{h}$ being equivalent.

In order to determine the scalar curvature of the connected components of $\mathrm{hyp}_1(h)$ with respect to the centro-affine metric g, we will construct a left-invariant orthonormal frame $\{X,Y\}$ of $T\mathcal{H}$ . We will work in our initial linear coordinates, so that

\begin{equation*} h=z^{\tau-2k}\left(xz-\frac{1}{2}y^2\right)^k\end{equation*}

as in (3·14). In these coordinates, it follows from (3·16) that the point $q\,:\!=\,(1,0,1)^\mathrm{T}$ with $h(q)=1$ is hyperbolic. We calculate

\begin{equation*} a_1q=\left(\begin{array}{c}0\\1\\0\end{array}\right),\quad Bq=\left(\begin{array}{c}b-1\\0\\b+1\end{array}\right),\end{equation*}

where we recall that $b= ({-\tau+2k})/{\tau}$ , and verify that $-\partial^2h_q(a_1q,Bq)=0$ . We further check that

\begin{equation*} g_q(a_1q,a_1q)=\frac{k}{\tau},\quad g_q(Bq,Bq)=\frac{4k(\tau-k)}{\tau^2},\end{equation*}

and thus define X and Y to be the left-invariant extensions of $X_q\,:\!=\,\sqrt{{\tau}/{k}}a_1q$ and $Y_q\,:\!=\, {\tau}/{2\sqrt{k(\tau-k)}}Bq$ , respectively. For their Lie bracket we obtain using $[a_1,B]=a_1$ that

\begin{equation*} [X,Y]=\frac{\tau}{2\sqrt{k(\tau-k)}}X.\end{equation*}

Using Koszul’s formula, we find for the Levi–Civita connection $\nabla$ of $(\mathcal{H},g)$

\begin{equation*} \nabla_X X= -\frac{\tau}{2\sqrt{k(\tau-k)}}Y,\quad \nabla_X Y= \frac{\tau}{2\sqrt{k(\tau-k)}}X,\quad \nabla_Y X= 0,\quad \nabla_Y Y= 0.\end{equation*}

Hence, the non-vanishing terms of the Riemannian curvature tensor R are determined by

\begin{equation*} R(X,Y)X=\frac{\tau^2}{4k(\tau-k)}Y,\quad R(X,Y)Y=-\frac{\tau^2}{4k(\tau-k)}X,\end{equation*}

and the scalar curvature of $(\mathcal{H},g)$ is given by

\begin{equation*} S=-\frac{\tau^2}{2k(\tau-k)}.\end{equation*}

Note that this result coincides with [ Reference Lindemann11 , Proposition 5·12] for the special case $\tau=3$ , $k=1$ . We further observe that the assignment $k\mapsto S$ for $1\leq k\leq \left\lfloor{\tau}/{2}\right\rfloor$ is injective, which follows from $\partial_k(k(\tau-k))>0$ for all $k\in\left(0,\left\lfloor {\tau}/{2}\right\rfloor\right)$ . This also presents an alternative way to see that the polynomials h in (3·2·1) are inequivalent for different values of k.

Lastly, we study the case $a_6$ . The equation $[a_6,B]=a_6$ is equivalent to B being of the form

\begin{equation*} B = \left(\begin{array}{c@{\quad}c@{\quad}c} B_{11} & -B_{12} & 0\\ B_{12} & B_{11}-1 & 0\\ 0 & 0 & B_{33} \end{array}\right).\end{equation*}

After adding $-B_{12}a_6$ to B, we might assume that $B_{12}=0$ . If $B_{33}=0$ , both the induced actions of $a_6$ and B restrict to any plane orthogonal to $(0,0,1)^\mathrm{T}$ , which contradicts the hyperbolicity of the considered polynomial. Hence, we can assume without loss of generality that B is given by

\begin{equation*} B=\left(\begin{array}{c@{\quad}c@{\quad}c} b & 0 & 0\\ 0 & b-1 & 0\\ 0 & 0 & b+c \end{array}\right),\quad b,c\in\mathbb{R},\end{equation*}

so that the set $\{b,b-1,b+c\}$ contains at least one negative and at least one positive number, which follows from Lemma 2·12. When studying one-dimensional stabilisers, we have seen that h being invariant under $a_6$ requires it to be of the form (3·2). We now calculate and find that ${\,\mathrm{d}}h_{\left(\begin{smallmatrix}x\\y\\z\end{smallmatrix}\right)}\left(B\cdot\left(\begin{smallmatrix}x\\y\\z\end{smallmatrix}\right)\right)=0$ is equivalent to

\begin{equation*} c_{\ell}\left((b\tau+c(\tau-2\ell))x^2 + (b\tau-2\ell+c(\tau-2\ell))y^2\right)=0\end{equation*}

for all $0\leq \ell \leq \left\lfloor{\tau}/{2}\right\rfloor$ , where $c_\ell$ is used as in equation (3·2). Hence the only potentially non-zero coefficient $c_\ell$ is $c_0$ . But $h=c_0z^\tau$ is not a hyperbolic polynomial. This excludes the generator $a_6$ and finishes the case of $\mathfrak{g}$ being non-abelian.

3·2·2. $\mathfrak{g}$ abelian

Now suppose that $\mathfrak{g}$ is abelian. For $B=(B_{ij})\in\mathfrak{gl}(3)$ we have to study $[a_i,B]=0$ for all $1\leq i\leq 6$ and then determine which of these pairs $(a_i,B)$ are linearly independent. In the following, (x, y, z) denote linear coordinates on $\mathbb{R}^3$ . We will first classify all two-dimensional abelian Lie subalgebras of $\mathfrak{gl}(3)$ that potentially act transitively on a special homogeneous surface, and in the second step classify the respective hyperbolic polynomials.

For $a_1$ , we obtain $[a_1,B]=0$ if and only if B is of the form

\begin{equation*} B=\left(\begin{array}{c@{\quad}c@{\quad}c} B_{11} & B_{12} & B_{13}\\ 0 & B_{11} & B_{12}\\ 0 & 0 & B_{11} \end{array}\right).\end{equation*}

Hence, B has one eigenvalue, namely $B_{11}$ . For $B_{11}\ne 0$ , the origin in in the closure of every orbit of the induced action of B and this case can thus be excluded by Lemma 2·12. For $B_{11}=0$ and $B\ne0$ , $a_1$ and B being linearly independent implies that B must be equivalent to $\left(\begin{array}{c@{\quad}c@{\quad}c} 0 & 1 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{array}\right)$ . This case is excluded by Lemma 2·11. Hence, there exists no two-dimensional abelian Lie subalgebra of $\mathfrak{gl}(3)$ acting on a special homogeneous surface with one generator being $a_1$ .

For $a_2$ , $[a_2,B]=0$ requires B to be of the form

(3·17) \begin{equation} B=\left(\begin{array}{c@{\quad}c@{\quad}c} B_{11} & B_{12} & 0\\ 0 & B_{11} & 0\\ 0 & 0 & B_{33} \end{array}\right).\end{equation}

We thus obtain three cases up to scale for B, depending on the sign of $B_{11}$ and $B_{12}$ ,

\begin{equation*} b_{21}=\left(\begin{array}{c@{\quad}c@{\quad}c} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & b \end{array}\right),\ b< 0,\quad b_{22}=\left(\begin{array}{c@{\quad}c@{\quad}c} 1 & 1 & 0\\ 0 & 1 & 0\\ 0 & 0 & b \end{array}\right),\ b< 0,\quad b_{23}=\left(\begin{array}{c@{\quad}c@{\quad}c} 0 & 1 & 0\\ 0 & 0 & 0\\ 0 & 0 & b \end{array}\right),\ b\ne 0.\end{equation*}

The respective conditions for b are, again, implied by Lemmas 2·11 and 2·12. Independent of the choice for b in $b_{23}$ , that case can be excluded by noting that either $a_2$ and $b_{23}$ are linearly independent, or the studied hyperbolic polynomial must be invariant under $a_2-b_{23}=\left(\begin{array}{c@{\quad}c@{\quad}c} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & \mu-b\end{array}\right)$ , which contradicts Lemma 2·11. Furthermore, after replacing $b_{22}\to b_{22}-a_2$ , we are left with only one case, namely $b_{21}$ . This is a special case for the first generator being $a_4$ which we will keep in mind. Since $b<0$ in the definition of $b_{21}$ , h is invariant under $a_2-\frac{\mu}{b}b_{21}$ , which is up to scale equivalent to one of

(3·18) \begin{equation} \left(\begin{array}{c@{\quad}c@{\quad}c} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 0 \end{array}\right),\quad \left(\begin{array}{c@{\quad}c@{\quad}c} 1 & 1 & 0\\ 0 & 1 & 0\\ 0 & 0 & 0 \end{array}\right),\quad \left(\begin{array}{c@{\quad}c@{\quad}c} 0 & 1 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{array}\right).\end{equation}

But all of the above cases are in contradiction to Lemma 2·11 or Lemma 2·12. We conclude that there exists no two-dimensional abelian Lie subalgebra of $\mathfrak{gl}(3)$ acting on a special homogeneous surface with one generator being $a_2$ .

Next consider $a_3$ . For B to fulfil $[a_3,B]=0$ , it must, as for $a_2$ , be of the form (3·17). With an analogous argument as for the case $a_2$ we obtain that, up to a linear transformation, the corresponding hyperbolic polynomial would be invariant under one of the matrices in (3·18), contradicting Lemma 2·11 or Lemma 2·12. Hence, we can also exclude $a_3$ as the first generator for our desired abelian Lie subalgebra of $\mathfrak{gl}(3)$ .

Next, consider the generator $a_4$ . Then $[a_4,B]=0$ is equivalent to B being of the form

\begin{equation*} B=\left(\begin{array}{c@{\quad}c@{\quad}c} B_{11} & 0 & 0\\ 0 & B_{22} & 0\\ 0 & 0 & B_{33} \end{array}\right)\end{equation*}

for $\lambda\in\mathbb{R}_{\geq 0}\setminus\{1\}$ , and of the form

\begin{equation*} B=\left(\begin{array}{c@{\quad}c@{\quad}c} B_{11} & B_{12} & 0\\ B_{21} & B_{22} & 0\\ 0 & 0 & B_{33} \end{array}\right)\end{equation*}

for $\lambda=1$ . We start with the case $\lambda\in\mathbb{R}_{\geq 0}\setminus\{1\}$ . After changing $B\to B-B_{11}a_4$ , rescaling, and using either Lemma 2·11 or Lemma 2·12, we might assume that B is of the form

\begin{equation*} B=\left(\begin{array}{c@{\quad}c@{\quad}c} 0 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & b \end{array}\right),\ b<0.\end{equation*}

Note this excludes the case of the generators being $a_4$ and $a_2$ as discussed above. For $\lambda\in\mathbb{R}_{\geq 0}\setminus\{1\}$ we might use alternative generators for the spanned abelian Lie subalgebra of $a_4$ and B by replacing $a_4\to a_4-\lambda B$ , namely B and $A\,:\!=\,\left(\begin{array}{c@{\quad}c@{\quad}c} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & a\end{array}\right)$ for some $a<0$ . The latter condition comes, again, from Lemma 2·11 and Lemma 2·12. For $\lambda=0$ , we can simply relabel $\mu=a$ and obtain the same set of generators.

Next, consider $\lambda=1$ . Since for $\lambda=1$ , $a_4$ is invariant under the adjoint action of $\mathrm{GL}(2)\times\mathrm{Id}_{z}$ and under switching the variables x and y of the ambient space, we can, after rescaling, assume without loss of generality that B is either diagonal, or of one the forms

\begin{align*} & b_{41}=\left(\begin{array}{c@{\quad}c@{\quad}c} 1 & 1 & 0\\ 0 & 1 & 0\\ 0 & 0 & b \end{array}\right),\ b< 0,\quad b_{42}=\left(\begin{array}{c@{\quad}c@{\quad}c} 0 & 1 & 0\\ 0 & 0 & 0\\ 0 & 0 & b \end{array}\right),\ b\ne 0,\\ & b_{43}=\left(\begin{array}{c@{\quad}c@{\quad}c} 1 & -c & 0\\ c & 1 & 0\\ 0 & 0 & b \end{array}\right),\ c>0,\ b<0.\end{align*}

Note that we have again used Lemmas 2·11 and 2·12 to obtain the respective ranges for b and c in the above equation. The diagonal case has already been covered in our study of the related $\lambda\in\mathbb{R}_{\geq 0}\setminus\{1\}$ . Since, up to relabelling of the variables, $b_{41}-a_4=a_3$ for $\lambda=1$ , and that case has already been shown not to admit an abelian Lie subalgebra fulfilling our hyperbolicity conditions, we can exclude the second generator being $b_{41}$ . Similarly, the case $b_{42}$ can be excluded. For $b_{43}$ , we can replace $b_{43}\to b_{43}-a_4$ and after rescaling obtain the generator $B\,:\!=\,\left(\begin{array}{c@{\quad}c@{\quad}c} 0 & -1 & 0\\ 1 & 0 & 0\\ 0 & 0 & b\end{array}\right)$ with $b<0$ . The abelian Lie subalgebra $\mathrm{span}\{a_4,B\}$ of $\mathfrak{gl}(3)$ is also a special case of the upcoming case with first generator being $a_5$ .

Suppose lastly that the first generator is either $a_5$ or $a_6$ . For $[a_5,B]=0$ , respectively $[a_6,B]=0$ , to hold, the upper left $2\times2$ -block of B must correspond to a complex number, that is B must be of the form

\begin{equation*} B=\left(\begin{array}{c@{\quad}c@{\quad}c} B_{11} & B_{12} & 0\\ -B_{12} & B_{11} & 0\\ 0 & 0 & B_{33} \end{array}\right).\end{equation*}

For $B_{12}=0$ , it follows that $B_{11}\ne 0$ by either Lemma 2·11 or Lemma 2·12, and we are in one of the cases with first generator being $a_4$ studied above. For $B_{12}\ne 0$ , we might replace $B\to B+B_{12}a_5$ , respectively $B\to B+B_{12}a_6$ , and are precisely in the latter case again.

Summarising, we have shown the following. Up to isomorphism, there exist two two-parameter families of abelian Lie subalgebras of $\mathfrak{gl}(3)$ that at this point can not be excluded to act transitively on a special homogeneous surface, namely

(3·19) \begin{align} &\mathfrak{g}_1\,:\!=\,\mathrm{span}\left\{A_1,B_1\right\},\ A_1=\left(\begin{array}{c@{\quad}c@{\quad}c} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & a\end{array}\right),\ B_1=\left(\begin{array}{c@{\quad}c@{\quad}c} 0 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & b\end{array}\right),\ a<0,\ b<0,\\[5pt] &\mathfrak{g}_2\,:\!=\,\mathrm{span}\left\{A_2,B_2\right\},\ A_2=\left(\begin{array}{c@{\quad}c@{\quad}c} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & a\end{array}\right),\ B_2=\left(\begin{array}{c@{\quad}c@{\quad}c} 0 & -1 & 0\\ 1 & 0 & 0\\ 0 & 0 & b\end{array}\right),\ a<0,\ b<0.\notag\end{align}

For both families we now need to determine the values of a and b, so that there exists a hyperbolic polynomial that is invariant under $\mathfrak{g}_1$ or $\mathfrak{g}_2$ , respectively. We then need to classify all such hyperbolic polynomials up to equivalence.

We start with $\mathfrak{g}_1$ . Note that for all monomials M in (x, y, z), both ${\,\mathrm{d}} M\left(A_1\cdot\left(\begin{smallmatrix}x\\y\\z\end{smallmatrix}\right)\right)$ and ${\,\mathrm{d}} M\left(B_1\cdot\left(\begin{smallmatrix}x\\y\\z\end{smallmatrix}\right)\right)$ are contained in $\mathbb{R}\cdot M$ . Hence, a homogeneous polynomial h of degree $\tau\geq 3$ is invariant under $\mathfrak{g}_1$ if and only if all its monomial terms are. Let $M=m z^{\tau-k}x^{k-\ell}y^\ell$ for some $m\ne0$ , $k\in\{0,\ldots,\tau\}$ , $\ell\in\{0,\ldots,k\}$ . We find that the monomial M is invariant under $\mathfrak{g}_1$ if and only if

(3·20) \begin{equation} a(\tau-k)+k-\ell=0,\quad b(\tau-k)+\ell=0.\end{equation}

The above two equations do not admit a solution with $k=0$ , since then by assumption $\ell=0$ and consequently $a=b=0$ , contradicting the definition of $\mathfrak{g}_1$ . For $k=\tau$ , the second equation implies $\ell=0$ , thereby contradicting the first of the two equations. Hence, (3·20) has a solution if and only $k\in\{1,\ldots,\tau-1\}$ , which is given by

(3·21) \begin{equation} a=\frac{\ell-k}{\tau-k},\quad b=-\frac{\ell}{\tau-k}.\end{equation}

Note that $a<0$ and $b<0$ imply that $\ell\in\{1,\ldots,k-1\}$ , which allows us to further restrict the range of k to $\{2,\ldots,\tau-1\}$ . Assume that a and b are as above for a chosen triple of $(\tau,k,\ell)$ . We will show that for $\tau$ fixed there exist no other such triple leading to the same values of a and b. Suppose that

(3·22) \begin{equation} \frac{\ell-k}{\tau-k}=\frac{L-K}{\tau-K},\quad -\frac{\ell}{\tau-k}=-\frac{L}{\tau-K},\end{equation}

for triples $(\tau,k,\ell),(\tau,K,L)\in\mathbb{N}_{\geq3}\times\{2,\ldots,\tau-1\}\times\{1,\ldots,k-1\}$ . By substituting the second equation in (3·22) into the first, we obtain $- {k}/({\tau-k})=- {K}/({\tau-K})$ which is easily shown to hold if and only if $- {\tau}/({\tau-k})=- {\tau}/({\tau-K})$ . Hence, $k=K$ , and this implies $\ell=L$ . Summarising, we have shown that a homogeneous polynomial h is invariant under $\mathfrak{g}_1$ if and only if a and b fulfil (3·21) and, up to positive rescaling, $h=\pm z^{\tau-k}x^{k-\ell}y^\ell$ for a choice of $(\tau,k,\ell)\in\mathbb{N}_{\geq3}\times\{2,\ldots,\tau-1\}\times\{1,\ldots,k-1\}$ . We can further restrict to

\begin{equation*} h=z^{\tau-k}x^{k-\ell}y^\ell,\end{equation*}

either by $(x,y,z)\to ({-}x,-y,-z)$ , or by noting that if $\tau-k$ , $k-\ell$ , and $\ell$ are all even, $-z^{\tau-k}x^{k-\ell}y^\ell$ is non-positive, thus excluding hyperbolicity. We now need to study which of the polynomials of the form $h=z^{\tau-k}x^{k-\ell}y^\ell$ are hyperbolic. It will turn out that, in fact, all of them are. To show this it suffices that in each case, $p=(1,1,1)^\mathrm{T}$ is a hyperbolic point of h. The latter point it contained in $\{h=1\}$ , and we further know by Euler’s theorem for homogeneous functions that

\begin{equation*} -\partial^2 h_{p}\left(p,p\right)=-\tau(\tau-1)<0.\end{equation*}

Recall that $T_{p}\{h=1\}$ is orthogonal to $\mathbb{R}\cdot p$ with respect to $-\partial^2h_{p}$ . Thus, in order to show that p is a hyperbolic point it suffices to show that $\det\big({-}\partial^2h_{p}\big)<0$ and that for some $v\in T_{p}\{h=1\}=\ker({\,\mathrm{d}} h_p)$ we have $-\partial^2h_{p}(v,v)>0$ . We check that

\begin{equation*} \det\big({-}\partial^2h_{p}\big)=-\ell(\tau-1)(\tau-k)(k-\ell),\end{equation*}

which is negative for the allowed values of $\tau,k,\ell$ . We further calculate ${\,\mathrm{d}} h_p=(k-\ell){\,\mathrm{d}} x + \ell{\,\mathrm{d}} y + (\tau-k){\,\mathrm{d}} z$ , and choose $v=(0, \tau-k, -\ell)^\mathrm{T}$ . We obtain

\begin{equation*} -\partial^2 h_p(v,v)=\ell(\tau-k)(\tau-k+\ell)>0\end{equation*}

as required. Hence, h is indeed hyperbolic.

It remains to describe the connected components of $\mathrm{hyp}_1(h)$ and the automorphism group $G^h$ for each allowed triple $(\tau,k,\ell)$ . Note that $\{h=0\}=\{x=0\}\cup\{y=0\}\cup\{z=0\}$ , meaning that $\{h\ne 0\}$ has eight connected components. For $\tau$ odd, either 1 or 3 out of the exponents $\tau-k,k-\ell,\ell$ must be odd. In any of these cases, we see that $\{h>0\}$ has four connected components and, in fact, coincides with $\mathrm{hyp}_1(h)$ . The connected components are pairwise isometric via combinations of sign flips in the coordinates x, y, z. For $\tau$ even, either 0 or 2 out of the exponents $\tau-k,k-\ell,\ell$ must be odd. In the first case we find that $\mathrm{hyp}_1(h)$ has eight isometric connected components, and in the latter case $\mathrm{hyp}_1(h)$ again has four isometric connected components. In order to find the automorphism group $G^h$ of h we need to find the stabiliser of p and determine which of its elements are not obtained from combinations of sign flips in the coordinates x, y, z. To do so, we write

(3·23) \begin{equation} h=x^a y^b z^c,\quad a,b,c\in\mathbb{N},\quad a+b+c=\tau,\quad a\leq b\leq c.\end{equation}

Every element s of $\mathrm{stab}^h_p$ must preserve the union of the pairwise intersections of $\{x=0\}$ , $\{y=0\}$ , and $\{z=0\}$ , since s is in particular a continuous map. Since s is linear, it can thus be written as the composition of a diagonal linear transformation and a permutation of coordinates. Since any variable switch maps p to itself and p is by assumption an eigenvector of s with eigenvalue 1, we find that s must be, up to switching coordinates, the identity transformation. Thus, $\mathrm{stab}^h_p$ is either trivial if $a<b<c$ , isomorphic to $\mathbb{Z}_2$ if $a\leq b<c$ or $a<b\leq c$ , or isomorphic to the group of permutations of a set with three elements $\sigma_3$ . Since no combinations of sign flips in x, y, z that is not the identity preserves any connected component of $\mathrm{hyp}_1(h)$ and elements of $\mathfrak{g}_1$ are diagonal, we find that for a, b, c even

(3·24) \begin{equation} G^h\cong\begin{cases} a<b<c: & \mathbb{R}^2\times\mathbb{Z}_2^3,\\[4pt] a=b<c\ \text{or}\ a<b=c: & \left(\mathbb{R}^2\times\mathbb{Z}_2^3\right)\ltimes\mathbb{Z}_2,\\[4pt] a=b=c: & \left(\mathbb{R}^2\times\mathbb{Z}_2^3\right)\ltimes\sigma_3, \end{cases}\end{equation}

and for at least one of a, b, c odd

(3·25) \begin{equation} G^h\cong\begin{cases} a<b<c: & \mathbb{R}^2\times\mathbb{Z}_2^2,\\[4pt] a=b<c\ \text{or}\ a<b=c: & \left(\mathbb{R}^2\times\mathbb{Z}_2^2\right)\ltimes\mathbb{Z}_2,\\[4pt] a=b=c: & \left(\mathbb{R}^2\times\mathbb{Z}_2^2\right)\ltimes\sigma_3. \end{cases}\end{equation}

It remains to show that the polynomials h as in (3·23) are pairwise inequivalent. Similarly to the calculation of the stabiliser of p above, we note that any supposed transformation relating two such polynomials must be compositions of switching variables and a diagonal transformation. The condition $a\leq b\leq c$ thus excludes that two distinct polynomials of the form (3·23) are equivalent.

Next consider $\mathfrak{g}_2$ and suppose that a hyperbolic polynomial h is invariant under it. Then h is in particular invariant under $C\,:\!=\,B_2- ({b}/{a})A_2$ , which is of the form

\begin{equation*} C=\left(\begin{array}{c@{\quad}c@{\quad}c} c & -1 & 0\\ 1 & c & 0\\ 0 & 0 & 0 \end{array}\right),\quad c<0.\end{equation*}

Hyperbolicity is an open condition and, hence, there exists a hyperbolic point p of h that is not contained in $\mathrm{span}\left\{\left(\begin{smallmatrix}0\\0\\1\end{smallmatrix}\right)\right\}$ , and h is constant along $f(t)\,:\!=\,e^{tC} p$ , $t\in\mathbb{R}$ . Since $c\ne 0$ , it follows that f is not periodic. The image of f is contained in the affine plane $E\,:\!=\,p+\mathrm{span}\left\{\left(\begin{smallmatrix}1\\0\\0\end{smallmatrix}\right),\left(\begin{smallmatrix}0\\1\\0\end{smallmatrix}\right)\right\}$ , and every line L therein intersects the image of f in countably infinitely many distinct points. Hence, by being a polynomial, h must be constant along L, which is a contradiction to Lemma 2·11. We conclude that there exists no hyperbolic polynomial h that is invariant under $\mathfrak{g}_2$ .

Next, we show that the scalar curvature of every special homogeneous surfaces $\mathcal{H}$ contained in the level set of $h=x^ay^bz^c$ as in (3·23) with respect to the centro-affine fundamental form g vanishes. Using a right-invariant, with respect to g orthonormal, frame $X_1,X_2$ of $T\mathcal{H}$ , we obtain from $\mathfrak{g}_1$ (3·19) being abelian and Koszul’s formula that $\nabla^g_{X_i}X_j=0$ for all $i,j\in\{1,2\}$ . Hence, the curvature of $(\mathcal{H},g)$ vanishes. Alternatively, we can consider a left-invariant orthonormal co-frame $\theta=\left(\begin{smallmatrix}e_1\\ e_2\end{smallmatrix}\right)$ of $T^*\mathcal{H}$ , so that

\begin{equation*} 0=\!{\,\mathrm{d}} \theta=-\omega_{\mathrm{LC}}\wedge\theta,\end{equation*}

where $\omega_{\mathrm{LC}}$ denotes the connection form of the Levi-Civita connection $\nabla^g$ . The above equation follows from $\mathfrak{g}_1$ being abelian and, hence, ${\,\mathrm{d}} e_i(X,Y)=-e_i\left([X,Y]\right)=0$ for $i\in\{1,2\}$ and all $X,Y\in\mathfrak{X}(\mathcal{H})$ . Thus, $\omega_{\mathrm{LC}}=0$ , showing that the curvature form $\Omega_{\mathrm{LC}}={\,\mathrm{d}}\omega_{\mathrm{LC}}+\omega_{\mathrm{LC}}\wedge\omega_{\mathrm{LC}}$ must also vanish identically.

This finishes the proof of Theorem 1·1 and Proposition 1·3.