2.1 Setup

Within the whole section, we work in a Minkowski space $(\mathbb {R}^n,F)$ with $n\ge 3$ . We denote $S_F=\{ y\in \mathbb {R}^n \mid F(y)=1\}$ the indicatrix of F, and g the Hessian metric $\tfrac {1}{2}{\textrm {d}}^2F^2$ of F on $\mathbb {R}^n\backslash \{0\}$ or its restriction to $S_F$ (and other submanifolds). We assume that F is invariant with respect to the standard block diagonal action of $SO(k)\times SO(n-k)$ , with $1\leq k\leq n/2$ .

We start with the following simple observation.

Note that $SO(1) =\{e\}$ , so the action of $O(n-1)=SO(1)\times O(n-1)$ is just that by the orthogonal matrices of the form $\mathrm {diag}(1,A)$ with $A\in O(n-1)$ .

2.2 Proof of Theorem 1.3 for $k=1$

We consider the indicatrix $S_F$ with the restriction of the Hessian metric g. Let $G_0$ be the connected isometry group for $(\mathbb {R}^n\backslash \{0\},g)$ , then it is also the connected isometry group for $(S_F,g)$ . We assume that F is invariant with respect to the standard block diagonal action of $SO(n-1)$ . It implies that $G_0$ naturally contains the group $SO(n-1)$ as a subgroup.

If $G_0$ coincides with $SO(n-1)$ , there is nothing to prove. The next lemma shows that if $G_0$ does not coincide with $SO(n-1)$ , then $(S_F, g)$ is isometric to the standard unit sphere.

Proof Let us assume that $G_0$ does not coincide with $SO(n-1)$ , i.e., $\dim G_0\geq \tfrac {(n-1)(n-2)}{2}+1$ .

We first prove that $(S_F, g)$ is a homogeneous Riemannian sphere. Here, we apply a proof of this claim for all $n\geq 3$ , which is similar to that of [Reference Yano51, Theorem 1] (see also [Reference Ishihara28, Section 4]). Notice that when $n\neq 5$ , [Reference Yano51, Theorem 1] provides an alternative approach. Indeed, we can also see that $(S_F,g)$ has constant sectional curvature, by [Reference Obata40, Theorem 10] and [Reference Ishihara28, Theorem 5] when $n\neq 5$ and $n=5$ , respectively, although it would not be needed in latter argument.

Consider the “pole” $y_0= (a_0,0,\ldots ,0)\in S_F$ . It is a fixed point for the $SO(n-1)$ -action. Consider its $G_0$ -orbit

$$ \begin{align*}G_0\cdot y_0= \{\Phi(y_0)\mid \Phi\in G_0\}.\end{align*} $$

Let $H\subset G_0$ be the stabilizer of $y_0$ . It is known that the stabilizer of a point with respect to an isometric action on an $(n-1)$ -dimensional manifold is at most $\tfrac {(n-1)(n-2)}{2}$ -dimensional, so we have $\dim G_0>\tfrac {(n-1)(n-2)}{2}\geq \dim H$ , i.e., there exists $y\in G_0\cdot y_0$ with $y\neq y_0$ . The orbit $G_0\cdot y_0$ is connected, so we can find a curve $\unicode{x3b3} \subset G_0\cdot y_0$ connecting y and $y_0$ . Then, $G_0\cdot y_0\supset SO(n-1)\cdot \unicode{x3b3} $ contains an $SO(n-1)$ -invariant neighborhood $U_0$ of $y_0$ in $S_F$ . By its homogeneity, $G_0\cdot y_0$ is an open subset of $S_F$ . On the other hand, it is closed, because $G_0$ is a compact Lie group. So we must have $G_0\cdot y_0=S_F$ , i.e., $(S_F,g)$ is a homogeneous sphere.

Next, we prove that the Hessian metric g on $\mathbb {R}^n\backslash \{0\}$ is flat, and its restriction to $S_F$ has constant curvature 1.

The Cartan tensor at $y=(x_1,\ldots ,x_n)\in \mathbb {R}^n\backslash \{0\}$ is defined as

$$ \begin{align*} C(u,v,w)=\tfrac{1}{4} \tfrac{\partial^3}{\partial s_1\partial s_2 \partial s_3}_{|s_1=s_2=s_3=0}F(y+s_1u+s_2v+s_3w)^2, \end{align*} $$

for any $u,v,w\in \mathbb {R}^n=T_y\mathbb {R}^n$ (so its $(ijk)$ -component is $C_{ijk}=\tfrac {1}{4}\tfrac {\partial ^3 (F^2)}{\partial x_i\partial x_j \partial x_k}$ ).

Now, we show the Cartan tensor vanishes at $y_0=(a_0,0,\ldots ,0)\in S_F$ .

Clearly, it is multiple linear and totally symmetric. By the positive $1$ -homogeneity of F, at every point $y\in \mathbb {R}^n\setminus \{0\}$ and for every vectors $u,v\in \mathbb {R}^n$ , we have $C(y, u, v)=0$ at y. So we only need to show, for each vector v with zero $x_1$ -coordinate (i.e., $v\in T_{y_0}S_F$ ), we have $C(v, v, v)=0$ at $y_0$ . Cartan’s trick can be applied to avoid direct calculation. The group $SO(n-1)$ acts transitively on the unit g-sphere in $T_{y_0}S_F$ . So there exists $A\in SO(n-1)$ with $Av=-v$ . That means, the linear isometry induced by A fixes $y_0$ and has a tangent map at $y_0$ mapping v to $-v$ . It preserves the Cartan tensor as well, so we have

$$ \begin{align*}C(v,v,v)= C(-v, -v, -v)=-C(v,v,v)\end{align*} $$

at $y_0$ , which implies $C=0$ there.

Now, we use the following well-known fact in Hessian geometry.

Fact 2.3 (e.g., Proposition 3.2 of [Reference Shima47])

Consider the Hessian metric generated by a (not necessary 2-homogeneous) function E, $g= {\textrm {d}}^2 E$ . Then, its curvature tensor $R_{ijk\ell }$ is given by

(2.5) $$ \begin{align} R_{ijk\ell} = \tfrac{1}{4} \sum_{s,r} \left(\frac{\partial^3 E}{\partial x_j \partial x_\ell \partial x_s} g^{sr} \frac{\partial^3 E}{\partial x_k \partial x_i \partial x_r}- \frac{\partial^3 E}{\partial x_i \partial x_\ell \partial x_s} g^{sr} \frac{\partial^3 E}{\partial x_k \partial x_j \partial x_r}\right),\end{align} $$

where $g^{rs}$ denote the components of the matrix inverse to $(g_{rs})$ .

If $E= \tfrac {1}{2} F^2$ for a Minkowski norm F, the curvature formula (2.5) is reduced to

(2.6) $$ \begin{align} R_{ijk\ell} = \sum_{s,r} \left(C_{i\ell s} g^{sr} C_{jk r} - C_{iks} g^{sr} C_{j\ell r} \right). \end{align} $$

As we explained above, at $y_0$ , every $C_{ijk}$ vanishes, so we have $R_{ijk\ell } =0$ , $\forall i,j,k,\ell $ . In particular, the sectional curvature of $(\mathbb {R}^n\backslash \{0\},g)$ vanishes at $y_0$ .

As we recalled in Section 1.1, the Hessian metric $g=({\textrm {d}}F)^2+F^2 g_{|S_F}$ on $\mathbb {R}^n\backslash \{0\}$ is the cone metric over its restriction to $S_F$ . Then, by the Gauss–Codazzi equation, the sectional curvature of $(S_F,g)$ equals to $1$ at $y_0$ . Because $(S_F,g)$ is homogeneous by assumptions, $(S_F,g_{S_F})$ has constant sectional curvature 1 at every point, i.e., it is isometric to the standard unit sphere. Then, the metric g is flat as we claimed.▪

The next lemma finishes the proof of Theorem 1.3 for $k=1$ .

Proof We first prove Lemma 2.4 when $n=3$ .

We consider the spherical coordinates $(r, \theta , \phi )\in \mathbb {R}_{>0}\times (0,\pi )\times (\mathbb {R}/(2\mathbb {Z}\pi ))$ on $\mathbb {R}^3$ determined by

$$ \begin{align*}x_1=r \cos\theta , \ \ x_2= r \sin\theta \cos\phi, \ \ x_3= r \sin\theta \sin\phi.\end{align*} $$

The $SO(2)$ -action is the left multiplication on column vectors by matrices of the form

$$ \begin{align*}\begin{pmatrix} 1 & 0 & 0\\ 0 & \cos s & \sin s\\ 0 & -\sin s & \cos s\end{pmatrix}, \end{align*} $$

i.e., it fixes the r- and $\theta $ -coordinates and shifts the $\phi $ -coordinate. By its $SO(2)$ -invariancy and homogeneity, the function $E= \tfrac {1}{2}F^2$ can be presented as

(2.7) $$ \begin{align} E= r^2 f(\theta). \end{align} $$

By the symmetry $(x_1,x_2,x_3)\mapsto (x_1, -x_2, -x_3)$ for E, the function $f(\theta )$ on $(0,\pi )$ can be extended to and will be viewed as an even positive smooth function on $\mathbb {R}$ with the period $2\pi $ , i.e., the restriction of E to the circle $\{(x_1,x_2,0)\mid x_1= \cos s, x_2= \sin s, \forall s\in \mathbb {R}\}$ .

Let us now calculate the Hessian metric g and the Cartan tensor C of F in the spherical coordinates. We use subscripts and superscripts r, $\theta $ , and $\phi $ , for example, $g_{r\theta }=g(\tfrac {\partial }{\partial r},\tfrac {\partial }{\partial \theta })$ , and $C_{\theta \theta \phi }=C(\tfrac {\partial }{\partial \theta },\tfrac {\partial }{\partial \theta },\tfrac {\partial }{\partial \phi })$ .

By its definition, g is the second covariant derivative of E with respect to the Levi-Civita connection of the standard flat metric on $\mathbb {R}^3$ , so we have

(2.8) $$ \begin{align} g(X,Y)=X(Y(E))-(\tilde{\nabla}_XY)(E), \end{align} $$

for any smooth tangent vector fields X and Y on $\mathbb {R}^3\backslash \{0\}$ , where $\tilde {\nabla }$ is the Levi-Civita connection for the standard flat metric

$$ \begin{align*}\tilde g:= {\textrm{d}}x_1^2+ {\textrm{d}}x_2^2 + {\textrm{d}}x_3^2= {\textrm{d}}r^2+r^2 {\textrm{d}}\theta^2+r^2\sin^2\theta {\textrm{d}}\phi^2.\end{align*} $$

Direct calculation gives

(2.9) $$ \begin{align} \begin{array}{ll} \tilde{\nabla}_{\tfrac{\partial}{\partial r}}{\tfrac{\partial}{\partial r}} =0, & \tilde{\nabla}_{\tfrac{\partial}{\partial \theta}}{\tfrac{\partial}{\partial r}} =\tilde{\nabla}_{\tfrac{\partial}{\partial r}}{\tfrac{\partial}{\partial \theta}} =\tfrac{1}{r}\tfrac{\partial}{\partial\theta},\\ \tilde{\nabla}_{\tfrac{\partial}{\partial \phi}}{\tfrac{\partial}{\partial r}} = \tilde{\nabla}_{\tfrac{\partial}{\partial r}}{\tfrac{\partial}{\partial \phi}} =\tfrac{1}{r}\tfrac{\partial}{\partial\phi}, & \tilde{\nabla}_{\tfrac{\partial}{\partial \theta}}{\tfrac{\partial}{\partial \theta}} =-r\tfrac{\partial}{\partial\phi},\\ \tilde{\nabla}_{\tfrac{\partial}{\partial \theta}}{\tfrac{\partial}{\partial \phi}} = \tilde{\nabla}_{\tfrac{\partial}{\partial \phi}}{\tfrac{\partial}{\partial \theta}}=\tfrac{\cos\theta}{\sin\theta} \tfrac{\partial}{\partial\phi}, & \tilde{\nabla}_{\tfrac{\partial}{\partial \phi}}{ \tfrac{\partial}{\partial \phi}} =-r\sin^2\theta\tfrac{\partial}{\partial r}-\sin\theta\cos\theta \tfrac{\partial}{\partial\theta}. \end{array} \end{align} $$

Combining (2.7) and (2.8), we obtain all components $g_{ab}$ , $a,b\in \{r,\theta ,\phi \}$ , for $g={\textrm {d}}^2 E$ . With the specified order $(r,\theta ,\phi )$ , they can be presented as the following matrix:

(2.10) $$ \begin{align} \left[ \begin{array}{ccc} 2 f (\theta) &r{ \frac {\textrm{d}}{{\textrm{d}}\theta }}f (\theta) &0 \\ r{\frac{\textrm{d}}{{\textrm{d}}\theta }}f (\theta) & 2r ^{2}f(\theta)+r^2 \frac{{\textrm{d}}^{2}}{{\textrm{d}}\theta^{2}}f (\theta) &0\\ 0&0&2r^2\sin^2\theta f(\theta)+r^2\sin\theta\cos\theta \frac{\textrm{d}}{{\textrm{d}}\theta}f (\theta)\end{array} \right]. \nonumber\\\end{align} $$

For further use, let us observe that the matrix (2.10) is block diagonal, so its inverse matrix is block diagonal as well, i.e., $g^{r\phi }=g^{\theta \phi }=0$ and $g^{\theta \theta }> 0$ .

To calculate the Cartan tensor $C_{abc}$ with $a,b,c\in \{r,\theta ,\phi \}$ , we can proceed analogically:

(2.11) $$ \begin{align} C(X,Y,Z)=\tfrac{1}{2}\left(Z(g(X,Y))-g(\tilde{\nabla}_ZX,Y) -g(X,\tilde{\nabla}_ZY)\right). \end{align} $$

Using (2.9) and (2.10), we see that the only possibly nonzero components of the Cartan tensor are

(2.12) $$ \begin{align} \begin{array}{rcl}C_{\theta\theta\theta} &=& 2r^2 \tfrac{{\textrm{d}}}{{\textrm{d}}\theta}f(\theta)+\tfrac12 r^2 \tfrac{{\textrm{d}}^3}{{\textrm{d}}\theta^3}f(\theta),\\ C_{\theta\phi\phi}&=&-\tfrac12 r^2\cos 2\theta\tfrac{{\textrm{d}}}{{\textrm{d}}\theta}f(\theta)+\tfrac14 r^2\sin2\theta \tfrac{{\textrm{d}}^2}{{\textrm{d}}\theta^2}f(\theta) \end{array}\end{align} $$

(of course, $C_{\theta \phi \phi }= C_{\phi \theta \phi }= C_{\phi \phi \theta }$ , because C is symmetric). Note that it is clear in advance that every component of the form $C_{r\cdot \cdot }=C(\tfrac {\partial }{\partial r},\cdot ,\cdot )$ is zero, because $\tfrac {\partial }{\partial r}$ is the Euler vector field annihilating C. It is also clear by Cartan’s trick that the component $C_{\theta \theta \phi }$ is zero, because the mapping given by $\phi \mapsto -\phi + \text {const}$ is, in fact, a linear isometry which from one side changes the sign for $C_{\theta \theta \phi }$ and from the other side preserves it.

In the case $n=3$ , the only curvature component we need to consider is

(2.13) $$ \begin{align} R_{\theta\phi\phi\theta}=g(R(\tfrac{\partial}{\partial \theta},\tfrac{\partial}{\partial\phi})\tfrac{\partial}{\partial\phi}, \tfrac{\partial}{\partial\theta}) =\sum_{a,b\in\{r,\theta,\phi\}}(C_{\theta\theta a}g^{ab} C_{\phi\phi b}-C_{\theta\phi a}g^{ab}C_{\phi\theta b}). \end{align} $$

Plugging (2.10) and (2.12) into (2.13) and using the vanishing of $g^{\theta \phi }$ , $C_{r\cdot \cdot }$ , and $C_{\theta \theta \phi }$ , we get

$$ \begin{align*} R_{\theta\phi\phi\theta}=C_{\theta\theta\theta}g^{\theta\theta} C_{\theta\phi\phi}-C_{\theta\phi\phi}g^{\phi\phi} C_{\theta\phi\phi}. \end{align*} $$

So the vanishing of the Riemann curvature implies

(2.14) $$ \begin{align} C_{\theta\theta\theta}g^{\theta\theta}C_{\theta\phi\phi} -C_{\theta\phi\phi}g^{\phi\phi}C_{\theta\phi\phi}=0.\end{align} $$

Note that the $\theta $ -derivative of $C_{\theta \phi \phi }$ is $\tfrac {1}{2} \sin 2\theta \ C_{\theta \theta \theta }$ . Indeed,

$$ \begin{align*} \tfrac{{\textrm{d}}}{{\textrm{d}}\theta}(\tfrac{C_{\theta\phi\phi}}{r^2}) &=\tfrac{{\textrm{d}}}{{\textrm{d}}\theta}\left(-\tfrac12\cos 2\theta\tfrac{{\textrm{d}}}{{\textrm{d}}\theta}f(\theta)+\tfrac14 \sin2\theta \tfrac{{\textrm{d}}^2}{{\textrm{d}}\theta^2}f(\theta)\right)\\ &=\tfrac12\sin 2\theta\left(2\tfrac{{\textrm{d}}}{{\textrm{d}}\theta}f(\theta)+\tfrac12\tfrac{{\textrm{d}}^3}{{\textrm{d}}\theta^3}f(\theta) \right)=\tfrac{1}{2}\sin2\theta\cdot\tfrac{C_{\theta\theta\theta}}{r^2}. \end{align*} $$

Thus, $h(\theta )=\left (\tfrac {C_{\theta \phi \phi }}{r^2}\right )^2$ is the solution of the following ODE:

(2.15) $$ \begin{align} \tfrac{{\textrm{d}}}{{\textrm{d}}\theta}h(\theta)=\tfrac{ g^{\phi\phi}\sin2\theta}{g^{\theta\theta}}\cdot h(\theta)=\tfrac{\cos\theta \left(4f(\theta)^2+2f(\theta)\tfrac{{\textrm{d}}^2}{{\textrm{d}}t^2}f(\theta)-\left(\tfrac{{\textrm{d}}}{{\textrm{d}}t}f(\theta)\right)^2\right)}{f(\theta)\left(2\sin\theta f(\theta)+\cos\theta \tfrac{{\textrm{d}}}{{\textrm{d}}t}f(\theta)\right)}\cdot h(\theta) \end{align} $$

on $(0,\pi )$ .

From (2.10), we see that $g_{\phi \phi }$ at the points of $S_F$ , i.e., when $r^2=\tfrac {1}{2f(\theta )}$ , is given by

$$ \begin{align*} \sin^2\theta+\tfrac{\sin\theta\cos\theta}{2f(\theta)} \tfrac{{\textrm{d}}}{{\textrm{d}}\theta}f(\theta). \end{align*} $$

In particular, we have $g_{\phi \phi }=1$ at $x\in S_F$ with $\theta $ -coordinate equal to $\pi /2$ . So the g-arc length of the curve $\{\theta =\pi /2\}$ on $S_F$ is $2\pi $ . When we identify $(S_F,g)$ with a standard $S^2(1)$ , the $SO(2)$ -action which shifts the $\phi $ -coordinates on $S_F$ coincides with a standard linear $SO(2)$ -action on $S^2(1)$ which orbits are the latitude lines. The curve $\theta =\pi /2$ on $S_F$ corresponds to the equator which has the maximal length among all latitude lines. So we have

$$ \begin{align*} \tfrac{{\textrm{d}}}{{\textrm{d}}\theta}\left(g_{\phi\phi}{}_{|r=(2f(\theta))^{-1/2}} \right)_{|\theta=\pi/2} &=\tfrac{{\textrm{d}}}{{\textrm{d}}\theta}\left(\sin^2\theta+\tfrac{\sin\theta\cos\theta}{2f(\theta)} \tfrac{{\textrm{d}}}{{\textrm{d}}\theta}f(\theta)\right)_{|\theta=\pi/2}\\ &= -\tfrac{1}{2f(\pi/2)}\tfrac{{\textrm{d}}}{{\textrm{d}}\theta}f(\theta)_{|\theta=\pi/2}=0, \end{align*} $$

i.e., $\tfrac {{\textrm {d}}}{{\textrm {d}}\theta }f(\theta )_{|\theta =\pi /2}=0$ . Plugging it into the formula of $C_{\theta \phi \phi }$ in (2.12), we see $C_{\theta \phi \phi }=0$ when $\theta =\pi /2$ and then $h(\pi /2)=0$ . By the theory of ODE, $h(\theta )$ is the zero solution of (2.15). So F must be euclidean, because it has zero Cartan tensor everywhere. Lemma 2.4 for $n=3$ is proved.

Remark 2.5 The equality (2.14) follows from (and, in fact, is equivalent to)

(2.16) $$ \begin{align} C_{\theta\theta\theta}g^{\theta\theta}-C_{\theta\phi\phi}g^{\phi\phi}=0 \end{align} $$

everywhere on $\mathbb {R}^3\backslash \{0\}$ . This is a third-order ODE for $f(\theta )$ , and has a three-parameter family of local solutions. Among these local solutions, $c_1+c_2\cos 2\theta $ with appropriate constants $c_1$ and $c_2$ corresponds to the euclidean norms. So we may generically perturb it among local solutions of (2.16), and use the resulting $f(\theta )$ to construct a flat Hessian metric $g={\textrm {d}}^2 E$ for $E=r^2 f(\theta )$ in some conic open subset of $\mathbb {R}^3\backslash \{0\}$ . Local Hessian isometries can be constructed between g and the Hessian metric for an euclidean norm. These local Hessian isometries are not linear. Note also that in Asanov’s examples, the sectional curvature of the restriction of the Hessian metric to the indicatrix is not one, so we obtain another ODE on the function f.

Let us now prove Lemma 2.4 when $n>3$ . Let $y_0\neq 0$ be any point fixed by the action of $O(n-1)$ , and $V_0$ any 3-dimensional vector subspace containing $y_0$ . We can find an involution in $O(n-1)$ , such that $V_0$ is its fixed point set. Indeed, we can find suitable orthonormal coordinates $(x_1,\ldots ,x_n)$ on $\mathbb {R}^n$ , such that $V_0$ consists of all vectors $(x_1,x_2,x_3,0,\ldots ,0)$ and $y_0$ is presented by $(a_0,0,\ldots ,0)$ . Then, $V_0$ is the fixed point set of the mapping $(x_1,\ldots ,x_n)\mapsto (x_1,x_2,x_3,-x_4,\ldots ,-x_n)$ in $O(n-1)$ .

The restriction $F_0=F_{|V_0}$ is invariant with respect to the standard block diagonal action of $O(2)=SO(1)\times O(2)$ . Its Hessian metric $g_0=\tfrac 12{\textrm {d}}^2F_0^2=g_{|V_0\backslash \{0\}}$ is flat, because it is the restriction of the ambient metric g which is flat to an automatically totally geodesic fixed points set. Then, $F_0$ is euclidean. By the $O(n-1)$ -invariance of F, we see that F is euclidean as well.▪

2.3 Proof of Theorem 1.3 for $2\leq k \leq n/2$

Assume now the Minkowski norm F on $\mathbb {R}^n$ is invariant with respect to the standard block diagonal action on $O(k)\times O(n-k)$ with $2\leq k\leq n/2$ . We denote by $G_0$ the connected isometry group for $(\mathbb {R}^n\backslash \{0\},g)$ and for $(S_F,g_{|S_F})$ .

We first consider the case when $(S_F, g_{|S_F})$ is a homogeneous Riemannian sphere. As in the previous section, let us apply Cartan’s trick to prove that the Cartan tensor vanishes at the point $ y_0= (a_0,0,\ldots ,0)\in S_F $ . Let $v\in \mathbb {R}^n$ be any vector contained in the tangent space $T_{y_0}S_F$ , then its $x_1$ -coordinate vanishes. The linear isometry $ (x_1,\ldots ,x_n)\mapsto (x_1,-x_2,-x_3,\dots , -x_n) $ in $O(k)\times O(n-k)$ fixes $y_0$ and its tangent map at $y_0$ sends $v\in T_{y_0}S_F$ to $-v$ . It preserves the Cartan tensor, so we have

$$ \begin{align*}C(v, v, v)= C(-v, -v, -v)=-C(v,v,v)\end{align*} $$

at $y_0$ for each $v\in T_{y_0}S_F$ , which implies $C=0$ there.

Using (2.6) and the same argument as for Lemma 2.2, we see $(S_F,g_{|S_F})$ has constant curvature 1 and $(\mathbb {R}^n\backslash \{0\},g)$ is flat. By $2\leq k\leq n/2$ , we have $n\geq 4$ , and the absolute 1-homogeneity for the $SO(k)\times SO(n-k)$ -invariant Minkowski norm F. By Theorem 1.1, we obtain that F is an euclidean norm, which ends the proof of Theorem 1.3 when $(S_F,g_{|S_F})$ is a homogeneous Riemannian sphere.

Next, we consider the case when $(S_F,g_{|S_F})$ is not a homogeneous Riemannian sphere. Because the $SO(k)\times SO(n-k)$ -action on $S_F$ has cohomogeneity one, $G_0$ must preserve each $SO(k)\times SO(n-k)$ -orbit. Then, the $G_0$ -action maps normal geodesics on $(S_F,g_{|S_F})$ (i.e., geodesics on $(S_F,g_{|S_F})$ which are orthogonal to all the $SO(k)\times SO(n-k)$ -orbits) to normal geodesics on $(S_F,g_{|S_F})$ . So each $\Phi \in G_0$ is determined by its restriction to any principal orbit $\mathcal {O}=(SO(k)\times SO(n-k))\cdot x$ , which results in an injective Lie group homomorphism from $G_0$ to the isometry group for $(\mathcal {O},g_{|\mathcal {O}})$ .

The restriction of the Hessian metric g to the principal orbit

$$ \begin{align*} \mathcal{O}&=(SO(k)\times SO(n-k))\cdot x =(SO(k)\times SO(n-k))/(SO(k-1)\times SO(n-k-1))\\ &= (SO(k)/SO(k-1))\times (SO(n-k)/SO(n-k-1)) \end{align*} $$

is isometric to the Riemannian product of two standard spheres, with dimensions $k-1$ and $n-k-1$ , respectively. The isometry group for $(\mathcal {O},g)$ has the Lie algebra $so(k)\oplus so(n-k)$ (see, e.g., [Reference Eschenburg and Heintze23, Corollary 1], so we have $\dim G_0\leq \dim SO(k)\times SO(n-k)$ . On the other hand, $G_0$ contains all the linear $SO(k)\times SO(n-k)$ -actions. Thus, we have $G_0=SO(k)\times SO(n-k)$ also in this case. Theorem 1.3 is proved.

3.1 Spherical coordinates presentations for local Hessian isometries

Assume the integers k and n satisfy $n\geq 3$ and $1\leq k\leq n/2$ .

The subgroup $O(k)\times O(n-k)$ of $O(n)$ , consisting of $\mathrm {diag}(A,B)$ for all $A\in O(k)$ and $B\in O(n-k)$ , has the standard block diagonal action on the euclidean $\mathbb {R}^n$ of column vectors, with respect to which we have the orthogonal linear decomposition $\mathbb {R}^n=V'\oplus V"$ , where $V'$ and $V"$ are k- and $(n-k)$ -dimensional $O(k)\times O(n-k)$ -invariant subspaces, respectively. For simplicity, if not otherwise specified, orbits are referred to $O(n-1)$ -orbits (which are the same as $SO(1)\times O(n-1)$ - and $SO(n-1)$ -orbits) when $k=1$ , and $O(k)\times O(n-k)$ -orbits (which are the same as $SO(k)\times SO(n-k)$ - and $SO(k)\times O(n-k)$ -orbits) when $k>1$ .

With the marking point $y\in \mathbb {R}^n\backslash \{0\}$ fixed, the orthonormal coordinates $(x_1,\ldots ,x_n)^T$ can and will be chosen such that

1. (1) $V'$ and $V"$ are represented by $x_{k+1}=\cdots =x_n=0$ and $x_1=\cdots =x_k=0$ , respectively;

2. (2) The marking point y has coordinates $(y_1,0,\ldots ,0,y_{k+1},0,\ldots ,0)^T$ with $y_1\geq 0$ and $y_{k+1}\geq 0$ .

Denote by

$$ \begin{align*} S'&=\{(x_1,\ldots,x_k)^T| x_1^2+\cdots+x_k^2=1\}\quad\mbox{and}\\ S"&=\{(x_{k+1},\ldots,x_n)^T| x_{k+1}^2+\cdots+x_n^2=1\} \end{align*} $$

the $(k-1)$ - and $(n-k-1)$ -dimensional standard unit spheres, respectively. Then, we set the spherical coordinates as the following.

If $k=1$ , the spherical coordinates $(r,\theta ,\xi )\in \mathbb {R}_{>0}\times (0,\pi )\times S"$ are determined by

$$ \begin{align*}x_1=r\cos\theta\quad\mbox{and}\quad (x_2,\ldots,x_n)^T=r\sin\theta\cdot \xi,\end{align*} $$

which are well defined on $\mathbb {R}^n\backslash V'$ . The action of $A\in O(n-1)$ (i.e., $\mathrm {diag}(1,A)\in SO(1)\times O(n-1)\subset O(n)$ ) fixes r and $\theta $ and changes $\xi $ to $A\xi $ .

If $k>1$ , the spherical coordinates $(r,\theta ,\xi ',\xi ")\in \mathbb {R}_{>0}\times (0,\pi /2)\times S'\times S"$ are determined by

$$ \begin{align*}(x_1,\ldots,x_k)^T=r\cos\theta\cdot \xi' \quad\mbox{and}\quad (x_{k+1},\ldots,x_n)^T=r\sin\theta\cdot \xi",\end{align*} $$

which are well defined on $\mathbb {R}^n\backslash (V'\cup V")$ . The action of $\mathrm {diag}(A',A")\in O(k)\times O( n-k)$ fixes r and $\theta $ , and changes $\xi '$ and $\xi "$ to $A'\xi '$ and $A"\xi "$ , respectively.

Let us now consider two $SO(k)\times SO(n-k)$ -invariant Minkowski norms $F_1$ and $F_2$ on $\mathbb {R}^n$ , and denote their Hessian metrics by $g_1=g_1(\cdot ,\cdot )$ and $g_1=g_2(\cdot ,\cdot )$ , respectively. To distinguish the different norms or Hessian metrics, we use t to denote the $\theta $ -coordinate where $F_1$ or $g_1$ is concerned, but still call it the $\theta $ -coordinate. By the homogeneity and $SO(k)\times O(n-k)$ -invariancy, $E_i=\tfrac 12F_i^2$ can be presented by spherical coordinates as

$$ \begin{align*} E_1=r^2 f(t)\quad \mbox{and}\quad E_2=r^2 h(\theta), \end{align*} $$

respectively. Although t and $\theta $ belong to $(0,\pi )$ or $(0,\pi /2)$ , $f(t)$ and $h(\theta )$ can be periodically extended to even positive smooth functions on $\mathbb {R}$ , with the period $2\pi $ or $\pi $ , when $k=1$ or $k>1$ , respectively.

Without loss of generality, we will further assume $y\in S_{F_1}$ . The $SO(k)\times O(n-k)$ -action on $(S_{F_i},g_{i})$ is of cohomogeneity one. The normal geodesics on $(S_{F_i},g_i)$ are those which intersect orbits orthogonally. Using fixed point set technique and similar Cartan’s trick as in the proof of Lemma 2.2, it is easy to see that around any principal orbit, normal geodesics are characterized by the following equations for spherical coordinates, $\xi \equiv \mathrm {const}$ when $k=1$ , or $(\xi ',\xi ")\equiv \mathrm {const}$ when $k>1$ .

Now, we assume y satisfies (1.2) in Theorem 1.4, i.e.,

$$ \begin{align*} g_1(v',v")\neq0\mbox{ at }y,\quad\mbox{for some }v'\in V'\mbox{ and }v"\in V". \end{align*} $$

Applying Cartan’s trick to those $\mathrm {diag}(\pm 1,\ldots ,\pm 1)\in O(k)\times O(n-k)$ which preserves $F_1$ and fix y, we see easily:

1. (1) When $k=1$ , we have $y\notin V'$ , and when $k>1$ , $y\notin V'\cup V"$ . So the spherical coordinates of y are well defined.

2. (2) The Hessian matrix $(a_{ij})=(g_1(\tfrac {\partial }{\partial x_i}, \tfrac {\partial }{\partial x_j}))$ is blocked-diagonal. To be precise, we have at y

   (3.17) $$ \begin{align} g_1(\tfrac{\partial}{\partial x_i},\tfrac{\partial}{\partial x_j})=0,\quad \mbox{when }i\neq j\mbox{ and } \{i,j\}\neq\{1,k+1\}. \end{align} $$

Using the spherical coordinates, the assumption (1.2) can be interpreted as the following.

Proof Because of (3.17) at y, (1) and (2) in Lemma 3.1 are equivalent. Further discussion can be reduced to the 3-dimensional subspace V given by $x_2=\cdots =x_k=x_{k+2}=\cdots =x_{n-1}=0$ . By similar calculation as for (2.10), we get

$$ \begin{align*} g_1(\tfrac{\partial}{\partial x_1},\tfrac{\partial}{\partial x_{k+1}})&=\pm g_1(\cos t\tfrac{\partial}{\partial r}-\tfrac{1}{r}\sin t \tfrac{\partial}{\partial t},\sin t\tfrac{\partial}{\partial r}+ \tfrac{\cos t}{r}\tfrac{\partial}{\partial t})\\ &=\pm\left(-\cos t\sin t\tfrac{{\textrm{d}}^2}{{\textrm{d}}t^2}f(t)+(\cos^2t-\sin^2t) \tfrac{{\textrm{d}}}{{\textrm{d}}t}f(t)\right). \end{align*} $$

Then, the equivalence between (2) and (3) in Lemma 3.1 follows immediately. Finally, we compare (3.18) with the formula for $C_{\theta \phi \phi }$ in (2.12), and we see the Cartan tensor does not vanish at y when (1.2) is satisfied. So $F_1$ is not locally euclidean there.▪

Let us consider a local Hessian isometry $\Phi $ from $F_1$ to $F_2$ which is defined on an $SO(k)\times O(n-k)$ -invariant conic neighborhood of y, and maps orbits to orbits. Notice that $\Phi $ satisfies the positive 1-homogeneity and preserves the norm.

The following spherical coordinates presentations of $\Phi $ are crucial for proving Theorems 1.4 and 1.5.

Lemma 3.2 When $k=1$ , the local Hessian isometry $\Phi $ can be presented by spherical coordinates as

(3.19) $$ \begin{align} (r,t,\xi)\mapsto \left(\tfrac{f(t)^{1/2}}{h(\theta(t))^{1/2}}\cdot r,\theta(t),A\xi\right) \end{align} $$

in some $O(n-1)$ -invariant conic neighborhood of y, where $A\in O(n-1)$ and $\theta (t)$ is a smooth function with nonzero derivatives everywhere.

Proof By the homogeneity of $\Phi $ , to prove (3.19), we only need to discuss $\Phi (x)$ for $x\in S_{F_1}$ . When x is sufficiently close to $y\in S_{F_1}$ , $x\notin V'$ , so its spherical coordinates $(r,t,\xi )=((2f(t))^{-1/2},t,\xi )$ are well defined. Because $\Phi $ maps principal orbits on $S_{F_1}$ to principal orbits on $S_{F_2}$ , and each principal orbit is characterized by constant $\theta $ -coordinates, we see that the $\theta $ -coordinate of $\Phi (x)$ only depends on t. So we may denote it as $\theta (t)$ , which smoothness is obvious. Because $F_1(x)=F_2(\Phi (x))=1$ , the r-coordinate of $\Phi (x)$ is $(2h(\theta (t)))^{-1/2}=\tfrac {f(t)^{1/2}}{h(\theta (t))^{1/2}}\cdot r$ .

Denote $\mathcal {O}_1=O(n-1)\cdot x$ the principal orbit in $S_{F_1}$ passing x. When endowed with the Hessian metric, it is a homogeneous Riemannian sphere $O(n-1)/O(n-2)$ , which is isometric to a radius R standard sphere (i.e., its perimeter is $2\pi R$ when $n=3$ or it has constant curvature $R^{-2}$ when $n>3$ ). For $\mathcal {O}_2=O(n-1)\cdot \Phi (x)$ in $S_{F_2}$ , we have a similar claim. Because the local Hessian isometry $\Phi $ maps $\mathcal {O}_1$ onto $\mathcal {O}_2$ , $(\mathcal {O}_2,g_2)$ is also isometric to a radius R standard sphere. Denote $g_{\mathrm {st}}$ the standard unit sphere metric on $S"$ , then the $O(n-1)$ -equivariant diffeomorphism $\Phi _1:(\mathcal {O}_1,g_1)\rightarrow (S",R^2 g_{\mathrm {st}})$ , mapping $x'\in \mathcal {O}_1$ to its $\xi $ -coordinate, is an isometry. Similarly, we have another homothetic correspondence $\Phi _2:(\mathcal {O}_2,g_2)\rightarrow (S",R^2g_{\mathrm {st}})$ . The composition

$$ \begin{align*}\Psi=\Phi_2\circ\Phi\circ\Phi_1^{-1}: (S",R^2g_{\mathrm{st}})\rightarrow (S",R^2g_{\mathrm{st}}),\end{align*} $$

which characterizes how the local Hessian isometry $\Phi $ changes the $\xi $ -coordinates, is an isometry. So $\Psi $ must be of the form $\xi \mapsto A\xi $ for some $A\in O(n-1)$ .

Because $\Phi $ maps orbits on $S_{F_1}$ to orbits on $S_{F_2}$ , it also maps normal geodesics to normal geodesics. Normal geodesics have constant $\xi $ -coordinates around each principal orbit. So the matrix $A\in O(n-1)$ in the presentation of $\Psi $ does not depend on t.

The above argument proves the spherical coordinates presentations of $\Phi $ in (3.19). Then, we prove $\theta (t)$ has nonzero derivatives everywhere.

We use (3.19) to calculate the tangent map $\Phi _*$ at x, which can be presented as the following Jacobi matrix:

$$ \begin{align*}\left( \begin{array}{ccc} \tfrac{f(t)^{1/2}}{h(\theta(t))^{1/2}} & \tfrac{h(\theta(t))-f(t)\tfrac{{\textrm{d}}}{{\textrm{d}}t}\theta(t)}{2f(t)(2h(\theta(t)))^{3/2}} & 0 \\ 0 & \tfrac{{\textrm{d}}}{{\textrm{d}}t}\theta(t) & 0 \\ 0 & 0 & A \\ \end{array} \right). \end{align*} $$

Because $\Phi $ is a local diffeomorphism, its Jacobi matrix must have nonzero determinant, which requires $\tfrac {{\textrm {d}}}{{\textrm {d}}t}\theta (t)\neq 0$ .▪

Lemma 3.3 When $k>1$ , the local Hessian isometry $\Phi $ can be presented by spherical coordinates either as

(3.20) $$ \begin{align} (r,t,\xi',\xi")\mapsto \left(\tfrac{f(t)^{1/2}}{h(\theta(t))^{1/2}}\cdot r,\theta(t),A'\xi',A"\xi"\right) \end{align} $$

or as

(3.21) $$ \begin{align} (r,t,\xi',\xi")\mapsto \left(\tfrac{f(t)^{1/2}}{h(\theta(t))^{1/2}}\cdot r,\theta(t),A"\xi", A'\xi'\right) \end{align} $$

in some $O(k)\times O(n-k)$ -invariant conic neighborhood of y, where $A'\in O(k)$ , $A"\in O(n-k)$ , $\theta (t)$ is a smooth function with nonzero derivatives everywhere, and (3.21) may happen only when $n=2k$ .

Proof We only need to discuss the spherical coordinates of $\Phi (x)$ for $x\in S_{F_1}$ sufficiently close to y. Denote the orbits

$$ \begin{align*} & \mathcal{O}_1=(O(k)\times O(n-k))\cdot x,\ \mathcal{O}^{\prime}_1=(O(k)\times\{e\})\cdot x,\ \mathcal{O}^{\prime\prime}_1=(\{e\}\times O(n-k))\cdot x,\\ & \mathcal{O}_2=(O(k)\times O(n-k))\cdot\Phi(x),\ \mathcal{O}^{\prime}_2=(O(k)\times\{e\})\cdot \Phi(x),\\ &\mathcal{O}^{\prime\prime}_2=(\{e\}\times O(n-k))\cdot \Phi(x). \end{align*} $$

When endowed with the restriction of $g_1$ , $\mathcal {O}_1=(O(k)\times O(n-k))/(O(k-1)\times O(n-k-1))$ is the Riemannian product of the two homogeneous Riemannian spheres, i.e., $\mathcal {O}^{\prime }_1=O(k)/O(k-1)$ , which is isometric to a radius $R^{\prime }_1$ standard sphere, and $\mathcal {O}^{\prime \prime }_1=O(n-k)/O(n-k-1)$ , which is isometric to a radius $R^{\prime \prime }_1$ standard sphere. Denote $g^{\prime }_{\mathrm {st}}$ and $g^{\prime \prime }_{\mathrm {st}}$ the standard unit sphere metrics on $S'$ and $S"$ , respectively, and $g_{R^{\prime }_1,R^{\prime \prime }_1}$ the product metric of ${R^{\prime }_1}^2g^{\prime }_{\mathrm {st}}$ and ${R^{\prime \prime }_1}^2g^{\prime \prime }_{\mathrm {st}}$ on $S'\times S"$ . Then, the $O(k)\times O(n-k)$ -equivariant diffeomorphism $\Phi _1:(\mathcal {O}_1,g_1)\rightarrow (S'\times S",g_{R^{\prime }_1,R^{\prime \prime }_1})$ is an isometry. Similarly, $(\mathcal {O}^{\prime }_2,g_2)$ and $(\mathcal {O}^{\prime \prime }_2,g_2)$ are isometric to standard spheres with radii $R^{\prime }_2$ and $R^{\prime \prime }_2$ , respectively, and we have another isometry

$$ \begin{align*}\Phi_2:(\mathcal{O}_2,g_2)\rightarrow (S'\times S",g_{R^{\prime}_2,R^{\prime\prime}_2}).\end{align*} $$

Because the local Hessian isometry $\Phi $ maps $\mathcal {O}_1$ onto $\mathcal {O}_2$ , the composition

$$ \begin{align*}\Psi=\Phi_2\circ\Phi\circ\Phi_1^{-1}:(S'\times S",g_{R^{\prime}_1,R^{\prime\prime}_1})\rightarrow (S'\times S",g_{R^{\prime}_2,R^{\prime\prime}_2}),\end{align*} $$

which characterizes how the local isometry $\Phi $ changes the $\xi '$ - and $\xi "$ -coordinates, is an isometry. The isometries on the Riemannian product of two standard spheres are completely known (see, e.g., [Reference Eschenburg and Heintze23, Corollary 1] or [Reference Helgason27]). Then, all possibilities for $\Psi $ are the following:

1. (1) $\Psi (\xi ',\xi ")=(A'\xi ',A"\xi ")$ for some $A'\in O(k)$ and $A"\in O(n-k)$ , $R^{\prime }_1=R^{\prime }_2$ and $R^{\prime }_2=R^{\prime }_1$ .

2. (2) $n=2k$ , $\Psi (\xi ',\xi ")=(A"\xi ",A'\xi ')$ for some $A',A"\in O(k)$ , $R^{\prime }_1=R^{\prime \prime }_2$ and $R^{\prime }_2=R^{\prime \prime }_1$ .

For each possibility, $\Psi $ represents a distinct homotopy class, which does not change when we move x continuously. Furthermore, $A'$ and $A"$ in the presentation of $\Psi $ are independent of t, because $\Phi $ maps normal geodesics on $S_{F_1}$ to those on $S_{F_2}$ , and normal geodesics on $S_{F_i}$ have constant $\xi '$ - and $\xi "$ -coordinates.

The remaining arguments are similar to those for Lemma 3.2, so we skip them.▪

3.2 Equivariant Hessian isometries

Analyzing the spherical coordinates presentations (3.19), (3.20), and (3.21) in Lemmas 3.2 and 3.3, we see immediately that a local Hessian isometry $\Phi $ mapping orbits to orbits can be decomposed as $\Phi =\Phi _1\circ \Phi _2$ , in which $\Phi _1$ is a linear isometry mapping orbits to orbits, and $\Phi _2$ is a local Hessian isometry fixing all $\xi $ -coordinates when $k=1$ , or fixing all $\xi '$ - and $\xi "$ -coordinates when $k>1$ . For example, when $n=2k$ and $\Phi $ is presented by spherical coordinates as in (3.21), i.e., $(r,t,\xi ',\xi ")\mapsto \left(\tfrac {f(t)^{1/2}r}{h(\theta (t))^{1/2}},\theta (t),A"\xi ",A'\xi '\right)$ , $\Phi _1$ is the action of $\left ( \begin {array}{cc} 0 & A" \\ A' & 0 \\ \end {array} \right ) $ in $O(n)$ . It maps orbits to orbits, exchanging the curvature constants of the two product factors in the orbit, and it induces a new $O(k)\times O(n-k)$ norm $F_3=F_2\circ \Phi _1$ . The composition $\Phi _2=\Phi _1^{-1}\circ \Phi $ is local Hessian isometry from $F_1$ to $F_3$ fixing $\xi '$ - and $\xi "$ -coordinates.

For simplicity, we call $\Phi $ equivariant if it is equivariant with respect to the $O(n-1)$ -action or the $O(k)\times O(n-k)$ -action, when $k=1$ or $k>1$ , respectively. Practically, we will only use those equivariant $\Phi $ which fix all $\xi $ -coordinates or all $\xi '$ - and $\xi "$ -coordinates.

Summarizing the above observations, we have the following theorem.

The following examples of global equivariant Hessian isometries are crucial for the proofs of Theorem 1.4.

Example 3.5 Let $F_1$ be any $SO(k)\times O(n-k)$ -invariant Minkowski norm on $\mathbb {R}^n$ , and $\Phi $ a linear map

(3.22) $$ \begin{align} (x_1,\ldots,x_k,x_{k+1},\ldots,x_n)\mapsto(ax_1,\ldots,ax_k,bx_{k+1}, \ldots,bx_{n}), \end{align} $$

with the parameter pair $(a,b)\in \mathbb {R}_{\neq 0}\times \mathbb {R}_{>0}$ when $k=1$ , or $(a,b)\in \mathbb {R}_{>0}\times \mathbb {R}_{>0}$ when $k>1$ . Then, $\Phi $ induces another $SO(k)\times O(n-k)$ -invariant Minkowski norm $F_2=F_1\circ \Phi ^{-1}$ , such that $\Phi $ is an equivariant Hessian isometry from $F_1$ to $F_2$ which fixes all $\xi $ - or all $\xi '$ - and $\xi "$ -coordinates. We will simply call it the linear example with the parameter pair $(a,b)$ .

If $k=1$ , the function $\theta (t)$ in the spherical coordinates presentations for the linear example with the parameter pair $(a,b)$ is

(3.23) $$ \begin{align} \theta(t)=\arccos\left(\tfrac{a\cos t}{(a^2\cos^2 t+b^2\sin^2 t)^{1/2}}\right), \end{align} $$

for $t\in (0,\pi )$ . It satisfies

(3.24) $$ \begin{align} \tfrac{{\textrm{d}}}{{\textrm{d}}t}\theta(t)=\tfrac{\sin\theta(t)\cos\theta(t)}{\sin t\cos t}, \end{align} $$

when $t\neq \pi /2$ .

If $k>1$ , the function $\theta (t)$ satisfies (3.23) and (3.24) for $t\in (0,\pi /2)$ .

Example 3.6 Let $F_1$ be any $SO(k)\times O(n-k)$ -invariant Minkowski norm on $\mathbb {R}^n$ , and $\Phi :\mathbb {R}^n\backslash \{0\}\rightarrow \mathbb {R}^n\backslash \{0\}$ the diffeomorphism

(3.25) $$ \begin{align} (x_1,\ldots,x_k,x_{k+1},\ldots,x_n) \mapsto (a\tfrac{\partial E_1}{\partial x_1},\ldots,a\tfrac{\partial E_1}{\partial x_k}, b\tfrac{\partial E_1}{\partial x_{k+1}}, \ldots,b\tfrac{\partial E_1}{\partial x_n}), \end{align} $$

where $E_1=\tfrac 12F_1^2=r^2f(t)$ in spherical coordinates, and the requirement for the parameter pair $(a,b)$ is the same as in Example 3.5. Then, $\Phi $ induces another $SO(k)\times O(n-k)$ -invariant Minkowski norm $F_2=F_1\circ \Phi ^{-1}$ , such that $\Phi $ is an equivariant Hessian isometry from $F_1$ to $F_2$ which fixes $\xi $ - or all $\xi '$ - and $\xi "$ -coordinates. We will simply call it the Legendre example with the parameter pair $(a,b)$ , because it is the composition between the Legendre transformation of $F_1$ , from $F_1$ to $\hat {F}_1$ , and a linear isometry from $\hat {F}_1$ to $F_2$ .

If $k=1$ , the function $\theta (t)$ in the spherical coordinates presentations for the Legendre example with the parameter pair $(a,b)$ is

(3.26) $$ \begin{align} &\theta(t) = \arccos\\ & \quad \left(\tfrac{2\cos t f(t)-\sin t \tfrac{{\textrm{d}}}{{\textrm{d}}t}f(t)}{\left[4\left(\cos^2 t+\tfrac{b^2}{a^2}\sin^2 t\right)f(t)^2 +4\left(\tfrac{b^2}{a^2}-1\right)\cos t\sin t f(t)\tfrac{{\textrm{d}}}{{\textrm{d}}t}f(t)+ \left(\sin^2 t+\tfrac{b^2}{a^2}\cos^2 t\right)\left(\tfrac{{\textrm{d}}}{{\textrm{d}}t}f(t)\right)^2\right]^{1/2}}\right),\nonumber \end{align} $$

for $t\in (0,\pi )$ . It satisfies

(3.27) $$ \begin{align} \tfrac{{\textrm{d}}}{{\textrm{d}}t}\theta(t)=\tfrac{ \left(2f(t)\tfrac{{\textrm{d}}^2}{{\textrm{d}}t^2}f(t)-\left(\tfrac{{\textrm{d}}}{{\textrm{d}}t}f(t)\right)^2+4f(t)^2\right)\sin\theta(t)\cos\theta(t)}{ \left(\cos t \tfrac{{\textrm{d}}}{{\textrm{d}}t}f(t)+2\sin t f(t)\right)\left(-\sin t \tfrac{{\textrm{d}}}{{\textrm{d}}t}f(t)+2\cos t f(t)\right)}, \end{align} $$

when $\cos t \tfrac {{\textrm {d}}}{{\textrm {d}}t}f(t)+2\sin t f(t)\neq 0$ and $-\sin t\tfrac {{\textrm {d}}}{{\textrm {d}}t}f(t)+2\cos t f(t)\neq 0$ .

By the strong convexity of $F_1$ , nonvanishing of $\cos t \tfrac {{\textrm {d}}}{{\textrm {d}}t}f(t)+2\sin t f(t)\neq 0$ is always guaranteed for $t\in (0,\pi )$ . In particular, when $n=3$ , $\cos t \tfrac {{\textrm {d}}}{{\textrm {d}}t}f(t)+2\sin t f(t)\neq 0$ is a product factor in $g_{\phi \phi }$ (see (2.10)). Meanwhile, by the calculation

$$ \begin{align*}\tfrac{\partial E_1}{\partial x_1}=r\left(-\sin t \tfrac{{\textrm{d}}}{{\textrm{d}}t}f(t)+2\cos t f(t)\right),\end{align*} $$

we see that $-\sin t \tfrac {{\textrm {d}}}{{\textrm {d}}t}f(t)+2\cos t f(t)$ vanishes iff $\tfrac {\partial E_1}{\partial x_1}=0$ . By the strong convexity and $O(n-1)$ -invariancy of $F_1$ , the equation $-\sin t \tfrac {{\textrm {d}}}{{\textrm {d}}t}f(t)+2\cos t f(t)=0$ has a unique solution $t'$ in $(0,\pi )$ . In particular, when $f(t)\equiv f(\pi -t)$ for $t\in (0,\pi )$ , $t'=\pi /2$ .

If $k>1$ , the corresponding function $\theta (t)$ satisfies (3.26) and (3.27) for all $t\in (0,\pi /2)$ .

3.3 Proof of Theorem 1.4: reduction to $n=3$

In the following two subsections, we prove Theorem 1.4. In this subsection, we explain why and how we can reduce the proof to the case $n=3$ . Then, in the next subsection, we prove Theorem 1.4 when $n=3$ .

Let $C(U_1)$ be any $SO(k)\times SO(n-k)$ -invariant connected conic open subset in $\mathbb {R}^n$ which satisfies (1.2) everywhere, and $\Phi $ a local Hessian isometry from $F_1$ to $F_2$ which is defined on $C(U_1)$ and maps orbits to orbits. By Theorem 3.4, we only need to prove Theorem 1.4 assuming that $\Phi $ fixes all $\xi $ - or all $\xi '$ - and $\xi "$ -coordinates. Then, $\Phi $ preserves the 3-dimensional subspace V given by $x_2=\cdots =x_k=x_{k+2}=\cdots =x_{n-1}=0$ . Furthermore, when $k>1$ , $\Phi $ preserves the subset $V_{x_1>0}\subset V$ with positive $x_1$ -coordinates. The restrictions ${F_i}_{|V}$ are Minkowski norms on V which are invariant with respect to the subgroup $O(2)\subset O(n-1)$ fixing each point of $V^\perp $ given by $x_1=x_{k+1}=x_n=0$ . Denote $C(U^{\prime }_1)$ the following $SO(2)$ -invariant connected conic open subset of V. When $k=1$ , $C(U^{\prime }_1)=C(U)\cap V$ , and when $k>1$ , $C(U^{\prime }_1)=C(U_1)\cap V_{x_1>0}$ . The restrictions ${g_i}_{|V}$ coincide with the Hessian metrics for ${F_i}_{|V}$ , so the restriction $\Phi _{|C(U^{\prime }_1)}$ is a local Hessian isometry from ${F_1}_{|V}$ to ${F_2}_{|V}$ . Each $SO(2)$ -orbit in $C(U^{\prime }_1)$ is the intersection of an $SO(k)\times O(n-k)$ -orbit with V or $V_{x_1>0}$ . So $\Phi _{|C(U^{\prime }_1)}$ maps $O(2)$ -orbits to $O(2)$ -orbits. By (3.17) and Lemma 3.1, we have

$$ \begin{align*}g_1\left(\tfrac{\partial}{\partial x_{k+1}},\tfrac{\partial}{\partial x_{n}}\right)=g_1\left(\tfrac{\partial}{\partial x_1},\tfrac{\partial}{\partial x_n}\right)=0\mbox{ and } g_1\left(\tfrac{\partial}{\partial x_1},\tfrac{\partial}{\partial x_k}\right)\neq0\end{align*} $$

at any $y=(y_1,0,\ldots ,0,y_{k+1},0,\ldots ,0)\in C(U^{\prime }_1)$ . So to summarize, we have the following observation.

Restricting to V, the following spherical $(r,\theta ,\phi )$ -coordinates are more convenient for calculation:

(3.28) $$ \begin{align} x_1=r\cos\theta,\quad x_{k+1}=r\sin\theta\cos\phi,\quad x_n=r\sin\theta\sin\phi, \end{align} $$

with $(r,\theta ,\phi )\in \mathbb {R}_{>0}\times (0,\pi )\times (\mathbb {R}/(2\mathbb {Z}\pi ))$ . Similarly, we use t to denote the $\theta $ -coordinate where ${F_1}_{|V}$ or ${g_1}_{|V}$ is concerned. It is easy to check that $\Phi _{|C(U^{\prime }_1)}$ fixes all $\phi $ -coordinates.

When $k=1$ , the $(r,\theta ,\phi )$ -coordinates are related to the $(r,\theta ,\xi )$ -coordinates in Section 3.1 by

$$ \begin{align*}(r,\theta,\phi)\leftrightarrow(r,\theta,\xi)=(r,\theta,(\cos\phi,0,\ldots,0,\sin\phi)^T).\end{align*} $$

The functions $f(t)$ , $h(\theta )$ , and $\theta (t)$ in the $(r,\theta ,\xi )$ -coordinates presentations are completely inherited by the $(r,\theta ,\phi )$ -coordinates presentations when restricted to V, i.e.,

$$ \begin{align*}{E_1}_{|V}=r^2 f(t),\quad {E_2}_{|V}=r^2 h(\theta), \quad \Phi_{|C(U^{\prime}_1)}:(r,t,\phi)\mapsto(\tfrac{f(t)^{1/2}r}{h({\theta(t)})^{1/2}}, \theta(t),\phi).\end{align*} $$

When $k>1$ , we can still use the spherical coordinates $(r,\theta ,\phi )$ in (3.28) on V, which is related to the $(r,\theta ,\xi ',\xi ")$ -coordinates by

$$ \begin{align*} & (r,\theta,\phi)\leftrightarrow (r,\theta,(1,0,\ldots,0)^T,(\cos\phi,0,\ldots,0,\sin\phi)^T),\quad \forall\theta\in(0,\pi/2), \\ & (r,\theta,\phi)\leftrightarrow (r,\pi-\theta,(-1,0,\ldots,0)^T,(\cos\phi,0,\ldots,0,\sin\phi)^T),\quad \forall\theta\in(\pi/2,\pi). \end{align*} $$

Because, in this case, $C(U^{\prime }_1)\subset V_{x_1>0}$ has positive $x_1$ -coordinates, i.e., its $\theta $ -coordinates range in $(0,\pi /2)$ , the functions $f(t)$ , $h(\theta )$ , and $\theta (t)$ in the $(r,\theta ,\xi ',\xi ")$ -coordinates presentations, which are originally defined on $(0,\pi /2)$ , can still be applied to the discussion for ${\Phi }_{|C(U^{\prime }_1)}$ . So to summarize, we have the following observation.

We see from the next subsection that the key steps in the proof, i.e., using the spherical $(r,\theta ,\phi )$ -coordinates to deduce and analyze the ODE system for $\theta (t)$ and $h(\theta )$ , and calculating the fundamental tensor for the linear and Legendre examples, are only relevant to the $x_1$ -, $x_{k+1}$ -, and $x_n$ -coordinates. So they are all contained in the proof of Theorem 1.4 when $n=3$ . The functions $\theta (t)$ in (3.23) and (3.26) for the linear example and the Legendre example, respectively, are irrelevant to the dimension. So to summarize, we have the following conclusion.

Conclusion: With some minor changes, the argument in the next subsection proves Theorem 1.4 generally.

3.4 Proof of Theorem 1.4 when $n=3$

Let $F_1$ , $F_2$ be two Minkowski norms on $\mathbb {R}^3$ , which are invariant with respect to (the same) standard block diagonal action of $O(2)$ generated by the matrices of the form $\mathrm {diag}(1,A)$ with $A\in O(2)$ . Their Hessian metrics are denoted as $g_1=g_1(\cdot ,\cdot )$ and $g_2=g_2(\cdot ,\cdot )$ , respectively.

We fix the orthonormal coordinates $(x_1,x_2,x_3)$ such that the $SO(2)$ -action fixes each point on the line $V'$ presented by $x_2=x_3=0$ and rotates the plane $V"$ presented by $x_1=0$ . We further require the marking point $y\in \mathbb {R}^3\backslash \{0\}$ has coordinates $(y_1,y_2,y_3)$ with $y_1\geq 0$ , $y_2\geq 0$ and $y_3=0$ .

In this subsection, we will only use the spherical coordinates $(r,\theta ,\phi )\in \mathbb {R}_{>0}\times (0,\pi )\times (\mathbb {R}/2\mathbb {Z}\pi )$ determined by

$$ \begin{align*}x_1=r\cos\theta,\quad x_2=r\sin\theta\cos\phi,\quad x_3=r\sin\theta\sin\phi.\end{align*} $$

Then, the $SO(2)$ -action fixes r and $\theta $ and shifts $\phi $ . We use t to denote the $\theta $ -coordinate and still call it the $\theta $ -coordinate where $F_1$ or $g_1$ is concerned.

By the homogeneity and $SO(2)$ -invariancy, $E_i=\tfrac 12F_i^2$ can be presented as

$$ \begin{align*} E_1=r^2 {f(t)}\quad \mbox{and}\quad E_2=r^2 {h(\theta)}, \end{align*} $$

respectively, in which $f(t)$ and $h(\theta )$ are some even positive smooth functions on $\mathbb {R}$ with the period $2\pi $ .

We have previously observed $y\notin V_1$ , so we have $y_1\geq 0$ and $y_2>0$ for $y=(y_1,y_2,0)$ , i.e., the $\theta $ -coordinate of y is contained in $(0,\pi /2]$ , and the $\phi $ -coordinate of y is $0\in \mathbb {R}/(2\mathbb {Z}\pi )$ . Without loss of generality, we assume $y\in S_{F_1}$ . So its spherical coordinates can be presented as $(r_0,t_0,\phi _0)=(f(t_0)^{-1/2},t_0,0)$ . By (3.17) and Lemma 3.1, we have the following at y:

(3.29) $$ \begin{align} & g_1(\tfrac{\partial}{\partial x_1},\tfrac{\partial}{\partial x_3}) =g_1(\tfrac{\partial}{\partial x_2},\tfrac{\partial}{\partial x_3})=0, \end{align} $$

(3.30) $$ \begin{align} & g_1(\tfrac{\partial}{\partial x_1},\tfrac{\partial}{\partial x_2})\neq0,\mbox{ and} \end{align} $$

(3.31) $$ \begin{align} & -\cos t_0\sin t_0\tfrac{{\textrm{d}}^2}{{\textrm{d}}t^2}f(t_0)+(\cos^2 t_0-\sin^2 t_0)\tfrac{{\textrm{d}}}{{\textrm{d}}t}f(t_0)\neq0. \end{align} $$

Let $\Phi $ be the local Hessian isometry from $F_1$ to $F_2$ , which is defined on some $SO(2)$ -invariant conic neighborhood of y and maps orbits to orbits. By Theorem 3.4 and Lemma 3.2, we only need to consider the situation that $\Phi $ fixes the $\phi $ -coordinates, and we can present it by spherical coordinates as

(3.32) $$ \begin{align} (r,t,\phi)\mapsto \left(\tfrac{f(t)^{1/2}}{h(\theta(t))^{1/2}}\cdot r,\theta(t),\phi\right). \end{align} $$

We will first discuss the situation that $t_0\neq \pi /2,t'$ , where $t'$ is the unique solution of $\sin t \tfrac {{\textrm {d}}}{{\textrm {d}}t}f(t)-2\cos tf(t)=0$ in $(0,\pi )$ .

Let $y(t)$ be a normal geodesic on $(S_{F_1},g_1)$ passing y, parameterized by the $\theta $ -coordinate. Using spherical coordinates, $y(t)$ can be locally presented as $((2f(t))^{-1/2}, t,0)$ around y, with its tangent vector field $\tfrac {{\textrm {d}}}{{\textrm {d}}t}y(t)=\tfrac {\partial }{\partial t}-\tfrac {1}{(2f(t))^{3/2}}\tfrac {{\textrm {d}}}{{\textrm {d}}t}f(t)\tfrac {\partial }{\partial r}$ . By (2.10),

(3.33) $$ \begin{align} g_1\left(\tfrac{{\textrm{d}}}{{\textrm{d}}t}y(t),\tfrac{{\textrm{d}}}{{\textrm{d}}t}y(t)\right) =\tfrac{1}{2f(t)}\tfrac{{\textrm{d}}^2}{{\textrm{d}}t^2}f(t)-\tfrac{1}{4f(t)^2}\left(\tfrac{{\textrm{d}}}{{\textrm{d}}t}f(t)\right)^2+1. \end{align} $$

The $\Phi $ -image $\unicode{x3b3} $ of the curve $y(t)$ is a curve on $S_{F_2}$ with constant $\phi $ -coordinate $0$ . When $\unicode{x3b3} =\unicode{x3b3} (\theta )$ is parameterized by the $\theta $ -coordinate, we similarly have

(3.34) $$ \begin{align} g_2(\tfrac{{\textrm{d}}}{{\textrm{d}}\theta}\unicode{x3b3}(\theta), \tfrac{{\textrm{d}}}{{\textrm{d}}\theta}\unicode{x3b3}(\theta)) =\tfrac{1}{2h(\theta)} \tfrac{{\textrm{d}}^2}{{\textrm{d}}\theta^2}h(\theta)- \tfrac{1}{4h(\theta)^2} \left(\tfrac{{\textrm{d}}}{{\textrm{d}}\theta}h(\theta)\right)^2+1. \end{align} $$

Because

$\Phi _*(\tfrac {{\textrm {d}}}{{\textrm {d}}t}y(t))=f'(t)\tfrac {{\textrm {d}}}{{\textrm {d}}\theta }\unicode{x3b3} (\theta (t))$ , and $\Phi $ is a local isometry around $y=y(t_0)$ , we have

(3.35) $$ \begin{align} &\tfrac{1}{2f(t)}\tfrac{{\textrm{d}}^2}{{\textrm{d}}t^2}f(t)-\tfrac{1}{4f(t)^2}\left(\tfrac{{\textrm{d}}}{{\textrm{d}}t}f(t)\right)^2+1\nonumber\\&= \left(\tfrac{{\textrm{d}}}{{\textrm{d}}t}\theta(t)\right)^2\cdot \left(\tfrac{1}{2h(\theta)} \tfrac{{\textrm{d}}^2}{{\textrm{d}}\theta^2}h(\theta(t))- \tfrac{1}{4h(\theta)^2} \left(\tfrac{{\textrm{d}}}{{\textrm{d}}\theta}h(\theta(t))\right)^2+1\right). \end{align} $$

On the other hand, the equivariancy of $\Phi $ implies that $\Phi _*(\tfrac {\partial }{\partial \phi })=\tfrac {\partial }{\partial \phi }$ , so by the isometric property of $\Phi $ and (2.10), we get

(3.36) $$ \begin{align} \sin^2 t+\tfrac{\cos t\sin t}{2f(t)}\tfrac{{\textrm{d}}}{{\textrm{d}}t}f(t) =\sin^2\theta(t)+ \tfrac{\cos \theta(t)\sin \theta(t)}{2h(\theta(t))} \tfrac{{\textrm{d}}}{{\textrm{d}}\theta}h(\theta(t)). \end{align} $$

We view (3.35) and (3.36) as an ODE system for the functions $\theta (t)$ and $h(\theta )$ . We first determine $\theta (t)$ . Rewrite (3.36) as

(3.37) $$ \begin{align} \tfrac{1}{h(\theta(t))}\tfrac{{\textrm{d}}}{{\textrm{d}}\theta}h(\theta(t)) =\left({2\sin^2 t+\tfrac{\cos t\sin t}{f(t)}\tfrac{{\textrm{d}}}{{\textrm{d}}t}f(t)}\right){\csc\theta(t)\sec\theta(t)}-2\tan\theta(t),\nonumber\\ \end{align} $$

and differentiate (3.37) with respect to t, then we get

(3.38) $$ \begin{align} & \tfrac{{\textrm{d}}}{{\textrm{d}}t}\theta(t)\cdot \left(\tfrac{1}{h(\theta(t))} \tfrac{{\textrm{d}}^2}{{\textrm{d}}\theta^2}h(\theta(t))- \tfrac{1}{h(\theta(t))^2}\left( \tfrac{{\textrm{d}}}{{\textrm{d}}\theta}h(\theta(t)) \right)^2 \right)\nonumber\\ &=\tfrac{{\textrm{d}}}{{\textrm{d}}t}\theta(t)\cdot {\left(2\sin^2 t+\tfrac{\cos t\sin t}{f(t)} \tfrac{{\textrm{d}}}{{\textrm{d}}t}f(t)\right) \left(\sec^2\theta(t)-\csc^2\theta(t)\right)} -2\tfrac{{\textrm{d}}}{{\textrm{d}}t}\theta(t)\cdot\sec^2\theta(t)\nonumber\\ & +\left({4\cos t\sin t+\tfrac{\cos^2t-\sin^2t}{f(t)}\tfrac{{\textrm{d}}}{{\textrm{d}}t}f(t) -\tfrac{\cos t\sin t}{f(t)^2}\left(\tfrac{{\textrm{d}}}{{\textrm{d}}t}f(t)\right)^2 +\tfrac{\cos t\sin t}{f(t)}\tfrac{{\textrm{d}}^2}{{\textrm{d}}t^2}f(t) }\right)\nonumber\\ & \cdot{\csc\theta(t)\sec\theta(t)}.\nonumber\\ \end{align} $$

We plug (3.37) and (3.38) into the right side of (3.35) to erase $h(\theta (t))$ and its derivatives, then we get a formal quadratic equation for $\tfrac {{\textrm {d}}}{{\textrm {d}}t}\theta (t)$ ,

(3.39) $$ \begin{align} A\left(\tfrac{{\textrm{d}}}{{\textrm{d}}t}\theta(t)\right)^2+B\left( \tfrac{{\textrm{d}}}{{\textrm{d}}t}\theta(t)\right)+C=0, \end{align} $$

in which

$$ \begin{align*} A&=\frac{\cos t\sin t\left(\cos t \tfrac{{\textrm{d}}}{{\textrm{d}}t}f(t)+2\sin t f(t)\right)\left(\sin t \tfrac{{\textrm{d}}}{{\textrm{d}}t}f(t)-2\cos t f(t)\right)}{2f(t)^2\cos^2\theta(t)\sin^2\theta(t)},\\ B&=\frac{\tfrac{\cos t\sin t}{f(t)}\tfrac{{\textrm{d}}^2}{{\textrm{d}}t^2}f(t)-\tfrac{\cos t\sin t}{f(t)^2}\left(\tfrac{{\textrm{d}}}{{\textrm{d}}t}f(t)\right)^2+\tfrac{\cos^2t-\sin^2t}{f(t)}\tfrac{{\textrm{d}}}{{\textrm{d}}t}f(t) +4\cos t\sin t }{\cos\theta(t)\sin\theta(t)},\\ C&=-\tfrac{1}{f(t)}\tfrac{{\textrm{d}}^2}{{\textrm{d}}t^2}f(t) +\tfrac{1}{2f(t)^2}\left(\tfrac{{\textrm{d}}}{{\textrm{d}}t}f(t)\right)^2-2. \end{align*} $$

By (3.36), $\theta _0=\theta (t_0)\in (0,\pi )$ equals $\pi /2$ iff $t_0=\pi /2$ or $t'$ , which has been excluded. So the denominators in above calculation do not vanish. Meanwhile, we see the coefficient A in (3.39) does not vanish for each value of t (when it is sufficiently close to $t_0$ ).

Direct calculation shows that for each t, the two solutions of (3.39) are

(3.40) $$ \begin{align} \frac{\cos\theta(t)\sin\theta(t)}{\cos t\sin t}\quad\mbox{and}\quad \tfrac{\left(-2f(t)\tfrac{{\textrm{d}}^2}{{\textrm{d}}t^2}f(t)+\left(\tfrac{{\textrm{d}}}{{\textrm{d}}t}f(t)\right)^2-4f(t)^2\right)\cos\theta(t)\sin\theta(t)}{ \left(\cos t\tfrac{{\textrm{d}}}{{\textrm{d}}t}f(t)+2\sin t f(t)\right)\left(\sin t\tfrac{{\textrm{d}}}{{\textrm{d}}t}f(t)-2\cos t f(t)\right)}. \end{align} $$

The discriminant of (3.39) is

(3.41) $$ \begin{align} B^2-4AC=\left(\frac{\cos t\sin t \tfrac{{\textrm{d}}^2}{{\textrm{d}}t^2}f(t)+(\sin^2 t-\cos^2t)\tfrac{{\textrm{d}}^2}{{\textrm{d}}t}f(t)}{\cos\theta(t)\sin\theta(t)}\right)^2. \end{align} $$

By the inequality (3.31), the discriminant is strictly positive when $t=t_0$ . By continuity, we have immediately the following lemma.

Now, we are ready to prove the following description for $\Phi $ .

Proof We first prove Lemma 3.8 with the assumption that the $\theta $ -coordinate $t_0$ of y satisfies $t_0\in (0,\pi )\backslash \{ \pi /2,t'\}$ .

In each case of Lemma 3.7, the local Hessian isometry $\Phi $ can be determined around y for any given pair of $\theta _0=\theta (t_0)\neq \pi /2$ and $h_0=h(\theta _0)>0$ . For example, in the case (1), we can use the ODE (3.42) and its initial value condition $\theta (t_0)=\theta _0$ to uniquely determine the function $\theta (t)$ , and then use the ODE (3.36) and its initial value condition $h(\theta _0)=h_0$ to uniquely determine $h(\theta )$ . Then, $\Phi $ is determined by (3.32) around y. Meanwhile, we see the ODE (3.42) coincides with (3.24), i.e., it is satisfied by the linear examples in Example 3.5. With the parameter pair $(a,b)$ suitably chosen, both initial value conditions can be met. So, in this case, $\Phi $ is a linear isometry in some $SO(2)$ -invariant conic neighborhood of y. In the case (2), the ODE (3.43) coincides with (3.27), i.e., it is satisfied by the Legendre examples in Example 3.6. We can suitably choose the parameter pair $(a,b)$ to meet both the initial value conditions. So, in this case, $\Phi $ coincides with a Legendre example in some $SO(2)$ -invariant conic neighborhood of y.

Let us now prove Lemma 3.8 when $t_0=\pi /2$ or $t'$ .

By (3.18), for $t\neq t_0$ sufficiently close to $t_0$ , we have $t\neq \pi /2,t'$ and

$$ \begin{align*}(\cos^2t-\sin^2t)\tfrac{{\textrm{d}}}{{\textrm{d}}t}f(t)-\cos t\sin t \tfrac{{\textrm{d}}^2}{{\textrm{d}}t^2}f(t)\neq0.\end{align*} $$

Previous arguments indicate $\Phi $ is either a linear example or a Legendre example, when restricted to each side $t<t_0$ and $t>t_0$ , respectively. When the restrictions of $\Phi $ to both sides are of the same type, by the smoothness of $\Phi $ , the parameter pairs $(a,b)$ for both sides must coincide. The proof ends immediately in this case.

Finally, we prove that it cannot happen that the restrictions of $\Phi $ to the two sides of $t_0$ have different types. Assume conversely that it happens. For example, when $t<t_0$ (or $t>t_0$ ), $\Phi $ is the linear example with the parameter pair $(a_1,b_1)$ , and when $t>t_0$ (or $t<t_0$ , respectively), $\Phi $ is the Legendre example with the parameter pair $(a_2,b_2)$ . In addition to $b_1>0$ and $b_2>0$ , we also have $a_1^{-1}a_2>0$ , because $a_1$ and $a_2$ have the same sign as $\tfrac {{\textrm {d}}}{{\textrm {d}}t}\theta (t_0)$ . Using the linear example to calculate the fundamental tensor $(b_{ij})=(g_2(\tfrac {\partial }{\partial x_i},\tfrac {\partial }{\partial x_j}))$ at $\Phi (y)$ , we get

(3.44) $$ \begin{align} \left( \begin{array}{cc} b_{11} & b_{12} \\ b_{21} & b_{22} \\ \end{array} \right)= \left( \begin{array}{cc} a_1^{-1} & 0 \\ 0 & b_1^{-1} \\ \end{array} \right)\left( \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{array} \right) \left( \begin{array}{cc} a_1^{-1} & 0 \\ 0 & b_1^{-1} \\ \end{array} \right), \end{align} $$

in which $(a_{ij})=(g_1(\tfrac {\partial }{\partial x_i},\tfrac {\partial }{\partial x_j}))$ is the fundamental tensor of $F_1$ at y. Using the Legendre example to calculate $(b_{ij})$ at $\Phi (y)$ , we get

(3.45) $$ \begin{align} \left( \begin{array}{cc} b_{11} & b_{12} \\ b_{21} & b_{22} \\ \end{array} \right)= \left( \begin{array}{cc} a_2^{-1} & 0 \\ 0 & b_2^{-1} \\ \end{array} \right)\left( \begin{array}{cc} a^{11} & a^{12} \\ a^{21} & a^{22} \\ \end{array} \right) \left( \begin{array}{cc} a_2^{-1} & 0 \\ 0 & b_2^{-1} \\ \end{array} \right), \end{align} $$

where $(a^{ij})_{1\leq i,j\leq 3}$ is the inverse matrix of $(a_{ij})_{1\leq i,j\leq 3}$ . Notice that $(a_{ij})_{1\leq i,j\leq 3}$ is blocked-diagonal by (3.29), so

(3.46) $$ \begin{align} \left( \begin{array}{cc} a^{11} & a^{12} \\ a^{21} & a^{11} \\ \end{array} \right)= \left( \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{array} \right)^{-1}. \end{align} $$

Summarizing (3.44)–(3.46), we get

$$ \begin{align*} & \left( \begin{array}{cc} a_1^{-2}a_2^2a_{11}^2+a_1^{-1}a_2b_1^{-1}b_2 a_{12}a_{21} & a_1^{-2}a_2^2a_{11}a_{12}+ a_1^{-1}a_2b_1^{-1}b_2a_{12}a_{22} \\ a_1^{-1}a_2b_{1}^{-1}b_2a_{11}a_{21}+b_1^{-2}b_2^2 a_{21}a_{22}& a_1^{-1}a_2b_1^{-1}b_2a_{12}a_{21}+b_1^2b_2^2 a_{22}^2 \\ \end{array} \right)\\ &=\left[\left( \begin{array}{cc} a_1^{-1}a_2 & 0 \\ 0 & b_1^{-1}b_2 \\ \end{array} \right) \left( \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{array} \right) \right]^2=\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right), \end{align*} $$

from which we see

$$ \begin{align*} a_1^{-2}a_2a_{11}a_{12}+a_1^{-1}a_2b_1^{-1}b_2a_{12}a_{22} =a_1^{-1}a_2a_{12}(a_1^{-1}a_2a_{11}+b_1^{-1}b_2a_{22})=0. \end{align*} $$

Because $a_1^{-1}a_2>0$ , $b_1>0$ , $b_2>0$ , $a_{11}>0$ , and $a_{22}>0$ , we get $a_{12}=g_1(\tfrac {\partial }{\partial x_1},\tfrac {\partial }{\partial x_2})=0$ at y. This is a contradiction to (3.30).▪