Self-similar curve shortening flow in hyperbolic 2-space

We are indebted to K Tenenblat for comments on the preprint version of this paper, which helped us to improve our presentation (especially, to correct an error in Proposition 2.3), and for bringing reference [10] to our attention. EW is grateful to the organizers and audience of the Clark University Geometric Analysis seminar of 13 Nov 2020, where these results were presented, for their interest and insightful comments. We thank to Michael Yi Li and Yingfei Yi for organizing the 2019 International Undergraduate Summer Enrichment Programme (IUSEP) at the University of Alberta, during which this work was begun. EW is supported by NSERC Discovery Grant RGPIN–2017–04896.

We review some basic facts of the hyperboloidal model, which represents hyperbolic 2-space ${\mathbb{H}}^{2}$ as the $z\geq 1$ sheet of the hyperboloid $x^{2}+y^{2}-z^{2}=-1$ in Minkowski $3$-space ${\mathbb{M}}^{3}:=\left({\mathbb{R}}^{3},\eta\right)$ where $\eta$ is the quadratic form $\operatorname{diag}(1,1,-1)$. We will adopt the convention that our coordinates are enumerated as $x^{i}=(x^{1},x^{2},x^{3})=(x,y,z)$ with $\eta(\partial_{x},\partial_{x})=\eta(\partial_{y},\partial_{y})=+1$ so that $\partial_{x}$ and $\partial_{y}$ are spacelike and $\eta(\partial_{z},\partial_{z})=-1$ so that $\partial_{z}$ is timelike in the terminology common in physics.

The hyperboloid modelling ${\mathbb{H}}^{2}$ is a spacelike hypersurface in ${\mathbb{M}}^{3}$; i.e., any vector tangent to this surface is spacelike. However, any vector $X$ from the origin to a point on ${\mathbb{H}}^{2}$ is timelike and future-directed ($\eta(X,\partial_{z})<0$), and has $\eta(X,X)=-1$. In fact, any such vector is a future-directed timelike unit normal field for the surface.

The group $G$ that preserves the quadratic form $\eta$ is the orthogonal group $\operatorname{O}(2,1)$. The proper orthochronous subgroup $G$ that preserves spatial orientation and time orientation will preserve the hyperboloid sheet $z=\sqrt{1+x^{2}+y^{2}}$ and the orientation of bases for its tangent spaces (as well as preserving the choice of future and past). This subgroup lies in the connected component of the identity in the isometry group of ${\mathbb{H}}^{2}$. The one-parameter subgroups of $G$ can be classified as compositions of boosts in the $x$-direction

boosts in the $y$-direction

and rotations about the $z$-axis

The corresponding Lie algebra is spanned by

The boosts in the $x$-direction preserve $\partial_{y}$ and each of the two null planes in which it lies. Likewise, the boosts in the $y$-direction preserve $\partial_{x}$ and the two null planes in which it lies. Using this, one can see that a general $G$-transformation mapping any orthonormal basis of ${\mathbb{M}}^{3}$ to any other (preserving orientation and time orientation) can be constructed from a product $A_{1}(\zeta)A_{2}(\xi)A_{3}(\theta)$, whose parameters $\zeta$, $\xi$, and $\theta$ play the role of “Euler angles”. We will sketch the argument. Consider two $\eta$-orthonormal basis (ONB) sets $\{e_{i}\}$ and $\{{\tilde{e}}_{i}\}$, where $\{e_{i}\}$ is the coordinate basis defined by the $x^{i}$ coordinates, with $e_{3}=\partial_{z}$ future-timelike and $\{e_{1},e_{2}\}$ right-handed (from here on, we simply say an oriented basis). Likewise, ${\tilde{e}}_{3}$ will be future-timelike as well. The span of $\{{\tilde{e}}_{1},{\tilde{e}}_{3}\}$ is a timelike plane $\Pi$. It’s a simple matter to find a (normalized spacelike) vector, call it $e_{1}^{\prime}$, that lies in $\Pi$ and is orthogonal to $e_{3}$. It’s also a simple exercise in linear algebra to find a rotation $A_{3}(\theta)$ about $e_{3}$ such that $A_{3}(\theta)e_{1}=e_{1}^{\prime}$. This obviously leaves $e_{3}$ invariant, but maps $e_{2}$ to $e_{2}^{\prime}=A_{3}(\theta)e_{2}$. Now apply a boost in the plane spanned by $e_{2}^{\prime}$ and $e_{3}$. Such a boost $A_{2}(\xi)$ will leave $e_{1}^{\prime}$ invariant, but we can choose $\xi$ such that $e_{3}^{\prime}:=A_{2}(\xi)e_{3}$, which remains timelike under a boost, lies in $\Pi$. This boost, incidentally, acts on $e_{2}^{\prime}$ to produce $e_{2}^{\prime\prime}:=A_{2}(\xi)e_{2}^{\prime}$. The plane $\Pi$ is now a coordinate plane for the ONB $\{e_{1}^{\prime},e_{2}^{\prime\prime},e_{3}^{\prime}\}$, with $\Pi=\operatorname{Span}\{e_{1}^{\prime},e_{3}^{\prime}\}=\operatorname{Span}\{{\tilde{e}}_{1},{\tilde{e}}_{3}\}$, and $e_{2}^{\prime\prime}$ is normal (i.e., $\eta$-normal) to this plane. A final boost $A_{1}(\zeta)$ in the plane $\Pi$ preserves $e_{2}^{\prime\prime}$ and can be chosen such that ${\tilde{e}}_{1}=e_{1}^{\prime\prime}:=A_{1}(\zeta)e_{1}^{\prime}$. Since $e_{3}^{\prime\prime}:=A_{1}(\zeta)e_{3}^{\prime}$ must be orthogonal to $e_{1}^{\prime\prime}$, it follows that $\{e_{1}^{\prime\prime},e_{2}^{\prime\prime},e_{3}^{\prime\prime}\}=\{{\tilde{e}}_{1},{\tilde{e}}_{2},{\tilde{e}}_{3}\}$.

Then a curve of isometries may be written as

If this curve passes through the identity isometry at $t=0$, we may write its tangent there as

For any curve of isometries containing the identity at $t=0$, differentiation at $t=t_{0}$ can be accomplished using that $A(t_{0}+\epsilon)=A(t_{0}+\epsilon)A^{-1}(t_{0})A(t_{0})=:B(\epsilon)A(t_{0})$ where $B(\epsilon):=A(t_{0}+\epsilon)A^{-1}(t_{0})$ so $B(0)=\operatorname{id}$. But $B(\epsilon)$ can be written as a product $B(\epsilon)=A_{1}(\zeta(\epsilon))A_{2}(\xi(\epsilon))A_{3}(\theta(\epsilon))$, and so

with ${\dot{B}}$ given by the right-hand side of (2.6).

Given a hypersurface $S$ in a Riemannian manifold and vectors $U$ and $V$ tangent to it, a standard formula in Riemannian geometry gives

where $\nabla$ is the connection on the ambient manifold, $D$ is the connection induced on the hypersurface, and $K$ is the vector-valued second fundamental form of the hypersurface. In the case of present interest, $\nabla$ is the Levi-Civita connection of the Minkowski metric $\eta=\langle\cdot,\cdot\rangle_{\eta}$. Since the vector $X$ from the origin to the unit hyperboloid is a unit future-timelike vector orthogonal to the hyperboloid, we may write a general vector $V$ as

where $P$ is orthogonal projection to the tangent space of the hyperboloid. The negative sign in the second term arises because $X$ is timelike. The second fundamental form can then be written as

where $X_{\flat}:=\langle\cdot,X\rangle_{\eta}$ is the 1-form metric-dual to $X$. If we write the first fundamental form of the hyperboloid as $h$ then the hyperboloid is umbilic in ${\mathbb{M}}^{3}$ such that $K=Xh$ (so that $\langle X,K(V,V)\rangle_{\eta}=-1$ for any unit vector $V$ tangent to the hyperboloid).

Now consider a (smooth) unit-speed curve $X(s)$ on the unit hyperboloid. The unit tangent vector is $T(s):=\frac{dX}{ds}$. It follows from the above formulas that

Since $T$ is a unit vector, $\nabla_{T}T$ lies in its orthogonal complement, from which we see that $D_{T}T$ does as well. But $D_{T}T$ must be tangent to the hyperboloid and so orthogonal to $X$, so we write

where $N$ is called the principal normal vector to the curve $s\mapsto X(s)$ and $\kappa_{g}$ is the geodesic curvature. The sign choices are such that $\{T,N,X\}$ is an orthonormal oriented basis oriented such that $X$ is future-pointing and the vector $\kappa_{g}N$ points to the concave side of the curve, viewed as a curve in the hyperboloid, whenever $\kappa_{g}$ is not zero. Now denote $\eta(\nabla_{T}T,\nabla_{T}T)=:\epsilon\kappa^{2}$ where $\epsilon=1$ if $\nabla_{T}T$ is spacelike, $0$ if it’s null, and $-1$ if it’s timelike. Then from (2.11) we can relate the curvature $\kappa$ of $X$ as a space curve in $({\mathbb{M}}^{3},\eta)$ to its geodesic curvature $\kappa_{g}$ in the hyperboloid by

Note that $\langle T,\nabla_{T}N\rangle_{\eta}=-\langle N,\nabla_{T}T\rangle_{\eta}=-\langle N,D_{T}T\rangle_{\eta}=-\kappa_{g}$. By similar reasoning, we see that $\langle X,\nabla_{T}N\rangle_{\eta}=0$, and of course since $N$ is a unit vector then $\langle N,\nabla_{T}N\rangle_{\eta}=0$. Thus $\nabla_{T}N=-\kappa_{g}T$. Collecting our results, we have that the $\{T,N,X\}$ basis evolves along the curve according to the Frenet-Serret equations in ${\mathbb{H}}^{2}$, which are

Here we follow closely the work of [12] for the ${\mathbb{S}}^{2}$ case, making changes as necessary. Choose an arbitrary vector ${\tilde{v}}\in{\mathbb{M}}^{3}\backslash\{0\}$, which can be either timelike, spacelike, or null. It is convenient to define ${\tilde{v}}=av$ where $a=\sqrt{|\eta({\tilde{v}},{\tilde{v}})|}$ when $v$ is not null. For presentation purposes, we will introduce the scale factor even when ${\tilde{v}}$ is null, but in that case $a\neq 0$ will not be determined and $v$, being null, cannot be normailzed. Then $\eta({\tilde{v}},{\tilde{v}})=\epsilon=0,\pm 1$ depending on whether $v$ is null ($0$), spacelike ($+1$), or timelike ($-1$).

Fix a unit speed curve $X:I\to{\mathbb{M}}^{3}$ on the unit hyperboloid, and define the functions

Then we can write

and

The next result shows that if the geodesic curvature of the curve $X(s)$ takes the form $\kappa_{g}(s)=a\tau(s)$, a choice corresponding to a flow by isometries under CSF as we will see in the next subsection, then $\tau$, $\nu$, and $\mu$ obey a certain autonomous system of differential equations along $X(s)$. The analogous result for curves in ${\mathbb{S}}^{2}$ is found in [12, Proposition 3.1].

Next we show that solutions of the system (2.18) are always realized by actual curves.

Having defined a moving basis along an arbitrary curve in the hyperboloid, we can now define the curve shortening flow in the hyperboloid to be a flow $X_{t}(s)=X(t,s)$ of smooth curves $X_{t}(s)=\left(x_{t}(s),y_{t}(s),z_{t}(s)\right)$ with principal normal $N_{t}$, for $t\in J\subset{\mathbb{R}}$ some connected interval about $t=0$, such that

The last two equations are equivalent to the conditions $\left\langle X_{t},X_{t}\right\rangle_{\eta}=-1$ and $X_{0}=\left(x_{0},y_{0},z_{0}\right)$.

We are interested in those solutions of (2.20) that are of the form

where $A(t):=A_{1}(\zeta(t))A_{2}(\xi(t))A_{3}(\theta(t))$ with $\zeta(0)=\xi(0)=\theta(0)=0$, with the $A_{i}$ defined in equations (2.1)–(2.3). Then $t\mapsto X_{t}$ is called a flow by isometries. Motion by an isometry preserves the curvature, so $\kappa_{g}(s,t)=\kappa)g(s)$ (i.e., it is independent of $t$).

As we saw in Proposition 2.1, if there is a (normalized or null) vector $v$ such that $a\langle T,v\rangle_{\eta}=\kappa_{g}$, then a system of three scalar equations governs the moving frame. The next result shows that if a curve shortening flow is also a flow by isometries, there is always such a $v$ which is constant with respect to the flow parameter $t$. For convenience, we work with the unrescaled vector ${\tilde{v}}=av$ ($a(s)$ is used for another purpose in the proof). We follow the proof of the analogous result for curves in ${\mathbb{S}}^{2}$ [12, Theorem 2.2].

These are solutions of (2.18) with $\tau\equiv 0$, so $\mu\equiv 0$ as well, and $\nu$ is constant. They are fixed points of (3.1). Since $\mu=0$, geodesics are plane curves lying in the plane orthogonal to $v$.

We may view the system (2.18) as an autonomous third-order system

where $\Phi=(\tau,\nu,\mu)$, $a>0$ is a constant, and $F_{a}:{\mathbb{R}}^{3}\to{\mathbb{R}}^{3}:(\tau,\nu,\mu)\mapsto(a\tau\nu+\mu,-a\tau^{2},\tau)$. We now show that at fixed points, the differential of $F_{a}$ has real eigenvalues of all three signs (positive, negative, and zero), so geodesics are unstable fixed points but attract in one direction.

Specifically, when $a=1$ and $(\tau,\nu,\mu)=(0,1,0)$ (respectively, $(\tau,\nu,\mu)=(0,1,0)$), then $\lambda=0,\frac{1+\sqrt{5}}{2},\frac{1-\sqrt{5}}{2}$ (respectively, $\lambda=0,\frac{-1+\sqrt{5}}{2},\frac{-1-\sqrt{5}}{2}$).