Shape Analysis of Surfaces Using General Elastic Metrics

Abstract

In this article, we introduce a family of elastic metrics on the space of parametrized surfaces in 3D space using a corresponding family of metrics on the space of vector-valued one-forms. We provide a numerical framework for the computation of geodesics with respect to these metrics. The family of metrics is invariant under rigid motions and reparametrizations; hence, it induces a metric on the “shape space” of surfaces. This new class of metrics generalizes a previously studied family of elastic metrics and includes in particular the Square Root Normal Field (SRNF) metric, which has been proven successful in various applications. We demonstrate our framework by showing several examples of geodesics and compare our results with earlier results obtained from the SRNF framework.

Appendices

Appendix A: The Geodesic Equation

In the following, we give the geodesic equation on the space of immersions ${\text {Imm}}(M, {\mathbb {R}}^3)$ with respect to the pullback of the metric (3) on the space of 1-forms. In this “Appendix,” we will assume that the domain M is a compact orientable surface without boundary, because we will need to use the Hodge decomposition. We will view $\alpha \in \varOmega ^1_+(M,{\mathbb {R}}^3)$ as a vector-valued 1-form with components $( \alpha ^1, \alpha ^2,\alpha ^3)$, where each $\alpha ^i$ is a 1-form on M in the usual sense. Then, metric (3) can be rewritten as

$$\begin{aligned} G_{\alpha }(\xi , \xi )&= \int _M {{\,\mathrm{tr}\,}}(\xi \varLambda _{\alpha }\xi ^T) \, \varphi _{\alpha } \, \mu \\&= \sum _{i=1}^3 \int _M \langle \xi _x^i, \varLambda _{\alpha } \xi _x^i\rangle \, \varphi _{\alpha } \, \mu \end{aligned}$$

where $\xi = (\xi ^1, \xi ^2,\xi ^3)\in T_{\alpha }\varOmega ^1_+(M,{\mathbb {R}}^3)$, $\varLambda _{\alpha } = (\alpha ^T\alpha )^{-1}$ is the induced Riemannian metric on 1-forms on M, and $\varphi _{\alpha } = \sqrt{\det {(\alpha ^T\alpha )}}$ is the induced volume form on M. As such, all computations can be done one component at a time.

If $F = (f^1, f^2, f^3)$ is a vector-valued function with each $f^i:M\rightarrow {\mathbb {R}}$ real-valued, then $\beta = dF$ is a vector-valued 1-form with $\beta ^i = df^i$. The Hodge decomposition tells us that every 1-form $\xi$ may be written as

$$\begin{aligned} \xi = df + \gamma , \end{aligned}$$

where $\delta \gamma = 0$ and $\delta :\varOmega ^1(M, {\mathbb {R}})\rightarrow C^{\infty }(M)$ is the codifferential operator.

The space ${\text {Imm}}(M,{\mathbb {R}}^3)$ is formally a submanifold of $\varOmega ^1_+(M,{\mathbb {R}}^3)$, and thus by general submanifold geometry, we know that the geodesic equation on ${\text {Imm}}(M,{\mathbb {R}}^3)$ will be given by

$$\begin{aligned} \frac{D}{dt} \frac{d}{dt} \alpha = \gamma , \qquad \alpha = d\varPhi , \qquad \delta \gamma = 0. \end{aligned}$$

Since $\delta \gamma = 0$, we know that $\star \gamma$ is an exact form, where $\star$ denotes the Hodge star operator. Then, there is a function p, unique up to a constant, such that $dp = \star \gamma$. We obtain

$$\begin{aligned} \varDelta p = \delta dp = \delta \!\star \!\gamma = \star d\left( \frac{D}{dt} \frac{d}{dt} \alpha \right) . \end{aligned}$$

In coordinates (u, v) on M, the operator $\star d$ is given by:

$$\begin{aligned} \star d (f \, du + g \, dv) = \frac{g_u - f_v}{\varphi }, \end{aligned}$$

where $\varphi$ is the volume form on M. From the geodesic equation on $\varOmega ^1_+(M,{\mathbb {R}}^3)$ with respect to the metric (3), as calculated in our previous paper [7], we know that the covariant derivative is given by

$$\begin{aligned}&\frac{D}{dt} \frac{d\alpha }{dt} = \alpha _{tt} - \alpha _t(\alpha ^T\alpha )^{-1} \alpha ^T_t \alpha - \alpha _t \alpha ^+\alpha _t + (\alpha _t\alpha ^+)^T\alpha _t \\&\quad - \tfrac{1}{2} {{\,\mathrm{tr}\,}}(\alpha _t(\alpha ^T\alpha )^{-1}\alpha ^T_t)\alpha + {{\,\mathrm{tr}\,}}(\alpha _t\alpha ^+) \alpha _t. \end{aligned}$$

Since $d\alpha _{tt} = 0$, we obtain

$$\begin{aligned} \varDelta p= & {} \star d\Big (-\alpha _t(\alpha ^T\alpha )^{-1} \alpha ^T_t \alpha \Big ) - \alpha _t \alpha ^+\alpha _t + (\alpha _t\alpha ^+)^T\alpha _t \\&- \tfrac{1}{2} {{\,\mathrm{tr}\,}}(\alpha _t(\alpha ^T\alpha )^{-1}\alpha ^T_t)\alpha + {{\,\mathrm{tr}\,}}(\alpha _t\alpha ^+) \alpha _t\Big ). \end{aligned}$$

Let $L = \alpha _t \alpha ^+$. Applying the Hodge star operator to both sides, we see that $\varPhi$ is a geodesic on ${\text {Imm}}(M, {\mathbb {R}}^3)$ if and only if we have

$$\begin{aligned} \star \varDelta p = \star \star d(\varOmega d\varPhi ) = d\varOmega \wedge d\varPhi \end{aligned}$$

where $\varOmega = LL^T + L^2 - L^T L +\frac{1}{2}{{\,\mathrm{tr}\,}}(L^TL) - {{\,\mathrm{tr}\,}}(L)L$. Here, we emphasize that p and $\varPhi$ are actually vector-valued functions, so these computations are done componentwise for each $i\in \{1,2,3\}$. In other words, we have

$$\begin{aligned} \star \varDelta p_i = \sum _{j=1}^3 d\varOmega _{ij} \wedge d\varPhi _j, \qquad i\in \{1,2,3\}. \end{aligned}$$

Appendix B: Proofs

Proof of Theorem 2

In the following, we prove the correspondence between our split metric on the space $\varOmega _+^1(M,{\mathbb {R}}^3)$ and the SRNF metric on the space of surfaces. Let $M_+(3,2)$ be the space of $3\times 2$ matrices with rank 2. Using the pointwise property of our metric, we will focus on the corresponding split metric on the matrix space $M_+(3,2)$. For $a\in M_+(3,2)$ and $v\in T_mM_+(3,2)$, we decompose v into four parts

$$\begin{aligned} v = v_m + \frac{1}{2}{{\,\mathrm{tr}\,}}(a^+v)a + v^{\perp } + v_0, \end{aligned}$$

where

$$\begin{aligned} v_m&= \frac{1}{2}a(a^Ta)^{-1}(a^Tv + v^Ta) - \frac{1}{2}{{\,\mathrm{tr}\,}}(a^+v)a \\ v^{\perp }&= v - a(a^Ta)^{-1}a^Tv \\ v_0&= \frac{1}{2}a(a^Ta)^{-1}(a^Tv - v^Ta). \end{aligned}$$

The corresponding split metric on $M_+(3,2)$ is then of the form:

$$\begin{aligned}&G^{{\mathfrak {a}},{\mathfrak {b}},{\mathfrak {c}},{\mathfrak {d}}}_{a}(v, v) \nonumber \\&\quad = {\mathfrak {a}}\langle v_m, v_m\rangle _a + {\mathfrak {b}}\left\langle \frac{1}{2}{{\,\mathrm{tr}\,}}(a^+v)a, \frac{1}{2}{{\,\mathrm{tr}\,}}(a^+v)a\right\rangle _a \nonumber \\&\qquad + {\mathfrak {c}}\langle v^{\perp }, v^{\perp }\rangle _a + {\mathfrak {d}}\langle v_0, v_0\rangle _a, \end{aligned}$$

(13)

Now consider the projection $\pi :M_+(3,2)\rightarrow {\text {Sym}}_+(2), \ a\mapsto a^Ta$. This projection is a Riemannian submersion, where $M_+(3,2)$ carries metric (13) with choices of constants (1, 1, 1, 1) and the space ${\text {Sym}}_+(2)$ is equipped with the following metric:

$$\begin{aligned} \langle h, k\rangle _g^{{\text {Sym}}} = \frac{1}{4}{{\,\mathrm{tr}\,}}(g^{-1}hg^{-1}k)\sqrt{\det (g)}. \end{aligned}$$

The horizontal bundle with respect to the projection $\pi$ is given by

$$\begin{aligned} {\mathcal {H}}_a = \{u\in M(3,2)\,|\,ua^+\in {\text {Sym}}(n)\}, \end{aligned}$$

and the differential $d\pi$ induces an isometry

$$\begin{aligned} d\pi _a: {\mathcal {H}}_a \rightarrow T_{\pi (a)}{\text {Sym}}_+(m). \end{aligned}$$

It is easy to check that $v_m$ and $\frac{1}{2}{{\,\mathrm{tr}\,}}(a^+v)a$ are horizontal vectors.

Let $g = \pi (a) =a^Ta$. By computation, we have

$$\begin{aligned} {{\,\mathrm{tr}\,}}(a^+v) = \frac{1}{2}{{\,\mathrm{tr}\,}}(g^{-1}d\pi _av) \end{aligned}$$

and

$$\begin{aligned} d\pi _a(v_m)&= a^Tv_m + v_m^Ta = a^Tv + v^Ta - {{\,\mathrm{tr}\,}}(a^+v)a^Ta\\&= d\pi _av - \frac{1}{2}{{\,\mathrm{tr}\,}}\left( g^{-1}d\pi _a v\right) g. \end{aligned}$$

Therefore, the first term in (13) becomes

$$\begin{aligned}&\langle v_m, v_m\rangle _a\\&\quad = \left\langle d\pi _a(v_m), d\pi _a(v_m)\right\rangle _{\pi (a)}^{{\text {Sym}}}\\&\quad =\left\langle d\pi _av - \frac{1}{2}{{\,\mathrm{tr}\,}}\left( g^{-1}d\pi _a v\right) g, d\pi _av - \frac{1}{2}{{\,\mathrm{tr}\,}}\left( g^{-1}d\pi _a v\right) g\right\rangle _g^{{\text {Sym}}} \\&\quad = \left\langle d\pi _av, d\pi _av\right\rangle _g^{{\text {Sym}}} - {{\,\mathrm{tr}\,}}\left( g^{-1}d\pi _a v\right) \left\langle d\pi _av, g\right\rangle _g^{{\text {Sym}}}\\&\qquad + \frac{1}{4}{{\,\mathrm{tr}\,}}^2\left( g^{-1}d\pi _a v\right) \left\langle g, g\right\rangle _g^{{\text {Sym}}} \\&\quad = \frac{1}{4}{{\,\mathrm{tr}\,}}\left( g^{-1}d\pi _avg^{-1}d\pi _av\right) \sqrt{\det (g)}\\&\qquad -\frac{1}{8}{{\,\mathrm{tr}\,}}^2\left( g^{-1}d\pi _a v\right) \sqrt{\det (g)} \end{aligned}$$

and the second term becomes

$$\begin{aligned} \left\langle \frac{1}{2}{{\,\mathrm{tr}\,}}(a^+v)a, \frac{1}{2}{{\,\mathrm{tr}\,}}(a^+v)a\right\rangle _a&= \frac{1}{2}{{\,\mathrm{tr}\,}}^2(a^+v) \sqrt{\det (a^Ta)} \\&= \frac{1}{8}{{\,\mathrm{tr}\,}}^2(g^{-1}d\pi _av)\sqrt{\det (g)}. \end{aligned}$$

For the third term in (13), we consider the corresponding unit normal map on the space of matrices given by

$$\begin{aligned} n: M_+(3,2)&\rightarrow {\mathbb {R}}^3\\ a&\mapsto \frac{a_1\times a_2}{|a_1\times a_2|} = \frac{a_1\times a_2}{\sqrt{\det (a^Ta)}}, \end{aligned}$$

where $a_1$ and $a_2$ are the first and the second columns of a, respectively. For any tangent vector $u = \begin{pmatrix} u_1&u_2 \end{pmatrix}$ at a, the differential of n at a is

$$\begin{aligned} dn_a(u) = \frac{u_1\times a_2+ a_1\times u_2 - (a_1\times a_2){{\,\mathrm{tr}\,}}(a^+u)}{\sqrt{\det (a^Ta)}}. \end{aligned}$$

It is easy to check that $aa^+v$ is in the kernel of the differential $dn_a$, i.e.,

$$\begin{aligned} dn_a(v) = dn_a(v^{\perp } + aa^+v) = dn_a(v^{\perp }). \end{aligned}$$

Note that ${{\,\mathrm{tr}\,}}(a^+v^{\perp }) = 0$, $aa^+a_1 = a_1$ and $aa^+a_2 = a_2$. Using the following identity for three-dimensional vectors b, c, d, e:

$$\begin{aligned} (b\times c)\cdot (d\times e) = b^Tdc^Te - b^Tec^Td \end{aligned}$$

and the formula for the inverse of $a^Ta$:

$$\begin{aligned} (a^Ta)^{-1} = \frac{1}{\det (a^Ta)}\begin{pmatrix} a_2^Ta_2 &{} -a_1^Ta_2\\ a_1^Ta_2 &{} a_1^Ta_1 \end{pmatrix}, \end{aligned}$$

we have

$$\begin{aligned}&\langle dn_av, dn_av\rangle _{{\mathbb {R}}^3} = \langle dn_av^{\perp }, dn_av^{\perp }\rangle _{{\mathbb {R}}^3}\\&\quad = \frac{1}{\det (a^Ta)}\langle v^{\perp }_1\times a_2+ a_1\times v^{\perp }_2, v^{\perp }_1\times a_2+ a_1\times v^{\perp }_2\rangle _{{\mathbb {R}}^3}\\&\quad = \frac{1}{\det (a^Ta)}\big [\left( v_1^Tv_1 - v_1^Taa^+v_1\right) a_2^Ta_2\\&\qquad - 2\left( v_1^Tv_2 - v_1^Taa^+v_2\right) a_1^Ta_2 + \left( v_2^Tv_2 - v_2^Taa^+v_2\right) a_1^Ta_1\big ], \end{aligned}$$

where $v^{\perp }_1, v^{\perp }_2$ are the first and the second columns of $v^{\perp }$ and $v_1, v_2$ are the first and the second columns of v, respectively. It follows that

$$\begin{aligned}&\left\langle v^{\perp }, v^{\perp }\right\rangle _a\sqrt{\det (a^Ta)} = {{\,\mathrm{tr}\,}}(v^{\perp }(a^Ta)^{-1}(v^{\perp })^T)\det (a^Ta)\\&\quad = \left( {{\,\mathrm{tr}\,}}(v(a^Ta)^{-1}v^T) - {{\,\mathrm{tr}\,}}(aa^+v(a^Ta)^{-1}v^T)\right) \det (a^Ta)\\&\quad = \left( {{\,\mathrm{tr}\,}}(v^Tv(a^Ta)^{-1}) - {{\,\mathrm{tr}\,}}(v^Taa^+v(a^Ta)^{-1})\right) \det (a^Ta)\\&\quad = {{\,\mathrm{tr}\,}}\left( \begin{pmatrix} v_1^T(I - aa^+)v_1 &{} v_1^T(I - aa^+)v_2\\ v_1^T(I - aa^+)v_2 &{} v_2^T(I - aa^+)v_2 \end{pmatrix}\begin{pmatrix} a_2^Ta_2 &{} -a_1^Ta_2\\ a_1^Ta_2 &{} a_1^Ta_1 \end{pmatrix}\right) \\&\quad = \left( v_1^Tv_1 - v_1^Taa^+v_1\right) a_2^Ta_2 - 2\left( v_1^Tv_2 - v_1^Taa^+v_2\right) a_1^Ta_2 \\&\qquad + \left( v_2^Tv_2 - v_2^Taa^+v_2\right) a_1^Ta_1\\&\quad = \langle dn_av, dn_av\rangle _{{\mathbb {R}}^3}\det (a^Ta), \end{aligned}$$

that is,

$$\begin{aligned} \left\langle v^{\perp }, v^{\perp }\right\rangle _a = \langle dn_av, dn_av\rangle _{{\mathbb {R}}^3}\sqrt{\det (g)}. \end{aligned}$$

Therefore, the split metric (13) on $M_+(3,2)$ can be rewritten as

$$\begin{aligned}&G^{{\mathfrak {a}},{\mathfrak {b}},{\mathfrak {c}},{\mathfrak {d}}}_{a}(v, v)\\&\quad = {\mathfrak {a}}\left( \frac{1}{4}{{\,\mathrm{tr}\,}}\left( g^{-1}d\pi _avg^{-1}d\pi _av\right) -\frac{1}{8}{{\,\mathrm{tr}\,}}^2\left( g^{-1}d\pi _a v\right) \right) \sqrt{\det (g)}\\&\qquad + \frac{{\mathfrak {b}}}{8}{{\,\mathrm{tr}\,}}^2\left( g^{-1}d\pi _a v\right) \sqrt{\det (g)}+ {\mathfrak {c}}\langle dn_av, dn_av\rangle _{{\mathbb {R}}^3}\sqrt{\det (g)}\\&\qquad + {\mathfrak {d}}\langle v_0, v_0\rangle _a. \end{aligned}$$

Now it is easy to see that the first three terms give rise to the formula of the full elastic metric on the space of surfaces for $A = {\mathfrak {a}}/4, B = ({\mathfrak {b}}-{\mathfrak {a}})/8, C = {\mathfrak {c}}$ and the SRNF metric corresponds to the split metric (4) with constants $(0 ,\frac{1}{2}, 1, 0)$. $\square$

Proof of Theorem 3

We first perform the computation in spherical coordinates $(\theta , \phi )\in [0, 2\pi ]\times [0, \pi ]$. Denote the usual spherical coordinate orthonormal basis by

$$\begin{aligned} e_1&= \langle \sin {\phi } \cos {\theta }, \sin {\phi } \sin {\theta }, \cos {\phi }\rangle , \\ e_2&= \langle \cos {\phi } \cos {\theta }, \cos {\phi } \sin {\theta }, -\sin {\phi }\rangle , \\ e_3&= \langle -\sin {\theta }, \cos {\theta }, 0\rangle . \end{aligned}$$

We have the following formulas for the partial derivatives:

$$\begin{aligned} \partial _{\phi } e_1&= e_2, \qquad \partial _{\phi } e_2 = -e_1, \qquad \partial _{\phi } e_3 = 0, \nonumber \\ \partial _{\theta } e_1&= \sin {\phi } e_3, \qquad \partial _{\theta } e_2 = \cos {\phi } e_3, \nonumber \\ \partial _{\theta } e_3&= -\sin {\phi } e_1 - \cos {\phi } e_2. \end{aligned}$$

(14)

We also note that the covariant derivatives are given by

$$\begin{aligned} \nabla _{e_2}e_2&= 0,\quad&\nabla _{e_2}\ e_3&= 0\nonumber \\ \nabla _{e_3}e_2&= \cot {\phi }\, e_3,\quad&\nabla _{e_3}e_3&= -\cot {\phi }\, e_2. \end{aligned}$$

(15)

Write

$$\begin{aligned} U(\theta , \phi ) = u(\theta , \phi ) e_2(\theta , \phi ) + v(\theta , \phi ) e_3(\theta , \phi ). \end{aligned}$$

For a real parameter t, we consider the following map $W:S^2 \rightarrow {\mathbb {R}}^3$ given in coordinates by

$$\begin{aligned} W(\theta , \phi )&= e_1(\theta , \phi ) + tU(\theta , \phi )\\&= e_1(\theta , \phi ) + t u(\theta , \phi ) e_2(\theta , \phi ) + t v(\theta , \phi ) e_3(\theta , \phi ). \end{aligned}$$

Then, $\eta = W/|W|$.

Note that in order for $\eta$ to be a diffeomorphism, we require that the Jacobian determinant be nonzero; it is given by

$$\begin{aligned} {\text {Jac}}(\eta ) = \frac{1}{\sin \phi }\left|\frac{\partial \eta }{\partial \phi }\times \frac{\partial \eta }{\partial \theta }\right|. \end{aligned}$$

Observe that

$$\begin{aligned} \eta _{\phi }&= \frac{1}{|W|} \left( W_{\phi } - \frac{W\cdot W_{\phi }}{|W|^2}\, W\right) = \frac{1}{|W|} P_{W^{\perp }}(W_{\phi }),\\ \eta _{\theta }&= \frac{1}{|W|} \left( W_{\theta } - \frac{W\cdot W_{\theta }}{|W|^2}\, W\right) = \frac{1}{|W|} P_{W^{\perp }}(W_{\theta }). \end{aligned}$$

Since $\eta _{\phi }$ and $\eta _{\theta }$ are both perpendicular to W, we know that $\eta _{\phi }\times \eta _{\theta }$ is parallel to W; thus, we obtain the formula

$$\begin{aligned} {\text {Jac}}(\eta )&= \frac{1}{\sin \phi |W|^2}\big |P_{W^{\perp }}(W_\phi )\times P_{W^{\perp }}(W_\theta )\big |\\&= \frac{1}{\sin \phi |W|^3}\left|W\cdot \big (P_{W^{\perp }}(W_\phi )\times P_{W^{\perp }}(W_\theta )\big )\right|\\&= \frac{1}{\sin \phi |W|^3}\left|W\cdot \big (W_\phi \times W_\theta \big )\right|, \end{aligned}$$

using the cyclic invariance of the scalar triple product and the fact that $W\times P_{W^{\perp }}(V) = W\times V$ for any vector V.

Since $W = e_1+tU$ for the vector field $U = ue_2+ve_3$, it is straightforward to compute using (14)-(15) that

$$\begin{aligned} W_{\phi }&= e_2+tU_{\phi } = e_2 +t\nabla _{e_2}U - tue_1,\\ \frac{W_{\theta }}{\sin \phi }&= e_3+\frac{t}{\sin \phi }U_{\theta } = e_3 +t\nabla _{e_3}U - tve_1, \end{aligned}$$

Let $m_1 = u_{\phi }, m_2 = v_{\phi }, m_3 = \frac{u_{\theta }-v\cos {\phi }}{\sin {\phi }}, m_4= \frac{v_{\theta }+u\cos {\phi }}{\sin {\phi }}$. We have by (15) that

$$\begin{aligned} \nabla _{e_2}U = m_1e_2+m_2e_3,\qquad \nabla _{e_3}U = m_3e_2+m_4e_3, \end{aligned}$$

which we abbreviate by

$$\begin{aligned} M :=\nabla U = \begin{pmatrix} m_1 &{} m_2\\ m_3 &{} m_4 \end{pmatrix}. \end{aligned}$$

Thus, the Jacobian is nonzero if and only if the following determinant is nonzero:

$$\begin{aligned} D = \left|W\cdot \big (W_\phi \times W_\theta \big )\right|= \begin{vmatrix} 1&tu&tv\\ -tu&1+ tm_1&tm_2\\ -tv&tm_3&1+tm_4 \end{vmatrix}. \end{aligned}$$

(16)

Expanding the determinant (16) along the first column, then it is given by

$$\begin{aligned} D = \det (1+tM) + t^2\langle JU, (1+tM)JU\rangle , \end{aligned}$$

where $J = \begin{pmatrix} 0 &{} -1\\ 1 &{} 0 \end{pmatrix}$.

Let ${\overline{M}} = \tfrac{1}{2} (M+M^T)$ denote the symmetrization of M, and let $\lambda _1\le \lambda _2$ denote the real eigenvalues of ${\overline{M}}$. Then, ${{\,\mathrm{tr}\,}}{M} = {{\,\mathrm{tr}\,}}{{\overline{M}}}$ and $\det {M} = \det {{\overline{M}}} + \tfrac{1}{4} (m_2-m_3)^2$, so that

$$\begin{aligned} \det (1+tM) \ge \det (1+t{\overline{M}}) = (1+\lambda _1t)(1+\lambda _2t). \end{aligned}$$

Since J is a rotation, we have

$$\begin{aligned} \langle JU, (1+tM)JU\rangle&= \langle JU, (1+t{\overline{M}})JU\rangle \\&\ge (1+\lambda _1t)|JU|^2\\&= (1+\lambda _1t)|U|^2. \end{aligned}$$

Thus,

$$\begin{aligned} D\ge (1+\lambda _1t)(1+\lambda _2t+|U|^2t^2). \end{aligned}$$

For sufficiently small t, we know $(1+\lambda _1t)$ is positive, and since $\lambda _1\le \lambda _2$, we obtain

$$\begin{aligned} D\ge (1+\lambda _1t)^2 \end{aligned}$$

Thus, $1+\lambda _1t>0$ is a sufficient condition for positivity of D, and this happens as long as $|t|< \frac{1}{|\lambda _1|}$. It is easy to compute that

$$\begin{aligned} \lambda _1 = \frac{m_1+m_4 - \sqrt{(m_1-m_4)^2 + (m_2 + m_3)^2}}{2}. \end{aligned}$$

In particular, $m_1+m_4 = {{\,\mathrm{tr}\,}}{(\nabla U)} = {\text {div}}{U}$, and by the divergence theorem, we know the integral of $m_1+m_4$ over $S^2$ is zero, and in particular $m_1+m_4$ is either identically zero or changes sign on $S^2$. Since t is nonnegative, we therefore are concerned about the most negative that $\lambda _1(x)$ can be:

$$\begin{aligned} 1 + \lambda _1(x) t \ge 1 + t\inf _{x\in S^2} \lambda _1(x) = 1 - t \sup _{p\in S^2}(-\lambda _1(x)) \ge 0, \end{aligned}$$

which is equivalent to

$$\begin{aligned} t < \frac{2}{\sup \limits _{p\in S^2} -(m_1+m_4) + \sqrt{(m_1-m_4)^2 + (m_2+m_3)^2}}. \end{aligned}$$

This is clearly (9). $\square$