Appendix
1.1 A.1 Formulas in \(\mathbb {R}^{3}\)
Let \({\Phi } : \mathbb {D} \rightarrow \mathbb {R}^{3}\) be a smooth conformal immersion. Let \(\vec {n} = \frac {{\Phi }_{z} \times {\Phi }_{\bar {z}} }{ i \left | {\Phi }_{z} \right |^{2} }\) denote its Gauss map (with × the classical vectorial product in \(\mathbb {R}^{3}\)), \(\lambda = \frac {1}{2} \log \left (2 \left | {\Phi }_{z} \right |^{2} \right )\) its conformal factor and \(H = \left \langle \frac { {\Phi }_{z \bar {z}}}{ \left | {\Phi }_{z} \right |^{2}} , \vec {n} \right \rangle \) its mean curvature. Its tracefree curvature is defined as follows
$${\Omega}:= 2\left\langle {\Phi}_{zz}, \vec{n} \right\rangle. $$
Then
$$ \vec{n}_{z} = - H {\Phi}_{z} - {\Omega} e^{-2 \lambda} {\Phi}_{\bar{z}}, $$
(60)
$$ {\Phi}_{z \bar{z}} = H \frac{e^{2\lambda}}{2} \vec{n} , $$
(61)
$$ {\Phi}_{z z} = 2 \lambda_{z} {\Phi}_{z} + \frac{\Omega}{2} \vec{n} $$
(62)
and Gauss-Codazzi can be written
$$ {\Omega}_{\bar{z}} e^{-2\lambda} = H_{z} . $$
(63)
Further if we write , \( A = \langle \nabla ^{2} {\Phi } , \vec {n} \rangle \), the second fundamental form of Φ, such that
$$ e^{-2\lambda} A= \left( \begin{array}{cccc} \epsilon & \varphi \\ \varphi & \gamma \end{array}\right),$$
then
$$ H= \frac{ \epsilon + \gamma }{2},$$
$$ \mathring{A} = \left( \begin{array}{cccc} \frac{\epsilon - \gamma }{2} & \varphi \\ \varphi & \frac{\gamma - \epsilon}{2} \end{array}\right),$$
with Å the tracefree second fundamental form defined in Eq. 9 and
$$ \begin{array}{llll} \nabla \vec{n} &= - H \nabla {\Phi} - \mathring{A} \nabla {\Phi} = \left( \begin{array}{cccc} \epsilon {\Phi}_{x} + \varphi {\Phi}_{y} \\ \varphi {\Phi}_{x} + \gamma {\Phi}_{y} \end{array}\right) \\ &= - H \nabla {\Phi} - \left( \begin{array}{cccc} \frac{\epsilon - \gamma }{2} {\Phi}_{x} + \varphi {\Phi}_{y} \\ \varphi {\Phi}_{x} - \frac{\epsilon - \gamma }{2} {\Phi}_{y} \end{array}\right). \end{array} $$
(64)
We can check
$$ \begin{array}{llll} \vec{n} \times \mathring{A} \nabla {\Phi} &= \vec{n} \times \left( \begin{array}{cccc} \frac{\epsilon - \gamma }{2} {\Phi}_{x} + \varphi {\Phi}_{y} \\ \varphi {\Phi}_{x} - \frac{\epsilon - \gamma }{2} {\Phi}_{y} \end{array}\right) \\ &= \left( \begin{array}{cccc} \frac{\epsilon - \gamma }{2} {\Phi}_{y} - \varphi {\Phi}_{x} \\ \varphi {\Phi}_{y} + \frac{\epsilon - \gamma }{2} {\Phi}_{x} \end{array}\right) \end{array}$$
and notice
$$ \nabla^{\perp} \vec{n} = - H \nabla^{\perp} {\Phi} - \vec{n} \times \mathring{A} \nabla {\Phi}. $$
(65)
1.2 A.2 Formulas in \(\mathbb {S}^{3}\)
Let \({\Phi } : \mathbb {D} \rightarrow \mathbb {R}^{3}\) be a smooth conformal immersion and \(X = \pi ^{-1} \circ {\Phi } : \mathbb {D} \rightarrow \mathbb {S}^{3}\). Let \({\Lambda } := \frac {1}{2} \log \left (2 \left | X_{z} \right |^{2} \right )\) be its conformal factor, \(\vec {N}\) such that \(\left (X, e^{-{\Lambda }} X_{x}, e^{- {\Lambda }} X_{y}, \vec {N} \right )\) is a direct orthonormal basis of \(\mathbb {R}^{4}\) its Gauss map, \( h = \left \langle \frac {X_{z\bar {z}}}{\left | X_{z} \right |^{2}}, \vec {N} \right \rangle \) its mean curvature and \(\omega := 2\left \langle X_{zz}, \vec {n} \right \rangle \) its tracefree curvature. Then
$$ X := \frac{1}{1+ |{\Phi}|^{2}} \left( \begin{array}{cccc} 2{\Phi} \\ |{\Phi}|^{2}-1 \end{array}\right) $$
(66)
which yields
$$ \begin{array}{llll} X_{z} &= d\pi^{-1} \left( {\Phi}_{z} \right)= \frac{2}{1+|{\Phi}|^{2} } \left( \begin{array}{cccc} {\Phi}_{z} \\ 0 \end{array}\right) -\frac{ 4 \langle {\Phi}_{z}, {\Phi} \rangle_{3}}{ \left( 1+ |{\Phi}|^{2} \right)^{2} } \left( \begin{array}{cccc} {\Phi} \\ -1 \end{array}\right). \end{array} $$
(67)
Since π is conformal, \(\left \langle d \pi ^{-1} \left ({\Phi }_{z} \right ), d \pi ^{-1} \left (\vec {n} \right ) \right \rangle = \left \langle {\Phi }_{z}, \vec {n} \right \rangle =0\). Then \( \vec {N} = \frac { d \pi ^{-1} \left (\vec {n} \right )}{ \left | d \pi ^{-1} \left (\vec {n} \right ) \right |}\) and thus
$$ \begin{array}{llll} \vec{N} &=\left( \begin{array}{cccc} \vec{n} \\ 0 \end{array}\right) - \frac{2 \langle \vec{n}, {\Phi} \rangle }{1+ |{\Phi}|^{2} } \left( \begin{array}{cccc} {\Phi} \\ -1 \end{array}\right). \end{array} $$
(68)
Using the corresponding definitions we successively deduce
$$ \begin{array}{llll} e^{2 {\Lambda}} &=2 \left\langle X_{z}, X_{\bar{z}} \right\rangle = \frac{4}{\left( 1+ |{\Phi}|^{2} \right)^{2} } e^{2\lambda} , \end{array} $$
(69)
$$ \begin{array}{llll} h &= \left\langle \frac{X_{z \bar{z}}}{ \left| X_{z} \right|^{2}}, \vec{N} \right\rangle = \frac{|{\Phi}|^{2} +1}{2}H + \langle \vec{n} , {\Phi} \rangle_{3} \end{array} $$
(70)
$$ \begin{array}{llll} \omega &= 2 \left\langle X_{zz}, \vec{N} \right\rangle= \frac{2{\Omega}}{1+ | {\Phi}|^{2} }. \end{array} $$
(71)
Then one can compute
$$ \begin{array}{llll} h \left( \begin{array}{cccc} X \\1 \end{array}\right) + \left( \begin{array}{cccc} \vec{N} \\ 0 \end{array}\right) &= \left( \frac{|{\Phi}|^{2} +1}{2}H + \langle \vec{n} , {\Phi} \rangle \right) \left( \begin{array}{cccc} \frac{2 {\Phi}}{1+ |{\Phi}|^{2}} \\ \frac{|{\Phi}|^{2}-1}{1+ |{\Phi}|^{2}} \\ 1 \end{array}\right)+ \left( \begin{array}{cccc} \vec{n} - \frac{2 \langle \vec{n}, {\Phi} \rangle }{1+ |{\Phi}|^{2} } {\Phi} \\ \frac{2 \langle \vec{n}, {\Phi} \rangle }{1+ |{\Phi}|^{2} } \\ 0 \end{array}\right) \\ &= \left( \begin{array}{cccc} H {\Phi} + \frac{2 \langle \vec{n}, {\Phi} \rangle }{1+ |{\Phi}|^{2} } {\Phi} \\ H \frac{ | {\Phi} |^{2} - 1 }{2} + \langle \vec{n} , {\Phi} \rangle \frac{|{\Phi}|^{2}-1}{1+ |{\Phi}|^{2}} \\ H \frac{ | {\Phi}|^{2} +1}{2} + \langle \vec{n}, {\Phi} \rangle \end{array}\right) + \left( \begin{array}{cccc} \vec{n} - \frac{2 \langle \vec{n}, {\Phi} \rangle }{1+ |{\Phi}|^{2} } {\Phi} \\ \frac{2 \langle \vec{n}, {\Phi} \rangle }{1+ |{\Phi}|^{2} } \\ 0 \end{array}\right) \\ &= H \left( \begin{array}{cccc} {\Phi} \\ \frac{ | {\Phi}|^{2} -1}{2} \\ \frac{ | {\Phi} |^{2} +1}{2} \end{array}\right) + \left( \begin{array}{cccc} \vec{n} \\ \langle \vec{n} ,{\Phi} \rangle \\ \langle \vec{n} , {\Phi} \rangle \end{array}\right). \end{array} $$
(72)
This shows that
$$ Y = h \left( \begin{array}{cccc} X \\ 1 \end{array}\right) + \left( \begin{array}{cccc} N \\ 0 \end{array}\right). $$
(73)
One may wish to compute in \(\mathbb {S}^{3}\) without going through Φ. The relevant formulas then are
$$ \vec{N}_{z} = - h X_{z} - \omega e^{-2 {\Lambda}} X_{\bar{z}}, $$
(74)
$$ X_{z \bar{z}} = h \frac{e^{2{\Lambda}}}{2} \vec{N} - \frac{e^{2{\Lambda}}}{2} X , $$
(75)
$$ X_{z z} = 2 {\Lambda}_{z} X_{z} + \frac{\omega}{2} \vec{N}, $$
(76)
and Gauss-Codazzi can be written
$$ \omega_{\bar{z}} e^{-2{\Lambda}} = h_{z} . $$
(77)
1.3 A.3 Mean Curvature of a Sphere in \(\mathbb {S}^{3}\)
Let σ be a sphere in \(\mathbb {S}^{3}\). Up to an isometry of \(\mathbb {S}^{3}\) σ can be assumed to be a sphere centered on the south pole S of radius \(r \le \frac {\pi }{2}\). Then π ∘ σ is a sphere of \(\mathbb {R}^{3}\) centered on the origin of radius R ≤ 1. It can be conformally parametrized over \(\mathbb {R}^{2} \cup \infty \) by \({\Phi } (x,y) = \frac {R}{1+ x^{2} +y^{2}} \left (\begin {array}{cccc} 2x \\2 y \\ x^{2} +y^{2}-1 \end {array}\right )\), of constant mean curvature \(H = \frac {1}{R}\). Then σ is conformally parametrized by
$$X = \frac{ 1}{1+ R^{2}} \left( \begin{array}{cccc} \frac{2R}{1+ x^{2} +y^{2}} \left( \begin{array}{cccc} 2x \\2 y \\ x^{2} +y^{2}-1 \end{array}\right) \\ R^{2} -1 \end{array}\right).$$
One can easily compute using basic trigonometry the tangent of r and find
$$ \begin{array}{llll} \tan \left( r \right) &= \frac{2R }{1-R^{2}}. \end{array}$$
Computing h at any point (x, y) using (70) yields with \(H= \frac {1}{R}\), \(\vec {n} = - \frac {\Phi }{R}\) (Fig. 3)
$$h= \frac{R^{2} +1}{2R} - R = \frac{1}{\tan (r)}$$
for any (x, y).
Since neither h nor r change under the action of isometries, any sphere σ of \(\mathbb {S}^{3}\) of radius r has constant mean curvature
$$ h = \text{cotan} (r). $$
(78)
1.4 A.4 Formulas in \(\mathbb {H}^{3}\)
Let \({\Phi } : \mathbb {D} \rightarrow \mathbb {R}^{3}\) be a smooth conformal immersion and \(Z = {\tilde \pi }^{-1} \circ {\Phi } : \mathbb {D} \rightarrow \mathbb {H}^{3}\). Then
$$ Z := \frac{1}{1- |{\Phi}|^{2}} \left( \begin{array}{cccc} 2{\Phi} \\ |{\Phi}|^{2}+1 \end{array}\right) $$
(79)
which yields
$$ \begin{array}{llll} Z_{z} &= \frac{2}{1- | {\Phi}|^{2} } \left( \begin{array}{cccc} {\Phi}_{z} \\ 0 \end{array}\right) + \frac{4 \langle {\Phi}_{z} , {\Phi} \rangle }{\left( 1- | {\Phi}|^{2} \right)^{2}} \left( \begin{array}{cccc} {\Phi} \\ 1 \end{array}\right) . \end{array} $$
(80)
Since \(\tilde \pi \) is conformal, \(\left \langle d \tilde \pi ^{-1} \left (\vec {n} \right ) ,Z_{z}\right \rangle = \left \langle {\Phi }_{z} , \vec {n} \right \rangle = 0\). Then \( \vec {n}^{Z} = \frac {d \tilde \pi ^{-1} (\vec {n})}{ \left | d \tilde \pi ^{-1} \left (\vec {n} \right ) \right |}\) and thus
$$ \begin{array}{llll} \vec{n}^{Z} &= \left( \begin{array}{cccc} \vec{n} \\ 0 \end{array}\right) + \frac{2 \langle \vec{n}, {\Phi} \rangle }{1- |{\Phi}|^{2}} \left( \begin{array}{cccc} {\Phi} \\ 1 \end{array}\right). \end{array} $$
(81)
Using the corresponding definition, we successively deduce
$$ \begin{array}{llll} e^{2\lambda^{Z}} &= \frac{4}{\left( 1- |{\Phi}|^{2} \right)^{2} } e^{2\lambda} , \end{array} $$
(82)
$$ \begin{array}{llll} H^{Z} &= \frac{1- |{\Phi}|^{2}}{2} H - \langle \vec{n} , {\Phi} \rangle , \end{array} $$
(83)
$$ \begin{array}{llll} {\Omega}^{Z} &= \frac{2{\Omega}}{1-|{\Phi}|^{2}}. \end{array} $$
(84)
Then one can compute
$$ \begin{array}{llll} H^{Z} \left( \begin{array}{cccc} Z_{h} \\ -1 \\ Z_{4} \end{array}\right) + \left( \begin{array}{cccc} {\vec{n}^{Z}_{h}} \\ 0 \\ {\vec{n}^{Z}_{4}} \end{array}\right) &= \left( \frac{1- |{\Phi}|^{2}}{2} H - \langle \vec{n} , {\Phi} \rangle \right) \left( \begin{array}{cccc} \frac{2 {\Phi}}{1- |{\Phi}|^{2}} \\ -1 \\ \frac{|{\Phi}|^{2}+1}{1-|{\Phi}|^{2}} \end{array}\right) + \left( \begin{array}{cccc} \vec{n} + \frac{2 \langle \vec{n}, {\Phi} \rangle }{1- |{\Phi}|^{2} } {\Phi} \\ 0 \\ \frac{2 \langle \vec{n},{\Phi} \rangle }{1- |{\Phi}|^{2} } \end{array}\right) \\ &= \left( \begin{array}{cccc} H {\Phi} - \frac{2 \langle \vec{n}, {\Phi} \rangle }{1- |{\Phi}|^{2} } {\Phi} \\- \frac{1- |{\Phi}|^{2}}{2} H + \langle \vec{n} , {\Phi} \rangle \\ H \frac{ | {\Phi} |^{2} +1 }{2} - \langle \vec{n}, {\Phi} \rangle \frac{|{\Phi}|^{2}+1}{1-|{\Phi}|^{2}} \end{array}\right) + \left( \begin{array}{cccc} \vec{n} + \frac{2 \langle \vec{n}, {\Phi} \rangle }{1- |{\Phi}|^{2} } {\Phi} \\ 0 \\ \frac{2 \langle \vec{n}, {\Phi} \rangle }{1- |{\Phi}|^{2} } \end{array}\right) \\ &= H \left( \begin{array}{cccc} {\Phi} \\ \frac{ | {\Phi}|^{2} -1}{2} \\ \frac{ | {\Phi} |^{2} +1}{2} \end{array}\right) + \left( \begin{array}{cccc} \vec{n} \\ \langle \vec{n} ,{\Phi} \rangle \\ \langle \vec{n} , {\Phi} \rangle \end{array}\right). \end{array} $$
(85)
Which shows that
$$ Y = H^{Z} \left( \begin{array}{cccc} Z_{h} \\ -1 \\ Z_{4} \end{array}\right) + \left( \begin{array}{cccc} {\vec{n}_{h}^{Z}} \\ 0 \\ {\vec{n}_{4}^{Z}} \end{array}\right). $$
(86)
1.5 A.5 Computations for the Conformal Gauss Map
Let \({\Phi } : \mathbb {D} \rightarrow \mathbb {R}^{3}\) be a smooth conformal immersion of representation X in \(\mathbb {S}^{3}\) and of conformal Gauss map Y.
Let us first use the expression (16). Then
$$ \begin{array}{llll} Y_{z} &= H_{z} \left( \begin{array}{cccc} {\Phi} \\ \frac{ | {\Phi}|^{2} -1}{2} \\ \frac{ | {\Phi} |^{2} +1}{2} \end{array}\right) + H \left( \begin{array}{cccc} {\Phi}_{z} \\ \langle {\Phi}_{z} , {\Phi} \rangle \\ \langle {\Phi}_{z} , {\Phi}_{z} \rangle \end{array}\right) + \left( \begin{array}{cccc} \vec{n}_{z} \\ \langle \vec{n}_{z} ,{\Phi} \rangle \\ \langle \vec{n}_{z} , {\Phi} \rangle \end{array}\right) \end{array} $$
and using (60)
$$ Y_{z}= H_{z} \left( \begin{array}{cccc} {\Phi} \\ \frac{ | {\Phi}|^{2} -1}{2} \\ \frac{ | {\Phi} |^{2} +1}{2} \end{array}\right) - {\Omega} e^{-2\lambda} \left( \begin{array}{cccc} {\Phi}_{\bar{z}} \\ \langle {\Phi}_{\bar{z}}, {\Phi} \rangle \\ \langle {\Phi}_{\bar{z}} , {\Phi} \rangle \end{array}\right). $$
(87)
Using (63) and (62) we compute
$$ \begin{array}{llll} Y_{z \bar{z}} &= H_{z \bar{z}} \left( \begin{array}{cccc} {\Phi} \\ \frac{ | {\Phi}|^{2} -1}{2} \\ \frac{ | {\Phi} |^{2} +1}{2} \end{array}\right) - \frac{ \left| {\Omega} \right|^{2} }{2}e^{-2\lambda} \left( \begin{array}{cccc} \vec{n} \\ \langle \vec{n}, {\Phi} \rangle \\ \langle \vec{n} , {\Phi}\rangle \end{array}\right) \\ &= \mathcal{W}({\Phi}) \left( \begin{array}{cccc} {\Phi} \\ \frac{ | {\Phi}|^{2} -1}{2} \\ \frac{ | {\Phi} |^{2} +1}{2} \end{array}\right) - \frac{\left| {\Omega} \right|^{2} e^{-2\lambda} }{2} Y \end{array} $$
(88)
where
$$ \mathcal{W}({\Phi}) = H_{z \bar{z}} + \frac{ \left| {\Omega} \right|^{2} e^{-2\lambda} }{2} H \in \mathbb{R} . $$
(89)
On the other hand
$$ \begin{array}{llll} Y_{zz} &= H_{zz} \left( \begin{array}{cccc} {\Phi} \\ \frac{ | {\Phi}|^{2} -1}{2} \\ \frac{ | {\Phi} |^{2} +1}{2} \end{array}\right) + H_{z} \left( \begin{array}{cccc} {\Phi}_{z} \\ \langle {\Phi}_{z}, {\Phi} \rangle \\ \langle {\Phi}_{z} , {\Phi} \rangle \end{array}\right) - \left( {\Omega} e^{-2 \lambda} \right)_{z} \left( \begin{array}{cccc} {\Phi}_{\bar{z}} \\ \langle {\Phi}_{\bar{z}}, {\Phi} \rangle \\ \langle {\Phi}_{\bar{z}} , {\Phi} \rangle \end{array}\right) \\& - {\Omega} \left( \frac{H}{2} \left( \begin{array}{cccc} \vec{n} \\ \langle \vec{n}, {\Phi} \rangle \\ \langle \vec{n} , {\Phi}\rangle \end{array}\right) + \frac{1}{2} \left( \begin{array}{cccc} 0 \\ 1 \\ 1 \end{array}\right) \right) \end{array} $$
(90)
using (61). Then if we define Bryant’s functional as \(\mathcal {Q} = \left \langle Y_{zz} , Y_{zz} \right \rangle \) we find
$$ \begin{array}{llll} \mathcal{Q} &=H_{zz}{\Omega} - H_{z} \left( {\Omega} e^{-2 \lambda} \right)_{z} e^{2\lambda} + {\Omega} \frac{H^{2}}{4} \\ &= \left( {\Omega}_{\bar{z}} e^{-2\lambda} \right)_{z} {\Omega} - {\Omega}_{\bar{z}} \left( {\Omega} e^{-2 \lambda} \right)_{z} + {\Omega} \frac{H^{2}}{4} \text{using (63)} \\ &= \left( {\Omega}_{z\bar{z}} {\Omega} - {\Omega}_{z} {\Omega}_{\bar{z}} \right) e^{-2\lambda} + {\Omega} \frac{H^{2}}{4} \\ &= {\Omega}^{2} e^{-2 \lambda} \left( \frac{{\Omega}_{z}}{\Omega} \right)_{\bar{z}} + {\Omega} \frac{H^{2}}{4} = {\Omega}^{2} e^{-2 \lambda} \left( \frac{{\Omega}_{\bar{z}}}{\Omega} \right)_{z} + {\Omega} \frac{H^{2}}{4}. \end{array} $$
(91)
We will now do the computations with the \(\mathbb {S}^{3}\) formalism. Then
$$ \begin{array}{llll} Y_{z} &= h_{z} \left( \begin{array}{cccc} X \\ 1 \end{array}\right) + h \left( \begin{array}{cccc} X_{z} \\1 \end{array}\right) + \left( \begin{array}{cccc} \vec{N}_{z} \\ 0 \end{array}\right) \end{array} $$
and using (74)
$$ Y_{z}= h_{z} \left( \begin{array}{cccc} X \\ 1 \end{array}\right) - \omega e^{-2{\Lambda}} \left( \begin{array}{cccc} X_{\bar{z}} \\0 \end{array}\right). $$
(92)
Using (77) and (76) we compute
$$ \begin{array}{llll} Y_{z \bar{z}} &= h_{z \bar{z}} \left( \begin{array}{cccc} X \\ 1 \end{array}\right) - \frac{ \left| \omega \right|^{2} }{2}e^{-2{\Lambda}} \left( \begin{array}{cccc} \vec{N} \\ 0 \end{array}\right) \\ &= \mathcal{W}_{\mathbb{S}^{3}}(X) \left( \begin{array}{cccc}X \\1 \end{array}\right) - \frac{\left| \omega \right|^{2} e^{-2{\Lambda}} }{2} Y \end{array} $$
(93)
where
$$ \mathcal{W}_{\mathbb{S}^{3}}(X) = h_{z \bar{z}} + \frac{ \left| \omega \right|^{2} e^{-2{\Lambda}} }{2} h \in \mathbb{R} . $$
(94)
Using (69), (70) and (71) yields
$$ \begin{array}{llll} \mathcal{W}_{\mathbb{S}^{3}}(X) &= \left( \frac{ |{\Phi}|^{2} +1}{2} H + \langle \vec{n} , {\Phi} \rangle \right)_{z \bar{z}} + \left( \frac{ |{\Phi}|^{2} +1}{2} H + \langle \vec{n} , {\Phi} \rangle \right) \frac{|{\Omega}|^{2}e^{-2 \lambda}}{2} \\ &=\left( \frac{ |{\Phi}|^{2} +1}{2} H_{z} + \langle {\Phi}_{z} , {\Phi} \rangle H + \langle \vec{n}_{z} , {\Phi} \rangle \right)_{\bar{z}} + \frac{ |{\Phi}|^{2} +1}{2} H \frac{ |{\Omega}|^{2} e^{-2\lambda}}{2}\\&+ \langle \vec{n}, {\Phi} \rangle \frac{ |{\Omega}|^{2} e^{-2\lambda}}{2} \\ &= \left( \frac{ |{\Phi}|^{2} +1}{2} H_{z} - {\Omega} e^{-2 \lambda} \langle {\Phi}_{\bar{z}} , {\Phi} \rangle \right)_{\bar{z}} + \frac{ |{\Phi}|^{2} +1}{2} H \frac{ |{\Omega}|^{2} e^{-2\lambda}}{2}\\&+ \langle \vec{n}, {\Phi} \rangle \frac{ |{\Omega}|^{2} e^{-2\lambda}}{2} \\ &= \frac{ |{\Phi}|^{2} +1}{2} \mathcal{W}({\Phi} ) + \langle {\Phi}_{\bar{z}} , {\Phi} \rangle H_{z} - {\Omega}_{\bar{z}} e^{-2\lambda} \langle {\Phi}_{\bar{z}}, {\Phi} \rangle - \frac{ |{\Omega}|^{2} e^{-2\lambda} }{2} \langle \vec{n}, {\Phi} \rangle\\& + \langle \vec{n}, {\Phi} \rangle \frac{ |{\Omega}|^{2} e^{-2\lambda}}{2} \\ &= \frac{ |{\Phi}|^{2} +1}{2} \mathcal{W}({\Phi} ), \end{array} $$
(95)
where we have used (60) to obtain the third equality and (77) to conclude. On the other hand
$$ \begin{array}{llll} Y_{zz} &= h_{zz} \left( \begin{array}{cccc} X \\ 1 \end{array}\right) + h_{z} \left( \begin{array}{cccc} X_{z} \\ 0 \end{array}\right) - \left( \omega e^{-2 {\Lambda}} \right)_{z} \left( \begin{array}{cccc} X_{\bar{z}} \\ 0 \end{array}\right) - \omega \left( \frac{h}{2} \left( \begin{array}{cccc} \vec{N} \\ 0 \end{array}\right) - \frac{1}{2} \left( \begin{array}{cccc} X \\ 0 \end{array}\right) \right) \end{array} $$
(96)
using (61). Then if we define \(\mathcal {Q} = \left \langle Y_{zz} , Y_{zz} \right \rangle \) we find, once more by applying (77),
$$ \begin{array}{llll} \mathcal{Q} &=h_{zz}\omega - h_{z} \left( \omega e^{-2 {\Lambda}} \right)_{z} e^{2{\Lambda}} + \omega^{2} \frac{h^{2}+1}{4} \\ &= \left( \omega_{\bar{z}} e^{-2{\Lambda}} \right)_{z} \omega - \omega_{\bar{z}} \left( \omega e^{-2 {\Lambda}} \right)_{z} + \omega^{2} \frac{h^{2}+1}{4} \\ &= \left( \omega_{z\bar{z}} \omega - \omega_{z} \omega_{\bar{z}} \right) e^{-2{\Lambda}} + \omega^{2} \frac{h^{2}+1}{4} \\ &= \omega^{2} e^{-2 {\Lambda}} \left( \frac{\omega_{z}}{\omega} \right)_{\bar{z}} + \omega^{2} \frac{h^{2}+1}{4} = \omega^{2} e^{-2 {\Lambda}} \left( \frac{\omega_{\bar{z}}}{\omega} \right)_{z} + \omega^{2} \frac{h^{2}+1}{4}. \end{array} $$
(97)
1.6 A.6 Formulas in \(\mathbb {S}^{4,1}\)
This section is devoted to computations for spacelike immersions in \(\mathbb {S}^{4,1}\) without relying on their being the conformal Gauss map of a given immersion.
Let \(Y : D \rightarrow \mathbb {S}^{4,1}\) be a smooth-spacelike conformal immersion, that is Y satisfies
$$\left\langle Y_{z} , Y_{z} \right\rangle = 0$$
and
$$\left\langle Y_{z} , Y_{\bar{z}} \right\rangle =: \frac{e^{2 \mathcal{L}}}{2} >0.$$
Let \(\nu , \nu ^{*} \in \mathcal {C}^{4,1}\) such that \(e = \left (Y, Y_{z} ,Y_{\bar {z}}, \nu , \nu ^{*} \right )\) is an orthogonal frame of \(\mathbb {R}^{4,1}\), that is
$$\left\langle Y, \nu \right\rangle = \left\langle Y_{z}, \nu \right\rangle = \left\langle Y_{\bar{z}}, \nu \right\rangle = \left\langle \nu, \nu \right\rangle =0$$
and
$$\left\langle Y, \nu^{*} \right\rangle = \left\langle Y_{z}, \nu^{*} \right\rangle = \left\langle Y_{\bar{z}}, \nu^{*} \right\rangle = \left\langle \nu^{*}, \nu^{*} \right\rangle =0.$$
We define successively the tracefree curvature in the direction ν
$$ {\Omega}_{\nu} =2 \left\langle Y_{zz}, \nu \right\rangle, $$
(98)
the tracefree curvature in the direction ν∗
$$ {\Omega}_{\nu^{*}} =2 \left\langle Y_{zz}, \nu^{*} \right\rangle, $$
(99)
the mean curvature in the direction ν
$$ H_{\nu} = 2 e^{-2 \mathcal{L} } \left\langle Y_{z \bar{z}}, \nu \right\rangle, $$
(100)
and the mean curvature in the direction ν∗
$$ H_{\nu^{*}} = 2 e^{-2 \mathcal{L} } \left\langle Y_{zz}, \nu^{*} \right\rangle. $$
(101)
Then,
$$ Y_{zz} = 2 \mathcal{L}_{z} Y_{z} + \frac{ {\Omega}_{\nu}}{2 \langle \nu , \nu^{*} \rangle } \nu^{*} + \frac{ {\Omega}_{\nu^{*}}}{2 \langle \nu , \nu^{*} \rangle } \nu, $$
(102)
and
$$ Y_{z\bar{z}}= \frac{H_{\nu} e^{2 \mathcal{L} } }{2 \langle \nu , \nu^{*} \rangle } \nu^{*} + \frac{H_{\nu^{*}}e^{2 \mathcal{L} }}{2 \langle \nu , \nu^{*} \rangle } \nu - \frac{e^{2 \mathcal{L} }}{2} Y. $$
(103)
Further
$$ \left\langle \nu_{z}, Y \right\rangle = \left( \left\langle \nu , Y \right\rangle \right)_{z} - \left\langle \nu , Y_{z} \right\rangle = 0, $$
(104)
and with Eq. 102,
$$ \begin{array}{llll} \left\langle \nu_{z}, Y_{z}\right\rangle &= \left( \left\langle \nu , Y_{z} \right\rangle \right)_{z} - \left\langle \nu , Y_{zz} \right\rangle \\ &= -2 \mathcal{L}_{z} \left\langle \nu , Y_{z} \right\rangle - \frac{ {\Omega}_{\nu}}{2 \langle \nu , \nu^{*} \rangle } \langle \nu , \nu^{*} \rangle - \frac{ {\Omega}_{\nu}^{*}}{2 \langle \nu , \nu^{*} \rangle } \langle \nu , \nu \rangle \\ &= - \frac{ {\Omega}_{\nu}}{2}. \end{array} $$
(105)
On the other hand, with Eq. 103 one has
$$ \begin{array}{llll} \left\langle \nu_{z}, Y_{\bar{z}}\right\rangle &= \left( \left\langle \nu , Y_{\bar{z}} \right\rangle \right)_{z} - \left\langle \nu , Y_{z \bar{z}} \right\rangle \\ &= -\frac{ H_{\nu} e^{2 \mathcal{L} } }{2 \langle \nu , \nu^{*} \rangle } \langle \nu , \nu^{*} \rangle \\ &= - \frac{ H_{\nu} e^{2 \mathcal{L} }}{2}, \end{array} $$
(106)
and
$$ \left\langle \nu_{z}, \nu \right\rangle = \left( \langle \nu , \nu \rangle \right)_{z} - \langle \nu , \nu_{z} \rangle, $$
meaning
$$ \left\langle \nu_{z}, \nu \right\rangle = 0. $$
(107)
Combining (104), (105), (106) and (107) yields
$$ \nu_{z}= - \left\langle \nu_{z}, \nu^{*} \right\rangle \nu - H_{\nu} Y_{z} - {\Omega}_{\nu} e^{-2 \mathcal{L} } Y_{\bar{z}}. $$
(108)
Similarly
$$ \left\langle \nu^{*}_{z}, Y \right\rangle = \left( \left\langle \nu^{*} , Y \right\rangle \right)_{z} - \left\langle \nu^{*} , Y_{z} \right\rangle = 0, $$
(109)
and with (102),
$$ \begin{array}{llll} \left\langle \nu^{*}_{z}, Y_{z}\right\rangle &= \left( \left\langle \nu^{*} , Y_{z} \right\rangle \right)_{z} - \left\langle \nu^{*} , Y_{zz} \right\rangle \\ &= -2 \mathcal{L}_{z} \left\langle \nu^{*} , Y_{z} \right\rangle - \frac{ {\Omega}_{\nu^{*}}}{2 \langle \nu , \nu^{*} \rangle } \langle \nu , \nu^{*} \rangle - \frac{ {\Omega}_{\nu}}{2 \langle \nu , \nu^{*} \rangle } \langle \nu^{*} , \nu^{*} \rangle \\ &= - \frac{ {\Omega}_{\nu^{*}}}{2}, \end{array} $$
(110)
while with (103)
$$ \begin{array}{llll} \left\langle \nu^{*}_{z}, Y_{\bar{z}}\right\rangle &= \left( \left\langle \nu^{*} , Y_{\bar{z}} \right\rangle \right)_{z} - \left\langle \nu^{*} , Y_{z \bar{z}} \right\rangle \\ &= -\frac{ H_{\nu^{*}} e^{2 \mathcal{L} } }{2 \langle \nu , \nu^{*} \rangle } \langle \nu , \nu^{*} \rangle \\ &= - \frac{ H_{\nu^{*}} e^{2 \mathcal{L} }}{2}, \end{array} $$
(111)
$$ \langle \nu^{*}_{z} \nu^{*} \rangle = 0, $$
(112)
$$ \nu^{*}_{z}= - \left\langle \nu^{*}_{z}, \nu \right\rangle \nu^{*} - H_{\nu^{*}} Y_{z} - {\Omega}_{\nu^{*}} e^{-2 \mathcal{L} } Y_{\bar{z}}. $$
(113)
Then
$$ \begin{array}{llll} \left\langle \nu_{z}, \nu_{z} \right\rangle &= H_{\nu} {\Omega}_{\nu} \\ \left\langle \nu^{*}_{z}, \nu^{*}_{z} \right\rangle &= H_{\nu^{*}} {\Omega}_{\nu^{*}}. \end{array} $$
(114)
