Spinorial Representation of Surfaces in Lorentzian Homogeneous Spaces of Dimension $\varvec{3}$

Abstract

We find a spinorial representation of a Riemannian or Lorentzian surface in a Lorentzian homogeneous space of dimension 3. We in particular obtain a representation theorem for surfaces in $\mathbb {L}(\kappa ,\tau )$ spaces. We then recover the Calabi correspondence between minimal surfaces in $\mathbb {R}^3$ and maximal surfaces in $\mathbb {R}_1^3$, and obtain a new Lawson type correspondence between CMC surfaces in $\mathbb {R}_1^3$ and in the 3-dimensional pseudo-hyperbolic space $\mathbb {H}_1^{3}.$

Appendices

Appendix A: Identification of Spinor Bundles

We consider a simply connected pseudo-Riemmanian surface M and the trivial bundle $E=M\times \mathbb {R}\longrightarrow M$ with metric ${\mp } d\nu ^2$ depending on whether M is Riemannian or Lorentzian and let us denote the spin structures of TM and E by ${\widetilde{Q}}_M$ and ${\widetilde{Q}}_E$ respectively. We identify here the spinor bundle $\Sigma M$ with a subbundle of the bundle $\Sigma =\left( {\widetilde{Q}}_{_M} \times _M {\widetilde{Q}}_{_E}\right) \times _\rho C\!\ell _{1,2}$ defined by (6).

Let us begin with the Riemannian case. We define the complexified quaternions as

$$\begin{aligned} \mathbb {H}^{\mathbb {C}}:=\mathbb {H}\otimes _{\mathbb {R}}\mathbb {C}=\{z_0+z_1I+z_2J+z_3K : z_i\in \mathbb {C}\}. \end{aligned}$$

This algebra is endowed with the quadratic form $\langle q,q\rangle =z_0^2+z_1^2+z_2^2+z_3^2$, where $q=z_0+z_1I+z_2J+z_3K\in \mathbb {H}^{\mathbb {C}}$.

Remark 6

Let $(e_0,e_1,e_2)$ be the standard basis of $\mathbb {R}^{1,2}$, which is such that $\langle e_i, e_j\rangle =0$ if $i\ne j$ and $\langle e_0, e_0\rangle =-1=-\langle e_1, e_1\rangle =-\langle e_2, e_2\rangle$. The map obtained by linearity from $e_0\longmapsto iI$, $e_1\longmapsto J$, $e_2\longmapsto JI=-K$ is a Clifford application which induces an isomorphism of algebras between $C\!\ell _{1,2}$ and $\mathbb {H}^{\mathbb {C}}$. Using this isomorphism, the even Clifford algebra $C\!\ell _{1,2}^{0}$ is isomorphic to

$$\begin{aligned} \{q_0+q_1I+iq_2J+iq_3K : q_i\in \mathbb {R}\}. \end{aligned}$$

We consider the following representations of the group $\mathrm {Spin}(0,2)$ given by the left-multiplication:

$$\begin{aligned} \rho _1:\mathrm {Spin}(0,2)\longrightarrow \mathrm {GL}(C\!\ell _{0,2}),&\rho _2:\mathrm {Spin}(0,2)\subset \mathrm {Spin}(1,2)\longrightarrow \mathrm {GL}(C\!\ell _{1,2}^{0}). \end{aligned}$$

Here and below we use the models

$$\begin{aligned} C\!\ell _{0,2}=\mathbb {H}\quad {and}\quad \mathrm {Spin}(0,2)=\{q_0+q_1I: q_i\in \mathbb {R},\ q_0^2+q_1^2=1\}. \end{aligned}$$

The representation $\rho _1$ is the standard spin representation (if we identify $\mathbb {H}\simeq \mathbb {C}^2$). The following lemma states that the representations $\rho _1$ and $\rho _2$ are equivalent.

Lemma 10

The isomorphism of vector spaces

$$\begin{aligned} f :\ \ \ C\!\ell _{0,2}\longrightarrow & {} C\!\ell _{1,2}^{0} \\ \ \ \ \ \ \ q_0+q_1I+J(q_2-Iq_3)\longmapsto & {} q_0+q_1I+iJ(q_2-Iq_3) \end{aligned}$$

induces a $\mathbb {C}$-linear isomorphism between the representations $\rho _1$ and $\rho _2$, where the complex structure in both spaces is given by the right multiplication by I.

Proof

A computation shows that $f(gq)=gf(q)$ $\forall g\in \mathrm {Spin}(0,2),\forall q\in C\!\ell _{0,2}.$ $\square$

Identifying $e_0$ with iI (see Remark 6), the isomorphism f satisfies the following:

Lemma 11

If $x\in \mathbb {R}^2$ and $q \in \mathbb {H}$ then $f(x\cdot q)=ie_0\cdot x\cdot f(q)$, where i is the complex structure in $C\!\ell _{1,2}^0$ given by the right multiplication by I.

Since E is trivial the orthonormal frame bundle is $Q_E=M\times \{1\}$ and ${\widetilde{Q}}_{E}=M\times \{\pm 1\}$ is also trivial. We consider the global section ${\widetilde{s}}_E:M\longrightarrow \{\pm 1\}$ given by $m\longmapsto +1$. Considering the inclusion ${\widetilde{Q}}_M\longrightarrow {\widetilde{Q}}_M\times {\widetilde{Q}}_E$, ${\widetilde{s}}_M\longmapsto ({\widetilde{s}}_M,{\widetilde{s}}_E)$ and the isomorphism f of Lemma 11 we get the bundle isomorphism

$$\begin{aligned} \Sigma M:= \overset{\sim }{Q}_{_M}\times _\rho \Sigma _2\longrightarrow & {} \Sigma _0:=\left( {\overset{\sim }{Q}_{_M}\times _{_M}\overset{\sim }{Q}_{_E}}\right) \times _\rho C\!\ell _{1,2}^{ ^\circ } \\ \psi :=\big [ \overset{\sim }{s}_{_M},q\big ]\longmapsto & {} \psi ^*:=\big [ (\overset{\sim }{s}_{_M},\overset{\sim }{s}_{_E}),f(q)\big ]; \end{aligned}$$

it satisfies

$$\begin{aligned} (\nabla _{_X}\psi )^*=\nabla _{_X}\psi ^*,\quad (X\cdot _M\psi )^*=iN\cdot X\cdot \psi ^*\quad {and}\quad |\psi ^+|^2-|\psi ^-|^2=\langle \langle \psi ^*,\psi ^*\rangle \rangle \end{aligned}$$

for all $X\in TM,$ where $N=[(\overset{\sim }{s}_M,\overset{\sim }{s}_E), e_0]$ (see [32, Prop. 3.4.9] for more details).

Let us see now the Lorentzian case. Here the spinor bundle of M is $\Sigma M={\widetilde{Q}}_M\times _{\rho _1} C\!\ell _{_{1,1}},$ where $\rho _1:\mathrm {Spin}(1,1)\longrightarrow \mathrm {GL}(C\!\ell _{1,1})$ is the representation given by left-multiplication. As above we have ${\widetilde{Q}}_E=M\times \{\pm 1\},$ the section $\overset{\sim }{s}_{_E}: m\mapsto +1$ of ${\widetilde{Q}}_E,$ and the inclusion $\overset{\sim }{Q}_{_M}\longrightarrow \overset{\sim }{Q}_{_M}\times \overset{\sim }{Q}_{_E}$, $\overset{\sim }{s}_{_M}\longmapsto (\overset{\sim }{s}_{_M},\overset{\sim }{s}_{_E}).$ Using the isomorphism $C\!\ell _{1,1}\simeq C\!\ell _{1,2}^{0}$ (induced by the Clifford application $\mathbb {R}^{1,1}\longrightarrow C\!\ell _{12}^{0}$, $x\longmapsto x\cdot e_2$, where $(e_0,e_1,e_2)$ is the standard basis of $\mathbb {R}^{1,2}$) we obtain a bundle isomorphism

$$\begin{aligned} \overset{\sim }{Q}_{_M}\times _{\rho _1} C\!\ell _{11}\longrightarrow & {} \left( {\overset{\sim }{Q}_{_M}\times _{_M}\overset{\sim }{Q}_{_E}}\right) \times _\rho C\!\ell _{1,2}^{0}\nonumber \\ \psi\longmapsto & {} \psi ^*; \end{aligned}$$

it satisfies the properties

$$\begin{aligned} (\nabla _{_X}\psi )^*=\nabla _{_X}\psi ^*,\quad (X\cdot _M\psi )^*=X\cdot N \cdot \psi ^*\quad {and}\quad |\psi ^+|^2-|\psi ^-|^2=\langle \langle \psi ^*,\psi ^*\rangle \rangle \end{aligned}$$

for all $X\in TM,$ where $N=[(\overset{\sim }{s}_M,\overset{\sim }{s}_E), e_2]$ (see [32, Prop. 3.4.16] for more details).

Appendix B: Bivectors and Linear Operators

We prove that a skew-symmetric operator $u:\mathbb {R}^{r,s}\longrightarrow \mathbb {R}^{r,s}$ is identified with a bivector ${\underline{u}}\in \Lambda ^2(\mathbb {R}^{r,s})$. Let us consider the following bracket in the Clifford algebra $C\!\ell _{r,s}$

$$\begin{aligned}{}[a,b]=\frac{1}{2}(a\cdot b-b\cdot a), \end{aligned}$$

(68)

for all $a,b\in C\!\ell _{r,s}$.

Lemma 12

Let $u:\mathbb {R}^{r,s}\longrightarrow \mathbb {R}^{r,s}$ be a skew-symmetric operator. The bivector that represents u is

$$\begin{aligned} {\underline{u}}=\frac{1}{2}\sum _{j=1}^{r+s}\varepsilon _je_j\cdot u(e_j),&\varepsilon _j=\langle e_j,e_j\rangle =\pm 1, \end{aligned}$$

(69)

and for all $\xi \in \mathbb {R}^{r,s}$ we have

$$\begin{aligned}{}[{\underline{u}},\xi ]=u(\xi ). \end{aligned}$$

(70)

Proof

Let us consider the linear application $u:\mathbb {R}^{r,s}\longrightarrow \mathbb {R}^{r,s}$ given by $e_i\longmapsto \varepsilon _j e_j$ and $e_j\longmapsto -\varepsilon _i e_i$ if $i<j$ and $e_k\longmapsto 0$ if $k\ne i,j$ which corresponds to $\varepsilon _i e_i\wedge \varepsilon _j e_j \in \Lambda ^2 \mathbb {R}^{r,s}$. This map is skew-symmetric and ${\underline{u}}=\varepsilon _i\varepsilon _je_i\cdot e_j=\frac{1}{2}\varepsilon _i\varepsilon _j\left( e_i\cdot e_j-e_j\cdot e_i\right)$ satisfies

$$\begin{aligned}{}[{\underline{u}},e_k]= & {} \frac{1}{2}\varepsilon _i\varepsilon _j(e_i\cdot e_j\cdot e_k-e_k\cdot e_i\cdot e_j), \end{aligned}$$

which yields for $k=i$

$$\begin{aligned}{}[{\underline{u}},e_i]=\frac{1}{2}\varepsilon _i\varepsilon _j(\varepsilon _i e_j+\varepsilon _i e_j)=\varepsilon _j e_j=u(e_i); \end{aligned}$$

similarly we can prove that $[{\underline{u}},e_j]=u(e_j)$ and also readily see that $[{\underline{u}},e_k]=0$ for $k\ne i, j$. Equality (70) is a consequence of linearity. $\square$

Table 1 $\mathbb {L}(\kappa , \tau )$ spaces