Clifford structures, bilegendrian surfaces, and extrinsic curvature

Abstract

We use Clifford algebras to construct a unified formalism for studying constant extrinsic curvature immersed surfaces in Riemannian and semi-Riemannian 3-manifolds in terms of immersed bilegendrian surfaces in their unitary bundles. As an application, we provide full classifications of both complete and compact immersed bilegendrian surfaces in the unit tangent bundle ${\text {U}}\mathbb {S}^3$ of the 3-sphere.

A. Calculating the curvature

Let W denote the bundle defined in Sect. 3.3. We now determine certain components of its curvature. The calculation is rather technical and uninformative, and for this reason we have chosen to relegate it to this appendix.

Lemma A.1

At every point (x, y) of M, and for every vector $\xi \in W_{(x,y)}$ such that $\pi _1(\xi )$ and $\pi _2(\xi )$ are colinear,

$$\begin{aligned} \omega _i(R_{X,J^\eta X}X,J^{\eta } X) = \frac{-\epsilon \Vert (x,y)\Vert _g^2}{2\Vert x\Vert _{{{\textbf{b}}}}^2\Vert y\Vert _{{{\textbf{b}}}}^2}{\hat{g}}(X,I^\eta X){\hat{g}}(X,X). \end{aligned}$$

(A.1)

Proof

Indeed, let $\lambda _x$ and $\lambda _y$ denote respectively the ${{\textbf{b}}}$-lengths of x and y. The normal bundle of W is generated by the vector fields

$$\begin{aligned} N_1:= \frac{1}{\lambda _x}(x,0),\ N_2:= \frac{1}{\lambda _y}(y,0),\ N_3:= \frac{1}{\lambda _x}(0,x),\ \text {and}\ N_4:= \frac{1}{\lambda _y}(0,y). \end{aligned}$$

For X and Y vector fields taking values in W, the respective shape operators of these normals are

$$\begin{aligned} B_1(X,Y)&= -\frac{1}{\lambda _x}{{\textbf{b}}}(\pi _1(X),\pi _1(Y)) ,\\ B_2(X,Y)&= -\frac{1}{\lambda _y}{{\textbf{b}}}(\pi _2(X),\pi _1(Y)) ,\\ B_3(X,Y)&= -\frac{\eta }{\lambda _x}{{\textbf{b}}}(\pi _1(X),\pi _2(Y)) ,\ \text {and}\\ B_4(X,Y)&= -\frac{\eta }{\lambda _y}{{\textbf{b}}}(\pi _2(X),\pi _2(Y)). \end{aligned}$$

We now suppose that $\pi _1(X)$ and $\pi _2(X)$ are colinear. Furthermore, we recall that

$$\begin{aligned} \pi _1\circ J^\eta&= \eta A\circ \pi _2 ,\\\pi _1\circ K&= A\pi _1,\\\pi _2\circ J^\eta&= A\circ \pi _1 ,\ \text {and}\\\pi _2\circ K&= -A\pi _2. \end{aligned}$$

Recall also that

$$\begin{aligned} A^2 = -\epsilon \text {Id}, \end{aligned}$$

where $\epsilon$ here denotes the sign of ${{\textbf{b}}}$. For each i, let $\sigma _i$ denote the component of curvature arising from $B_i$. We obtain

$$\begin{aligned} \sigma _1(X,J^\eta X,X,KX)&=-\frac{1}{\Vert x\Vert _{{\textbf{b}}}^2}{{\textbf{b}}}(\pi _1(X),\pi _1(X)){{\textbf{b}}}(\pi _1(J^\eta X),\pi _1(KX))\\&=-\frac{\eta }{\Vert x\Vert _{{\textbf{b}}}^2}{{\textbf{b}}}(\pi _1(X),\pi _1(X)){{\textbf{b}}}(A\pi _2(X),A\pi _1(X))\\&=-\frac{\epsilon \eta }{\Vert x\Vert _{{\textbf{b}}}^2}{{\textbf{b}}}(\pi _1(X),\pi _1(X)){{\textbf{b}}}(\pi _1(X),\pi _2(X)). \end{aligned}$$

In a similar manner, we obtain

$$\begin{aligned} \sigma _2(X,J^\eta ,X,KX)&= -\frac{\epsilon }{\Vert y\Vert _{{\textbf{b}}}^2}{{\textbf{b}}}(\pi _1(X),\pi _1(X)){{\textbf{b}}}(\pi _2(X),\pi _1(X)),\\ \sigma _3(X,J^\eta ,X,KX)&= \frac{\epsilon }{\Vert x\Vert _{{\textbf{b}}}^2}{{\textbf{b}}}(\pi _1(X),\pi _2(X)){{\textbf{b}}}(\pi _2(X),\pi _2(X)),\ \text {and}\\ \sigma _4(X,J^\eta ,X,KX)&= \frac{\epsilon \eta }{\Vert y\Vert _{{\textbf{b}}}^2}{{\textbf{b}}}(\pi _2(X),\pi _2(X)){{\textbf{b}}}(\pi _1(X),\pi _1(X)). \end{aligned}$$

Combining these identities yields

$$\begin{aligned} g(R_{X,J^\eta X}X,KX)&= {{\textbf{b}}}(\pi _1(X),\pi _2(X))\bigg (-\frac{\epsilon \eta }{\Vert x\Vert _{{\textbf{b}}}^2} {\hat{g}}(X,X)-\frac{\epsilon }{\Vert y\Vert _{{\textbf{b}}}^2}{\hat{g}}(X,X)\bigg )\\&= \frac{-\epsilon \Vert (x,y)\Vert _g^2}{2\Vert x\Vert _{{\textbf{b}}}^2\Vert y\Vert _{{\textbf{b}}}^2}{\hat{g}}(X,I^\eta X){\hat{g}}(X,X), \end{aligned}$$

as desired.$\square$