CONSTANT ANGLE SURFACES IN PRODUCT SPACESThanks: This research was supported by Research Grant G.0432.07 of the Research Foundation-Flanders (FWO)

Let $M^{2}(c_{1})\times M^{2}(c_{2})$ be the product of two $2$-dimensional space forms of constant sectional curvature $c_{1}$ and $c_{2}$ with the standard product metric $\widetilde{g}$, with $c_{1}$ and $c_{2}$ not both $0$. Denote by $\widetilde{\nabla}$ the Levi-Civita connection of $(M^{2}(c_{1})\times M^{2}(c_{2}),\widetilde{g})$ and by $F$ the product structure of $M^{2}(c_{1})\times M^{2}(c_{2})$, see [16]. This is the $(1,1)$-tensor of $M^{2}(c_{1})\times M^{2}(c_{2})$ defined by

$$F(X_{1}+X_{2})=X_{1}-X_{2},$$

for any vector field $X=X_{1}+X_{2}$, where $X_{1}$ and $X_{2}$ denote the parts of $X$ tangent to the first and second factors, respectively. By definition of the product structure $F$, we see that $\frac{I+F}{2}(X)$ is the projection of the vector field $X$ on the first component and that the $(1,1)$-tensor $\frac{I+F}{2}$ has rank $2$ everywhere. Analogously we have that $\frac{I-F}{2}(X)$ is the projection of the vector field $X$ on the second component and that the $(1,1)$-tensor $\frac{I-F}{2}$ has rank $2$ everywhere. We note that the product structure has the following properties:

(1)		$$\displaystyle F^{2}=I\>(F\neq I),$$
(2)		$$\displaystyle\widetilde{g}(FX,Y)=\widetilde{g}(X,FY),$$

and

$$(\widetilde{\nabla}_{X}F)(Y)=0,$$

for any vector field $X$ and $Y$ of $M^{2}(c_{1})\times M^{2}(c_{2})$. The Riemann-Christoffel curvature tensor $\widetilde{R}$ of $(M^{2}(c_{1})\times M^{2}(c_{2}),\widetilde{g})$ is given by

$$\widetilde{R}(X,Y)Z=c_{1}\left(\frac{I+F}{2}(X)\wedge\frac{I+F}{2}(Y)\right)Z+c_{2}\left(\frac{I-F}{2}(X)\wedge\frac{I-F}{2}(Y)\right)Z,$$

where $\wedge$ associates to two tangent vectors $v,w\in T_{p}(M^{2}(c_{1})\times M^{2}(c_{2}))$ the endomorphism defined by

$$(v\wedge w)u=\widetilde{g}(w,u)v-\widetilde{g}(v,u)w,$$

for every $u\in T_{p}(M^{2}(c_{1})\times M^{2}(c_{2}))$.

Let us now consider a surface $M^{2}$ immersed in $M^{2}(c_{1})\times M^{2}(c_{2})$. We will denote by $X,Y,Z,\dots$ tangent vector fields and by $\xi,\xi_{1},\xi_{2},\dots$ vector fields normal to $M^{2}$ in $M^{2}(c_{1})\times M^{2}(c_{2})$. We can now let the product structure $F$ of $M^{2}(c_{1})\times M^{2}(c_{2})$ act on a tangent vector field $X$ or on a normal vector field $\xi$. We can consider the decomposition of $FX$ and $F\xi$ into a tangent component and a normal component as

	$$\displaystyle FX=fX+hX,$$
	$$\displaystyle F\xi=s\xi+t\xi,$$

where $f:TM^{2}\rightarrow TM^{2}$, $h:TM^{2}\rightarrow T^{\perp}M^{2}$, $s:T^{\perp}M^{2}\rightarrow TM^{2}$ and $t:T^{\perp}M^{2}\rightarrow T^{\perp}M^{2}$ are $(1,1)$-tensors on $M^{2}$. It can be easily deduced from equations $(\ref{F^2})$ and $(\ref{Sym})$ that

(3)		$$\displaystyle f\textrm{ is a symmetric }(1,1)\textrm{-tensor field on }M^{2}\textrm{ such that }f^{2}X=X-shX,$$
(4)		$$\displaystyle t\textrm{ is a symmetric }(1,1)\textrm{-tensor field on }M^{2}\textrm{ such that }t^{2}\xi=\xi-hs\xi,$$
(5)		$$\displaystyle\widetilde{g}(hX,\xi)=g(X,s\xi),$$
(6)		$$\displaystyle fs\xi+st\xi=0\textrm{ and }hfX+thX=0,$$

for every $X\in TM^{2}$ and every $\xi\in T^{\perp}M^{2}$. If we denote by $R$ the Riemann-Christoffel curvature tensor of $M^{2}$, then with the previous notations we obtain that Gauss, Codazzi and Ricci equations are written as follows in terms of $f$ and $h$:

(7)

$$R(X,Y)Z=S_{\sigma(Y,Z)}X-S_{\sigma(X,Z)}Y+a((X\wedge Y)Z+\\ (fX\wedge fY)Z)+b(f(X\wedge Y)Z+(X\wedge Y)fZ),$$

(8)

$$(\nabla\sigma)(X,Y,Z)-(\nabla\sigma)(Y,X,Z)=a(g(fY,Z)hX-g(fX,Z)hY)+\\ b(g(Y,Z)hX-g(X,Z)hY),$$

(9)

$$R^{\perp}(X,Y)\xi=a(\widetilde{g}(hY,\xi)hX-\widetilde{g}(hX,\xi)hY)-\sigma(S_{\xi}X,Y)+\sigma(S_{\xi}Y,X),$$

where $a=\frac{c_{1}+c_{2}}{4}$ and $b=\frac{c_{1}-c_{2}}{4}$. Since $M^{2}$ is a surface immersed in $M^{2}(c_{1})\times M^{2}(c_{2})$, we have that equation (7) is equivalent to the fact that the Gaussian curvature $K$ is equal to

$$\det(S_{\xi_{1}})+\det(S_{\xi_{2}})+c_{1}\det(\frac{I+f}{2})+c_{2}\det(\frac{I-f}{2}),$$

where $\{\xi_{1},\xi_{2}\}$ is an orthonormal basis of $T^{\perp}M^{2}$. Moreover we have the following proposition that we can prove using the formulas of Gauss and Weingarten and the fact that $\widetilde{\nabla}F=0$.

We remark that the $(1,1)$-tensor $s$ is, in some sense, a kind of transpose of the $(1,1)$-tensor $h$, because of equation (5). We can easily see that equations (11) and (13) are equivalent because of equation (5). Analogously we can see that the two equations of (6) are equivalent.

The equations (7), (8), (9), (10), (11) and (12) are called the compatibility equations of surfaces in $M^{2}(c_{1})\times M^{2}(c_{2})$. The following theorems follow from more general results proven in [12].

In this short subsection we will discuss curves in $2$-dimensional space forms $M^{2}(c)$ with $c\neq 0$. It is known that $M^{2}(c)$ is isometric to the $2$-dimensional sphere $\mathbb{S}^{2}(c)$ of radius $\frac{1}{\sqrt{c}}$ if $c>0$, i.e.

$$\mathbb{S}^{2}(c)=\{(p_{1},p_{2},p_{3})\in\mathbb{E}^{3}\;|\;p_{1}^{2}+p_{2}^{2}+p_{3}^{2}=\frac{1}{c}\},$$

endowed with the induced metric of $\mathbb{E}^{3}$. The tangent space $T_{p}\mathbb{S}^{2}(c)$ in every point $p$ is given by

$$\mathbb{S}^{2}(c)=\{v\in T_{p}\mathbb{E}^{3}\;|\;\langle p,v\rangle=0\}.$$

Using the cross-product $\times$ in $\mathbb{E}^{3}$, we define a complex structure $J$ on $T\mathbb{S}^{2}(c)$ by

$$J:T\mathbb{S}^{2}(c)\rightarrow T\mathbb{S}^{2}(c):v_{p}\mapsto\sqrt{c}(p\times v)_{p}.$$

It is easy to see that if $v\in T_{p}\mathbb{S}^{2}(c)$ and $\|v\|^{2}=1$, then $\{v,Jv\}$ is an orthonormal basis of $T_{p}\mathbb{S}^{2}(c)$.

We can define in a similar manner a complex structure when $c<0$. It is known that $M^{2}(c)$ is isometric to the hyperbolic plane $\mathbb{H}^{2}(c)$ if $c<0$. We use here the Minkowski or the hyperboloid model of the hyperbolic plane. Denote by $\mathbb{R}^{3}_{1}$ the Minkowski $3$-space with standard coordinates $p_{1},p_{2}$ and $p_{3}$, endowed with the Lorentzian metric

$$\langle.,.\rangle_{1}=-dp_{1}^{2}+dp_{2}^{2}+dp_{3}^{2}.$$

The hyperbolic plane $\mathbb{H}^{2}(c)$ can be constructed as the upper sheet ($p_{1}>0$) of the hyperboloid

$$\{(p_{1},p_{2},p_{3})\in\mathbb{R}^{3}_{1}\;|\;-p_{1}^{2}+p_{2}^{2}+p_{3}^{2}=\frac{1}{c}\},$$

endowed with the induced metric of $\mathbb{R}^{3}_{1}$. The tangent space $T_{p}\mathbb{H}^{2}(c)$ in every point $p$ is given by

$$T_{p}\mathbb{H}^{2}(c)=\{v\in T_{p}\mathbb{R}^{3}_{1}\;|\;\langle p,v\rangle_{1}=0\}.$$

Using the Lorentzian cross-product $\boxtimes$ in $\mathbb{R}^{3}_{1}$ (see for example [7]), we define a complex structure $J$ on $T\mathbb{H}^{2}(c)$ by

$$J:T\mathbb{H}^{2}(c)\rightarrow T\mathbb{H}^{2}(c):v_{p}\mapsto\sqrt{-c}(p\boxtimes v)_{p}.$$

It is easy to see that if $v\in T_{p}\mathbb{H}^{2}(c)$ and $\|v\|^{2}=1$, then $\{v,Jv\}$ is an orthonormal basis of $T_{p}\mathbb{H}^{2}(c)$. In the following we will denote $J$ as the complex structure of $M^{2}(c)$.

Let $\alpha:I\rightarrow M^{2}(c)$ be an arc-length parameterized curve in $M^{2}(c)$. Denote by $T(s)\in T_{\alpha(s)}M^{2}(c)$ the tangent unit vector $\alpha^{\prime}(s)$ and by $N(s)\in T_{\alpha(s)}M^{2}(c)$ the normal vector $JT(s)$. By direct calculations, one can show that

$$\displaystyle T^{\prime}=D_{T}T=\kappa N-c\alpha,N^{\prime}=D_{T}N=-\kappa T,$$

where $D$ is the Levi-Civita connection of $\mathbb{E}^{3}$ or of $\mathbb{R}^{3}_{1}$. We call $\kappa$ the geodesic curvature of $\alpha$ in $M^{2}(c)$. We will need the geodesic curvature of a curve in $M^{2}(c)$ in order to state our classification results of constant angle surfaces.

Let $\widetilde{J}$ and $\overline{J}$ be complex structures on $M^{2}(c_{1})\times M^{2}(c_{2})$ defined by

$$\widetilde{J}v=\widetilde{J}(v_{1},v_{2})=(J_{1}v_{1},J_{2}v_{2})=(J_{1}\frac{I+F}{2}v,J_{2}\frac{I-F}{2}v)$$

and

$$\overline{J}v=\overline{J}(v_{1},v_{2})=(J_{1}v_{1},-J_{2}v_{2})=(J_{1}\frac{I+F}{2}v,-J_{2}\frac{I-F}{2}v),$$

respectively, where $J_{1}$ and $J_{2}$ denote the standard complex structures on $M^{2}(c_{1})$ and $M^{2}(c_{2})$. We obtain the following connection between the angle functions and the complex structures $\widetilde{J}$ and $\overline{J}$.

By direct computations, one can now easily prove the following theorem.

We show now that totally geodesic surfaces in $M^{2}(c_{1})\times M^{2}(c_{2})$ are constant angle surfaces in $M^{2}(c_{1})\times M^{2}(c_{2})$.

In this section we will give a local classification of totally geodesic surfaces for which $\lambda_{1}=\lambda_{2}=-\frac{b}{a}$ with $c_{1}c_{2}>0$. The other cases will be treated in the next sections and will appear as special cases of constant angle surfaces. We classify the totally geodesic surfaces of $M^{2}(c_{1})\times M^{2}(c_{2})$ in Theorem 7. Suppose that $c_{1},c_{2}>0$, the other case can be treated analogously and the result of the second case is stated together with the first case in Proposition 4.

We can immerse $M^{2}(c_{1})\times M^{2}(c_{2})$ as a submanifold of codimension $2$ in the Euclidean space $\mathbb{E}^{6}$. We also remark that we obtain, by using the equation of Gauss, that the surface $M^{2}$ has constant Gaussian curvature $\frac{c_{1}c_{2}}{c_{1}+c_{2}}$. So let $\psi:M^{2}\rightarrow M^{2}(c_{1})\times M^{2}(c_{2})$ be a totally geodesic surface in $M^{2}(c_{1})\times M^{2}(c_{2})$ with $\lambda_{1}=\lambda_{2}=-\frac{b}{a}$. Let us fix a point $p$ in open set $U$ of $M^{2}$ and let $(u,v)$ be Fermi coordinates of $U$ in $M^{2}$, there always exist such coordinates on an open set of a surface (see for example [11]). The metric $g$ of $M$ then has the form

$$du^{2}+G(u,v)dv^{2}$$

on the open set $U$ of $M$, with $G(0,v)=1$ and $\frac{\partial}{\partial_{u}}G(0,v)=0$ for every $v$, in terms of the Fermi coordinates $(u,v)$. Since $M^{2}$ has constant Gaussian curvature, we have that $G$ is uniquely determined by the partial differential equation

$$\frac{\partial^{2}}{\partial^{2}u}\sqrt{G}=-\frac{c_{1}c_{2}}{c_{1}+c_{2}}\sqrt{G}.$$

So we find that $G$ is given by

(18)

$$\cos^{2}(\sqrt{\frac{c_{1}c_{2}}{c_{1}+c_{2}}}u),$$

because of the initial conditions $G(0,v)=1$ and $\frac{\partial}{\partial_{u}}G(0,v)=0$ for every $v$. Let us now consider $M^{2}$ as a surface of codimension $4$ immersed in $\mathbb{E}^{6}$. The formulas of Gauss are then given by

(19)		$$\displaystyle D_{\partial_{u}}\partial_{u}=-\frac{c_{1}c_{2}}{c_{1}+c_{2}}\overrightarrow{x},$$
(20)		$$\displaystyle D_{\partial_{u}}\partial_{v}=D_{\partial_{v}}\partial_{u}=-\sqrt{\frac{c_{1}c_{2}}{c_{1}+c_{2}}}\tan(\sqrt{\frac{c_{1}c_{2}}{c_{1}+c_{2}}}u)\partial_{v},$$
(21)		$$\displaystyle D_{\partial_{v}}\partial_{v}=\sqrt{\frac{c_{1}c_{2}}{c_{1}+c_{2}}}\cos(\sqrt{\frac{c_{1}c_{2}}{c_{1}+c_{2}}}u)\sin(\sqrt{\frac{c_{1}c_{2}}{c_{1}+c_{2}}}u)\partial_{u}-\frac{c_{1}c_{2}}{c_{1}+c_{2}}\cos^{2}(\sqrt{\frac{c_{1}c_{2}}{c_{1}+c_{2}}}u)\overrightarrow{x},$$

where $\overrightarrow{x}$ is the position vector of $M^{2}$ in $\mathbb{E}^{6}$. Solving equations (19) and (20) we find that $\psi$ is locally given by

$$(\cos(\sqrt{\frac{c_{1}c_{2}}{c_{1}+c_{2}}}u)\widetilde{f}_{1}(v)+\sin(\sqrt{\frac{c_{1}c_{2}}{c_{1}+c_{2}}}u)\widetilde{g}_{1},\dots,\cos(\sqrt{\frac{c_{1}c_{2}}{c_{1}+c_{2}}}u)\widetilde{f}_{3}(v)+\sin(\sqrt{\frac{c_{1}c_{2}}{c_{1}+c_{2}}}u)\widetilde{g}_{3},\\ \cos(\sqrt{\frac{c_{1}c_{2}}{c_{1}+c_{2}}}u)\overline{f}_{1}(v)+\sin(\sqrt{\frac{c_{1}c_{2}}{c_{1}+c_{2}}}u)\overline{g}_{1},\dots,\cos(\sqrt{\frac{c_{1}c_{2}}{c_{1}+c_{2}}}u)\overline{f}_{3}(v)+\sin(\sqrt{\frac{c_{1}c_{2}}{c_{1}+c_{2}}}u)\overline{g}_{3}),$$

where $\widetilde{g}=(\widetilde{g}_{1},\widetilde{g}_{2},\widetilde{g}_{3})$ and $\overline{g}=(\overline{g}_{1},\overline{g}_{2},\overline{g}_{3})$ are constant vectors in $\mathbb{R}^{3}$. Moreover we have the following conditions

	$$\displaystyle g(\psi_{u},\psi_{u})=1,\;g(\psi_{u},\psi_{v})=0,\;g(\psi_{v},\psi_{v})=\cos^{2}(\sqrt{\frac{c_{1}c_{2}}{c_{1}+c_{2}}}u),$$
	$$\displaystyle g(\psi_{u},h\partial_{u})=0,\;g(\psi_{u},h\partial_{v})=0,\;g(h\partial_{u},h\partial_{u})=\frac{4c_{1}c_{2}}{(c_{1}+c_{2})^{2}},\;g(h\partial_{u},h\partial_{v})=0,$$
	$$\displaystyle g(\psi_{v},h\partial_{v})=0,\;g(\psi_{v},h\partial_{u})=0,\;g(h\partial_{v},h\partial_{v})=\frac{4c_{1}c_{2}}{(c_{1}+c_{2})^{2}}\cos^{2}(\sqrt{\frac{c_{1}c_{2}}{c_{1}+c_{2}}}u),$$
	$$\displaystyle g(\psi_{u},\widetilde{\xi})=0,\;g(\psi_{v},\widetilde{\xi})=0,\;g(\widetilde{\xi},\widetilde{\xi})=\frac{1}{c_{1}},$$
	$$\displaystyle g(\psi_{u},\overline{\xi})=0,\;g(\psi_{v},\overline{\xi})=0,\;g(\overline{\xi},\overline{\xi})=\frac{1}{c_{2}},$$
	$$\displaystyle g(h\partial_{u},\widetilde{\xi})=g(h\partial_{v},\widetilde{\xi})=g(h\partial_{u},\overline{\xi})=g(h\partial_{v},\overline{\xi})=0,$$

in which $\widetilde{\xi}=(\psi_{1},\psi_{2},\psi_{3},0,0,0)$ and $\overline{\xi}=(0,0,0,\psi_{4},\psi_{5},\psi_{6})$. This conditions are equivalent to

	$$\displaystyle\sum_{i=1}^{3}\widetilde{f}_{i}^{2}=\sum_{i=1}^{3}\widetilde{g}_{i}^{2}=\frac{1}{c_{1}},$$
	$$\displaystyle\sum_{j=1}^{3}\overline{f}_{j}^{2}=\sum_{j=1}^{3}\overline{g}_{j}^{2}=\frac{1}{c_{2}},$$
	$$\displaystyle\sum_{i=1}^{3}\widetilde{f}_{i}\widetilde{g}_{i}=\sum_{i=1}^{3}\widetilde{f}^{\prime}_{i}\widetilde{g}_{i}=0,$$
	$$\displaystyle\sum_{j=1}^{3}\overline{f}_{j}\overline{g}_{j}=\sum_{j=1}^{3}\overline{f}^{\prime}_{j}\overline{g}_{j}=0,$$
	$$\displaystyle\sum_{i=1}^{3}(\widetilde{f}^{\prime}_{i})^{2}=\frac{c_{2}}{c_{1}+c_{2}},$$
	$$\displaystyle\sum_{j=1}^{3}(\overline{f}^{\prime}_{i})^{2}=\frac{c_{1}}{c_{1}+c_{2}}.$$

From the above equations we can conclude that $\widetilde{f}=(\widetilde{f}_{1},\widetilde{f}_{2},\widetilde{f}_{3})$ and $\overline{f}=(\overline{f}_{1},\overline{f}_{2},\overline{f}_{3})$ are curves in $M^{2}(c_{1})$ and $M^{2}(c_{2})$, respectively. Moreover we see that $\widetilde{f}$ and $\overline{f}$ are curves of speed $\sqrt{\frac{c_{2}}{c_{1}+c_{2}}}$ and $\sqrt{\frac{c_{1}}{c_{1}+c_{2}}}$, respectively. The constant vector $\widetilde{g}$ is perpendicular to the vectors $\widetilde{f}$ and $\widetilde{f}^{\prime}$ and the constant vector $\overline{g}$ is perpendicular to the vectors $\overline{f}$ and $\overline{f}^{\prime}$. Hence we obtain that $\widetilde{g}=\pm\sqrt{\frac{c_{1}+c_{2}}{c_{2}}}\widetilde{f}\times\widetilde{f}^{\prime}$ and $\overline{g}=\pm\sqrt{\frac{c_{1}+c_{2}}{c_{1}}}\overline{f}\times\overline{f}^{\prime}$. Since $\widetilde{g}$ and $\overline{g}$ are constant vectors we obtain that the curves $\widetilde{f}$ and $\overline{f}$ are circles of radius $\frac{1}{\sqrt{c_{1}}}$ and $\frac{1}{\sqrt{c_{2}}}$, respectively. We obtain the following proposition.