Almost Complex Surfaces in the Nearly Kähler SL(2,ℝ) × SL(2,ℝ)

Abstract

The space $SL(2,\mathbb{R})\times SL(2,\mathbb{R})$ admits a natural homogeneous pseudo-Riemannian nearly Kähler structure. We investigate almost complex surfaces in this space. In particular, we obtain a complete classification of the totally geodesic almost complex surfaces and of the almost complex surfaces with parallel second fundamental form.

1. Introduction

Almost complex structures were introduced by C. Ehresmann in 1947 [1] in order to study the problem of finding differentiable manifolds that admit a complex analytic structure related to the differentiable structure on the manifold. Approaching the same problem but by using different methods, H. Hopf has shown that the spheres $S^4$ and $S^8$ do not have any complex structure [2]. Moreover, A. Borel and J.P. Serre [3,4] have shown that $S^2$ and $S^6$ are the only spheres admitting almost complex structures. Up to today, it is still an open problem to determine whether, on $S^6$, there exists an integrable almost complex structure.

A manifold endowed with an almost complex structure is said to be an almost complex manifold. If the structure is compatible with the metric, the manifold is called an almost Hermitian manifold. We can consider different types of submanifolds of an almost Hermitian manifold (C. Ehresmann [5]). Two natural classes for instance are almost complex and totally real submanifolds, which are defined, as follows. A submanifold M of an almost complex manifold $\tilde{M}$ is called almost complex (resp. totally real) if each tangent space of M is mapped into itself (resp. into the normal space) by the almost complex structure of $\tilde{M}.$ An almost complex submanifold is endowed with an almost complex structure that is induced by that of the ambient manifold. C. Ehresmann showed (non-published results) that there exist always, for any dimension of the manifold, real submanifolds. A similar statement is not true for almost complex submanifolds if the dimension of the manifold is not 2. For example, $S^6$ does not admit almost complex submanifolds of dimension 4 and there does not exist almost complex functions on $S^6.$ S.S. Chern [6] also studied the problem of existence of complex analytic functions on a complex manifold.

We are interested in nearly Kähler manifolds, which are (pseudo-)Riemannian manifolds that are endowed with an almost complex structure compatible with a metric on the manifold satisfying one additional condition (Definition 1). These manifolds have been intensively studied by Gray [7]. In a recent work of J.-B. Butruille [8], it has been shown that the only homogeneous six-dimensional Riemannian nearly Kähler manifolds are the nearly Kähler six-sphere, $S^3\times S^3,$ the projective space $\mathbb{C}{P}^3$, and the flag manifold $SU\left(3\right)/U\left(1\right)\times U\left(1\right).$

In this paper, we study the analogue of the nearly Kähler space $S^3\times S^3$ in the pseudo-Riemannian case, which is $SL(2,\mathbb{R})\times SL(2,\mathbb{R})$. In particular, we study almost complex surfaces in the six-dimensional space $SL(2,\mathbb{R})\times SL(2,\mathbb{R})$, where a nearly Kähler structure exists naturally. We give this structure explicitly. Just as for its analogue $S^3\times S^3$, it is of course homogeneous, see [9]. However, in contrast to the Riemannian case, a complete classification of such spaces in the pseudo-Riemannian case does not yet exist. All that is known is that besides the analogues to the Riemannian spaces some exceptional examples do exist. In this paper, we investigate in details totally geodesic and parallel almost complex surfaces in $SL(2,\mathbb{R})\times SL(2,\mathbb{R})$.

2. Preliminaries

In this section, we will recall some basic definitions, properties, and formulas that will be used in what follows.

An almost Hermitian manifold is a (pseudo-)Riemannian manifold $(M,g,J)$, which admits an endomorphism J of the tangent bundle, such that

• $J^2=-Id$, i.e., J is an almost complex structure and
• J is compatible with the metric g, i.e., $g(JX,JY)=g(X,Y)$ for all vector fields X and $Y.$

We remark that the first condition requires the real dimension of M to be even.

Definition 1.

An almost Hermitian manifold is called a nearly Kähler manifold if the almost complex structure J verifies

$$\left({\tilde{\nabla }}_{X}J\right)X=0,$$

where $\tilde{\nabla }$ is the Levi–Civita connection that is associated with $g.$ This is equivalent to saying that the tensor G defined by $G(X,Y):=\left({\tilde{\nabla }}_{X}J\right)Y$ is skew-symmetric.

For all vector fields $X,Y,Z,$ the tensor G satisfies the following properties:

$$\left\{\begin{array}{cc}G(X,Y)+G(Y,X)\hfill & =0,\hfill \\ G(X,JY)+JG(X,Y)\hfill & =0,\hfill \\ g\left(G\right(X,Y),Z)+g\left(G\right(X,Z),Y)\hfill & =0.\hfill \end{array}\right.$$

(1)

Remark that, from the last two equations of (1), we can notice that the lowest dimension, in which non-Kähler (i.e., G does not vanish identically) nearly Kähler manifold can exist is 6.

There are two very natural types of submanifolds of nearly Kähler manifolds to be studied: almost complex submanifolds, for which the tangent spaces are invariant by the almost complex structure J (i.e., $JTM=TM$) and totally real submanifolds for which J interchanges tangent and normal vectors (i.e., $JTM\perp TM$).

Moreover, it will be convenient to mention some formulas that will be used through the calculation of the paper. First, the formula of Gauss that gives the decomposition of ${\nabla }_{X}Y$ for tangent vectors $X,Y$ on a submanifold M, where ∇ is the induced connection on $M.$ The Gauss formula is given by

$${\tilde{\nabla }}_{X}Y={\nabla }_{X}Y+h(X,Y),$$

where $\tilde{\nabla }$ is the Levi–Civita connection on the ambient space and h is the second fundamental form. Next, the formula of Weingarten that gives the decomposition along a tangent and a normal vector $\xi$, is given by

$${\tilde{\nabla }}_X\xi =-A_{\xi }X+{\nabla }_X^{\perp }\xi ,$$

where A is the shape operator that is given by the relation $g(h(X,Y),\xi )=g(A_{\xi }X,Y)$ and ${\nabla }^{\perp }$ is the normal connection on $M.$

We recall from [10] some useful relations that hold for an almost complex submanifold M and that follow from the Gauss and Weingarten formulas:

$$\begin{array}{cccc}\hfill {\nabla }_{X}JX& =J{\nabla }_{X}X,\hfill & \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hfill h(X,JY)& =Jh(X,Y),\hfill \\ \hfill A_{J\xi }X=J{A}_{\xi }X& =-A_{\xi }JX,\hfill & \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hfill G(X,\xi )& ={\nabla }_X^{\perp }J\xi -J{\nabla }_X^{\perp }\xi ,\hfill \end{array}$$

where $X,Y$ are tangent vectors and $\xi$ is a normal vector of $M.$

Finally, let us finish this section by the Gauss and Codazzi equations that are respectively given by:

$$\left\{\begin{array}{c}g(\tilde{R}(X,Y)Z,W)=g(R(X,Y)Z,W)+g(h(X,Z),h(Y,W))-g(h(Y,Z),h(X,W)),\hfill \\ {\left(\tilde{R}(X,Y)Z\right)}^{\perp }=\left(\tilde{\nabla }h\right)(X,Y,Z)-\left(\tilde{\nabla }h\right)(Y,X,Z),\hfill \end{array}\right.$$

where $\tilde{R}$ denotes the curvature tensor of the ambient space. Another useful property is the Ricci identity, which states that

$$\left({\nabla }^{2}h\right)(X,Y,Z,W)-\left({\nabla }^{2}h\right)(Y,X,Z,W)=R^{\perp }(X,Y)h(Z,W)-h(R(X,Y)Z,W)-h(X,R(Y,Z)W).$$

3. The Nearly Kähler Structure on SL(2,$\mathbb{R}$) × SL(2,$\mathbb{R}$)

To begin with, consider the nondegenerate indefinite inner product in ${\mathbb{R}}^4$ given by

$$⟨a,b⟩=-\frac{1}{2}(a_1{b}_4-a_2{b}_3-a_3{b}_2+a_4{b}_1),$$

(2)

where $a=(a_1,a_2,a_3,a_4)$, $b=(b_1,b_2,b_3,b_4)\in {\mathbb{R}}^4.$

By identifying the $2\times 2$ real matrices space, $M(2,\mathbb{R})$, with the four-dimensional vector space ${\mathbb{R}}^4$, the above inner product can be viewed on $M(2,\mathbb{R})$ as

$$⟨A,B⟩=-\frac{1}{2}\mathrm{Trace}\left({\left(\mathrm{adj}A\right)}^{T}B\right)\mathrm{for}A,B\in M(2,\mathbb{R}).$$

(3)

Therefore, the space of $2\times 2$-real matrices with a determinant 1, denoted by $SL(2,\mathbb{R}),$, is given by

$$SL(2,\mathbb{R})=\left\{A\in M(2,\mathbb{R})|⟨A,A⟩=-1\right\}.$$

The restriction of $⟨.,.⟩$ to the tangent spaces at the points of $SL(2,\mathbb{R})$ will also be denoted by $⟨.,.⟩$. We then consider $SL(2,\mathbb{R})$, equipped with the pseudo-Riemannian metric $⟨.,.⟩$.

Now, let us proceed by defining the tangent space of $SL(2,\mathbb{R})\times SL(2,\mathbb{R}).$ For that aim, let $(A,B)\in SL(2,\mathbb{R})\times SL(2,\mathbb{R})$. By the natural identification

$$T_{(A,B)}(SL(2,\mathbb{R})\times SL(2,\mathbb{R}))\cong T_{A}SL(2,\mathbb{R})\oplus T_{B}SL(2,\mathbb{R}),$$

we may write a tangent vector at $(A,B)$ as $Z(A,B)=\left(U\right(A,B),V(A,B\left)\right)$ or simply $Z=(U,V).$

Let $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in SL(2,\mathbb{R})$. The tangent vector fields $X_1,X_2$ and $X_3,$ given by

$$\begin{array}{ccc}\hfill X_1\left(A\right)& =& A\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)=\left(\begin{array}{cc}a& -b\\ c& -d\end{array}\right),\hfill \\ \hfill X_2\left(A\right)& =& A\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)=\left(\begin{array}{cc}b& a\\ d& c\end{array}\right),\hfill \\ \hfill X_3\left(A\right)& =& A\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)=\left(\begin{array}{cc}-b& a\\ -d& c\end{array}\right),\hfill \end{array}$$

form an orthogonal (semi-orthonormal) basis of $T_{A}SL(2,\mathbb{R})$ such that

$$⟨X_1,X_1⟩=1,⟨X_2,X_2⟩=1\mathrm{and}⟨X_3,X_3⟩=-1.$$

Hence, the tangent space of $SL(2,\mathbb{R})$ can be defined as

$$T_{A}SL(2,\mathbb{R})=\left\{A\alpha |\alpha \mathrm{is}\mathrm{a}2\times 2\mathrm{matrix}\mathrm{of}\mathrm{trace}0\right\}.$$

Consequently, the vector fields

$$\begin{array}{cccc}\hfill E_1(A,B)& =\left(A\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right),0\right),\hfill & \hspace{1em}\hspace{1em}\hfill F_1(A,B)& =\left(0,B\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\right),\hfill \\ \hfill E_2(A,B)& =\left(A\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right),0\right),\hfill & \hspace{1em}\hspace{1em}\hfill F_2(A,B)& =\left(0,B\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\right),\hfill \\ \hfill E_3(A,B)& =\left(A\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right),0\right),\hfill & \hspace{1em}\hspace{1em}\hfill F_3(A,B)& =\left(0,B\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)\right),\hfill \end{array}$$

are mutually orthogonal with respect to the usual product metric on $SL(2,\mathbb{R})\times SL(2,\mathbb{R})$, as given by:

$${⟨Z,Z^{\prime }⟩}_{SL(2,\mathbb{R})\times SL(2,\mathbb{R})}=⟨U,U^{\prime }⟩+⟨V,V^{\prime }⟩$$

for $Z=(U,V),Z^{\prime }=(U^{\prime },V^{\prime })$ tangential to $SL(2,\mathbb{R})\times SL(2,\mathbb{R}).$ We note that the usual product metric ${⟨.,.⟩}_{SL(2,\mathbb{R})\times SL(2,\mathbb{R})}$ will also be denoted by $⟨.,.⟩.$ Hence, the Lie brackets are

$$[E_i,E_j]=-2{ϵ}_{ijk}{\eta }_k{E}_k[F_i,F_j]=-2{ϵ}_{ijk}{\eta }_k{F}_k[E_i,F_j]=0$$

where $i,j\in \{1,2,3\}$, ${\eta }_i$ denotes the causal character of $E_i$ and $F_i$ and ${ϵ}_{ijk}$ is the Levi–Civita symbol.

Now, we define the almost complex structure J on $SL(2,\mathbb{R})\times SL(2,\mathbb{R})$ by

$$\begin{array}{ccc}\hfill J{E}_i& =& \frac{1}{\sqrt{3}}\left(2F_i+E_i\right),\hfill \\ \hfill J{F}_i& =& -\frac{1}{\sqrt{3}}\left(E_i+F_i\right),\hfill \end{array}$$

or more generally by

$$J(A\alpha ,B\beta )=\frac{1}{\sqrt{3}}\left(A(\alpha -2\beta ),B(2\alpha -\beta )\right),$$

for $\alpha ,\beta$ 2-dimensional matrices of trace 0 and therefore, ${\displaystyle (A\alpha ,B\beta )\in T_{(A,B)}(SL(2,\mathbb{R})\times SL(2,\mathbb{R})).}$

The metric on $SL(2,\mathbb{R})\times SL(2,\mathbb{R})$ that corresponds to the almost complex structure J is the metric g given as follows:

$$\begin{array}{c}\hfill g(E_i,F_j)=\left\{\begin{array}{ccccc}\hfill -\frac{1}{3}{\delta }_{ij}& \mathrm{for}\hfill & \hfill i& =\hfill & \hfill 1,2,\\ \hfill \frac{1}{3}{\delta }_{ij}& \mathrm{for}\hfill & \hfill i& =\hfill & \hfill 3,\end{array}\right.\end{array}$$

$$\begin{array}{c}\hfill g(E_i,E_j)=g(F_i,F_j)=\left\{\begin{array}{ccccc}\hfill \frac{2}{3}{\delta }_{ij}& \mathrm{for}\hfill & \hfill i& =\hfill & \hfill 1,2,\\ \hfill -\frac{2}{3}{\delta }_{ij}& \mathrm{for}\hfill & \hfill i& =\hfill & \hfill 3.\end{array}\right.\end{array}$$

More generally, on arbitrary tangent vectors in $T_{(A,B)}(SL(2,\mathbb{R})\times SL(2,\mathbb{R}))$, the metric g is given in terms of the usual product metric on $SL(2,\mathbb{R})\times SL(2,\mathbb{R})$ by

$$g((A\alpha ,B\beta ),(A\gamma ,B\delta ))=\frac{2}{3}⟨(A\alpha ,B\beta ),(A\gamma ,B\delta )⟩-\frac{1}{3}⟨(A\beta ,B\alpha ),(A\gamma ,B\delta )⟩$$

(4)

for any $\alpha ,\beta ,\gamma ,\delta \in M(2,\mathbb{R})$ of trace 0.

We can check that the almost complex structure J is compatible with the metric $g.$

Let us now consider the following lemma that will be used through the paper.

Lemma 1.

The Levi–Civita connection $\tilde{\nabla }$ on $SL(2,\mathbb{R})\times SL(2,\mathbb{R})$ with respect to the metric g is given by

$$\begin{array}{cc}\hfill {\tilde{\nabla }}_{E_i}{E}_j& =-{ϵ}_{ijk}{\eta }_k{E}_k,\hfill \\ \hfill {\tilde{\nabla }}_{F_i}{F}_j& =-{ϵ}_{ijk}{\eta }_k{F}_k,\hfill \\ \hfill {\tilde{\nabla }}_{E_i}{F}_j& =\frac{{ϵ}_{ijk}{\eta }_k}{3}(E_k-F_k),\hfill \\ \hfill {\tilde{\nabla }}_{F_i}{E}_j& =\frac{{ϵ}_{ijk}{\eta }_k}{3}(F_k-E_k),.\hfill \end{array}$$

Subsequently, the covariant derivative $\tilde{\nabla }J$ can be computed, as follows

$$\begin{array}{cc}\hfill \left({\tilde{\nabla }}_{E_i}J\right)E_j& =\frac{2}{3\sqrt{3}}{ϵ}_{ijk}{\eta }_k(E_k+2F_k),\hfill \\ \hfill \left({\tilde{\nabla }}_{F_i}J\right)F_j& =-\frac{2}{3\sqrt{3}}{ϵ}_{ijk}{\eta }_k(2E_k+F_k),\hfill \\ \hfill \left({\tilde{\nabla }}_{E_i}J\right)F_j& =\left({\tilde{\nabla }}_{F_i}J\right)E_j=\frac{2}{3\sqrt{3}}{ϵ}_{ijk}{\eta }_k(E_k-F_k).\hfill \end{array}$$

Let us set $G:=\tilde{\nabla }$. Afterwards, G is skew symmetric and it satisfies the following equations

$$G(X,JY)+JG(X,Y)=0,g\left(G\right(X,Y),Z)+g\left(G\right(X,Z),Y)=0,$$

for any vector fields $X,Y,Z\in T\left(SL\right(2,\mathbb{R})\times SL(2,\mathbb{R}\left)\right).$ Therefore, $SL(2,\mathbb{R})\times SL(2,\mathbb{R})$ endowed with g and J, becomes a nearly Kähler manifold.

In order to express the curvature tensor of the nearly Kähler $SL(2,\mathbb{R})\times SL(2,\mathbb{R})$, it is convenient to introduce an almost product structure P on which is defined as

$$\begin{array}{c}\hfill P(pU,qV)=(pV,qU),\forall Z=(pU,qV)\in T_{(p,q)}(SL(2,\mathbb{R})\times SL(2,\mathbb{R})).\end{array}$$

(5)

We can easily verify that P satisfies the following properties:

$$\begin{array}{cc}\hfill P^2& =Id,\hfill \\ \hfill PJ& =-JP,\hfill \\ \hfill g(PZ,P{Z}^{\prime })& =g(Z,Z^{\prime }),\mathrm{i}.\mathrm{e}.,P\mathrm{is}\mathrm{compatible}\mathrm{with}g,\hfill \\ \hfill g(PZ,Z^{\prime })& =g(Z,P{Z}^{\prime }),\mathrm{i}.\mathrm{e}.,P\mathrm{is}\mathrm{symmetric}.\hfill \end{array}$$

It then turns out that the Riemann curvature tensor $\tilde{R}$ on $\left(SL\right(2,\mathbb{R}\left)\right)\times SL(2,\mathbb{R}),g)$ is given by

$$\begin{array}{ccc}\hfill \tilde{R}(U,V)W& =& -\frac{5}{6}\left(g(V,W)U-g(U,W)V\right)\hfill \\ \hfill & & -\frac{1}{6}\left(g(JV,W)JU-g(JU,W)JV-2g(JU,V)JW\right)\hfill \\ \hfill & & -\frac{2}{3}\left(g(PV,W)PU-g(PU,W)PV+g(JPV,W)JPU-g(JPU,W)JPV\right),\hfill \end{array}$$

and the tensors $\tilde{\nabla }G$ and G satisfy

$$\begin{array}{ccc}\hfill \left(\tilde{\nabla }G\right)(X,Y,Z)& =& -\frac{2}{3}\left(g(X,Z)JY-g(X,Y)JZ-g(JY,Z)X\right),\hfill \end{array}$$

(6)

$$\begin{array}{ccc}\hfill g\left(G\right(X,Y),G(Z,W\left)\right)& =& -\frac{2}{3}\left(g(X,Z)g(Y,W)-g(X,W)g(Y,Z)\right.\hfill \\ \hfill & & +\left.g(JX,Z)g(JW,Y)-g(JX,W)g(JZ,Y)\right)\hfill \end{array}$$

(7)

From (7) and by using the fact that $g\left(G\right(X,Y),Z)=-g\left(G\right(X,Z),Y)$, we deduce the following equation

$$\begin{array}{ccc}\hfill G(X,G(Z,W\left)\right)& =& \frac{2}{3}\left(g(X,Z)W-g(X,W)Z+g(JX,Z)JW\right.\hfill \\ \hfill & & \left.-g(JX,W)JZ\right).\hfill \end{array}$$

(8)

Moreover, we can express the tensor G explicitly for any tangent vectors fields. For that aim, let us present the following proposition.

Proposition 1.

Let $X=(A\alpha ,B\beta ),Y=(A\gamma ,B\delta )\in T_{(A,B)}(SL(2,\mathbb{R})\times SL(2,\mathbb{R})).$ Subsequently,

$$G(X,Y)=\frac{2}{3\sqrt{3}}\left(A(-\alpha \times \gamma -\alpha \times \delta +\gamma \times \beta +2\beta \times \delta ),B(-2\alpha \times \gamma +\alpha \times \delta -\gamma \times \beta +\beta \times \delta )\right),$$

where × is a product that is similar to the vector product, which we define on the space of real matrices of dimension 2 and trace 0 by $\alpha \times \beta =\frac{1}{2}(\alpha \beta -\beta \alpha ).$

Proof. 

Let ${\alpha }_1,{\alpha }_2,{\alpha }_3$ be the coefficients of $\alpha$ in the basis $\left\{\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right),\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right),\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)\right\}$, similarly for $\beta .$ Then, we write

$$X=U_{\alpha }+V_{\beta },$$

where $U_{\alpha }={\alpha }_1{E}_1+{\alpha }_2{E}_2+{\alpha }_3{E}_3$ and $V_{\beta }={\beta }_1{F}_1+{\beta }_2{F}_2+{\beta }_3{F}_3$. Similarly,

$$Y=U_{\gamma }+V_{\delta }.$$

By using Lemma 1, we can compute

$$G(U_{\alpha },V_{\beta })=-\frac{2}{3\sqrt{3}}(U_{\alpha \times \beta }-V_{\alpha \times \beta }),G(U_{\alpha },U_{\beta })=-\frac{2}{3\sqrt{3}}(U_{\alpha \times \beta }+2V_{\alpha \times \beta }).$$

As $P{U}_{\alpha }=V_{\alpha },$, we obtain

$$G(V_{\alpha },V_{\beta })=\frac{2}{3\sqrt{3}}(V_{\alpha \times \beta }+2U_{\alpha \times \beta }).$$

Subsequently, by linearity we complete the proof of the proposition. ☐

Moreover, the almost product structure P can be expressed in terms of the usual product structure Q, given by $QZ=Q(A\alpha ,B\beta )=(-A\alpha ,B\beta )$ and vice versa:

$$\begin{array}{cc}\hfill QZ& =-\frac{1}{\sqrt{3}}(2PJZ-JZ),\hfill \end{array}$$

(9)

$$\begin{array}{cc}\hfill PZ& =-\frac{1}{2}(Z+\sqrt{3}JQZ).\hfill \end{array}$$

(10)

The metric g is expressed in terms of the metric $⟨.,.⟩$ by:

$$g(Z,Z^{\prime })=\frac{1}{4}\left(⟨Z,Z^{\prime }⟩+⟨JZ,J{Z}^{\prime }⟩\right).$$

Afterwards,

$$g(QZ,Q{Z}^{\prime })+g(Z,Z^{\prime })=\frac{4}{3}⟨Z,Z^{\prime }⟩,$$

so that the metric $⟨.,.⟩$ can be written in terms of the metric g:

$$⟨Z,Z^{\prime }⟩=2g(Z,Z^{\prime })+g(Z,P{Z}^{\prime }).$$

(11)

Let us end this section with the following.

Lemma 2.

The relation between the Levi-Civita connection $\tilde{\nabla }$ of the metric g and that of the usual product metric $⟨.,.⟩$, denoted ${\nabla }^E$, is

$${\nabla }_X^{E}Y={\tilde{\nabla }}_{X}Y+\frac{1}{2}\left(JG(X,PY)+JG(Y,PX)\right).$$

4. Totally Geodesic Almost Complex Surfaces in SL(2,$\mathbb{R}$) × SL(2,$\mathbb{R}$)

In this section, we start by studying almost complex surfaces in $SL(2,\mathbb{R})\times SL(2,\mathbb{R})$, which are totally geodesic. The reasoning in some cases is similar to that applied in [10] in the case of the nearly Kähler $S^3\times S^3.$ We note that the identities in Lemma 3.1. of [10] remain true in our case and we will always assume that the almost complex surface is a regular surface, i.e., the induced metric is non degenerate. As J is compatible with the metric, this either implies that the induced metric is positive definite or negative definite.

Proposition 2.

If M is a totally geodesic almost complex surface in SL(2,$\mathbb{R}$)×SL(2,$\mathbb{R}$), then either

(1) 

P maps the tangent space into the normal space and the Gaussian curvature K is $-\frac{4}{3}$ and

(2) 

P preserves the tangent space (and therefore also the normal space) and the Gaussian curvature is 0.

Proof. 

Let $(A,B)\in M$ be a point of M and v a tangent vector to M at $(A,B)$ such that $g(v,v)=\pm 1$.

Case 1: If $g(v,v)=1$. Using Codazzi’s equation, we have ${\left(\tilde{R}(v,Jv)v\right)}^{\perp }=0$, then $\tilde{R}(v,Jv)v$ is a multiple of $Jv$. By the Gauss equation, and using the fact that $h=0$, we have

$$R(v,Jv)v=-\frac{4}{3}\left(-Jv+g(PJv,v)Pv-g(Pv,v)PJv\right).$$

Because the metric g is positive definite on v, we can choose v such that $g(v,Pv)$ is maximal for all unit vectors in $(A,B)$, so that $g(Pv,Jv)=g(PJv,v)=0$ (since P is symmetric). Subsequently, the Gauss equation becomes

$$R(v,Jv)v=\frac{4}{3}\left(Jv+g(Pv,v)PJv\right).$$

Now, we will distinguish between two subcases.

If $g(Pv,v)=0$, then $K=g(R(v,Jv)Jv,v)=-\frac{4}{3}.$ In this case $Pv$ and $PJv$ are normal vectors because $g(Pv,v)=g(Pv,Jv)=0$ and $g(PJv,v)=0$, $g(PJv,Jv)=-g(JPv,Jv)=-g(Pv,v)=0$. We can verify in this case that the vectors $v,Jv,Pv$ and $PJv$ are of length 1 and $G(v,Pv)$, $JG(v,Pv)$ are of length $-2/3.$

If $g(Pv,v)\ne 0$, we have $g(PJv,Jv)=-g(Pv,v)\ne 0$, which implies that ${\displaystyle PJv=-g(Pv,v)Jv}$, a non-zero multiple of $Jv$, then $g(PJv,PJv)=-g(Pv,v)g(Jv,PJv).$ Therefore, ${\displaystyle g(Pv,v)=\pm 1}$ and $PJv=\pm Jv$. We assume that $PJv=-Jv$, then $Pv=v$ and $K=-\frac{4}{3}g(Jv+g(Pv,v)PJv,Jv)=-\frac{4}{3}(1-g{(Pv,v)}^2)=0.$

Case 2: If $g(v,v)=-1$. Using Codazzi’s equation we have $\tilde{R}(v,Jv)v$ is a multiple of $Jv$ and by Gauss equation, we deduce

$$R(v,Jv)v=-\frac{4}{3}\left(Jv+g(PJv,v)Pv-g(Pv,v)PJv\right).$$

We can suppose $g(PJv,v)=0$, then the Gauss equation simplifies to

$$R(v,Jv)v=-\frac{4}{3}\left(Jv-g(Pv,v)PJv\right).$$

In this case, the assumption $g(Pv,v)=0$ is not possible, since the vectors $Pv$ and $PJv$ are normal and so we have two tangent vectors (v and $Jv$) and two normal vectors $Pv$ and $PJv$ of negative length. Hence, $g(Pv,v)\ne 0$, as before, $g(Pv,v)=\pm 1$, and consequently, $PJv=\pm Jv$. We assume that $PJv=-Jv$, which implies $Pv=v$ and $K=0$. ☐

Let us now investigate in more detail the two subcases introduced in the previous proposition. We start with a flat surface for which P preserves both the tangent and normal space.

Explicit Examples

In this section, we explicitly find the submanifolds M, such that $PTM\subset TM$ with Gauss curvature $K=0$:

Let $(A,B)\in M$ be a point of M and v a tangent vector to M, such that $g(v,v)=1.$ The vectors v and $Jv$ form an orthonormal basis of $T_{(A,B)}M.$ Additionally, we have

$$Pv=v$$

(12)

and

$$PJv=-Jv.$$

(13)

Similar to Lemma 3.1. in [10], we know in this case that $\left({\nabla }_{X}P\right)Y=0$ for any tangent vectors $X,Y$, which leads to $P\left({\nabla }_{X}v\right)={\nabla }_{X}v$. On the other hand, ${\nabla }_{X}v$ is a multiple of $Jv$ which gives $P\left({\nabla }_{X}v\right)=-{\nabla }_{X}v$. Subsequently, for any tangent vector $X,$

$${\nabla }_{X}v=0,$$

similarly for

$${\nabla }_{X}Jv=0.$$

Consequently, $[v,Jv]=0$ and therefore we can find coordinates s and t on the surface, such that

$$v:=F_s\mathrm{and}Jv:=F_t,$$

where F denotes the immersion, i.e.,

$$F:M\to SL(2,\mathbb{R})\times SL(2,\mathbb{R}):(s,t)\mapsto \left(A\right(s,t),B(s,t\left)\right).$$

Similar to [10], we may assume that $F_t=J{F}_s$ and therefore there are $2\times 2$-real matrices $\alpha ,\beta ,\gamma ,\delta$ with vanishing trace, such that

$$A_s=A\alpha ,B_s=B\beta ,A_t=A\gamma ,B_t=B\delta .$$

The matrices $\alpha ,\beta ,\gamma ,\delta$ are such that

$$\alpha =\beta ,\gamma =-\delta =-\frac{\alpha }{\sqrt{3}}.$$

This comes from $P{F}_s=F_s,J{F}_s=F_t$, which implies the following equations

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}\hfill A_s& =A\alpha \hfill \\ \hfill A_t& =-\frac{1}{\sqrt{3}}A\alpha ,\hfill \end{array}\right.\mathrm{and}\left\{\begin{array}{cc}\hfill B_s& =B\alpha ,\hfill \\ \hfill B_t& =\frac{1}{\sqrt{3}}B\alpha .\hfill \end{array}\right.\end{array}$$

(14)

Now, we will distinguish between two subcases:

• If g is positive definite: From (4), we have

  $$⟨\alpha ,\alpha ⟩=⟨\beta ,\beta ⟩=\frac{3}{2}\mathrm{and}⟨\gamma ,\gamma ⟩=⟨\delta ,\delta ⟩=\frac{1}{2}.$$

Moreover, by (11), we have

$$⟨F_s,F_s⟩=3,⟨F_s,F_t⟩=0,⟨F_t,F_t⟩=1$$

and using (9), we have in general $⟨X,QY⟩=-\sqrt{3}g(X,PJY)$ so that

$$⟨F_s,Q{F}_s⟩=⟨F_t,Q{F}_t⟩=0,⟨F_s,Q{F}_t⟩=⟨F_t,Q{F}_t⟩=\sqrt{3}.$$

Additionally, the usual connection ∇ vanishes on the vectors $F_s$ and $F_t,$ then we deduce

$$F_{ss}=\frac{3}{2}F,F_{tt}=\frac{1}{2}F\mathrm{and}{F}_{st}=\frac{\sqrt{3}}{2}QF,$$

which is equivalent to say that the second derivatives of the components A and B satisfy the following equations

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}\hfill A_{ss}& =\frac{3}{2}A,\hfill \\ \hfill A_{st}& =-\frac{\sqrt{3}}{2}A,\hfill \\ \hfill A_{tt}& =\frac{1}{2}A,\hfill \end{array}\right.\mathrm{and}\left\{\begin{array}{cc}\hfill B_{ss}& =\frac{3}{2}B,\hfill \\ \hfill B_{st}& =\frac{\sqrt{3}}{2}B,\hfill \\ \hfill B_{tt}& =\frac{1}{2}B.\hfill \end{array}\right.\end{array}$$

(15)

On the other hand, the integrablity condition $F_{st}=F_{ts}$ implies that $\alpha$ is a constant matrix. Hence, by deriving (14), we get

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}\hfill A_{ss}& =A\alpha \alpha ,\hfill \\ \hfill A_{tt}& =\frac{1}{3}A\alpha \alpha ,\hfill \end{array}\right.\mathrm{and}\left\{\begin{array}{cc}\hfill B_{ss}& =B\alpha \alpha ,\hfill \\ \hfill B_{tt}& =\frac{1}{3}B\alpha \alpha .\hfill \end{array}\right.\end{array}$$

(16)

By identifying (15) and (16), we obtain

$$\alpha \alpha =\frac{3}{2}I,$$

where I is the identity matrix. The above equation is equivalent to

$$\left(\begin{array}{cc}{\alpha }_1^2+{\alpha }_2^2-{\alpha }_3^2& 0\\ 0& {\alpha }_1^2+{\alpha }_2^2-{\alpha }_3^2\end{array}\right)=\left(\begin{array}{cc}\frac{3}{2}& 0\\ 0& \frac{3}{2}\end{array}\right),$$

where ${\alpha }_1,{\alpha }_2,{\alpha }_3$ are the coefficients of $\alpha$ in the basis

$$\left\{\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right),\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right),\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)\right\}.$$

By using now that on the nearly Kähler $SL(2,\mathbb{R})\times SL(2,\mathbb{R})$, the map

$$(p,q)\mapsto (ApC,BqC)$$

is an isometry. We can assume that $\alpha =\left(\begin{array}{cc}\sqrt{\frac{3}{2}}& 0\\ 0& -\sqrt{\frac{3}{2}}\end{array}\right).$

Now we are able to solve equations (14).

Proposition 3.

A flat totally geodesic almost complex surface M of $SL(2,\mathbb{R})\times SL(2,\mathbb{R})$ with positive definite induced metric is isometric to the immersion $(s,t)\mapsto \left(A\right(s,t),B(s,t\left)\right)$ where

$$A(s,t)=\left(\begin{array}{cc}{e}^{\frac{1}{\sqrt{2}}\left(\sqrt{3}s-t\right)}& 0\\ 0& e^{-\frac{1}{\sqrt{2}}\left(\sqrt{3}s-t\right)}\end{array}\right)$$

and

$$B(s,t)=\left(\begin{array}{cc}{e}^{\frac{1}{\sqrt{2}}\left(\sqrt{3}s+t\right)}& 0\\ 0& e^{-\frac{1}{\sqrt{2}}\left(\sqrt{3}s+t\right)}\end{array}\right)$$

Proof. 

To determine the solution A of the first system of equations in (14), we may pose

$$x=\frac{\sqrt{3}s+t}{2}\mathrm{and}y=\frac{\sqrt{3}s-t}{2}.$$

The system is now equivalent to

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}\hfill A_x& =0\hfill \\ \hfill A_y& =\frac{2}{\sqrt{3}}A\alpha \hfill \end{array}\right.\end{array}$$

which has the solution

$$A(x,y)=k\left(\begin{array}{cc}{e}^{\frac{2}{\sqrt{2}}y}& 0\\ 0& e^{-\frac{2}{\sqrt{2}}y}\end{array}\right),$$

where k is a constant. By applying some isometries, we can suppose that $k=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right).$ Subsequently, we have the solution given in the proposition. In a similar way, we obtain the solution $B.$ ☐

• We now study the case, where g is negative definite

In this case,

$$⟨\alpha ,\alpha ⟩=⟨\beta ,\beta ⟩=-\frac{3}{2}\mathrm{and}⟨\gamma ,\gamma ⟩=⟨\delta ,\delta ⟩=-\frac{1}{2},$$

and from (11),

$$⟨F_s,F_s⟩=-3,⟨F_s,F_t⟩=0,⟨F_t,F_t⟩=-1,$$

also from (9),

$$⟨F_s,Q{F}_s⟩=⟨F_t,Q{F}_t⟩=0,⟨F_s,Q{F}_t⟩=⟨F_t,Q{F}_t⟩=-\sqrt{3}.$$

We proceed similarly as before to obtain that the second derivatives of the components A and B satisfy in this case

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}\hfill A_{ss}& =-\frac{3}{2}A,\hfill \\ \hfill A_{st}& =\frac{\sqrt{3}}{2}A,\hfill \\ \hfill A_{tt}& =-\frac{1}{2}A,\hfill \end{array}\right.\mathrm{and}\left\{\begin{array}{cc}\hfill B_{ss}& =-\frac{3}{2}B,\hfill \\ \hfill B_{st}& =-\frac{\sqrt{3}}{2}B,\hfill \\ \hfill B_{tt}& =-\frac{1}{2}B.\hfill \end{array}\right.\end{array}$$

(17)

and $\alpha$ is a constant matrix that satisfies

$$⟨\alpha ,\alpha ⟩=-\frac{3}{2}$$

hence, $\alpha =\left(\begin{array}{cc}0& \sqrt{\frac{3}{2}}\\ -\sqrt{\frac{3}{2}}& 0\end{array}\right)$

Now the equations (14) can be solved by posing, as before, $x=\frac{\sqrt{3}s+t}{2}$ and $y=\frac{\sqrt{3}s-t}{2}$, we will obtain the solution that is given by the following.

Proposition 4

A flat totally geodesic almost complex surface M of $SL(2,\mathbb{R})\times SL(2,\mathbb{R})$ with negative definite induced metric is isometric to the immersion $(s,t)\mapsto \left(A\right(s,t),B(s,t\left)\right)$ where

$$A(s,t)=\left(\begin{array}{cc}cos\left(\frac{1}{\sqrt{2}}\left(\sqrt{3}s-t\right)\right)& sin\left(\frac{1}{\sqrt{2}}\left(\sqrt{3}s-t\right)\right)\\ -sin\left(\frac{1}{\sqrt{2}}\left(\sqrt{3}s-t\right)\right)& cos\left(\frac{1}{\sqrt{2}}\left(\sqrt{3}s-t\right)\right)\end{array}\right)$$

and

$$B(s,t)=\left(\begin{array}{cc}cos\left(\frac{1}{\sqrt{2}}\left(\sqrt{3}s+t\right)\right)& sin\left(\frac{1}{\sqrt{2}}\left(\sqrt{3}s+t\right)\right)\\ -sin\left(\frac{1}{\sqrt{2}}\left(\sqrt{3}s+t\right)\right)& cos\left(\frac{1}{\sqrt{2}}\left(\sqrt{3}s+t\right)\right)\end{array}\right)$$

5. Parallel Almost Complex Surfaces in SL(2,$\mathbb{R}$) × SL(2,$\mathbb{R}$)

Now, we start the study of almost complex surfaces with parallel second fundamental form. As mentioned before, the properties of Lemma 3.1 of [10] remain valid for almost complex surfaces in SL(2,$\mathbb{R}$)×SL(2,$\mathbb{R})$.

Theorem 1.

If M is an almost complex surface in SL(2,$\mathbb{R}$)×SL(2,$\mathbb{R}$) with parallel second fundamental form, then either

(1)

P maps the tangent space into the normal space and, in this case, either M is totally geodesic with constant Gaussian curvature $-\frac{4}{3}$ or M has constant Gaussian curvature $-\frac{5}{9}$

(2)

P preserves the tangent space, in this case the Gaussian curvature is 0 and M is totally geodesic and, therefore, M is congruent to one of the two previous mentioned examples.

Proof. 

Let $(A,B)\in M$ be a point of M and v a tangent vector to M at $(A,B)$ such that $g(v,v)=\pm 1.$

-

Case 1. We suppose $g(v,v)=1.$

From Codazzi’s equation, we can see that $\tilde{R}(v,Jv)v$ is a multiple of $Jv.$ As we mentioned before, (in Theorem 2) we can suppose

$$g(Pv,Jv)=g(PJv,v)=0.$$

Subsequently, from Gauss equation, we have

$$R(v,Jv)v=\tilde{R}(v,Jv)v+2J{A}_{h(v,v)}v,$$

with

$$\tilde{R}(v,Jv)v=\frac{4}{3}\left(Jv+g(Pv,v)PJv\right).$$

We study the two subcases $g(Pv,v)=0$ and $g(Pv,v)\ne 0.$

If $g(Pv,v)\ne 0,$ then it follows from the Codazzi equation that $PJv$ is a tangent vector. Hence, P preserves the tangent space. Therefore we can take v, such that $g(Pv,v)=\pm 1,$ hence $PJv=\mp Jv.$ It follows then from the Gauss equation that $A_{h(v,v)}v=-\frac{1}{2}Kv$ and the Gauss curvature $K=-{2\left|\right|h(v,v)\left|\right|}^2.$ Accordingly, $\left|\right|A_{h(v,v)}{v\left|\right|}^2={\left|\right|h(v,v)\left|\right|}^4=\frac{K^2}{4}.$ On the other hand, as P preserves the tangent space, we have that $\nabla P=0$. Moreover, as P is symmetric, $P^2=I$ and P anticommutes with J, we can locally find a frame $\{e_1=v,e_2=Jv\}$, such that $P{e}_1=e_1$ and $P{e}_2=PJ{e}_1=-e_2$. It now follows immediately from $\nabla P=0$ that ${\nabla }_X{e}_1={\nabla }_X{e}_2=0$. Hence, we deduce that the surface is flat. As $K=-2\parallel h(v,v)\parallel$, it follows that $h(v,v)=0$. This implies that the surface is totally geodesic.

If $g(Pv,v)=0,$ then P maps tangent vectors into normal vectors and from the Gauss equation

$$K=-\frac{4}{3}-{2\left|\right|h(v,v)\left|\right|}^2.$$

Because $g(PJv,h(v,v\left)\right)=g(Pv,h(v,v\left)\right)=0,$ Ricci equation implies

$$\begin{array}{ccc}\hfill g(R^{\perp }(v,Jv)h(v,v),Jh(v,v))& =& \frac{1}{3}{\left|\right|h(v,v)\left|\right|}^2-{2\left|\right|h(v,v)\left|\right|}^4\hfill \\ \hfill & =& -\frac{K^2}{2}-\frac{3}{2}K-\frac{10}{9}.\hfill \end{array}$$

On the other hand, Ricci identity gives

$$\begin{array}{ccc}\hfill g(R^{\perp }(v,Jv)h(v,v),Jh(v,v))& =& 2g\left(h\right(R(v,Jv)v,v),Jh(v,v\left)\right)\hfill \\ \hfill & =& -2g\left(h\right(KJv,v),Jh(v,v\left)\right)\hfill \\ \hfill & =& -{2K\left|\right|h(v,v)\left|\right|}^2\hfill \\ \hfill & =& K^2+\frac{4}{3}K.\hfill \end{array}$$

By identification, we have the equation

$$\frac{3}{2}{K}^2+\frac{17}{6}K+\frac{10}{9}=0,$$

which has the solutions

$$K=-\frac{4}{3}K=-\frac{5}{9}.$$

-

Case 2. We suppose $g(v,v)=-1.$ In a similar way as we did in case 1, we have

$$R(v,Jv)v=-\frac{4}{3}\left(Jv-g(Pv,v)PJv\right)+2J{A}_{h(v,v)}v.$$

If $g(Pv,v)\ne 0,$ similarly as before we get that P preserves the tangent space. Moreover, $g(Pv,v)PJv=-Jv$ and

$${\left|\right|h(v,v)\left|\right|}^2=-\frac{K}{2}.$$

As $\nabla P=0$ and the eigenvalues of P are constant, namely $\pm 1$, it again follows immediately that the surface is flat and, therefore, totally geodesic.

If $g(Pv,v)=0,$ it follows that $G(v,Pv)$ is a normal with negative length. As the tangent space is negative definite in view of the index of the metric, we have that the normal space has to be positive definite. Hence, we obtain a contradiction. ☐

Now, in order to be able to exclude the case with constant constant Gaussian curvature $-\frac{5}{9}$ in the previous theorem and obtain an explicit expression of the (totally geodesic) almost complex surface with constant Gaussian curvature $-\frac{4}{3}$ we will study more generally almost complex submanifolds with arbitrary Gauss curvature in more detail. We will, in particular, focus on those for which P maps the tangent space into the normal space.

Let M be an almost complex surface of $SL(2,\mathbb{R})\times SL(2,\mathbb{R})$ defined by the almost complex immersion $F:M\to SL(2,\mathbb{R})\times SL(2,\mathbb{R}):(s,t)\mapsto \left(A\right(s,t),B(s,t\left)\right)$, where $(s,t)$ are the isothermal coordinates on $M.$ Similarly as in [10], we may assume $F_t=J{F}_s$ and there exist $2\times 2$-real matrices with vanishing trace $\tilde{\alpha },\tilde{\beta },\tilde{\gamma },\tilde{\delta }$, such that

$$A_s=A\tilde{\alpha },A_t=A\tilde{\beta },B_s=B\tilde{\gamma },B_t=B\tilde{\delta }.$$

(18)

Subsequently, the matrices $\tilde{\alpha },\tilde{\beta },\tilde{\gamma },\tilde{\delta }$ are such that

$$\begin{array}{ccc}\hfill \tilde{\gamma }& =& \frac{\tilde{\alpha }}{2}-\frac{\sqrt{3}}{2}\tilde{\beta }\hfill \end{array}$$

(19)

$$\begin{array}{ccc}\hfill \tilde{\delta }& =& \frac{\sqrt{3}}{2}\tilde{\alpha }+\frac{\tilde{\beta }}{2}\hfill \end{array}$$

(20)

Furthermore, by using the integrability conditions $A_{st}=A_{ts}$ and $B_{st}=B_{ts}$, we have the two equations

$$\begin{array}{ccc}\hfill {\tilde{\alpha }}_t-{\tilde{\beta }}_s& =& 2\tilde{\alpha }\times \tilde{\beta }\hfill \end{array}$$

(21)

$$\begin{array}{ccc}\hfill {\tilde{\alpha }}_s+{\tilde{\beta }}_t& =& -\frac{2}{\sqrt{3}}\tilde{\alpha }\times \tilde{\beta },\hfill \end{array}$$

(22)

where we define the product × by $\tilde{\alpha }\times \tilde{\beta }=\frac{1}{2}(\tilde{\alpha }\tilde{\beta }-\tilde{\beta }\tilde{\alpha }).$

Now by writing $\alpha =cos\theta \tilde{\alpha }+sin\theta \tilde{\beta }$ and $\beta =-sin\theta \tilde{\alpha }+cos\theta \tilde{\beta }$, where $\theta =\frac{\pi }{3}$, Equations (21) and (22) imply

$$\begin{array}{ccc}\hfill {\alpha }_s+{\beta }_t& =& -\frac{4}{\sqrt{3}}\alpha \times \beta \hfill \\ \hfill {\alpha }_t-{\beta }_s& =& 0\hfill \end{array}$$

Note that, in the special case $PTM\subset T^{\perp }M,$, we have that $\tilde{\alpha }$ and $\tilde{\beta }$ are orthogonal with respect to the induced Euclidean product metric and of the same length (this is also true for $\alpha$ and $\beta$).

Hence, from the relation between the nearly Kähler metric and the usual Euclidean product metric (see (4)), we have

$$g(F_s,F_s)=⟨\alpha ,\alpha ⟩=g(F_t,F_t)\mathrm{and}g(F_s,F_t)=0.$$

Because $s,t$ are the isothermal coordinates, we may assume $g(F_s,F_s)=g(F_t,F_t)=e^{2\omega }$ for a smooth positive function $\omega$ on M. Subsequently, the induced metric on the surface M is given by

$$g=e^{2\omega }d{s}^2+e^{2\omega }d{t}^2.$$

It follows that the Levi–Civita connection on the surface M is given by

$$\begin{array}{ccc}\hfill {\nabla }_{F_s}{F}_s& =& {\omega }_s{F}_s-{\omega }_t{F}_t,\hfill \\ \hfill {\nabla }_{F_t}{F}_t& =& {\omega }_t{F}_t-{\omega }_s{F}_s,\hfill \\ \hfill {\nabla }_{F_s}{F}_t={\nabla }_{F_t}{F}_s& =& {\omega }_t{F}_s+{\omega }_s{F}_t.\hfill \end{array}$$

From (18) and (19), we have that the components A and B of the immersion F verify the equations

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}\hfill A_{ss}& =A(e^{2\omega }I+{\tilde{\alpha }}_s),\hfill \\ \hfill A_{st}& =A(\tilde{\alpha }\tilde{\beta }+{\tilde{\beta }}_s),\hfill \\ \hfill A_{tt}& =A(e^{2\omega }I+{\tilde{\beta }}_t),\hfill \end{array}\right.\mathrm{and}\left\{\begin{array}{cc}\hfill B_{ss}& =B(e^{2\omega }I+\frac{1}{2}{\tilde{\alpha }}_s-\frac{\sqrt{3}}{2}{\tilde{\beta }}_s),\hfill \\ \hfill B_{st}& =B(\frac{1}{4}(\tilde{\alpha }\tilde{\beta }-3\tilde{\beta }\tilde{\alpha })+\frac{\sqrt{3}}{2}{\tilde{\alpha }}_s+\frac{1}{2}{\tilde{\beta }}_s),\hfill \\ \hfill B_{tt}& =B(e^{2\omega }I+\frac{\sqrt{3}}{2}{\tilde{\alpha }}_t+\frac{1}{2}{\tilde{\beta }}_t).\hfill \end{array}\right.\end{array}$$

(23)

So far, the above equations remained valid for any almost complex surface for which P maps to tangent space to the normal space. From now on we will assume that the surface is moreover totally geodesic. Subsequently, we have the following.

Proposition 5.

A totally geodesic almost complex surface M of $SL(2,\mathbb{R})\times SL(2,\mathbb{R})$ for which P maps tangent vectors into normal vectors is isometric to the immersion $(s,t)\mapsto \left(A\right(s,t),B(s,t\left)\right)$, where

$$A(s,t)=\left(\begin{array}{cc}\frac{\sqrt{3}{y}_1}{2}+\frac{1}{2}& \frac{1}{2}\sqrt{3}(y_2+y_3)\\ \frac{1}{2}\sqrt{3}(y_2-y_3)& \frac{1}{2}-\frac{\sqrt{3}{y}_1}{2}\end{array}\right)$$

and

$$B(s,t)=\left(\begin{array}{cc}\frac{1}{2}-\frac{\sqrt{3}{y}_1}{2}& -\frac{1}{2}\sqrt{3}(y_2+y_3)\\ -\frac{1}{2}\sqrt{3}(y_2-y_3)& \frac{\sqrt{3}{y}_1}{2}+\frac{1}{2}\end{array}\right),$$

where $(y_1,y_2,y_3)$ is a point of the hyperbolic quadric

$$y_1^2+y_2^2-y_3^2=-1$$

Proof. 

Using

$$D_{X}Y={\nabla }_X^{E}Y+\frac{1}{2}⟨X,Y⟩F+\frac{1}{2}⟨X,QY⟩QF,$$

which relates the pseudo Euclidean connection with the Levi–Civita connection of the nearly kähler metric and by supposing that M is totally geodesic and then by using Proposition 1 we have

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}\hfill A_{ss}& =A({\omega }_s\tilde{\alpha }-{\omega }_t\tilde{\beta }-\frac{1}{\sqrt{3}}\tilde{\alpha }\times \tilde{\beta }+e^{2\omega }I),\hfill \\ \hfill A_{st}& =A({\omega }_t\tilde{\alpha }+{\omega }_s\tilde{\beta }),\hfill \\ \hfill A_{tt}& =A(-{\omega }_s\tilde{\alpha }+{\omega }_t\tilde{\beta }-\frac{1}{\sqrt{3}}\tilde{\alpha }\times \tilde{\beta }+e^{2\omega }I),\hfill \end{array}\right.\end{array}$$

and

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}\hfill B_{ss}& =B({\omega }_s\tilde{\gamma }-{\omega }_t\tilde{\delta }-\frac{1}{\sqrt{3}}\tilde{\alpha }\times \tilde{\beta }+e^{2\omega }I),\hfill \\ \hfill B_{st}& =A({\omega }_t\tilde{\gamma }+{\omega }_s\tilde{\delta }),\hfill \\ \hfill B_{tt}& =B(-{\omega }_s\tilde{\gamma }+{\omega }_t\tilde{\delta }+\frac{1}{\sqrt{3}}\tilde{\alpha }\times \tilde{\beta }+e^{2\omega }I).\hfill \end{array}\right.\end{array}$$

Hence, by identification, we deduce that $\tilde{\alpha }$ and $\tilde{\beta }$ are determined by the following system of partial differential equations

$$\begin{array}{ccc}\hfill {\tilde{\alpha }}_s& =& -{\omega }_t\tilde{\beta }+{\omega }_s\tilde{\alpha }-\frac{1}{\sqrt{3}}\tilde{\alpha }\times \tilde{\beta },\hfill \\ \hfill {\tilde{\alpha }}_t& =& {\omega }_t\tilde{\alpha }+{\omega }_s\tilde{\beta }+\tilde{\alpha }\times \tilde{\beta },\hfill \\ \hfill {\tilde{\beta }}_s& =& {\omega }_t\tilde{\alpha }+{\omega }_s\tilde{\beta }-\tilde{\alpha }\times \tilde{\beta },\hfill \\ \hfill {\tilde{\beta }}_t& =& {\omega }_t\tilde{\beta }-{\omega }_s\tilde{\alpha }-\frac{1}{\sqrt{3}}\tilde{\alpha }\times \tilde{\beta },\hfill \end{array}$$

which in terms of $\alpha$ and $\beta$ become

$$\begin{array}{ccc}\hfill {\alpha }_s& =& -{\omega }_t\beta +{\omega }_s\alpha -\frac{2}{\sqrt{3}}\alpha \times \beta ,\hfill \\ \hfill {\alpha }_t& =& {\omega }_t\alpha +{\omega }_s\beta ,\hfill \\ \hfill {\beta }_s& =& {\omega }_t\alpha +{\omega }_s\beta ,\hfill \\ \hfill {\beta }_t& =& {\omega }_t\beta -{\omega }_s\alpha -\frac{2}{\sqrt{3}}\alpha \times \beta .\hfill \end{array}$$

Afterwards, locally there exists a matrix $\epsilon$, such that ${\epsilon }_s=\alpha$, ${\epsilon }_t=\beta$. The surface $\epsilon$ is determined by

$$\begin{array}{ccc}\hfill {\epsilon }_{ss}& =& -{\omega }_t{\epsilon }_t+{\omega }_s{\epsilon }_s-\frac{2}{\sqrt{3}}{\epsilon }_s\times {\epsilon }_t,\hfill \\ \hfill {\epsilon }_{st}& =& {\omega }_t{\epsilon }_s+{\omega }_s{\epsilon }_t,\hfill \\ \hfill {\epsilon }_{tt}& =& {\omega }_t{\epsilon }_t-{\omega }_s{\epsilon }_s-\frac{2}{\sqrt{3}}{\epsilon }_s\times {\epsilon }_t.\hfill \end{array}$$

(24)

A unit normal vector field on the surface M determined by the matrix $\epsilon$ is given by $\xi =-\frac{{\epsilon }_s\times {\epsilon }_t}{e^{2\omega }}$.

On the other hand, ${\epsilon }_{st}=D_{{\epsilon }_s}{\epsilon }_t={\nabla }_{{\epsilon }_s}{\epsilon }_t+\tilde{h}({\epsilon }_s,{\epsilon }_t)\xi$, by identification with the second equation of (24), we deduce that $h({\epsilon }_s,{\epsilon }_t)=0$, in a similar way $h({\epsilon }_s,{\epsilon }_s)=h({\epsilon }_t,{\epsilon }_t)=\frac{2}{\sqrt{3}}{e}^{2\omega }\xi .$ Hence, the surface determined by $\epsilon$ is totally umbilical with shape operator $S=\frac{2}{\sqrt{3}}I.$ Moreover, as $D_{\frac{\partial }{{\partial }_u}}(\epsilon +\frac{\sqrt{3}}{2}\xi )=D_{\frac{\partial }{{\partial }_v}}(\epsilon +\frac{\sqrt{3}}{2}\xi )=0$, we deduce that $\epsilon$ is a hyperbolic space of cartesian equation

$${\epsilon }_1^2+{\epsilon }_2^2-{\epsilon }_3^2=-\frac{3}{4}.$$

From this, we can now reverse engineer the original immersion. We can take a local isothermal parametrisation of this hypersurface by

$$\begin{array}{cc}\hfill & {\epsilon }_1=-\frac{\sqrt{3}s}{(s^2+t^2-1)},\hfill \\ \hfill & {\epsilon }_2=-\frac{\sqrt{3}t}{(s^2+t^2-1)},\hfill \\ \hfill & {\epsilon }_3=\frac{\sqrt{3}}{2}\frac{s^2+t^2+1}{s^2+t^2-1},\hfill \end{array}$$

where we look at $\epsilon$ as the matrix

$$\epsilon =\left(\begin{array}{cc}{\epsilon }_1& {\epsilon }_2+{\epsilon }_3\\ {\epsilon }_2-{\epsilon }_3& {\epsilon }_1\end{array}\right).$$

It then follows immediately that

$$\begin{array}{c}\hfill \alpha =\left(\begin{array}{cc}\frac{\sqrt{3}\left(s^2-t^2+1\right)}{{\left(s^2+t^2-1\right)}^2}& \frac{2\sqrt{3}s(t-1)}{{\left(s^2+t^2-1\right)}^2}\\ \frac{2\sqrt{3}s(t+1)}{{\left(s^2+t^2-1\right)}^2}& \frac{\sqrt{3}\left(-s^2+t^2-1\right)}{{\left(s^2+t^2-1\right)}^2}\end{array}\right),\\ \hfill \beta =\left(\begin{array}{cc}\frac{2\sqrt{3}st}{{\left(s^2+t^2-1\right)}^2}& \frac{\sqrt{3}\left({(t-1)}^2-s^2\right)}{{\left(s^2+t^2-1\right)}^2}\\ \frac{\sqrt{3}\left({(t+1)}^2-s^2\right)}{{\left(s^2+t^2-1\right)}^2}& -\frac{2\sqrt{3}st}{{\left(s^2+t^2-1\right)}^2}\end{array}\right).\end{array}$$

From this, we now deduce that

$$\begin{array}{cc}\hfill & \tilde{\alpha }=\left(\begin{array}{cc}\frac{\sqrt{3}{s}^2-6st-\sqrt{3}\left(t^2-1\right)}{2{\left(s^2+t^2-1\right)}^2}& \frac{3s^2+2\sqrt{3}s(t-1)-3{(t-1)}^2}{2{\left(s^2+t^2-1\right)}^2}\\ \frac{3s^2+2\sqrt{3}s(t+1)-3{(t+1)}^2}{2{\left(s^2+t^2-1\right)}^2}& \frac{-\sqrt{3}{s}^2+6st+\sqrt{3}\left(t^2-1\right)}{2{\left(s^2+t^2-1\right)}^2}\end{array}\right),\hfill \\ \hfill & \tilde{\beta }=\left(\begin{array}{cc}\frac{3s^2+2\sqrt{3}st-3t^2+3}{2{\left(s^2+t^2-1\right)}^2}& \frac{-\sqrt{3}{s}^2+6s(t-1)+\sqrt{3}{(t-1)}^2}{2{\left(s^2+t^2-1\right)}^2}\\ \frac{-\sqrt{3}{s}^2+6s(t+1)+\sqrt{3}{(t+1)}^2}{2{\left(s^2+t^2-1\right)}^2}& -\frac{3s^2+2\sqrt{3}st-3t^2+3}{2{\left(s^2+t^2-1\right)}^2}\end{array}\right),\hfill \\ \hfill & \tilde{\gamma }=\left(\begin{array}{cc}\frac{-\sqrt{3}{s}^2-6st+\sqrt{3}\left(t^2-1\right)}{2{\left(s^2+t^2-1\right)}^2}& \frac{3s^2-2\sqrt{3}s(t-1)-3{(t-1)}^2}{2{\left(s^2+t^2-1\right)}^2}\\ \frac{3s^2-2\sqrt{3}s(t+1)-3{(t+1)}^2}{2{\left(s^2+t^2-1\right)}^2}& \frac{\sqrt{3}{s}^2+6st-\sqrt{3}\left(t^2-1\right)}{2{\left(s^2+t^2-1\right)}^2}\end{array}\right),\hfill \\ \hfill & \tilde{\delta }=\left(\begin{array}{cc}\frac{3s^2-2\sqrt{3}st-3t^2+3}{2{\left(s^2+t^2-1\right)}^2}& \frac{\sqrt{3}{s}^2+6s(t-1)-\sqrt{3}{(t-1)}^2}{2{\left(s^2+t^2-1\right)}^2}\\ \frac{\sqrt{3}{s}^2+6s(t+1)-\sqrt{3}{(t+1)}^2}{2{\left(s^2+t^2-1\right)}^2}& \frac{-3s^2+2\sqrt{3}st+3t^2-3}{2{\left(s^2+t^2-1\right)}^2}\end{array}\right).\hfill \end{array}$$

Solving now the differential equations for A and B, we find that

$$\begin{array}{c}\hfill A=\left(\begin{array}{cc}\frac{1}{2}-\frac{\sqrt{3}s}{s^2+t^2-1}& \frac{\sqrt{3}\left(s^2+{(t-1)}^2\right)}{2\left(s^2+t^2-1\right)}\\ -\frac{\sqrt{3}\left(s^2+{(t+1)}^2\right)}{2\left(s^2+t^2-1\right)}& \frac{\sqrt{3}s}{s^2+t^2-1}+\frac{1}{2}\end{array}\right),\\ \hfill B=\left(\begin{array}{cc}\frac{\sqrt{3}s}{s^2+t^2-1}+\frac{1}{2}& -\frac{\sqrt{3}\left(s^2+{(t-1)}^2\right)}{2\left(s^2+t^2-1\right)}\\ \frac{\sqrt{3}\left(s^2+{(t+1)}^2\right)}{2\left(s^2+t^2-1\right)}& \frac{1}{2}-\frac{\sqrt{3}s}{s^2+t^2-1}\end{array}\right).\end{array}$$

Replacing now the coordinates by ${\epsilon }_i$ and rescaling completes the proof of the proposition. ☐

Finally, we will study the case when $K=-5/9$ and $PTM\subset T^{\perp }M.$ We consider the basis

$$e_1=V,e_2=JV,e_3=PV,e_4=JPV,e_5=G(V,PV),e_6=-G(JV,PV),$$

where V is a unit tangent vector. All of these vectors are mutually orthogonal, $e_1,e_2,e_3,e_4$ are of unit length and $e_5,e_6$ are of length $-2/3.$

By using the relation (8), we have

$$\begin{array}{cccccccccc}\hfill G(e_1,e_2)& =,0\hfill & \hfill G(e_1,e_3)& =e_5,\hfill & \hfill G(e_1,e_4)& =-e_6,\hfill & \hfill G(e_1,e_5)& =\frac{2}{3}{e}_3,\hfill & \hfill G(e_1,e_6)& =-\frac{2}{3}{e}_4,\hfill \\ \hfill G(e_2,e_3)& =-e_6,\hfill & \hfill G(e_2,e_4)& =-e_5,\hfill & \hfill G(e_2,e_5)& =-\frac{2}{3}{e}_4,\hfill & \hfill G(e_2,e_6)& =-\frac{2}{3}{e}_3,\hfill & \hfill G(e_3,e_4)& =0,\hfill \\ \hfill G(e_3,e_5)& =\frac{2}{3}{e}_1,\hfill & \hfill G(e_3,e_6)& =\frac{2}{3}{e}_2,\hfill & \hfill G(e_4,e_5)& =\frac{2}{3}{e}_2,\hfill & \hfill G(e_4,e_6)& =\frac{2}{3}{e}_1,\hfill & \hfill G(e_5,e_6)& =0.\hfill \end{array}$$

Let $\tilde{\nabla }$ denote the Levi-Civita connection on $SL(2,\mathbb{R})\times SL(2,\mathbb{R}).$ We will write ${\tilde{\nabla }}_{e_1}{e}_1={\nabla }_{e_1}{e}_1+h(e_1,e_1)=a_1{e}_2+a_2{e}_5+a_3{e}_6$ (the coefficients of $e_3$ and $e_4$ are 0 from page 6 in [10]), where $a_1,a_2,a_3$ are functions to determine and let $b_1=g({\tilde{\nabla }}_{e_2}{e}_1,e_2)$.

The Levi–Civita connection $\tilde{\nabla }$ is computed in term of $a_1,a_2,a_3$ and $b_1$, more precisely

$$\begin{array}{ccc}\hfill {\tilde{\nabla }}_{e_1}{e}_2& =& -a_1{e}_1-a_3{e}_5+a_2{e}_6,\hfill \\ \hfill {\tilde{\nabla }}_{e_1}{e}_3& =& -a_1{e}_4+a_2{e}_5+(\frac{1}{2}-a_3)e_6,\hfill \\ \hfill {\tilde{\nabla }}_{e_1}{e}_4& =& a_1{e}_3+(\frac{1}{2}+a_3)e_5+a_2{e}_6,\hfill \\ \hfill {\tilde{\nabla }}_{e_1}{e}_5& =& \frac{2}{3}{a}_2{e}_1-\frac{2}{3}{a}_3{e}_2+\frac{2}{3}{a}_2{e}_3+\frac{2}{3}(\frac{1}{2}+a_3)e_4,\hfill \\ \hfill {\tilde{\nabla }}_{e_1}{e}_6& =& \frac{2}{3}{a}_3{e}_1+\frac{2}{3}{a}_2{e}_2+\frac{2}{3}(\frac{1}{2}-a_3)e_3+\frac{2}{3}{a}_2{e}_4,\hfill \\ \hfill {\tilde{\nabla }}_{e_2}{e}_1& =& b_1{e}_2-a_3{e}_5+a_2{e}_6,\hfill \\ \hfill {\tilde{\nabla }}_{e_2}{e}_2& =& -b_1{e}_1-a_2{e}_5-a_3{e}_6,\hfill \\ \hfill {\tilde{\nabla }}_{e_2}{e}_3& =& -b_1{e}_4+(\frac{1}{2}-a_3)e_5-a_2{e}_6,\hfill \\ \hfill {\tilde{\nabla }}_{e_2}{e}_4& =& b_1{e}_3+a_2{e}_5-(\frac{1}{2}+a_3)e_6,\hfill \\ \hfill {\tilde{\nabla }}_{e_2}{e}_5& =& -\frac{2}{3}{a}_3{e}_1-\frac{2}{3}{a}_2{e}_2+\frac{2}{3}(\frac{1}{2}-a_3)e_3+\frac{2}{3}{a}_2{e}_4,\hfill \\ \hfill {\tilde{\nabla }}_{e_2}{e}_6& =& \frac{2}{3}{a}_2{e}_1-\frac{2}{3}{a}_3{e}_2-\frac{2}{3}(\frac{1}{2}+a_3)e_4-\frac{2}{3}{a}_2{e}_3.\hfill \end{array}$$

From the Gauss equation, we have that the Gauss curvature $K=⟨R(e_1,e_2)e_2,e_1⟩$ is given by

$$\begin{array}{ccc}\hfill K& =& ⟨\tilde{R}(e_1,e_2)e_2,e_1⟩-⟨h((e_1,e_2),h(e_1,e_2)⟩+⟨h(e_2,e_2),h(e_1,e_1)⟩\hfill \end{array}$$

(25)

which implies that

$$\begin{array}{ccc}\hfill -\frac{5}{9}& =& -\frac{4}{3}+\frac{4}{3}(a_2^2+a_3^2).\hfill \end{array}$$

Subsequently,

$$a_2^2+a_3^2=\frac{7}{12}.$$

(26)

Now, by applying that $\tilde{R}(e_1,e_2)X={\tilde{\nabla }}_{e_1}{\tilde{\nabla }}_{e_2}X-{\tilde{\nabla }}_{e_2}{\tilde{\nabla }}_{e_1}X-{\tilde{\nabla }}_{[e_1,e_2]}X$ for $X=e_1,\dots ,e_6$, with $[e_1,e_2]=-a_1{e}_1-b_1{e}_2$, we deduce the following equations

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}\hfill -3(a_1^2+b_1^2-e_2\left(a_1\right)+e_1\left(b_1\right))+\frac{5}{3}& =0,\hfill \\ \hfill 2(a_1{a}_3+a_2{b}_1)+e_1\left(a_2\right)-e_2\left(a_3\right)& =0,\hfill \\ \hfill 2(a_1{a}_2-a_3{b}_1)-e_2\left(a_2\right)-e_1\left(a_3\right)& =0.\hfill \end{array}\right.\end{array}$$

If the surface is parallel, $\nabla h=0$ gives the following equations

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}\hfill \frac{2}{3}{a}_2^2+\frac{1}{3}(1-2a_3)a_3& =0,\hfill \\ \hfill \frac{2}{3}{a}_2{a}_3+\frac{1}{3}(1+2a_3)a_2& =0,\hfill \\ \hfill 2a_1{a}_3+e_1\left(a_2\right)& =0,\hfill \\ \hfill -2a_1{a}_2+e_1\left(a_3\right)& =0.\hfill \end{array}\right.\end{array}$$

This system has a unique null solution, which is in contradiction with $\left(26\right)$. Thus, no solutions exist. Subsequently, Proposition 1 becomes

Theorem 2.

If M is an almost complex surface in $SL(2,\mathbb{R})\times SL(2,\mathbb{R})$ with parallel second fundamental form, then either

• P maps the tangent space into the normal space and, in this case, M is totally geodesic with constant Gaussian curvature $-{\displaystyle \frac{4}{3}}$. Moreover, M is totally geodesic and congruent to the example that is described in Proposition 5.
• P preserves the tangent space, in this case the Gaussian curvature is 0. Moreover, M is totally geodesic and congruent to either the example described in Proposition 3 or the example described in Proposition 4.

This means that every parallel almost complex surface in $SL(2,\mathbb{R})\times SL(2,\mathbb{R})$ is totally geodesic.