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abstract: 'We present a reduction of codimension theorem for surfaces with parallel mean curvature in symmetric spaces.' 
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**Parallel Mean Curvature Surfaces in**

**Symmetric Spaces**

.4in

**by**

.2in

**M. J. Ferreira and Renato Tribuzy**

.5in

[**I - Introduction**]{}

Surfaces with parallel mean curvature have been considered by many geometers.  Yau \[Y\] studied them when the ambient space is a space form.  He showed that such a surface must be, either minimal in a round sphere, or its image must be contained in a three dimensional totally geodesic, or totally umbilic submanifold.  This result was recently extended to maps into $E^n(c) \times \mathbb R$ by Alencar, do Carmo and Tribuzy \[A-C-T\], where $E^n(c)$ a space form with constant sectional curvature $c \neq 0$. They have proved that the image of the surface is either minimal in a totally umbilical hypersurface of $E^n(c)$, has constant mean curvature in a three dimensional totally geodesic, or totally umbilical, submanifold of $E^n(c)$, or lies in $E^4 \times \mathbb R$.

Fetcu \[F\] has shown that a surface immersed in $CP^n$ with parallel mean curvature must be either pseudo-umbilic and totally real or its image must be contained in $CP^5$. We recall that a immersion $\varphi: M \rightarrow N$ between two Riemannian manifolds is said to be pseudo-umbilic if the $H$-Weigarten operator $A_H$ is a multiple of the identity, where $H$ denotes the mean curvature. 

The aim of this work is to generalize the above results to immersions into symmetric spaces. 

From now on we let $S$ denote a Riemannian symmetric space and $R$ its curvature tensor.  For $p \in S$, we say that a subspace $K
\subset T_pS$ is invariant by the curvature tensor $R_p$, if $R_p(u,v)w \in K$, whenever $u,v,w \in K$. When $X \subset T_p S$, $R_p(X)$ will represent the least vector subspace of $T_p S$ invariant by the curvature tensor $R_p$ at $p$.

Let $\varphi : M \rightarrow S$ be an isometric immersion from a Riemann surface $M$. For each $x \in M$, the $n$-th osculating space of $\varphi$ at $x$ will be denoted by $O^n_x (\varphi)$.

Our main result is the following:

[**Theorem 1**]{}

If $\varphi$ has parallel mean curvature $H\neq 0$, one of the following conditions hold:

i\) $\varphi$ is pseudo-umbilic;

ii\) The dimension $d$ of $R_{\varphi(x)}(O^2_{x})$ is independent of $x \in M$ and there exists a totally geodesic $S^{\prime} \subset
S$, with dimension $d$, in such a way that $\varphi (M) \subset
S^{\prime}$ . 

[**II - Proof of the main theorem**]{}

The following theorem dealing with reduction of codimension of maps is fundamental to our purpose:

[**Theorem 2**]{}

Let $\varphi:M \rightarrow S$ be an isometric immersion from a Riemannian manifold into a symmetric space.  If there exists a parallel fiber bundle $L$, over $M$, such that $R(L)=L$ and $TM
\subset L$, then there exists a totally geodesic submanifold $S^{\prime}$ of $S$ with $\varphi (M) \subset S^{\prime}$ and $L_x =
T_{\varphi(x)}S^{\prime}$, $\forall x \in M$.

For a proof see \[E-T\]. 

Let $\alpha$ denote the second fundamental form of the immersion.  We remark that, for each $x \in M$, $O^2_x(\varphi)=T_x M + N_1(x)$, where $N_1(x) =\{\alpha(u,v): u,v \in T_x M\}$. We will show that either $H$ is a umbilical direction, or $L=R(O^2(\varphi))$ is a parallel fiber bundle. 

Let $U$ be an open subset of $M$ where $L_x = R_x (O^2_x (\varphi))$ has maximal dimension.  We will prove that $L$ is parallel on $U$ when $H$ is not a umbilical direction.  Using analiticity arguments we may conclude that its dimension is constant on $M$.

[**Lemma 1**]{}

Whenever $X,Y,W \in \Gamma(L)$ and $\nabla X, \nabla Y, \nabla W \in
\Gamma (L)$, $R(X,Y)W \in \Gamma (L)$.

[**Proof of Lemma 1:**]{}

It follows straightforward from the parallelism of $R$.

We remark that $L$ is constructed from $O^2(\varphi)$ by successive applications of $R$ to its elements.  Thus it is enough to show that $ \nabla \xi \in \Gamma(L)$, for every $\xi= \alpha (U,V)$, to conclude that $L$ is parallel. 

Let $\{\varepsilon_1, \varepsilon_2\}$ be a local orthonormal frame field defined on open subset $V$ of $U$. Notice that $R_x(\varepsilon_1,\varepsilon_2)H \in O^2_x$, $\forall x\in V$. Let $\pi_{N_1(x)}:T_{\varphi(x)} S \rightarrow N_1(x)$ denote the orthogonal projection and consider $n_1(x)=\pi_{N_1(x)}(R_x(\varepsilon_1,\varepsilon_2)H)$.

[**Lemma 2**]{}

Assume $n_1$ vanishes identically on $V$. Then $L$ is parallen on $V$.

[**Proof of Lemma 2:**]{}

From Ricci equations we get

$$0=<R^{\perp}(\varepsilon_1,\varepsilon_2)H,\eta>=-<[A_H,
A_\eta]\varepsilon_1,\varepsilon_2>,$$ so that $A_H$ commutes with $A_\eta$ for every section $\eta$ of the normal bundle.  Therefore without loss of generality $\{\varepsilon_1,\varepsilon_2\}$ may be chosen in such a way that $\alpha(\varepsilon_1,\varepsilon_2)=0$ and it is enough to show that $$\nabla_{\varepsilon_i} \alpha(\varepsilon_j,\varepsilon_j)\in
\Gamma(L)$$.

But this is equivalent to prove that

$$(\nabla_{\varepsilon_i} \alpha)(\varepsilon_j , \varepsilon_j) \in
\Gamma(L),$$ since

$$(\nabla_{\varepsilon_i} \alpha)(\varepsilon_j ,
\varepsilon_j)=\nabla_{\varepsilon_i}
\alpha(\varepsilon_j,\varepsilon_j)- 2
\alpha(\nabla_{\varepsilon_i}\varepsilon_j,\varepsilon_j).$$

From Codazzi equations we have:

$$(\nabla_{\varepsilon_i} \alpha)(\varepsilon_j ,
\varepsilon_j)=(\nabla_{\varepsilon_j} \alpha)(\varepsilon_i ,
\varepsilon_j)+(R(\varepsilon_i,\varepsilon_j)\varepsilon_j)^{\perp},$$ where $( ~~  )^{\perp}$ represents the orthogonal projection onto the normal bundle.  But $(R(\varepsilon_i,\varepsilon_j)\varepsilon_j)^{\perp}$ sits in $L$ since $$(R(\varepsilon_i,\varepsilon_j)\varepsilon_j)^{\perp}=R(\varepsilon_i,\varepsilon_j)\varepsilon_j
- A,$$ with $A=(R(\varepsilon_i,\varepsilon_j)\varepsilon_j)^{T} \in
\Gamma(L)$, where $(~~ )^T$ stands for the orthogonal projection onto $TV$.

Of course $\nabla_{\varepsilon_i}\alpha(\varepsilon_j,\varepsilon_j)
\in \Gamma(L)$ if $i\neq j$. When $i=j$, $$\nabla_{\varepsilon_i} \alpha(\varepsilon_i,\varepsilon_i)=
\nabla_{\varepsilon_i}(2H-\alpha(\varepsilon_j,\varepsilon_j))=-\nabla_{\varepsilon_i}
\alpha(\varepsilon_j,\varepsilon_j)+B \in \Gamma(L),$$ with $j\neq i$ and $B=2(\nabla_{\varepsilon_i}H)^T$.

We proceed now with the proof of theorem 1, considering the case where $n_1$ does not vanish identically. 

Let us consider an open subset $V$ of $U$ where $n_1 (x) \neq 0$, $\forall x \in V$.

We remark that $n_1$ and $H$ are linearly independent, $\nabla n_1
\in \Gamma(L)$ and $\nabla H \in \Gamma(L)$. Remember that, due to the parallelism of $R$, $\nabla n_1=R(\nabla
\varepsilon_1,\varepsilon_2)H+R(\varepsilon_1,\nabla \varepsilon_2)H
+ R(\varepsilon_1,\varepsilon_2)\nabla H$.

Let $Z$ denote the sub-bundle of the normal bundle spanned by $n_1$ and $H$. Consider a unitary section $\gamma$ orthogonal to $Z$ and sitting in $N_1$ and take an orthonormal frame field $\{\varepsilon_1,\varepsilon_2\}$ diagonalizing the Weingarten operator $A_{\gamma}$.

We observe that $\alpha(X,Y)=\alpha(X,Y)_{Z}+\alpha(X,Y)_{\gamma}$, where $\alpha_Z$ and $\alpha_\gamma$ are respectively the $Z$ and $<\gamma>$ components of $\alpha$; here $<\gamma>$ denotes the sub-bundle spanned by $\gamma$.

Now, to show that $\nabla \alpha(X,Y) \in \Gamma(L)$, it is enough to prove that $\nabla \alpha_\gamma(X,Y) \in \Gamma(L)$, since $\alpha_Z(X,Y) \in \Gamma(L)$. Therefore we get that, for any $i,j,k
\in \{1,2\}$, $\nabla_{\varepsilon_i} \alpha_\gamma(\varepsilon_j,
\varepsilon_k) \in \Gamma(L)$ if and only if $\nabla_{\varepsilon_i}
\alpha(\varepsilon_j, \varepsilon_k)$, which is equivalent to $(\nabla_{\varepsilon_i} \alpha)(\varepsilon_j, \varepsilon_k)\in
\Gamma(L)$.

Clearly $\nabla_{\varepsilon_i}\alpha_\gamma(\varepsilon_j,\varepsilon_k)=0$, whenever $j\neq k$.

When $j=k \neq i$, using Codazzi equations, we have $$(\nabla_{\varepsilon_i}\alpha)(\varepsilon_j,\varepsilon_j)=(\nabla_{\varepsilon_j}\alpha)(\varepsilon_i,\varepsilon_j)+A,$$ where $A\in \Gamma(L)$ and, since $(\nabla_{\varepsilon_j}\alpha_{\gamma})(\varepsilon_i,\varepsilon_j)
\in \Gamma(L)$, we obtain $(\nabla_{\varepsilon_i}\alpha)(\varepsilon_j,\varepsilon_j)\in
\Gamma(L)$.

If $i=j=k$,

$$\nabla_{\varepsilon_i}\alpha_\gamma(\varepsilon_i,\varepsilon_i)=\nabla_{\varepsilon_i}(2H_\gamma
-
\alpha_\gamma(\varepsilon_j,\varepsilon_j))=-\nabla_{\varepsilon_i}\alpha_\gamma(\varepsilon_j\varepsilon_j),$$ where $j \neq i$ and we have used the fact that $H_\gamma = 0$. Hence $\nabla_{\varepsilon_i}\alpha_\gamma(\varepsilon_i,\varepsilon_i)
\in \Gamma(L)$.

[**III - Remarks**]{}

[**1 -**]{}

When $S=E^n(c)$ is a space form of constant sectional curvature $c$, it follows directly from Ricci equation that $A_H$ is either a multiple of the identity, or there exists a basis diagonalizing, at each point, the second fundamental form.  Hence the second normal space has dimension less or equal than $4$. Of course the curvature tensor leaves this space invariant and we get that $\varphi(M)
\subset E^4(c)$ (\[Y\]). 

[**2 -**]{}

When $S=CP^n$ we observe that $$R(O^2(\varphi)) \subset O^2(\varphi) + J O^2(\varphi),$$ where $J$ denotes the standard complex structure of $CP^n$. Therefore, if $\varphi$ is not pseudo-umbilic, we must clearly have $\varphi(M) \subset CP^5$. In the case were $H$ is a umbilical direction, a direct computation gives $R(X,Y)H=0$, whenever $X,Y \in
\Gamma(TM)$. Thus, from the explicit formula of the curvature tensor, we obtain straightforward that $JTM$ is orthogonal to $TM$, i.e $\varphi$ is totally real.  This case was studied by Fetcu \[F\]. 

[**3 -**]{}

Assume now that the target manifold is a product $S_1 \times S_2$ of symmetric spaces.  We have:

[**Corolary 1 -** ]{} Let $\varphi :M \rightarrow S_1 \times
S_2$ an immersion from a Riemann surface with parallel mean curvature $H \neq 0$. Then, one of the following conditions hold:

i\) $\varphi$ is pseudo-umbilic;

ii\) $\varphi(M) \subset S_1^{\prime} \times S_2^{\prime}$, where, for each $i\in \{1,2\}$, $S_i^{\prime}\subset S_i$ is a symmetric space totally geodesic embedded in $S_i$ and $\dim S_i^{\prime}=dim
R_i((O^2(\varphi))^{T_i})$; here $R_i$ represents the curvature tensor of $S_i$ and $(~~)^{T_i}$ stands for the orthogonal projection onto $TS_i$.

[**Proof of Corollary 1:**]{}

Clearly $$R(O^2(\varphi) )\subset R_1((O^2(\varphi))^{TS_1}) +
R_2((O^2(\varphi))^{TS_2}).$$ Therefore it is enough to prove that each $R_i((O^2(\varphi))^{TS_i})$ is parallel.  Let us take an open subset of $M$ where the dimension of $O^2(\varphi)^{T_i}$ is maximal.  By Lemma 1, it suffices to show that $\nabla \mu_i \in \Gamma
(R((O^2(\varphi))^{T_i}))$, whenever $\mu_i \in
\Gamma(O^2(\varphi))$. This follows from the fact that there exists $\mu \in\Gamma(O^2(\varphi))$ such that $\mu_i = (\mu)^{T_i}$. Since $\nabla \mu = (\nabla \mu_1, \nabla \mu_2) \in R(O^2(\varphi))$ we obtain $\nabla \mu_i \in R_i(O^2(\varphi))$.

In the particular situation where $S_2=\mathbb R$, it follows from corollary 1 that, either $\varphi$ is pseudo-umbilic, or $\varphi(M)
\subset S^{\prime} \times \mathbb R$, where $\dim S^{\prime}=\dim
R((O^2(\varphi))^ {TS})$ and $R$ denotes the curvature tensor of $S$. For instance, if $S=E^n(c)$, clearly $S^{\prime}=E^4(c)$, like in \[A-C-T\]. 

When the target manifold is $E^n(c) \times E^m(d)$, either $\varphi$ is pseudo-umbilic, or $\varphi(M) \subset E^4(c) \times E^4(d)$.

[WW]{}

Alencar, H., do Carmo, M., Tribuzy, R., [*A Hopf Theorem for ambient spaces of dimension higher than three* ]{}, J. Diff.  Geom.  vol.  84, nº1 (2010), 1-17. 

Eschenburg, J., Tribuzy, R., [*Existence and uniqueness of maps in affine homogeneous spaces* ]{}, Rend.  Sem.  Mat.  Univ.  Padova,80 (1993), 11-18. 

Fetcu, Dorel, [*Surfaces with parallel mean curvature in complex space forms*]{}, arxiv.1011.5892v1\[math. DG\]. 

Yau, S. T., [*Subamnifolds with constant mean curvature I* ]{}, Am.  J. Mat.  96 (1974), 346–366. 

mjferr@ptmat.fc.ul.pt

tribuzy@pq.cnpq.br

Work supported by FCT (Fundação para a Ciência e Tecnologia) in Portugal and by CNPq and FAPEAM in Brasil
