Suppose $G$ is a non-compact, semi-simple Lie group, of rank at least two, with maximal compact subgroup $K$ and $G/K$ the corresponding Riemannian symmetric space. Let $\mathbb{H}^2(-c^2)$ be the 2-dimensional real hyperbolic space of sectional curvature $-c^2.$ Suppose, \begin{align} f:\mathbb{H}^2(-c^2)\rightarrow G/K, \end{align} is an isometric, minimal (i.e. vanishing mean curvature) immersion. Is $f$ necessarily totally geodesic? I'm primarily concerned in the case that G is a complex simple Lie group, but perhaps the statement is true in this broader context. Thanks for any help!


1 Answer 1


Below, I have added an answer to your question about the case of a complex semi-simple Lie group. It turns out that the answer is 'no' even in this case.

The answer to your general question is 'no'. A simple example is to take $G = \mathrm{SU}(1,2)$ with its maximal compact $K = \mathrm{U}(2)$. You can think of the homogeneous space $G/K$ as the unit ball $\mathbb{B}^2\subset \mathbb{C}^2$, but you can also think of it as the space of complex lines $L$ in $\mathbb{C}^3$ on which the Hermitian form $h =|z^0|^2-|z^1|^2-|z^2|^2$ is positive definite.

When endowed with its canonical $G$-invariant Kähler metric, $\mathbb{B}^2=G/K$ contains three mutually noncongruent, homogeneous copies of the hyperbolic disc that are minimal surfaces: The first, which is totally geodesic, consists of the $h$-positive complex lines $L$ that are complexifications of real lines in $\mathbb{R}^3$, i.e., those lines for which $L = \bar L$. The second, which is also totally geodesic, consists of the $h$-positive complex lines $L$ that lie in a fixed complex $2$-plane $P\subset\mathbb{C}^3$ of $h$-hermitian type $(1,1)$. The third, which is not totally geodesic but is minimal (because it is a complex curve in $\mathbb{B}^2$), is the set of $h$-positive lines $L$ that are null with respect to the complex inner product $q = (z^0)^2-(z^1)^2-(z^2)^2$, i.e., that satisfy $L\cdot L = 0$ (while $L\cdot \bar L > 0$). This curve is an orbit of the subgroup $\mathrm{SO}(1,2)$ and hence is $\mathrm{SO}(1,2)/\mathrm{SO}(2)$, i.e., the hyperbolic plane.

Here is where I have modified my original answer:

As for the case of a complex simple Lie group modulo its maximal compact, the natural first case to check would be $\mathrm{SL}(3,\mathbb{C})/\mathrm{SU}(3)$ (since the answer is 'yes' for the lowest possible case, $\mathrm{SL}(2,\mathbb{C})/\mathrm{SU}(2)\simeq H^3$). However, right away, the above example shows that the answer is 'no' for $\mathrm{SL}(3,\mathbb{C})/\mathrm{SU}(3)$. The reason is that the noncompact real form $\mathrm{SU}(1,2)\subset \mathrm{SL}(3,\mathbb{C})$ acts on the quotient $\mathrm{SL}(3,\mathbb{C})/\mathrm{SU}(3)$ with orbit $\mathbb{B}^2 = \mathrm{SU}(1,2)/\mathrm{U}(2)$ (since $\mathrm{SU}(1,2)\cap\mathrm{SU}(3)\simeq \mathrm{U}(2)$), and this orbit is embedded as a totally geodesic submanifold of $\mathrm{SL}(3,\mathbb{C})/\mathrm{SU}(3)$. Since we already constructed an isometric, minimal-but-not-totally-geodesic immersion of the hyperbolic disc into $\mathbb{B}^2$ (see above), it follows that this yields an isometric, minimal-but-not-totally-geodesic immersion of the hyperbolic disc into $\mathrm{SL}(3,\mathbb{C})/\mathrm{SU}(3)$. Thus, the answer is 'no' even in the special case you care about.

Remark: Meanwhile, the answer to your question is 'yes' for $G = \mathrm{SO}(1,n)$. This is in an old article of mine: Minimal surfaces of constant curvature in $S^n$, Trans. Amer. Math. Soc. 20 (1985), 259–271. (Don't be fooled by the title; at the end of the article, I treat the case of hyperbolic $n$-space $H^n$ as well, which is the case $G=\mathrm{SO}(1,n)$.)

  • $\begingroup$ Dear Robert, thank you for this excellent answer. Reading it, I realize I should have seen this example coming. I'm not sure if I should revise the question itself, but I'll ask the following refinement here first. What if we assume G is the split real form of a complex, simple Lie group. Perhaps to make things very concrete, SL(n,R)/SO(n) is my first example of interest. I'm new to mathoverflow, so if this refinement should be edited into the original question, I will do so. Thanks! $\endgroup$ Apr 7, 2014 at 22:36
  • $\begingroup$ @AndySanders: You're welcome. If you do want to edit the question, you should only add to it (and note the addition), not remove what's there. (Otherwise, you might make my answer become irrelevant.) I haven't thought about the split case, but I'll do so when I get a chance. If one does have rigidity there, then a proof will likely depend on techniques such as the ones I used in that TAMS article. $\endgroup$ Apr 8, 2014 at 8:51

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