11
$\begingroup$

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!

$\endgroup$

1 Answer 1

6
$\begingroup$

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 $G=\mathrm{SO}(1,n)$.)

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.