7
$\begingroup$

Let $M$ be a submanifold of a symmetric space $Q$. The normal bundle $NM$ is called abelian if $\exp(N_{p}M)$ is contained in some totally geodesic and flat submanifold of $Q$ for all $p \in M$; see Terng & Thorbergsson, "Submanifold geometry in symmetric spaces", J. Differential Geom. 42 (1995), 665–718.

It is clear that every codimension-one submanifold (i.e., hypersurface) of $Q$ has abelian normal bundle.

I am interested in the case where $Q$ is a Lie group $G$ equipped with a bi-invariant metric.

Questions: Are submanifolds with abelian normal bundle (and codimension greater than one) plentiful or rare in $G$? In particular, do two-dimensional examples exist?

$\endgroup$
8
  • 2
    $\begingroup$ In rank 1 Lie groups, clearly there are no other examples besides codimension one submanifolds. $\endgroup$
    – Ben McKay
    Aug 30, 2021 at 8:30
  • 2
    $\begingroup$ In the simplest rank 2 case, i.e., $Q=S^2\times S^2$, such manifolds are surfaces $M=C_1\times C_2$ where $C_i\subset S^2$ is a curve, so there are very many of them. Similarly for the product of any two rank 1 symmetric spaces. On the other hand, when $Q=\mathrm{SU}(3)$, regarded as an irreducible rank 2 symmetric space, such a submanifold $M^6$ has to satisfy a system of 10 first-order equations that is not involutive. One would have to prolong the system to determine how many such submanifolds there are (if any). $\endgroup$ Aug 30, 2021 at 10:11
  • 2
    $\begingroup$ You have examples by considering hyperpolar actions. In particular, principal orbits of the isotropy $K$ of some point $p \in Q$. See for example: jstor.org/stable/2693761 or arxiv.org/abs/1412.2626 $\endgroup$
    – Holonomia
    Aug 31, 2021 at 19:16
  • 1
    $\begingroup$ @MK7: A $2$-dimensional surface with abelian normal bundle could only happen in $G$ for which $\dim G - \mathrm{rank}\, G \le 2$. For compact connected Lie groups $G$, this only happens for $G = Z$ or $G = \mathrm{SU}(2)\times Z$, where $Z$ is the center of $G$, which is, itself, a torus. In both those cases, we know what the two-dimensional examples are. $\endgroup$ Sep 2, 2021 at 13:08
  • 2
    $\begingroup$ @MK7: You are welcome. Oh, and I realized that I should have also mentioned that one can take $\mathbb{Z}_2$-quotients of $G=\mathrm{SU}(2)\times Z$, such as $\mathrm{SO}(3)\times Z$ or $\mathrm{U}(2) =( \mathrm{SU}(2)\times S^1)/\{(\pm I,\pm 1)\}$. $\endgroup$ Sep 3, 2021 at 9:26

2 Answers 2

6
$\begingroup$

One class of examples in a compact, connected Lie group $G$ of rank $r$ are the conjugacy classes of codimension $r$. Taking an element $a\in G$ whose centralizer is a maximal torus (a generic condition), the conjugacy class of $a$ is a submanifold $C_a\subset G$ of codimension $r$ whose normal plane at $a$ is the tangent at $a$ to $Z_a$, the centralizer of $a$ in $G$, which is a flat, totally geodesic submanifold. By the $G$-homogeneity of $C_a = G/Z_a\subset G$, the fact that this holds at $a$ implies that it holds at all point of $C_a$.

This gives an $r$-parameter family of mutually noncogruent examples.

There are other examples: When $G=\mathrm{SU}(n)$, and $a = \mathrm{diag}(\lambda_1,\lambda_2,\ldots,\lambda_n)$ is a diagonal element for which the $\lambda_i^2$ are all distinct, the submanifold $$ M_a = \{\ g_1 a g_2\ |\ g_1,g_2\in\mathrm{SO}(n)\ \}\subset\mathrm{SU}(n) $$ is homogeneous under the isometry group of $\mathrm{SU}(n)$ (endowed with its biïnvariant metric), and its tangent plane at $a$ is orthogonal to the diagonal maximal torus $T\subset\mathrm{SU}(n)$, so, by homogeneity, its tangent plane at any point is orthogonal to a flat, totally geodesic submanifold of $\mathrm{SU}(n)$. Hence it has an abelian normal bundle.

This gives another $r=n{-}1$ parameter family of mutually noncongruent examples of codimension $r$, distinct from the conjugacy classes.

Using the methods of exterior differential systems, one can show that, when $n=3$, these two families account for all of the codimension $2$ submanifolds of $\mathrm{SU}(3)$ with abelian normal bundle, in the sense that any connected submanifold $M^6\subset\mathrm{SU}(3)$ with abelian normal bundle is, up to ambient isometry, an open subset of one of the examples listed above. The argument that I have written out is a calculation using exterior differential systems and the moving frame, but, when I have time, I can sketch the proof, if there is interest.

Addendum 1: I had a flight with some time to look at the other two compact simple rank 2 groups. It turns out that all of the connected codimension two submanifolds with abelian normal bundle in $\mathrm{SO}(5)$ and $\mathrm{G}_2$ are (open subsets of) homogeneous compact ones as well. In each case, there is one additional $2$-parameter family of examples beyond the principal conjugacy classes.

Addendum 2: I also checked the rank $r=3$ case $G = \mathrm{SU}(4)$, and found that every connected codimension $3$ submanifold of $G$ with abelian normal bundle is an open subset of one of the two types of homogeneous examples listed above. Since any codimension $2$ submanifold of $\mathrm{SU}(4)$ that is foliated by codimension $3$ submanifolds of $\mathrm{SU}(4)$ with abelian normal bundle will have abelian normal bundle, it follows that there are many non-homogeneous codimension $2$ submanifolds of $\mathrm{SU}(4)$ with abelian normal bundle.

In fact, it now seems likely that, for any compact simple group $G$ of rank $r>1$, there are two $r$-parameter families of homogeneous codimension $r$ submanifolds with abelian normal bundle and every connected codimension $r$ submanifold with abelian normal bundle is, up to ambient isometry, an open subset of a homogeneous one. The two families are as follows: The first family is the family of principal conjugacy classes in $G$, and the second family is the set of principal orbits of $K\times K$ acting by left and right multiplication in $G$ where $G/K$ is a symmetric space of rank $r$. For example, when $G=\mathrm{SU}(n)$ $(n\ge3)$, $K=\mathrm{SO}(n)$; when $G = \mathrm{SO}(n)$ $(n\ge 5)$, $K = \mathrm{SO}(p)\times\mathrm{SO}(n{-}p)$ where $p = \bigl[\tfrac12 n\bigr]$; when $G=\mathrm{Sp}(n)$ $(n\ge3)$, $K=\mathrm{U}(n)$; when $G=\mathrm{G}_2$, $K=\mathrm{SO}(4)$; when $G=\mathrm{F}_4$, $K=\mathrm{Sp}(3)\mathrm{SU}(2)$; when $G=\mathrm{E}_6$, $K=\mathrm{Sp}(4)$; when $G=\mathrm{E}_7$, $K=\mathrm{SU}(8)$, and when $G=\mathrm{E}_8$, $K=\mathrm{SO}'(16)$.

$\endgroup$
2
$\begingroup$

Let $M$ be a submanifold of a symmetric space $Q$, $G = I^0(Q)$, $\mathfrak{g} = \mathrm{Lie}(G)$. Take $p \in M$ and write $K$ for the isotropy subgroup of $G$ at $p$ and $s_p \in I(Q)$ for the geodesic symmetry at $p$. We have $\Theta = C_{s_p} \colon g \mapsto s_p g s_p$ an involutive automorphism of $G$ and $\theta = \Theta_*$ the corresponding automorphism of $\mathfrak{g}$. As usual, we decompose $\mathfrak{g} = \mathfrak{k} \oplus \mathfrak{p}$ into the $(+1)$- and $(-1)$-eigenspaces of $\theta$, where $\mathfrak{k} = \mathrm{Lie}(K)$. Note that $(G,K)$ is a Riemannian symmetric pair representing $Q$. We identify $\mathfrak{p}$ with $T_pQ$ in a standard way. Since $p \in M$, the normal space $N_pM$ becomes a subspace of $\mathfrak{p}$ and thus $\mathfrak{g}$.

First of all, a nice algebraic way to rephrase the condition that $\exp(N_pM)$ is contained in a flat immersed totally geodesic submanifold of $Q$ is to say that $N_pM$ is an abelian subspace of $\mathfrak{g} = \mathrm{Lie}(G)$. This follows from the standard correspondence between complete totally geodesic immersed submanifolds of $Q$ passing through $p$ and Lie triple systems in $\mathfrak{p}$ and the fact that flat such submanifolds correspond precisely to abelian Lie triple systems, i.e. just abelian subspaces of $\mathfrak{p}$. As noted in the comments, it is necessary for $M$ to be of codimension $\leqslant \mathrm{rk}(Q)$ if you want it to have an abelian normal bundle, simply by definition of the rank of a symmetric space.

As Robert Bryant noticed in his comment, one can easily find submanifolds with abelian normal bundles when $Q$ is reducible by taking products of submanifolds of the factors. In other words, if $Q = Q_1 \times \ldots \times Q_k$ is a Riemannian product of smaller symmetric spaces and $M_i \subseteq Q_i$ are submanifolds with abelian normal bundles, then $M_1 \times \ldots \times M_k$ has an abelian normal bundle as a submanifold of $Q$. As a special case, you can take $M_i$'s to be any hypersurfaces.

Now, the question of abundance and classification of such submanifolds is difficult in general, but it becomes easier in case $M$ is homogeneous. Recall that $M$ is called homogeneous if the subgroup of $I(Q)$ of isometries preserving $M$ acts transitively on $M$. An equivalent definition as that $M$ is homogeneous if it is an orbit of a closed connected subgroup of $G$ (here we assume that $M$ is properly embedded and connected). If $M$ is homogeneous and has an abelian normal bundle, then it is an orbit (maybe singular) of a hyperpolar action on $Q$. Somewhat conversely, if $H$ is a closed connected subgroup of $G$ acting on $Q$ hyperpolarly, then any nonsingular orbit of $H$ ($=$ of the maximal dimension) has an abelian normal bundle. Recall that the action of $H$ on $Q$ is called polar if it admits a section, i.e. a complete immersed submanifold $S \subseteq Q$ that intersects all $H$-orbits and does so orthogonally. It is called hyperpolar if it is polar and admits a flat section. Sections are always totally geodesic and their dimension is equal to the cohomogeneity of the action. So if you have a hyperpolar action of cohomogeneity greater than one, its nonsingular orbits are examples of submanifolds of codimension $> 1$ with an abelian normal bundle.

Classification of polar and hyperpolar actions on symmetric spaces has a long history and is quite hard in general. When $Q$ is irreducible of compact type (in particular, when it is a simple compact Lie group), a classification was obtained by A. Kollross. He explicitly classified hyperpolar actions and their special case of actions of cohomogeneity one here and then, in a series of papers, he showed (together with A. Lytchak) that polar actions are automatically hyperpolar on such spaces. So what you need is hyperpolar actions in Kollross's classification whose cohomogeneity is greater than one in case $Q$ is a simple compact Lie group. This is discussed in Section 3 of his paper. If I'm not mistaken and got the paper correctly, in his notation (since $Q$ is a Lie group, he denotes it by $G$), every such action is orbit equivalent to the action of a subgroup $H \times K \subseteq G \times G = I^0(G)$, where $H$ and $K$ are as in the following tables

enter image description here, enter image description here, enter image description here,

or the action of the isotropy subgroup $\Delta \subseteq G \times G$ at $e$ (i.e. the adjoint action of $G$ on itself; the cohomogeneity of this action equals the rank of $G$), or a so-called $\sigma$-action (whose construction is explained in the paper):

enter image description here.

Let me finally remark that if you're interested in such submanifolds in symmetric spaces other than compact Lie groups, there are some classification results as well. In this paper Berndt, Tamaru, and Diaz-Ramos classified hyperpolar actions without singular orbits on symmetric spaces of noncompact type (possibly reducible). Since it is very easy to compute the cohomogeneity of the actions they construct, their result gives a wealth of examples of submanifolds of noncompact symmetric spaces with abelian normal bundles.

$\endgroup$
4
  • 1
    $\begingroup$ Thanks for this detailed discussion. While you mention that the homogeneous case is easier than the non-homogeneous case, you only mention examples of non-homogeneous examples that essentially arise from hypersurfaces in rank 1 symmetric spaces. What examples do you know of irreducible Riemannian symmetric spaces of higher rank $r>1$ that contain nonhomogeneous codimension $r$ submanifolds $M$ with abelian normal bundle? $\endgroup$ Sep 1, 2021 at 13:40
  • $\begingroup$ @RobertBryant, I didn't say I possessed non-homogeneous examples. On the contrary, the point of my answer was to deal with the situation when $M$ is homogeneous. $\endgroup$ Sep 1, 2021 at 14:23
  • 1
    $\begingroup$ I'm sorry, I didn't mean to suggest that you had said that there were non-homogeneous examples in the (irreducible) higher rank case. In fact, you clearly didn't claim this, but I wondered whether you knew of any such examples. You only said that the homogeneous case was 'easier' (but still difficult). $\endgroup$ Sep 1, 2021 at 14:48
  • $\begingroup$ Oh I see, sorry, I used the word 'easier' a bit frivolously there :) May need to ponder over non-homogeneous examples a bit. $\endgroup$ Sep 1, 2021 at 15:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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