数理解析研究所講究録
第 1929 巻 2014 年 64-74

64

ON THE GEOMETRY OF SYMMETRIC$R$-SPACESPETER QUAST AND MAKIKO SUMI TANAKA
ABSTRACT. In this survey article we report on our recent work [9, 35], partially in collaboration with Jost-Hinrich Eschenburg, on geometric properties of symmetric -spaces andtheir submanifolds. But this paper also contains a new result, Theorem 12, on the convexityof certain reflective submanifolds in simply connected irreducible compact symmetric spacesof Dynkin type$R$$\mathfrak{a}.$1. INTRODUCTION AND PRELIMINARIES
1.1. Riemannian symmetric spaces. Important isometries of a euclidean space are reflections through aﬃne subspaces. They generate the full isometry group of . A particularthrough a point $p\in E$ . It reverses the orientation of oriented
reflection is the symmetry
straight lines (geodesics) emanating in . Analogously, if is a connected Riemannian manifold and a point in , an isometry of that fixes and reverses the orientations of all
geodesics emanating in is called (geodesic) symmetry of through . A generic Riemannian
manifold does, of course, not admit any geodesic symmetry.
(Riemannian) symmetric space is a connected Riemannian manifold such that for eachexists. Locally symmetric spaces are characterized
point $p\in S$ the geodesic symmetryby the property that their Riemannian curvature tensor is covariantly constant (parallel).
From this point of view symmetric spaces are generalizations of euclidean space. Symmetricspaces were introduced by \’Elie Cartan in the $1920s$ (see [3, Chap. IV] and [1, \S 6.7-\S 6.9] forinteresting historical accounts). Classical references on symmetric spaces include Sigurdur
Helgason’s monograph [16] and Ottmar Loos’ two volumes [24, 25]. We refer to these books
for further details and proofs.
To a symmetric space one associates two transitively acting closed subgroups of the fullisometry group $Iso(S)$ : the symmetry group $Sym(S)$ generated by all geodesic symmetries,
and the transvection group Trans (S) generated by compositions of two geodesic symmetries.
If is compact, the transvection group of is actually the identity component of $Iso(S)$ .
We choose a base point $0\in S$ and denote by $K$ the identity component of the isotropy. This action is calledgroup $\{g\in Iso(S)|g(0)=0\}$ . The group $K$ acts eﬀectively on
isotropy representation. A symmetric space is called irreducible, if its isotropy representation
is irreducible. Further we say that a symmetric space is of compact type, if its universal
Riemannian cover is still compact.
An importa tool for studying symmetric spaces are flats. flat $F$ of a symmetric space
is a maximal connected complete totally geodesic submanifold $F\subset S$ of vanishing sectionalcan be identified by an isometry ofcurvature. Any two flats of a symmetric space
$E$

$E$

$s_{p}$

$S$

$p$

$S$

$p$

$s_{p}$

$S$

$p$

$S$

$p$

$p$

$S$

$A$

$s_{p}$

$S$

$S$

$S$

$T_{o}S$

$S$

$A$

$1t$

$S$

2010 Mathematics Subject Classification. Primary $53C35$ ; Secondary $53C40.$$S.$

Key words and phrases. symmetric $R$-spaces, extrinsic symmetry, submanifolds, unit lattice, convexity.

65
PETER QUAST AND MAKIKO SUMI TANAKA

Thus the dimension of any two flats of coincide. This dimension is called the rank ofrank $(S)=\dim(F)$ .
If a symmetric space is compact, then any flat of is a flat torus. The unit lattice of acompact symmetric space is the unit lattice of one of its flats. Let $F$ be a flat of and let$o\in F$, then the unit lattice ofwith respect to $F$ and is
$S$

$S,$

$S$

$S$

$S$

$S$

$S$

$0$

$\Gamma=\Gamma(T_{o}F) :=\{X\in T_{o}F|Exp_{0}(X)=0\},$: $T_{o}Sarrow S$ denotes the Riemannian exponential map ofwith respect towhere
the point . We say that is rectangular (resp. cubic), if there exists an orthogonal (resp.
orthonormal) basis
ofsuch that$S$$Exp_{0}$$\Gamma$$0$$\{e_{1}, .

.

.

, e_{r}\}$$T_{o}F$$\Gamma=spann_{\mathbb{Z}}(e_{1}, \ldots, e_{r})=\{\sum_{j=1}^{r}\lambda_{j}e_{j}$$\lambda_{1}$,...,$\lambda_{r}\in \mathbb{Z}\}.$

1.2. Symmetric -spaces. Symmetric -spaces were introduced by Tadashi Nagano [31]and Masaru Takeuchi [39] in 1965 as compact symmetric spaces which are at the sametime -spaces. This means that they also admit a transitive action of a centre-free noncompact semisimple Lie group and the corresponding stabilizer of a point is a certain maximal
parabolic subgroup. For a geometric interpretation of this non-compact transformation groupof a symmetric -space we refer to [42] and [15]. We call a symmetric -space indecomposable,
if it is not (global) Riemannian product of two symmetric -spaces. Symmetric -spaces
appear in various geometric contexts.
Shoshichi Kobayashi and Tadashi Nagano classified symmetric -spaces in [18]. It turns
can be obtained as follows: Let beout that every indecomposable symmetric -spacea simply connected irreducible compact symmetric space and let $0\in S$ . We take an element$\xi\in T_{o}S$ such that the linear operator $T_{o}Sarrow T_{o}S,$ $X\mapsto R(\xi, X)\xi$ has precisely spectrum$\{0, -1\}$ . Those elements
are called extrinsically symmetric. The extrinsically symmetriccan be read oﬀ from the Satake diagram (see [18, Section 6]) or from theelements in
Dynkin diagram (see e.g. [28, Lemma 2.1]) of . Every connected component of the set of
all extrinsically symmetric elements in
is an orbit of the isotropy representation and an
indecomposable symmetric -space. Vice-versa, every indecomposable symmetric -space isobtained in this way (see [18, 19] and also [17,40
Using an algebraic description of symmetric -spaces in terms of so called compact Jordan
triple systems, Ottmar Loos characterized symmetric -spaces among all compact symmetricspaces as those whose metric on irreducible factors can be rescaled in such a way that theunit lattice gets cubic (see [26, 27
An important subclass of symmetric -spaces are the hermitian symmetric spaces of compact type. A hermitian symmetric space is a symmetric space that also carries a K\"ahler
structure such that all geodesic symmetries are holomorphic. Moreover, Masaru Takeuchi
showed that every real form (that is a totally real totally geodesic submanifold of half dimension) of a hermitian symmetric space of compact type is a symmetric -space and thatevery symmetric -space can be realized in such a way (see [41] and also [23], [45, proof of
Theorem 4.3] and [36]).
Raoul Bott used symmetric -spaces to prove his famous periodicity theorem for the stablehomotopy of unitary, orthogonal and symplectic groups (see [4] and [30, Sections 23, 24
$R$

$R$

$R$

$R$

$R$

$R$

$R$

$a$

$R$

$S$

$P$

$R$

$\xi$

$T_{o}S$$S$$T_{o}S$

$R$

$R$

$R$

$R$

$R$

$R$

$R$

$R$

66
ON THE GEOMETRY OF SYMMETRIC$R$-SPACES

There is also a periodicity result for some typical embeddings of symmetriceach other (see [29]).

$R$-spaces into

1.3. Extrinsically symmetric spaces. A connected submanifold $P\subset E$ of a euclideanisspace $E$ is called an extrinsically symmetric space if for all $p\in P$ the submanifold$P$$\rho_{p}\in
Iso(E)$$p+N_{p}P$
of
space
of
through the aﬃne normal
invariant under the reflections
atWhile Riemannian symmetric spaces are locally characterized by the parallelism of their
Riemannian curvature tensors, extrinsic symmetric spaces are characterized by the paral(w.r.t. the induced connection on the normal
lelism of their second fundamental form
bundle). Unlike in the case of Riemannian symmetric spaces, the parallelism of the second
fundamental form of $P\subset E$ characterizes extrinsically symmetric spaces globally, if one
assumes that is connected and complete, as shown by Wolf Str\"ubing (see [38] and [14]).
Moreover, for compact submanifolds of euclidean space this characterization is stable in thesense that compact submanifolds of euclidean space with almost parallel second fundamental
form are just small deformations of extrinsically symmetric ones (see [34]).
rlmetric spaces by $showi_{Il}g$ that every extrinsicallyDirk Fcrus classified cxtrinsicallysymmetric space is a product of a compact extrinsically symmetric space and an aﬃne subspace. Further, every compact extrinsically symmetric space is a symmetric -space, realized
as a connected component of the set of extrinsically symmetric elements in the tangent spaceof some symmetric space of compact type (see [12, 14] and also [7]). Vice-versa, every symmetric -space realized in this manner is extrinsically symmetric (see [11, 14
By the very definition of an extrinsically symmetric space $P\subset E$ , any element $f\in Sym(P)$is the restriction to $P$ of an isometry of $E$ . Recently Jost-Hinrich Eschenburg and theauthors have shown that this holds for any isometry of $P$ :
$P$

$p.$

$\alpha$$\alpha$$P$$sy_{I}$$R$$R$$\hat{f}$Theorem 1 ([8,10 Every isometryis the restriction of a linearl isometry$f$$\hat{f}$of a compact extrinsically symmetric spaceof$P\subset E$$E.$

Unfortunately, in the case of non-hermitian extrinsically symmetric spaces our proof uses
the classification of compact extrinsically symmetric spaces and a case-by-case verification.
A beautiful standard reference for extrinsically symmetric spaces and related topics is [2].
1.4. Intrinsically and extrinsically reflective submanifolds. A reflective submanifold
invoin a Riemannian manifold $N$ is a connected component of the fixed point set of
$N$
$M$
$N$
of
.
Reflective
isometrythroughwill be called reflection
lutive isometry of . This
submanifolds are automatically totally geodesic. Reflective submanifolds in symmetric spaces
havc bcen studied and clas ified by Dominic S. P. Leung in the series of papers [20, 21, 22, 23].
They include important totally geodesic submanifolds such as polars, meridians (see below)
and centrioles, which were introduced and extensively studied by Bang-Yen Chen and Tadashi
Nagano and their students (see e.g. [5, 6] or [32]).
A totally geodesic submanifold $M\subset N$ of a submanifold $N\subset E$ of a euclidean space $E$
is called extrinsically reflective, if $M$ is a connected component of the intersection of $N$ withthe fixed set of an involutive isometry of that leaves $N$ invariant.
$M$

$aI1$

$\sigma$$\sigma$$E$

1We may always assume that the barycentre of$P$

is the origin of $E.$

67
PETER QUAST AND MAKIKO SUMI TANAKA

Proposition 2 ([9, Theorem 2
An extrinsically reflective submanifold $M\subset P$ of an$P\subset
E$
extrinsically symmetric spaceis extrinsically symmetric in $E.$

Observation 3. In view of Theorem 1, every reflective submanifold of a compact extrinsicallysymmetric space is actually extrinsically reflective, and thus extrinsically symmetric. In otherwords, any reflective submanifold of a symmetric -space is a symmetric -space. This
generalizes a claim in [43, Lemma 3.1].
$R$

$R$

1.5. Meridians. Let be a compact symmetric space. We choose an origin $0\in S$ and afixed point $p\in S$ of the geodesic symmetry
with $p\neq 0$ Then the geodesic symmetries, is theandcommute. The meridian of corresponding to and , often denoted bythat contains
connected component of the fixed point set of the involutive isometry
The terminology (meridian’ was introduced by Bang-Yen Chen and Tadashi Nagano in [5].
$S$$s_{o}$$s_{o}$$S$$s_{p}$$0$$S_{-}$$p$$p.$$s_{o}os_{p}$Remark. If is an isolated fixed point of , then $s_{p}=s_{o}$ and $s_{o}os_{p}=$ id and thecorresponding meridian
coincides with . Unless is an isolated fixed point of we have$p$$s_{o}$$S$$S_{-}$$p$$s_{o}$$\dim(S_{-})<\dim(S)$ .

Proposition 4 ([5, Lemma 2.3]). The rank of a compact symmetric space coincides withrank(S). In particular, any flat of, that is rank(S-)
the rank of any of its meridians
is also a flat of$S$$S_{-}$$=$$S.$$S_{-}$

If is a compact extrinsically symmetric space in $E$ , then the involutive isometry
is full in $E$ , then
. If we assume thatis the restriction to of the linear isometryis has order two. With Proposition 2 we conclude:$S$$s_{o}os_{p}$

$S$

$S$

$\rho_{0}\circ\rho_{p}$$\rho_{0}\circ\rho_{p}$Observation 5. Any meridian

itself extrinsically symmetric in$P_{-}$

of a compact extrinsically symmetric space$P\subset E$is$E^{2}$

2. THE UNIT LATTICE OF COMPACT EXTRINSICALLY SYMMETRIC SPACES
In this section we report on our recent work [9] joint with Jost-Hinrich Eschenburg.
Using algebraic techniques Ottmar Loos [26, 27] proved that symmetric -spaces are precisely the compact symmetric spaces whose unit lattice is cubic, after a suitable rescalingof the metric on irreducible factors. In [9] Jost-Hinrich Eschenburg and the authors gave apurely diﬀerential geometric proof of the following statement originally due to Ottmar Loos:$R$

Theorem 6 ([26, 27
rectangular.

The unit lattice of a compact extrinsically symmetric space $P\subset E$ is

Although this fact is well known, we think that some methods used in [9] might still beinteresting. A first statement for which we provide a detailed and elementary proof concerns
extrinsically symmetric flat tori:
Theorem 7 ([14, Theorem 3], [9, Theorem 3 A full -dimensional extrinsically symmetricflat torus $F\subset E$ is an extrinsic product torus. This means that $F$ is a Riemannian product$d$

Although this isjust a special case of Observation 3, our proof does not use classification and case-by-caseverification.
$2$

68
ON THE GEOMETRY OF SYMMETRIC$R$-SPACES$i=1$ , . . . , , of possibly diﬀerent radii

in aﬃne 2, which are perpendicular to each otheP. In particular, thedimensional subspaces ofunit lattice of is rectangular.

of planar round circles$d$$S^{1}(r_{i})\subset E_{i},$$r_{i}$$E_{i}\subset E$$F$

Proof. This proof is a detailed elaboration of the arguments given [14] and [9].

be the isometric immersion given by the universal covering of . By
Let :. Similarly
denote the operators
anddenote the partial derivative operator
$f$

$F$

$\mathbb{R}^{d}arrow E$$\partial_{ijk}$$\partial_{ij}$$\frac{\partial}{\partial x_{l}}$we$\partial_{i}$$\frac{\partial^{2}f}{\partial x_{i}\partial_{x_{j}}}$is an orthonormal tangent frame
respectively. Notice thatare parallel and is an isometric immersion, theon . Since constant vector fields on
is a normal
on $F$ are parallel, too. Thus
tangent vectors fields
vector field on $F$ for any $i,j\in\{1, . . . , d\}.$As $F$ is extrinsically symmetric, its second fundamental form is parallel. By fullness,$i,j\in\{1, . . . , d\}$ . Moreoverthe normal space of $F$ is generated by theare tangent vector fields on
are parallel and commute with each other.
We observe that the linear endomorphisms$TF=E_{1}\oplus\cdots\oplus
E_{r},$ $r\leq d$ , of $TF$ into parallel
decomposition
Thus there is an orthogonal. We may assume that the parallel tangent vector fields
common eigendistributions of the. The. Thus we get
are common eigenvectors of thepartialendomorphisms
constant,
parallel.
are
the
the
becauseSince
areeigenvalues
must vanish, if at least two indicesderivatives commute, we see that the eigenvalues. We are left withare distinct. The only possible non-zero eigenvalues are therefore
for $i=1$ , . . . , . From
the diﬀerential equationsis
if $i\neq j$ , we conclude thatfor $i\neq j$ . In a similar way we observe that$i\neqj.$foreverywhere perpendicular to
Summing up we get the following system of diﬀerential equations
and$\{\partial_{\’{i}}f|i=1, .

$\frac{\partial^{3}f}{\partial x_{t}\partial x_{j}\partial x_{k}}$.

.

$\mathbb{R}^{d}$$F$, d\}$$f$$\partial_{ij}f=\alpha(\partial_{i}f, \partial_{j}f)=:\alpha_{ij}$$\partial_{i}f$$\alpha$$\partial_{ijk}f=-A_{\alpha_{lj}}\partial_{k}f$$\alpha_{ij},$$F.$$A_{\alpha_{ij}}$$A_{\alpha_{ij}}$$\partial_{i}f$$\partial_{ijk}f=-A_{\alpha_{tj}}\partial_{k}f=\lambda_{ijk}\partial_{k}f$$A_{\alpha_{ij}}$$A_{\alpha_{1j}}$$\lambda_{ijk}$$i,$ $j,$ $k$$\lambda_{ijk}$$\lambda_{i}$$d$$\partial_{iii}f=\lambda_{i}\partial_{i}f$$\langle\alpha_{ij},$$:=\lambda_{iii}$$\alpha_{ij}\rangle=\langle A_{\alpha}:j\partial_{i}f,$$\alpha_{ij}=\partial_{ij}f=0$$\partial_{j}f\rangle=0$$\alpha_{ii}$$\alpha_{jj}$

(1)
(2)

$\partial_{iii}f-\lambda_{i}\partial_{i}f$

$=$

$0$

$\partial_{ij}f$

$=$

$0$

for $i=1$ , . . . ,for $i\neq j.$$d,$Solving Equation (1) yields

if$c+c_{l}\cdot x_{i}c_{i1}\exp(\sqrt{\lambda_{i}’}x_{i})+c_{\iota 2}\exp(-\sqrt{\lambda_{i}}x_{i})$,$\partial_{i}f=\{$$c_{i1}\sin(\sqrt{-\lambda_{i}}x_{i})+c_{i2}\cos(\sqrt{-\lambda_{i}}x_{i})$:andwhere the functions
But, by Equation (2), the functions$\mathbb{R}^{d}arrow E$$c_{i1}$$c_{i2}$$\partial_{i}f$

:

:

if
, if

$\lambda_{i}=0$$\lambda_{i}>0$$\lambda_{i}<0$

do not depend on the i-th variableandareonly depend on . Thus$\mathbb{R}^{d}arrow E$$\mathbb{R}^{d}arrow E$$x_{i}.$$x_{i}$$c_{\iota 1}$$c_{i2}$constant functions.
Since
has everywhere length one, only the following two cases are possible:$\partial_{i}f$$\partial_{i}f=\{\begin{array}{ll}c_{\iota 1}, if \lambda_{i}=0c_{r1}\sin(\sqrt{-\lambda_{i}}x_{i})+c_{\iota 2}\cos(\sqrt{-\lambda_{i}}x_{i}) , if \lambda_{i}<0\end{array}$

Although the radii of the planar round circles can be distinct, such a torus is sometimes still called a
Cliﬀord torus.
$3$

69
PETER QUAST AND MAKIKO SUMI TANAKA

By integrating
occur, because

in the direction of we see that onlyis compact; that is$\partial_{i}f$$x_{i}$

for all $i\in\{1, . . . , d\}$ can$\lambda_{i}<0$$f(\mathbb{R}^{d})=F$$\partial_{i}f=c_{i1}\sin(\sqrt{-\lambda_{i}}x_{i})+c_{i2}\cos(\sqrt{-\lambda_{i}}x_{i})$.

In particular nohas zeros. Thus the set $\{\partial_{i}f|i=1, . . . , d\}\cup\{\partial_{ii}f|i=1, . . . , d\}$ is at
has constant lengtheach point an orthogonal basis of $E$ . In particular $\dim(E)=2d$ . Asone,
are unit vectors.
andRecall thatis everywhere perpendicular tofor $i\neq j$ . By taking appropriate values
for $i\neq j$ and $l\in\{1$ , 2 . Moreover,
for and , we observe thatis perpendicular to
since
and
$\alpha_{ii}$$\partial_{i}f$$c_{i2}$$c_{i1}$$\partial_{i}f$$x_{i}$$\partial_{j}f$$c_{\tau k}$$x_{j}$$k,$$c_{jl}$$\}$$\partial_{i}f$$\partial_{ii}f=c_{i1}\sqrt{-\lambda_{i}}\cos(\sqrt{-\lambda_{i}}x_{i})-c_{i2}\sqrt{-\lambda_{i}}\sin(\sqrt{-\lambda_{i}}x_{i})$, that the unit vectors
are everywhere perpendicular, we see, if we take $x_{i}=0$ and$\{c_{i1}|i=1,.
.
.
,
d\}\cup\{c_{i2}|i=1,. . . , d\}$ is anandare perpendicular. Summing up,$E.$orthonormal basis of
By integration we get$x_{i}= \frac{\pi}{2\sqrt{-\lambda_{l}}}$$c_{i2}$$c_{i1}$$f(x_{1}, \ldots, x_{d})=v+\sum_{i=1}^{d}\frac{1}{\sqrt{-\lambda_{i}}}(-\cos(\sqrt{-\lambda_{i}}x_{i})c_{i1}+\sin(\sqrt{-\lambda_{i}}x_{i})c_{i2})$for some $v\in E$ . This shows that

$F=f(\mathbb{R}^{d})$is an extrinsic product torus.

$\square$Sketch of proof of Theorem 6. We are now able to sketch our proof of Theorem 6. Using
Proposition 4, we lower the dimension of the submanifold while keeping the rank by theof . By Observation 5following iteration: Starting with $P\subset E$ , we take a meridian
is again extrinsically symmetric in $E$ . Next we consider a meridian of -and then take
$P_{-}$$P$$P$$P_{-}$

rneridia.ns again and again, until we reach a fixed point of this iteration scheme.

$T1_{1}is$

fixed

point must be a compact extrinsically symmetric space all of whose geodesic symmetries
only have isolated fixed points. In other words, we could have assumed right away that thegeodesic symmetries of our compact extrinsically symmetric space $P\subset E$ only have isolated
fixed points,
Considering a compact covering where all euclidean factors split oﬀ, we could show:
Proposition 8 ([9, Lemma 7 A compact symmetric space $P$ all of whose geodesic symmetries only have isolated fixed points is a Riemannian product of a simply connected symmetricspace $P’$ of compact type and possibly a flat torus , that is $P=P’\cross T.$$T$

If the root system of a simply connected irreducible symmetric space $P’$ of compact type
thathas not type , then there is a closed geodesic
in $P’$ admitting a Jacobi-field$0=\gamma(O)$
of
in
the
vanishes at the starting pointof but not at the antipodal point$Looki_{Il}g$circleat the variation of closed geodesics defined by , we see that the connectedcomponent of Fix
that contains has positive dimension. This yields:$J$$\alpha_{1}$$\gamma$$p$$\gamma$$0$$J$$\gamma.$$(s_{o})\subset P’$$p$

Theorem 9 ([9, Theorem 9
The only simply connected irreducible symmetric spaces ofcompact type whose geodesic symmetries only have isolated fixed points are round spheres.4
was already known before as a consequence of the classification of polars in compact symmetricspaces (these are connected components of the fixed point set of a geodesic symmetry) due to Bang-Yen
Chen and Tadashi Nagano (see [5, 6, 32 But our proof is purely conceptional.
$4_{This}$

70
ON THE GEOMETRY OF SYMMETRIC$R$-SPACESAt this point we may assume that our compact extrinsically symmetric space $P\subset E$ isintrinsically a Riemannian product of round spheres , . . . , and possibly a flat torus
that is$k$$\mathbb{S}_{1}$$T,$$\mathbb{S}_{k}$$P=\mathbb{S}_{1}\cross\cdots\cross \mathbb{S}_{k}\cross T.$

Obviously, a maximal torus of$P$

has the form
$F=C_{1}\cross\cdots\cross C_{k}\cross T,$

for $j=1$ , . . . , , and it is a reflective submanifold of
where
is a great circle in
Theorem 6 follows directly from Theorem 7 if $k=0$ . If contains even-dimensional spheres,has even dimension and set $P’$
we split them oﬀ as follows: Assume w.l. . that...
lies in the. The geodesic symmetryofat. We choose a pointlies in the identity component oftransvection group of . Therefore the involution$Iso(P)$ , which is Trans (P) . Thus
extends to an extrinsic reflection, and the connected$\{x_{1}\}\crossP’\congP’$is extrinsically symmetric by Proposition 2.
componentof FixWe are left to show that $P’$ has a rectangular unit lattice. Applying the above argument
recursively, we may assume that all sphere factors of $P’$ have odd dimensions.
In other words, we may assume that our extrinsically symmetric space $P\subset E$ is intrinsically a Riemannian product$C_{j}$$P^{5}$$k$$\mathbb{S}_{j}$$P$$:=\mathbb{S}_{2}\cross$$\mathbb{S}_{1}$$g$$x_{1}\in \mathbb{S}_{1}$$\cross \mathbb{S}_{k}\cross T$$\mathbb{S}_{1}$$x_{1}$$s_{x_{1}}$$s_{x_{1}}\cross id_{P’}$$\mathbb{S}_{1}$$s_{x_{1}}\cross id_{P’}$$(s_{x_{1}}\cross id_{P’})$$P=\mathbb{S}_{1}\cross\cdots\cross \mathbb{S}_{k}\cross T$of odd dimensional round spheres , . . . ,and perhaps a flat torus . For each $j\in$throughofis ain . The reflection
we now choose a great circle
transvection of , since has odd dimension. Thus the involutive isometry
is a transvection of $P$ and therefore extends to an extrinsic involutive isometry. With Proposition 2 we conclude that the maximal torus $F=C_{1}\cross\cdots\cross C_{k}\cross T$ ofis extrinsicallysymmetric and has rectangular unit lattice by Theorem 7.
$\{1, .

.

.

, k\}$$C_{j}$$S_{j}$$T$$\mathbb{S}_{k}$$\mathbb{S}_{1}$$C_{j}$$\mathbb{S}_{j}$$\mathbb{S}_{j}$$r_{j}$$r_{1}\cross\cdots\cross r_{k}\cross id_{T}$$\mathbb{S}_{j}$$P$

Question. At RIMS Workshop we were asked by Professor Yoshihiro Ohnita if our method
can be adapted to show that the unit lattice of an indecomposable symmetric -space isactually cubic. Unfortunately, we cannot answer this question.6$R$

3. CONVEXITY OF REFLECTIVE SUBMANIFOLDS OF SYMMETRIC$R$-SPACESIn this section we give an overview of our work published in [35].
A geodesically complete submanifold $M\subset N$ in a Riemannian manifold $N$ is called$m_{2}\in M$ mea(geodesically) convex, if the Riemannian distance between any two points$M$sured within coincides with the Riemannian distance between
andmeasured within$N$ , or, equivalently, if any shortest geodesic arc in $M$ is still shortest in $N$ . One might thinkof convexity as a ‘global version’ of being totally geodesic.
Reflective submanifolds certainly form a very important class of totally geodesic subman$m_{1},$$m_{1}$$m_{2}$

ifolds of compact symmetric spaces. One may therefore wonder whether reflective submani-folds in compact symmetric spaces are geodesically convex. Already in the fairly easy example
one uses Observation 3 (for which we unfortunately only know a proof using classification and whichis therefore not in the spirit of [9]), then Theorem 6 follows directly from Theorem 7.
$6At$
this point we recall that Dirk Ferus has shown in [13] that the shape operator in the direction of themean curvature vector field of a compact indecomposable extrinsically symmetric space is a multiple of theidentity.
$5If$

71
PETER QUAST AND MAKIKO SUMI TANAKA, whose unit lattice is rhombic and not rectangular, the long diagonalof a flat 2-torus
is an example of a reflective but non-convex submanifold. This is bad news! Indeed, adapting
results of Takashi Sakai [37] on the cut locus of compact symmetric spaces, Hiroyuki Tasaki
showed that convexity is already detected on the level of flats.
$\Gamma$$\mathbb{R}^{2}/\Gamma$

Proposition 10 ([46, Lemma 2.2]). Let $M\subset S$ be a reflective submanifold in a compact, that isa flat of containing
be a flat of $M$ andsymmetric space S. Let$M$
is convex in if and only if
is convex in
Then$F_{M}$$S$$F_{S}$$S$$F_{M}$$F_{M}\subset F_{S}.$$F_{M}$$F_{S}.$

It turns out that the situation described above appears in the symmetric spaceof the reflective submanifold $M$ which isomorphic
whose unit lattice is rhombic. The flat(see [35]). This example works in allinis the long diagonal intoiscomplexconjugation inbythedimensions: The reflective submanifold given$n\geq3$. On the other hand, Felix Platzer and the first author have shown
never convex forusing case-by-case arguments:$S=SU_{3}/\mathbb{Z}_{3},$$F_{M}$$SO_{3}$$S=SU_{3}/\mathbb{Z}_{3}$$F_{S}$$S=SU_{n}/\mathbb{Z}_{n}$Theorem 11 ([33]). Every reflective submanifold in a special unitary group$SU_{n}$is convex.

As a consequence we get:
Theorem 12. Let be a simply connected irreducible symmetric space of compact type and
Let $M\subset S$ be a reflective submanifold thatof rank $r\geq 2$ , whose root system is of typehas the following property:$(*)$
which is regular $w.r.t$ . the root system of
There exists
$S$

$7$

$\alpha_{r}.$$S^{8}$$\xi\in T_{O}F_{M}\subset T_{O}F_{S}$Then $M$ is convex in$S.$for short. Choose a maximal abelian subspace
the flat of the special unitary groupand denote byof the Lie algebra$:=SU_{r+1}$ satisfying
both have rankand. Since the symmetric spaces
and root systems of type , there exists (for a suitably scaled bi-invariant metric on ) an
that identifies the root system of with the root system oforthogonal linear map. Since the unit lattice of a simply connected irreducible symmetric space is generated bycoincides with
its system of inverse roots (see e.g. [25, pp. 25, 69, 77 which in the case$\iota(0)=e.$
with
the set of root vectors, induces an isometry :invariant (see [44, Lemma 3.1], [35, Observation
The reflection of through $M$ leaves
4 and its diﬀerential
restricted to $t=T_{o}F_{S}$ is an involutive orthogonal map thatleaves the root system of invariant. Consider

Proof. Set$t:=T_{o}F_{S}$and$\hat{t}$$\alpha$$:=T_{O}F_{M}\subset\{$$\mathfrak{s}u_{r+1}\cong T_{I}SU_{r+1}$$\hat{S}$$F_{\hat{S}}$$\hat{S}$$S$$\hat{t}=T_{I}F_{\hat{S}}$$r$$\hat{S}$$\alpha_{r}$$S$$\iota_{*}:tarrow\hat{t}$$\hat{S}$$\mathfrak{a}_{r}$$\iota$$\iota_{*}$$\sigma$$S$$F_{S}arrow F_{\hat{S}}$$F_{S}$$\sigma_{*}:tarrow t$$S$$\hat{\sigma}_{*}:=\iota_{*}\circ\sigma_{*}\circ\iota_{*}^{-1}:\hat{t}arrow\hat{t}.$fiesThis is an involutive orthogonal map that leaves the root system of invariant and$(*)$,. Moreover, by condition
fixes the regular element. Thusleaves the system of positiveinthat
contains
and therefore the Weyl chamber$\hat{S}$$sati_{i\supset}\backslash$$\hat{\alpha}$$\hat{\xi}$$:=Fix(\hat{\sigma}_{*})=\iota_{*}(\alpha)$$\hat{\sigma}_{*}$$\hat{t}^{+}$$\hat{t}$$:=\iota_{*}(\xi)\in\hat{\alpha}$$\hat{\xi}$$\hat{\sigma}_{*}$

7 This means that $S$ is $SU_{r+1}$ or $SU_{r+1}/SO_{r+1}$ or $SU_{2r+2}/Sp_{r+1}$ or $E_{6}/F_{4}$ (see e.g. [16, Chapter X

is called regular
we use the notation introduced in Proposition 10 with $0\in F_{M}$ . A vector
for all roots of . For details on root systems of symmetric spaces
w.r. . the root system of , if
we refer to the standard literature such as [25] or [16]. An equivalent formulation of condition $(*)$ is thatis the unique extension of
to a flat of$8_{Here}$$t$$\xi\in T_{o}F_{S}$$S$$\alpha(\xi)\neq 0$$\alpha$$S$$F_{S}$$F_{M}$$S.$

72
ON THE GEOMETRY OF SYMMETRIC$R$-SPACESdefining
simple roots ofinvariant. Since $\hat{S}=SU_{r+1}$ is a simply connected simple
induces an involutive Lie group automorphism of $\hat{S}=SU_{r+1}$ thatcompact Lie group,
leavesinvariant (see [25, Proposition 3.4, p. 128
Let
be the connected component of Fix (a) containing the identity . We claim
be a maximal abelian subspace ofis a flat of . Indeed, let
that
$(*)$
,
is a regular element w.r.
that contains . By condition
that contains
uniqueabeliansubsetofismaximalthethe root system of . Thus
But thethat contains
and therefore also the unique maximal abelian subset ofis a flat ofandis . Henceintersection of withis convex
is convex in $SU_{r+1}$ by Theorem 11, and thus
The reflective submanifold
in . Since is an isometry,is also convex in . Proposition 10 eventually implies theclaim.
$\hat{t}^{+}$$\mathfrak{s}u_{r+1}$$\hat{\sigma}$$\hat{\sigma}_{*}$$F_{\hat{S}}$$\hat{M}\subset\hat{S}$$I$$\hat{M}$$F_{\hat{M}}$$\hat{\mathfrak{a}}’$$:=\iota(F_{M})\subset F_{\hat{S}}$$T_{I}\hat{M}\subset \mathfrak{s}u_{r+1}$$\hat{\xi}:=\iota_{*}(\xi)\in\hat{\mathfrak{a}}$$\hat{\alpha}$$\hat{S}$$t.$$\hat{t}$$\mathfrak{s}u_{r+1}$$\hat{\xi}$$\alpha$$\mathfrak{s}u_{r+1}$$T_{I}\hat{M}$$t$$\hat{\mathfrak{a}}’=\hat{\mathfrak{a}}$$\hat{\mathfrak{a}}$$\hat{M}.$$\iota(F_{M})$$\hat{M}$$F_{\hat{M}}$$F_{\hat{S}}$$F_{M}$$\iota$$F_{S}$$\square$Now the following question arises:
Question. Can one drop the somewhat technical condition $(*)$ in Theorem 12?

More generally on may ask:
Question. How can one describe the class formed by the compact symmetric spaces all of
whose reflective submanifolds are convex?
The authors have shown that the symmetric$R$-spaces form a subclass of this class:

Theorem 13 ([35]). Every reflective submanifold in a symmetric$R$-space is convex.

be an involutive isometry of a symmetric -space and let $M$ be aofof its fixed point set. In view of Proposition 10 we take a flat
connected$M$ and a flat
. Recall that leaves
invariant (see [44, Lemma 3.1],
of withis convex in . For this we choose an origin[35, Observation 4 We have to show that$o\in F_{M}$ . Then we have to prove that

Outline of proof. Let$R$$\sigma$$P$$F_{M}$$c(Inpon(^{\lrcorner},nt$$F_{S}$$S$$F_{M}\subset F_{S}$$F_{S}$$\sigma$$F_{M}$$F_{S}$$d_{F_{M}}(0, x)=d_{F_{S}}(0, x)$anddenote the Riemannian distances inandholds for all $x\in F_{M}$ , where
respectively.
As a consequence of Theorem 6 one can show that there exists an orthogonal basis $B=$ofsuch thatthe unit lattice of is generated by , that is$d_{F_{M}}$$\{e_{1}, .

.

.

, e_{r}\}$$T_{o}F_{S}$

$B$

$S$

$\Gamma$$\bullet$$\Gamma=\{\sum_{j=1}^{r}\lambda_{j}e_{j}$$\bullet$$F_{M}$$d_{F_{S}}$$\lambda_{j}\in \mathbb{Z},$$j=1$ , . . . , $r\}$ ;there exists $p\in\{1, . . . , r\}$ and $q\in\{2p, . . . , r\}$ such that
$T_{O}F_{M}= \{\sum_{j=1}^{r}\lambda_{j}e_{j}$$\lambda_{2j-1}=\lambda_{2j}$for $1\leq j\leq p$ and $\lambda_{q+1}=\cdots=\lambda_{r}=0\}$$F_{S}$

73
PETER QUAST AND MAKIKO SUMI TANAKA(see [44, Proposition 3.3] for a diﬀerential geometric proof and [35, Proposition 5] for anelementary linear algebraic one).
Given $x\in F_{M}$ we take an element $X\in T_{O}F_{M}$ such that $x=Exp_{0}(X)$ , where $Exp_{0}$ denotes
the Riemannian exponential map of $P$ at . Then $d_{F_{S}}(0, x)= \min_{Y\in\Gamma}\Vert X+Y$ One easily
constructs an element $Z\in T_{O}F_{M}\cap\Gamma$ such that $\Vert X+Z\Vert=\min_{Y\in\Gamma}\Vert X+Y$ We conclude that$0$$d_{F_{M}}(0, x)=d_{Fs}(0, x)$.

$\square$ACKNOWLEDGEMENTS
The authors wish to thank Jost-Hinrich Eschenburg and Ernst Heintze for helpful discusslons.

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375-382(Quast) INSTITUT F\"UR MATHEMATIK, UNIVERSIT\"AT AUGSBURG, D 86135 AUGSBURG, GERMANY-mail address: peter.quast@math.uni-augsburg.de$E$(Tanaka) FACULTY OF SCIENCE AND TECHNOLOGy, TOKYO UNIVERSITY OF SCIENCE, NODA, CHIBA
278-8510, JAPAN-mail address: tanaka nakiko@ma.noda.tus.ac.jp$E$$\lrcorner$