Curvature-adapted submanifolds of semi-Riemannian groups
Margarida Camarinha Matteo Raffaelli
math.DG 53C40 (Primary) 53B25, 53C30 (Secondary)

Curvature-adapted submanifolds of Lie groups]Curvature-adapted submanifolds
of semi-Riemannian groups


CMUC

University of Coimbra

Department of Mathematics

3000-143 Coimbra

Portugal
mmlsc@mat.uc.pt
This work was partially supported by the Centre for Mathematics of the University of Coimbra – UIDB/00324/2020, funded by the Portuguese Government through FCT/MCTES.The second author was partially supported by Austrian Science Fund (FWF) project F 77 (SFB “Advanced Computational Design”).

CMUC

University of Coimbra

Department of Mathematics

3000-143 Coimbra

Portugal
Institute of Discrete Mathematics and Geometry

TU Wien

Wiedner Hauptstraße 8-10/104

1040 Vienna

Austria
matteo.raffaelli@tuwien.ac.at



[2020]Primary: 53C40; Secondary: 53B25, 53C30



We study semi-Riemannian submanifolds of arbitrary codimension in a Lie group G equipped with a bi-invariant metric. In particular, we show that, if the normal bundle of M ⊂ G is closed under the Lie bracket, then any normal Jacobi operator K of M equals the square of the associated invariant shape operator α. This permits to understand curvature adaptedness to G geometrically, in terms of left translations. For example, in the case where M is a Riemannian hypersurface, our main result states that the normal Jacobi operator commutes with the ordinary shape operator precisely when the left-invariant extension of each of its eigenspaces has first-order tangency with M along all the others. As a further consequence of the equality K = α^2, we obtain a new case-independent proof of a well-known fact: every three-dimensional Lie group equipped with a bi-invariant semi-Riemannian metric has constant curvature.

[
    Matteo Raffaelli
    May 12, 2023
====================







§ INTRODUCTION AND MAIN RESULT
 
Given a Riemannian manifold, it is natural to study submanifolds whose geometry is somehow adapted to that of the ambient space. This idea led to the concept of curvature-adapted submanifold <cit.>; a concept that, since its introduction, has attracted the interest of many geometers.

Here we define curvature adaptedness in a slightly more general setting, that of a semi-Riemannian manifold (Q,g)≡ Q.

Let M be an m-dimensional semi-Riemannian submanifold of Q, let N^1M be its unit normal bundle, and let R be the ambient curvature tensor. For (p,η)≡η∈ N^1_pM, the normal Jacobi operator

    K_η≡ K  T_pM    → T_pQ 
    
    v    ↦ R(η,v)η

of M with respect to η measures the curvature of the ambient manifold along η. On the other hand, denoting by N a unit normal local extension of η along M, and by ∇ the Levi-Civita connection of Q, the shape operator

    A_η≡ A  T_pM    → T_pM 
    
    v    ↦π^⊤_vN

of M with respect to η describes the curvature of M as a submanifold of Q; here π^⊤ denotes orthogonal projection onto T_pM.


One says that M is curvature adapted to Q at a point p∈ M if, for every η∈ N^1_pM,


  *  The normal Jacobi operator leaves T_pM invariant, i.e., K(T_pM) ⊂ T_pM;

  *  The operators A and K commute, i.e., K ∘ A = A ∘ K.


Consequently, one calls M curvature adapted if it is curvature adapted at p for all p ∈ M.



Condition <ref> in Definition <ref> is always satisfied for hypersurfaces.



Both A and K are self-adjoint with respect to the induced semi-Riemannian metric. Thus, if these operators are also diagonalizable, then M is curvature adapted at p precisely when they share a common orthonormal basis of eigenvectors (Lemma <ref>). Recall that diagonalizability is automatic if the metric is positive definite.



In general, it could also be interesting to consider a weaker notion of curvature adaptedness, obtained by replacing the operator K with π^⊤K in the definition. This modification would make any totally umbilic submanifold necessarily curvature adapted.



It is easy to see (e.g., using <cit.>) that every semi-Riemannian submanifold of a real semi-Riemannian space form is curvature adapted. However, for other ambient spaces, the definition is restrictive. For example, if Q is an (m+1)-dimensional nonflat complex space form with complex structure J, then A and K commute precisely when -Jη is an eigenvector of A. Also, if Q is an (m+1)-dimensional nonflat quaternionic space form with quaternionic structure ℑ, then A and K commute precisely when the maximal subspace of T_pM invariant under ℑ is also invariant under A; see <cit.> and <cit.>.

In a symmetric space of nonconstant curvature, the situation is more involved, yet many interesting results have been obtained. Among others (see for example <cit.>), the most important is arguably Gray's theorem <cit.>, which states that any tubular hypersurface around a curvature-adapted submanifold is itself curvature adapted.


Gray's theorem has been further generalized to the family of Riemannian manifolds such that, for every geodesic γ, the Jacobi operator R(γ̇,·)γ̇ is diagonalizable by a parallel orthonormal frame field along γ. It turns out that, in such spaces, the classification of the curvature-adapted submanifolds is fully determined by that of the curvature-adapted hypersurfaces <cit.>.

In this article, we shall examine the case where Q is a semi-Riemannian group, i.e., a Lie group G ≡ (G, ⟨·,·⟩) equipped with a bi-invariant semi-Riemannian metric ⟨·,·⟩. In particular, we will focus our attention on the class of (semi-Riemannian) submanifolds of G having closed normal bundle; here by closed we mean that each normal space of M corresponds, under the group's left action, to a subspace of 𝔤 that is closed under the Lie bracket, i.e., to a Lie subalgebra (Definition <ref>).


Examples of Lie groups G abounds <cit.>: every semisimple Lie group can be furnished with a bi-invariant metric, and every compact Lie group admits one that is Riemannian. Note that, if G is absolutely simple, then any bi-invariant metric on G is a scalar multiple of the Killing form of 𝔤. 

In order to explain our main result, we first set up some notation. Provided that K is diagonalizable, let (e_1, …, e_m) be an orthonormal basis of eigenvectors of K, that is, a basis of T_pM such that |⟨ e_j, e_h⟩| =δ_jh and K(e_j) = λ_j e_j for all j,h = 1,…,m; for each j, let e_j^L denote the left-invariant extension of e_j.

 
Let M be a semi-Riemannian submanifolds of a semi-Riemannian group G. Assume that K is diagonalizable and that the normal space of M at p is closed under the Lie bracket. Then the following statements are equivalent:


  * A and K commute.

  *  If λ_j and λ_h are distinct eigenvalues, then e_j(⟨ N, e_h^L⟩) = 0.

Moreover, if the induced metric is positive definite at p, then e_j(⟨ N, e_h^L⟩) = e_h(⟨ N, e_j^L⟩) whenever e_j and e_h are in distinct eigenspaces.



For all j =1,…,m, we have λ_j = -sec(e_j,η) ≤ 0; see section <ref>.





Roughly speaking, Theorem <ref><ref> expresses the condition that the left-invariant extension of each eigenspace of K is “orthogonal to first order" to N (at p) along all other eigenspaces; in other words, along any curve that starts at p and is tangent to a different eigenspace. 

In particular, in the case of hypersurfaces, the condition on the normal space is automatically satisfied. Specializing the theorem to that case, we obtain the result below.

If M is a hypersurface and K is diagonalizable, then the following statements are equivalent:


  * A and K commute.

  * The left-invariant extension of each eigenspace of K has first-order tangency with M along all other eigenspaces.



The main significance of Theorem <ref> lies in the fact that it permits to understand curvature adaptedness to G geometrically, in terms of left translations. More precisely, it reveals that, in order for a generic submanifold (with closed normal bundle) to be curvature adapted, its tangent bundle needs to behave reasonably well under left translations. Note that the tangent bundle of a Lie subgroup is fully left-invariant; conversely, if M is a closed, connected submanifold that contains the identity and has left-invariant tangent bundle, then it is a Lie subgroup.

The basic fact that allows us to prove Theorem <ref> is that the shape operator of M⊂ H with respect to η, being H any Lie group with a left-invariant semi-Riemannian metric, decomposes as the sum of two terms <cit.>: an invariant shape operator, which depends only on η, T_pM and H; plus a second term, here denoted by 𝒲, that is closely related to the Gauss map of M; see section <ref> for details. In particular, if the metric is bi-invariant and the normal bundle is closed, then the invariant shape operator commutes with K (Proposition <ref>), and so, by linearity, commutativity of A and K reduces to that of 𝒲 and K.

In fact, if M = 2 or M is Riemannian, then the nonzero eigenvalues of K have even multiplicities (Corollary <ref>), which leads us to the following conclusions.

Every two-dimensional surface with closed normal bundle is curvature adapted to G.



If M is a three-dimensional Riemannian submanifold of G with closed normal bundle, and if K ≠ 0 for all η∈ N^1_pM, then the following statements are equivalent:


  * M is curvature adapted in a neighborhood U of p.

  * For all η∈ N^1U, the 0-eigenvector of K is an eigenvector of A.



Proposition <ref> implies that, if G =3, then every Riemannian surface in G is curvature adapted.  This is by no means surprising, because every three-dimensional semi-Riemannian group has constant curvature. Indeed, it is well known that the Lie algebra of such a Lie group is either abelian or isomorphic to 𝔰𝔩(2,ℝ) or 𝔰𝔲(2) <cit.>; in each of the latter two cases, a direct computation would reveal that the curvature of the Killing form (which, up to scaling, coincides with the metric) vanishes. Alternatively, since the Ricci curvature of G is proportional to the Killing form, the statement follows from the fact that every three-dimensional Einstein manifold has constant curvature <cit.>; see Remark <ref>.

On the other hand, applying classical results of Cartan and Dajczer–Nomizu (which, for the reader's convenience, are included in section <ref>), we can prove the following statement.





 Suppose that Q is a Riemannian resp., Lorentzian manifold of dimension at least three. If, for some p∈ Q and each x ∈ T_pQ such that ⟨ x, x ⟩=1 resp., -1, the map from x^⊥ to x^⊥ defined by y ↦ R(x,y)x is a multiple of the identity, then Q has constant curvature.
 


Assume the hypothesis of the lemma. If x,y,z are orthonormal vectors in T_pQ (resp., if x,y,z are orthonormal vectors in T_pQ and ⟨ x, x ⟩ =-1), then ⟨ R(x,y)z,x⟩ = -⟨ R(x,y)x,z⟩=0. Applying <cit.> (resp., <cit.>) and Schur's lemma <cit.>, the claim follows.


Corollary <ref> thus yields a case-independent proof that, in dimension three, every metric Lie group has constant curvature.


An alternative case-independent proof of the same fact may be sketched as follows. First, by Levi decomposition and the absence of simple Lie algebras of dimension one and two, observe that a three-dimensional Lie algebra is either solvable or simple. However, if a nonabelian solvable Lie algebra of dimension three admits an ad-invariant bilinear form, then such form is necessarily degenerate <cit.>; in other words, a three-dimensional nonabelian Lie algebra admitting an ad-invariant metric is (absolutely) simple.

Suppose, thus, that G is absolutely simple. Then any semi-Riemannian metric on G is a scalar multiple of the Killing form of 𝔤, implying that G is an Einstein manifold. Hence, in dimension three, it has constant curvature <cit.>.


The remainder of the paper is organized as follows.
In the next section we briefly review some background material. In section <ref> we introduce the invariant shape operator and examine its properties. In section <ref> we then prove Theorem <ref> and Proposition <ref>. We conclude with Appendix <ref>, where, for the sake of illustration, we give a direct proof that condition <ref> in Theorem <ref> holds whenever p is an umbilical point of M.

Some final remarks about notation:


  * The indices j,h,i satisfy j,h ∈{1, …, m} and i ∈{1, …, m+n}; note that we always use Einstein summation convention.

  * For x ∈ T_pH, we denote the left-invariant extension of x by x^L.







§ PRELIMINARIES
 
Here we recall some basic results that are used throughout the paper; see e.g. <cit.> for further details about semi-Riemannian geometry and metric Lie groups.



 §.§ Semi-Euclidean vector spaces

Let V be a real vector space, of finite dimension d, equipped with a nondegenerate symmetric bilinear form f, and let L be an endomorphism of V that is self-adjoint with respect to f, i.e., such that f(L(x), y) = f(x,L(y)) for all x,y∈ V. Recall that a basis v_1, …, v_d of V is called orthonormal if | f(v_j,v_h) |= δ_jh for all j,h =1, …, d.


It is well known that, if f is positive definite, then L is diagonalizable by an orthonormal basis of eigenvectors of L. Moreover, two self-adjoint endomorphisms of V commute if and only if they share a common orthonormal basis of eigenvectors.

While it is not possible to fully extend these classic results beyond the positive definite case, something interesting can still be said.

 
	If L is diagonalizable, then there exists an orthonormal basis of eigenvectors. Moreover, any two  diagonalizable self-adjoint endomorphisms of V commute if and only if they share a common orthonormal basis of eigenvectors. 


Assume that L is diagonalizable. Since V is the direct sum of mutually orthogonal eigenspaces of L, each eigenspace must be nondegenerate, and so it has an orthonormal basis; by concatenating these bases, we obtain an orthonormal basis of V.

As for the second statement, one direction is obvious; for the other, it suffices to note that any two commuting linear maps on V preserve each other's eigenspaces.





 §.§ Semi-Riemannian geometry

Let (Q,g) be a semi-Riemannian manifold, and let ∇ be its Levi-Civita connection. The curvature endomorphism R 𝔛(Q)^3→𝔛(Q) of (Q,g) is the (1,3)-tensor field on Q defined by

    R(X,Y)Z= ∇_X∇_YZ-∇_Y∇_XZ-∇_[X,Y]Z.

(Caution: some authors define the curvature endomorphism as the negative of ours.)

Being R a tensor, if x,y,z are vectors in T_pQ, then the value R(x,y)z is independent of the extension of x,y,z and thus well-defined. 

Suppose that x ∈ T_pQ has unit length, i.e., that | g(x,x) | =1. Then the Jacobi operator of (Q,g) with respect to x is the linear map

    K_x x^⊥   → x^⊥
    
    	y    ↦ R(x,y)x.

Clearly, by the symmetry by pairs of the (0,4)-curvature tensor (obtained by lowering the last index of R), the operator K_x is self-adjoint with respect to g.

Next, suppose that x,y ∈ T_pQ are orthonormal. Then the sectional curvature sec(x,y) of the nondegenerate plane spanned by x and y is given by the formula

    sec(x,y)= g(R(x,y)y,x)=-g(K_x(y),y),

where the last equality follows from the skew-symmetry of the (0,4)-curvature tensor.

In the Riemannian setting, an important criterion for discerning whether a manifold has constant sectional curvature is provided by the following lemma.

 
	Suppose that Q is a Riemannian manifold, and that Q ≥ 3. If, at some point p ∈ Q, the curvature tensor satisfies g(R(x,y)z, x)=0 whenever x,y,z are orthonormal, then all sectional curvatures of Q at p are equal.


Several generalizations of Lemma <ref> to the semi-Riemannian setting have appeared <cit.>. Below we recall the one that is most relevant to our discussion.


	Let x,y ∈ T_pQ. We say that the pair (x,y) is orthonormal of signature (-,+) if g(x,x) =-1, g(y,y) =1, and g(x,y) =0.



	Suppose that Q ≥ 3. If, at some point p ∈ Q, the curvature tensor satisfies g(R(x,y)z,x) =0 whenever (x,y) is orthonormal of signature (-,+) and g(x,z) = g(y,z) =0, then all nondegenerate two-planes have the same sectional curvature.





 §.§ Semi-Riemannian groups

Let G be a Lie group equipped with a left- and right-invariant (i.e., bi-invariant) semi-Riemannian metric ⟨·,·⟩, and let 𝔤 be its Lie algebra, that is, the Lie algebra of left-invariant vector fields on G. As customary, we identify 𝔤 with the tangent space T_eG of G at the identity e.

Suppose that X, Y, Z ∈𝔛(G) are left-invariant, i.e., that X,Y,Z ∈𝔤. Then the Levi-Civita connection is given by

    ∇_XY = -∇_YX= 1/2[X,Y]

and the curvature endomorphism by

    R(X,Y)Z = 1/4[Z,[X,Y]].

In addition, the following equality holds:

    ⟨ [X,Y],Z ⟩ = ⟨ X,[Y,Z] ⟩.


Suppose that x,y ∈ T_pG are orthonormal; let x^L,y^L be their left-invariant extensions. The sectional curvature of the the two-plane spanned by x and y may be computed by

    sec(x,y) =1/4⟨ [x^L,y^L], [x^L,y^L] ⟩.

Note that sec(x,y) ≥ 0, with equality if and only if [x^L,y^L]=0.

Now, let M be a semi-Riemannian submanifold of G.


The normal space N_pM is said to be closed under the Lie bracket if dL_p^-1(N_pM) is a Lie subalgebra of 𝔤. Consequently, one calls the normal bundle of M closed under the Lie bracket if every normal space is closed.


It is clear that N_pM is closed exactly when exp(N_pM) is contained in a Lie subgroup of G.


Following <cit.>, the normal space N_pM is called abelian if exp(N_pM) is contained in a totally geodesic, flat submanifold of G. It is easy to see that N_pM is abelian if and only if dL_p^-1(N_pM) is an abelian subalgebra of 𝔤.




§ THE INVARIANT SHAPE OPERATOR
 
In this section we consider the general case of an orientable semi-Riemannian submanifold M of a Lie group H equipped with a left-invariant metric ⟨·,·⟩. Given a unit normal vector η of M at p, the invariant shape operator of M with respect to η is the map


    α T_pM    → T_pM 
    
    v    ↦π^⊤_vη^L,

where, as usual, π^⊤ is the orthogonal projection onto T_pM and η^L the left-invariant extension of η.

The significance of the invariant shape operator lies in the fact that it represents the deviation of the ordinary shape operator from the differential of the Gauss map of M, as we now explain.

Let N^1M be the unit normal bundle of M, and let 𝕊^m+n-1_e be the unit sphere inside the Lie algebra 𝔥 of H. The Gauss map of M is the map


    𝒢 N^1M    →𝕊^m+n-1_e
    
    (p,η)    ↦(L_p^-1)(η);

here L_p^-1 T_pH →𝔥 denotes left translation by p^-1.

Let N be a unit normal vector field along M such that N_p = η, and consider the map 𝒢̅ = 𝒢∘ N. Its differential at p is a linear map T_pM →𝒢(p,η)^⊥. Thus, since (L_p^-1) takes η to 𝒢(p,η) and is an isometry, it follows that

    𝒲 = π^⊤∘(L_p) ∘𝒢̅

is an endomorphism of T_pM.


The Gauss map of a hypersurface in a metric Lie group was first defined by Ripoll in <cit.>. It is worth pointing out that our definition can be extended to the case where the ambient manifold is parallelizable <cit.>, or even just Killing-parallelizable <cit.>.


Clearly, if H=ℝ^m+1, then 𝒢 is the classical Gauss map of M, whereas 𝒲 its shape operator. In our setting, the following result holds.


For any v ∈ T_pM,

    A(v)  = α(v) + 𝒲(v).



Let (b_1, …, b_m+n) be an orthonormal basis of T_pH such that b_1,…,b_m∈ T_pM and b_m+n=η. For each i, let b^L_i be the left-invariant extension of b_i, so that (b^L_1, …, b^L_m+n= η^L) is an orthonormal frame for H (and a basis of T_eH).

If q ∈ M—writing N^i as a shorthand for ⟨ N, b^L_i⟩—then

    𝒢̅(q) = (L_q^-1)(N_q) = (L_q^-1) ( N^i(q) . b^L_i|_q)=N^i(q) . b^L_i|_e.

Thus, if v ∈ T_pM, then

    𝒢̅(v) = dN^i(v) . b^L_i|_e = v (N^i) . b^L_i|_e.

Since

    (L_p)(𝒢̅(v))=v (N^i ) b_i,

it follows that

    π^⊤(L_p)(𝒢̅(v))=v(N^j) b_j.


On the other hand,

    A(v)    = π^⊤_vN^ib^L_i
       =π^⊤( N^i(p)_vb^L_i + v(N^i) b_i).

Since, by construction, N^1(p)= … =N^m+n-1(p) =0 and N^m+n(p)=1, we have

    A(v) =α(v) + v(N^j) b_j,

which, together with (<ref>), gives (<ref>).



Proposition <ref> shows that 𝒲 does not depend on the particular choice of normal vector field N but only on its value at p.



Using equation (<ref>), it is not difficult to see that statement <ref> in Theorem <ref> is nothing but the coordinate expression, with respect to the frame (e^L_1, …, e^L_m), of the condition

    π_j𝒲|_Λ_h = 0   for all j,h = 1, …, m such that λ_j≠λ_h,

where Λ_h is the eigenspace of K corresponding to the eigenvalue λ_h, and where π_j is the orthogonal projection onto Λ_j. Note that (<ref>) holds if and only if 𝒲 leaves the eigenspaces of K invariant.






A useful property of the invariant shape operator, which is crucial in proving Theorem <ref>, is contained in the following proposition.


If ⟨·,·⟩ is bi-invariant and the normal space N_pM is closed, then


  * K=α∘α, and so α and K commute;

  * K leaves T_pM invariant.



The proof will be based on a lemma.

 
Under the hypotheses of Proposition <ref>, α(v)=∇_vη^L.


Let ξ be a unit normal vector at p. Since N_pM is closed,

    [η^L,ξ^L] ∈ L_p^-1(N_pM),

under the usual identification of 𝔤 with T_eG. By the bi-invariance of the metric, it follows that

    ⟨[ v^L, η^L], ξ^L⟩ = ⟨ v^L, [η^L, ξ^L]⟩=0,

which implies 1/2[v^L,η^L]_p=∇_vη^L∈ T_pM, and so α(v)= π^⊤_vη^L= ∇_vη^L.



The converse of Lemma <ref> holds: if the metric is bi-invariant and α(v) = ∇_vη^L for all (v,η) ∈ T_pM× N_pM, then N_pM is closed. In other words, given a nondegenerate subspace S of 𝔤, the orthogonal complement S^⊥ of S is closed under the Lie bracket if and only if [S, S^⊥] ⊂ S.



Clearly, being the second assertion in the proposition a direct consequence of the first, we only need to prove the latter.

Let v ∈ T_pM. Since K is tensorial, the value K(v) may be computed in terms of the left-invariant extensions v^L and η^L of v and η:

    K(v) = R(η^L, v^L)η^L.


Assume that the metric is bi-invariant. Then, using (<ref>) and (<ref>), we have

    K(v)    = 1/4[ η^L, [η^L, v^L] ]
       = ∇_∇_vη^Lη^L.

From here the statement follows directly from Lemma <ref>.



Suppose that the induced semi-Riemannian metric on M is positive definite at p ∈ M. Then, under the hypotheses of Proposition <ref>, the nonzero eigenvalues of K are negative and have even multiplicities.


We deduce from equations (<ref>) and (<ref>) that the invariant shape operator α of G is skew-adjoint with respect to the semi-Riemannian metric. On the other hand, K is self-adjoint. Thus, if the induced metric on M is positive definite at p and K(T_pM) ⊂ T_pM, then the matrices of K and α with respect to any orthonormal basis of T_pM are symmetric and skew-symmetric, respectively. Since K =α∘α when the hypotheses of Proposition <ref> are fulfilled, the statement follows from <cit.>.




Dropping the assumption that the metric is positive definite, the matrix of α in any orthonormal basis (b_1, …, b_m) of T_pM becomes

    [                         0 -⟨α(b_1), b_2⟩⟨ b_1, b_1⟩                         ⋯ -⟨α(b_1), b_m⟩⟨ b_1, b_1⟩;  ⟨α(b_1), b_2⟩⟨ b_2, b_2⟩                         0                         ⋯ -⟨α(b_2), b_m⟩⟨ b_2, b_2⟩;                         ⋮                         ⋮                         ⋱                         ⋮;  ⟨α(b_1), b_m⟩⟨ b_m, b_m⟩  ⟨α(b_2), b_m⟩⟨ b_m, b_m⟩                         ⋯                         0 ].

Hence, when m=2, the operator α∘α is a multiple of the identity regardless of the signature of the metric.




§ PROOF OF THE MAIN RESULT
 
We are now ready to prove Theorem <ref> and Proposition <ref>.


It follows from equation (<ref>) that

    A(e_j)= α(e_j) + ∑_h=1^me_j(⟨ N,e^L_h⟩)e_h.

Hence, by linearity of K, we have

    K(A(e_j)) = K(α(e_j)) + ∑_h=1^m e_j(⟨ N,e^L_h⟩)K(e_h),

whereas

    A(K(e_j)) = α(K(e_j)) + ∑_h=1^m K(e_j)(⟨ N,e^L_h⟩)e_h.


Assume that N_pM is closed; this way, since G is equipped with a bi-invariant metric, K and α commute by Proposition <ref>. It follows that K(A(e_j)) = A(K(e_j)) exactly when

    ∑_h=1^mλ_h e_j(⟨ N,e^L_h⟩)e_h = ∑_h=1^mλ_j e_j(⟨ N,e^L_h⟩)e_h.

Being (e_1, …, e_m) a basis of T_pM, we conclude that A and K commute if and only if e_j(⟨ N, e^L_h⟩) = 0 for all j and h such that λ_j≠λ_h.

It remains to show that e_j(⟨ N, e^L_h⟩) = e_h(⟨ N, e^L_j⟩) when λ_j≠λ_h and the induced metric on M is positive definite at p. To this end, identify α, A, and K with their matrices in the basis (e_j)_j=1^m. The first is a skew-symmetric matrix, by equation (<ref>), whereas A is symmetric and K diagonal. Since α and K commute, the (j,h)-entries of α K and Kα are equal, and so we must have α_jhλ_h= λ_jα_jh.

Assume λ_j≠λ_h. Then α_jh=-α_hj=0 and so, by equation (<ref>),

    A_jh   = ⟨𝒲(e_j),e_h⟩,
    
    A_hj   = ⟨𝒲(e_h),e_j⟩,

implying ⟨𝒲(e_j),e_h⟩ = ⟨𝒲(e_h),e_j⟩ by symmetry of A.





Suppose that M is a three-dimensional Riemannian submanifold with closed normal bundle, and suppose that α≠ 0. It follows by Corollary <ref> that K has one zero eigenvalue, while the remaining two are equal. Without loss of generality, we may assume that λ_3=0. Since K ≠ 0, it is clear that λ_1=λ_2≠0.

Extend η to a unit normal vector field N along M. Then, by continuity, the multiplicity of λ_3 is locally constant, i.e., there exists a neighborhood U=U(N) of p in M such that the extension of K has two negative definite eigenvalues in U.



Assume that A and K commute, i.e., they share a common basis of eigenvectors. Since the 0-eigenspace of K is one-dimensional, it follows that e_3 is an eigenvector of A. Conversely, if e_3 is an eigenvector of A, then its other two eigenvectors lie in the λ_1-eigenspace of K, from which we infer that A and K commute.







§ 
 
Here we present a direct proof of the following obvious corollary of Theorem <ref>.



Suppose that N_pM is closed. If the shape operator of M with respect to η is a multiple of the identity, then e_j(⟨ N, e^L_h⟩) = 0 for all e_j, e_h in different eigenspaces.



Assume the hypotheses of the corollary. Since α commutes with K, each eigenspace of K is invariant under α. Moreover, being A a multiple of the identity, ⟨ A(e_j), e_h⟩ =0 for j ≠ h, and so equation (<ref>) implies 

    ⟨α(e_j), e_h⟩ = ± e_j(⟨ N,e^L_h⟩)  for j ≠ h.


Clearly, if λ_j≠λ_h, then e_j(⟨ N,e^L_h⟩)=0, because α(e_j) and e_h are in different (orthogonal) eigenspaces of K.




§ ACKNOWLEDGMENTS

We thank Marcos Salvai and the anonymous referees for their constructive comments, which contributed to improve the quality of the paper.

amsplain

