Curvature locus of a 3-manifold having a normal bundle with some flat direction

A well known result states that the curvature of the normal bundle of an n-manifold immersed in the Euclidean space, Rn+k\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathbb {R}}^{n+k}$$\end{document}, vanishes if and only if the curvature locus at any point is a convex polytope. In the case of 3-manifolds, the curvature locus is, generically, a triangle. In this paper we determine, among all the possible curvature locii of 3-manifolds immersed in R3+3\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathbb {R}}^{3+3}$$\end{document}, which ones can be projected on a triangle and we relate them with the curvature of the normal bundle.


Introduction
The study of the geometric properties of immersions of n-manifolds in Euclidean spaces can be done through the object called the second fundamental form. The image by the second fundamental form of all unitary tangent vectors at a point is a compact subset of the normal space. This compact set is called the curvature locus of the immersion at the point. In the case of immersions of surfaces the curvature locus is nothing but the curvature ellipse.
The notion of curvature locus for higher dimensional submanifolds was first given in [5]. In [2] it was shown that for 3-manifolds immersed in R 3+3 , all the possible substantial (i.e. non-planar) curvature locus models are Steiner surfaces of types 1, 3, 5, 6, ellipsoids and truncated cones.
In this paper we discuss the 3-manifolds immersed in R 3+3 for which the normal bundle is non totally flat. To be non totally flat means that there is at least some direction for which the curvature of the normal bundle vanishes.
The curvature of the normal bundle at a point is related with the commutators of the symmetric matrices, associated with an orthonormal basis of the tangent space, of the shape operators associated with an orthonormal basis of the normal space. This relationship trans-lates conditions on the curvature of the normal bundle into conditions on a family of matrices, as many as the dimension of the normal space.
As differential geometers know very well, the choice of a convenient orthonormal basis simplifies problems. The matrices associated with the shape operator, or equivalently, to the second fundamental form, are symmetric, therefore, they are diagonalizable. The problem is that we do not have just one matrix, but as many as the codimension of the immersion. Conjugation by an orthogonal matrix diagonalizing a matrix does not need to diagonalize the others simultaneously. The condition for having simultaneous diagonalization of two (diagonalizable) matrices is the vanishing of the commutator of the two matrices. This is how to relate the curvature of the normal bundle with triangular curvature locii. Indeed, if all the commutators of matrices associated with the shape operator vanish, then all the matrices are diagonalizable simultaneously. The eigenvalues of the matrices are the coordinates of the vertices of a polyhedron, and the curvature locus is the convex hull of such vertices. This is a well known result in [6].
Here we have obtained a generalization of this result for immersions whose normal bundle has some flat direction, this is, where not all the commutators vanish, but maybe only one. Once we have chosen the appropriate pair of orthonormal basis, one for the tangent space and one for the normal space, we classify the curvature locus thanks to a result in [1]. Among all the possible substantial (non-planar) curvature locus, we have determined that for this kind of immersions, Cross-Cap surfaces, Steiner Type 6 surfaces or ellipsoids are not allowed as curvature locus. The only possibilities are Steiner Roman surfaces, Steiner type 5 surfaces or truncated cones. Furthermore, the curvature locus must admit a projection on a triangle for some direction in the normal space. This is true for truncated cones, but not for the other two types of allowed curvature locus. Thus, we have called Steiner Roman (resp. type 5) surfaces with parallel sides to the subfamily of Steiner Roman (resp. Steiner Type 5), surfaces for which there is a projection on a triangle.

Curvature of the normal bundle of an immersion
Let M be a smooth immersed n-dimensional manifold in R n+k . We consider in M the Riemannian metric induced by the Euclidean metric of R n+k . Given a point p ∈ M, we have a decomposition where N p M = T p M ⊥ and the corresponding orthogonal projections : R n+k → T p M and ⊥ : R n+k → N p M.
Let us denote by∇ the Riemannian connection of R n+k . Given vector fields X , Y , locally defined along M, we can choose local extensions,X ,Ȳ , over R n+k , and define the Riemannian connection on M as ∇ X Y = ∇XȲ .
where ∇ is the Levi-Cività connection in R m+n and ⊥ denotes the projection on the normal bundle.
Let us recall that the second fundamental form is symmetric, α(X , Y ) = α(Y , X ), due to the fact that∇ is a Levi-Cività connection. Moreover, this bilinear symmetric map induces, for each p ∈ M and for each u ∈ N p M, u = 0, a bilinear symmetric form on the tangent space T p M given by wherev andw are tangent vector fields such thatv( p) = v andw( p) = w and · denotes the usual scalar product.

Definition 2
The shape operator. Let us denote by A : whereū is a local extension of u ∈ N p M over a neighborhood of p in R n+k .
For each normal vector u ∈ N p M, A u is the shape operator in the u-direction.
The shape operator satisfies for u ∈ N p M and X , Y ∈ T p M.

Definition 3
The curvature tensor of the normal bundle is defined as the map, R N : whereX ,Ȳ ,ū are local extensions of X , Y ∈ T p M and u ∈ N p M, respectively.
Recall that an immersion is said to have a flat normal bundle if the curvature of the normal bundle vanishes. As a generalization of this notion, let us introduce the concept of normal bundle with some flat direction.
Notice first that given normal vectors Definition 4 An immersion is said to have a normal bundle with some flat direction at a point p if the vector subspace is not of maximal dimension n 2 .
Let us recall the next result whose proof is included here for the sake of completeness.

Lemma 1 The curvature R N of the normal bundle at p satisfies
for any u, v ∈ N p M, X , Y ∈ T p M.
Proof The reference to the point p will be suppressed in the following computations and the first equality in Eq. (1) is proven because On the other hand, Therefore, we get the second equality in Eq. (1).
Notice that given a normal vector u ∈ N , then A u : T → T is an endomorphism of the tangent space. If {t i } n i=1 is an orthonormal basis of T , then we can obtain the matrix of the endomorphism A u in the basis {t i } n i=1 as The Veronese of curvature may also be defined as the restriction of η to the unit sphere in T p M. The image of the Veronese is called the curvature locus.
be an orthonormal basis of the tangent space at p ∈ M. Any vector in the upper hemisphere of S T can be parameterized as where u i ∈ R, i = 1, . . . , n − 1. Thus, using expression (2), we get Let us define The symmetry of the second fundamental form ( Thus, the result is a quadratic rational surface. Let us write the associated homogeneous parametrization (in coordinates u i for i = 1, . . . , n):

The associated matrices
The numerator in Eq. (3) can be written as where the entries of the symmetric matrix are vectors of the normal bundle, N p M. Therefore, if we take a basis on the normal bundle, , then each entry of the matrix in expression (4) can be written as where B i j ∈ R for = 1, 2, . . . , k and i, j = 1, 2, . . . , n. Thus, the matrix B i j n i, j=1 can be written in fact as k symmetric matrices: , of the tangent space T p M and given an orthonormal basis, where Notice that A is, up to a sign, nothing but the matrix of the shape operator A n : T p M → T p M in the basis of the tangent space {t i } n i=1 (see Def. 2). All matrices A , = 1, 2, . . . , k, are diagonalizable because they are symmetric. Nevertheless, they are not simultaneously diagonalizable. Recall that a pair of symmetric matrices, A and B, is simultaneously diagonalizable if and only if they commute (see, for instance, Th. 1.3.12 in [4]). Moreover, a family of diagonalizable matrices is simultaneously diagonalizable if and only if they commute pair by pair.

Two actions
The definition of the matrices A 's depends on the choice of a basis of the tangent space and on the choice of another basis of the normal space. Let us see what happens if one of the two basis changes.

Change of basis of the tangent space
A change of basis of the tangent space is given by an orthogonal matrix, M ∈ O(n). Thus, the new associated matrices are given as follows:

Change of basis of the normal space
Analogously, a change of basis of the normal space is given by another orthogonal matrix, P ∈ O(k). The action of the change of basis acts on the B i 's as usual: Therefore the result of the action on the matrices A i 's is Both actions are commutative.

Relationship with the curvature of the normal bundle
Next result is an interpretation of the commutator of two associated matrices, A r and A s in terms of the curvature of the normal bundle.
Proof It is just a consequence of Lemma 1. In fact, [A r , A s ] is the matrix associated with [A n r , A n s ] and, thanks to Eq. (1), Therefore, the curvature locus is the convex hull of the points In this case, the parametrization of the curvature locus can be written as where Therefore, we get the parametrization of the triangle with vertices B 11 , B 22 and B 33 . Let us go one step further in the analysis of the curvature locus of immersions of 3manifolds. Let us see how to characterize immersions having normal bundle with some flat direction in terms of the commutators of the associated matrices. We need a lemma first. Reciprocally, let A 1 , A 2 , A 3 be the matrices associated with an orthonormal basis of N p M, {n 1 , n 2 , n 3 }, and let us suppose that the family is not of maximal rank. Also, the family is linearly dependent. Then, there is a non null vector, n 3 = (n 3,1 , n 3,2 , n 3,3 ) = n 3,1 n 1 + n 3,2 n 2 + n 3,3 n 3 , in the normal space, such that (Notice the permutation of indexes 1 ↔ 3 and the minus sign in the second term. ) We can suppose that the vector n 3 is unitary and we can complete an oriented orthonormal basis of the normal space with two more vectors n 1 = (n 1,1 , n 1,2 , n 1,3 ) and n 2 = (n 2,1 , n 2,2 , n 2,3 ) such that n 3 = n 1 ∧ n 2 . This is, such that n 3,1 = n 1,2 n 2,3 − n 1,3 n 2,2 , n 3,2 = n 1,3 n 2,1 − n 1,1 n 2,3 , n 3,3 = n 1,1 n 2,2 − n 1,2 n 2,1 . The action of P on the matrices A 1 , A 2 and A 3 produces three new matrices A P 1 , A P 2 and A P 3 given by A P k = n k,1 A 1 + n k,2 A 2 + n k,3 A 3 , k = 1, 2, 3.
Let us compute, for k = 1 and = 2, the commutator of the new matrices A P 1 and A P 2 :

Diagonalizing two of the three matrices
Let us suppose that there is a basis of the normal space such that the first two associated matrices commute. Therefore, they can be simultaneously diagonalized, this is, there is a basis of the tangent space such that the first two matrices are diagonal. Therefore, we can suppose that In this case, the parametrization of the curvature locus can be written as where B 12 = (0, 0, q 12 ), B 13 = (0, 0, q 13 ), B 23 = (0, 0, q 23 ).
The first term in Eq. (8) is nothing but the parametrization of the triangle, T , with vertices B 11 , B 22 and B 33 .
The generic curvature locus in this situation is a Roman Steiner surface. To prove this, we will make use of the classification of the curvature locus of immersions parametrized in Monge form that can be found in [1]. Let us include here the statement. y, z, f 1 (x, y, z), f 2 (x, y, z), f 3 (x, y, z)), where f i , i = 1, 2, 3, have zero linear and constant parts and let us denote its quadratic part by j 2 f i . Consider the matrix ⎛

Theorem 1 (Th. 3.1 in [1]) Given a local parametrization in Monge form
and let V (M f ) be the space of zeros of the 3 × 3 minors of this matrix. Let us suppose finally that the curvature locus at p = 0 is non-planar. Then, Notice that if the points B 11 , B 22 and B 33 are collinear then, the first two coordinates of the parametrization of the curvature locus (Eq. (8)) are a parametrization of a segment. Therefore, the curvature locus is a planar figure. Moreover, the same happens if the projections on the x y-plane of the points B 11 , B 22 and B 33 are collinear. Therefore, we will add this condition as a hypothesis.

The curvature locus is a Roman Steiner surface if and only if V (M f ) is 6 real lines. 2. The curvature locus is a Cross-cap surface if and only if V (M f ) is 2 real lines and 4 complex lines. 3. The curvature locus is a type 5 surface if and only if V (M
As we will see in the following statement, some of the possible curvature locus we will obtain, are not generic Roman Steiner surfaces but a specific type of these. So let's first look at what this kind of Steiner Roman surfaces is. Notice first that any Steiner surface has an enveloping tetrahedron. The four faces of the tetrahedron are determined by the four tangent planes to the Steiner surface along the respective ellipses (see Fig. 1, left).

Definition 7
We will say that a Roman Steiner surface has parallel sides when one of the vertices of the enveloping tetrahedron is placed at an infinite point. See Fig. 1, right, for a Roman Steiner surface with parallel (and vertical) sides. vanish. 4. …a triangle when q 12 = q 13 = q 23 = 0.

Proposition 3 Let us suppose that the projections on the x y-plane of the points B
Proof 1) Let us see this fact as an application of Th. 1. Given the associated matrices A 1 , A 2 , A 3 , we will apply this result to a parametrization in Monge form such that This is, Fig. 1 The enveloping tetrahedron of a Roman Steiner surface. At the left, the general case. At the right, the case of a normal bundle with some flat direction: one of the vertices is a point at infinity Given the parametrization of the curvature locus given in Eq. (8), one has to compute zeros of the 3 × 3 minors of the matrix (the common factor 2 has been removed) The simplest one of the four 3 × 3 minors is It is easy to check that if (a 1 , a 2 ), (b 1 , b 2 ) and (c 1 , c 2 ) are collinear, then the curvature locus is planar. Conversely, if the three points (a 1 , a 2 ), (b 1 , b 2 ) and (c 1 , c 2 ) are not collinear, then, the only solution is that x = 0 or y = 0 or z = 0.
Let us suppose that z = 0. Substituting z = 0 in the matrix (10) we get ⎛ ⎜ ⎜ ⎝ The 3 × 3 minor made of the first three lines is equal to Therefore, there are two possible solutions: the first one, that one of the two variables, x or y, vanishes. The second solution is given by where m ∈ R. It is just a matter of computation to check that the points in the lines (x, 0, 0) or (−mq 23 , mq 13 , 0) are zeroes of the four 3 × 3 minors.
Analogously, if we start with x = 0 or with y = 0. Therefore, V (M) has six real lines: Therefore, according to the item 1) in Th. 1, we are in the case of a Roman Steiner surface. Now, let us check that the "sides" are parallel. Notice that Loc(u, 0) = 1 1 + u 2 B 11 u 2 + 2B 13 u + B 33 is the parametrization of an ellipse in the plane passing through the points B 11 , B 13 and B 33 . Indeed, if we write Loc tan t 2 , 0 = 1 2 ((B 11 + B 33 + ((B 33 − B 11 ) cos(t) + 2B 13 sin(t)) .
Therefore, the ellipse is centered at B 33 + B 11 and with oblique axes B 33 − B 11 and 2B 13 . Since B 13 = (0, 0, q 12 ) and B 11 and B 33 are two of the vertices of the triangle, then the ellipse lays in the vertical plane passing through the side B 11 B 33 of the triangle T . Moreover, this vertical plane is tangent to the curvature locus along the ellipse. Indeed, which is a vector parallel to (0, 0, 1). Analogously, Loc(0, v) is the parametrization of an ellipse in the vertical plane passing through the side B 22 B 33 of the triangle T .
And finally, Loc w , when w → 0, is the parametrization of an ellipse laying in the vertical plane passing through the side B 11 B 22 of the triangle T .
(2) Notice that until now we are supposing that q 12 q 13 q 23 = 0. Conversely, if for instance, q 13 = 0, then the solution (−mq 23 , mq 13 , 0) agrees with solution (x, 0, 0) and solution (0, −mq 13 , mq 12 ) agrees with solution (0, 0, z). Or, in other words, V (M) has four real lines, two of them with multiplicity 2 and two with multiplicity 1. Therefore, according to the item 3) in Th. 1, we are in the case of a Steiner type 5 surface. See Fig. 3, center. (3) If for instance, q 13 = q 23 = 0, then matrix in Eq. (10) reduces to Now, it is easy to check that the real line (0, 0, z) and the plane (x, y, 0) are zeroes of all the 3 × 3 minors. Thus, according to the item 6) in Th. 1, the curvature locus is a truncated cone. Moreover, it is easy to check that the base of the truncated cone is a vertical ellipse. See Fig. 3, right side.

Remark 1
An alternative proof of Prop. 3 can be obtained thanks to the classification of quadratically parametrizable surfaces shown in [3], Sect. 6. has not the maximal dimension. In the case of 3-manifolds immersed in R 3+3 the maximal dimension is 3 2 = 3. Therefore the possibilities for the dimension of Q are 0, 1 and 2. The basis-free statement of the classification result is as follows: Proof Recall that, thanks to Corollary 1, the dimension of the vector subspace Q is the same than the rank of the family

Theorem 2 Let M be a 3-manifold immersed in R 3+3 and let Q be the vector subspace
The commutator of two 3 × 3 symmetric matrices, A and B, is a skew-symmetric matrix. Moreover, the set of 3 × 3 skew-symmetric matrices is a vector space of dimension 3. The standard basis of this vector space is Since the rank of Q is less than 2, there is an orthonormal basis on N p M and an orthonormal basis of T p M such that the associated matrices are those in Eq. (7). For those matrices, If the projections on the x y-plane of the points B 11 , B 22 and B 33 are collinear then, the curvature locus is a planar figure. Therefore, we can suppose that the projections on the x y-plane of the points B 11 , B 22 and B 33 are not collinear. The 2 × 2 minors of Q are It is easy to check that the vanishing of one of the three determinants implies that the projections on the x y-plane of the points B 11 , B 22 and B 33 are collinear. Therefore, the three determinants are not zero. Thus, the rank of Q is 2 if and only if q 12 q 13 q 23 = 0 or just one of the three parameters q 12 , q 13 and q 23 vanishes. Therefore, according to item 1) and 2) of Th. 3, the curvature locus is a Roman Steiner surface or a type-5 surface, both with parallel sides. The rank of Q is 1 if and only if just two of the three parameters q 12 , q 13 and q 23 vanish. Therefore, according to item 3) of Th. 3, the curvature locus is a truncated cone.
Finally, the rank of Q is 0 if and only if the three parameters q 12 , q 13 and q 23 vanish. Therefore, according to item 4) of Th. 3, the curvature locus is a triangle.

The curvature locus of the Veronese immersion
We are going to illustrate Th. 2 with an example. We will compute the curvature locus of the Veronese immersion:

Proposition 4 At any regular point of the Veronese surface the curvature locus is always a truncated cone.
Proof First of all, notice that, since the expression of the Veronese immersion is through second order homogeneous monomials, then where x i denotes the partial derivative with respect to u i . Taking partial derivatives once more, Let {n 1 , n 2 , n 3 } be an orthonormal basis of the normal space. Therefore, if we multiply by any normal vector, n , = 1, 2, 3, we get This means that where α is the second fundamental form of the immersion. Instead of computing the rank of the commutators of the three matrices defined through the vectors α(t i , t j ), we are going to compute the rank of the matrices defined through the vectors α(x i , x k ). The difference is just a change of basis, so the rank does not change.
If we put b i j = α(x i , x j ), then Eq. (14) can be written as Since the only singular point of the immersion is x(0, 0, 0) = 0 ∈ R 6 , then we can suppose that (u 1 , u 2 , u 3 ) = 0. Let us take u 1 = 0. Therefore, we can write Therefore, the three matrices which we have to consider are the three components in the basis The rank of the commutators of the three matrices is the same than the rank of the commutator of the matrices ⎛ A simple computation shows that the matrix defined from the commutators is  Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article's Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder.