Compositions of isometric immersions in higher codimension

Marcos Dajczer & Luis A. Florit

An isometric immersion f: M^{n} → ℝ^{n+p} of a connected n-dimensional Riemannian manifold into euclidean space with codimension p is said to be rigid if any other isometric immersion into the same ambient space is congruent to f by an euclidean motion. But rigidity is lost once we allow new immersions to have higher codimension than the given one. In fact, for given q ≥ 1, an abundance of isometric immersions g: M^{n} → ℝ^{n+p+q} can be produced by composing f with isometric immersions into ℝ^{n+p+q} of open subsets V ⊂ ℝ^{n+p} so that f(M) ⊂ V. We recall that the study of the large set of local isometric immersions between euclidean spaces goes back to Cartan ([Ca]). Complete descriptions for codimensions one and two were given in [DG] and [DF], respectively.

In this paper we answer a rigidity question already considered in [DT] for the special case of hypersurfaces (p = 1). Namely, we find sufficient (generic) conditions which, for f and g as above, imply that g must be a composition in the sense of the following definition. Definition. Given an isometric immersion f: M^{n} → ℝ^{n+p} we say that an isometric immersion g: M^{n} → ℝ^{n+p+q}, q ≥ 0, is a composition when there is an isometric embedding f′: M^{n} ↪ N_{0}^{n+p} into a flat manifold N_{0}^{n+p}, an isometric immersion j: N_{0}^{n+p} → ℝ^{n+p} (i.e., a local isometry) satisfying f = j∘f′ and an isometric immersion h: N_{0}^{n+p} → ℝ^{n+p+q} such that g = h∘f′.

For any open subset U ⊂ M^{n} where f as in the definition is an embedding, it follows that there exists an isometric immersion h: V ⊂ ℝ^{n+p} → ℝ^{n+p+q} of a tubular neighborhood V of f(U) such that

g = h ∘ f.  (1)

In particular, (1) holds globally if f itself is an embedding.

Observe that for q = 0 being a composition just means that the two immersions are congruent. Hence, the notion of composition extends the one of rigidity. In fact, it seems to us that considering rigidity results within the more general setting of compositions leads to a deeper understanding of the theory.

Next we deal with second fundamental forms which carry the structure corresponding to compositions. Given f: M^{n} → ℝ^{n+p}, we say that the second fundamental form α_{g} of g: M^{n} → ℝ^{n+p+q} decomposes at x ∈ M^{n} if there are a subspace L^{p} ⊂ T_{g(x)}^{⊥}M and an isometry τ: T_{f(x)}^{⊥}M → L^{p} so that

α_{g}(x) = τ ∘ α_{f}(x) ⊕ γ,  (2)

where γ: T_{x}M × T_{x}M → L^{⊥}. If α_{g} decomposes at each point of M^{n}, we call the decomposition regular when the image S(γ) and kernel N(γ) of γ have both constant dimension.

Besides the aforementioned result in [DT] for hypersurfaces, our main result also generalizes the rigidity theorem in [CD]. There, it was shown that f: M^{n} → ℝ^{n+p} is rigid for p ≤ 5 if the s-nullity of f satisfies everywhere ν_{s}^{f} ≤ n − 2s − 1 for all 1 ≤ s ≤ p, where

ν_{s}^{f}(x) = max{dim N(α_{Γ^{s}})(x): Γ^{s} ⊂ T_{f(x)}^{⊥}M},

with α_{Γ^{s}} denoting the orthogonal projection of α_{f} onto the subspace Γ^{s}. Here we prove the following for compositions.

Theorem 1. Let f: M^{n} → ℝ^{n+p} and g: M^{n} → ℝ^{n+p+q}, q ≥ 0, be isometric immersions. Suppose p ≤ 6, and assume that f satisfies everywhere

ν_{s}^{f} ≤ n − q − 2s − 1, for all 1 ≤ s ≤ p.

When q ≥ 5, assume further that ν_{1}^{f} ≤ n − 2q + 1. Then α_{g} decomposes everywhere and g is a composition if α_{g} decomposes regularly.

A rather simple example due to Henke ([He]) shows that the above global result does not hold without the regularity assumption even if f is an embedding. Nevertheless, we always have regularity and that f is an embedding along connected components of an open dense subset of M^{n}. We conclude that (1) holds along each one of these components under assumptions on f only.

Application. By Theorem 1 an isometric immersion g: 𝕊^{r} × 𝕊^{k} → ℝ^{n+q+2}, for r ≤ k and n = r+k, must be a composition of 𝕊^{r} ×𝕊^{k} ↪ ℝ^{r+1}×ℝ^{k+1} = ℝ^{n+2} if either 3 ≤ r ≤ 7 and q ≤ r − 3 or r ≥ 8 and q ≤ (r + 1)/2. Of course, the same conclusion holds if spheres are replaced by convex hypersurfaces.

We also have the following rigidity result for minimal immersions.

Theorem 2. Let f: M^{n} → ℝ^{n+p} be a minimal isometric immersion which satisfies ν_{s}^{f}(x_{0}) ≤ n−q −2s−1 at x_{0} ∈ M^{n} for all 1 ≤ s ≤ p ≤ 6. Then, any minimal isometric immersion g: M^{n} → ℝ^{n+p+q} is congruent to f in ℝ^{n+p+q}.

A stronger result was obtained for hypersurfaces in [BDJ]. See also [Da] for the case q = 1 for both results.

The paper is organized as follows. In Section 1, we generalize in two directions by means of a simpler proof a result in [CD] on flat symmetric bilinear forms. In particular, also the conformal rigidity theorem there remains valid for an extra unit in the codimension. Moreover, we have good reasons to believe that this result will be crucial in the understanding of isometric deformations of submanifolds in low codimension. The proofs are given in Section 2.

§1 Flat bilinear forms.

Given a symmetric bilinear form β: V × V → W between finite dimensional real vector spaces, the kernel N(β) ⊂ V of β is defined as

N(β) = {X ∈ V: β(X, Y) = 0, Y ∈ V },

and the image S(β) ⊂ W of β is given by

S(β) = span{β(X, Y), X, Y ∈ V }.

A vector Y ∈ V is called a regular element of β if the linear map B_{Y}: V → W, defined as B_{Y} (X) = β(X, Y), satisfies

dim B_{Y} (V) = max{dim B_{Z}(V), Z ∈ V }.

It is easy to check that the subset RE(β) ⊂ V of regular elements is open and dense; see [Mo] or [D] for details.

We denote by W^{p,q} a (p + q)–dimensional vector space with a possible indefinite inner product of type (p, q), where q ≥ 0 is the maximal dimension of a negative definite subspace. We call a subspace U ⊂ W^{p,q} degenerate if the restriction of the metric of W^{p,q} to U is degenerate, and denote by rank U the rank of the metric induced on U. Thus, rank U = dim U − dim U ∩ U^{⊥} and U is nondegenerate if rank U = dim U. Finally, we say that U is null when rank U = 0, i.e., U ⊂ U^{⊥}.

Theorem 3. Let β: V^{n} × V^{n} → W^{p,q}, where p ≤ q and p + q < n, be a nonzero symmetric bilinear form which is flat, that is,

⟨β(X, Y), β(Z, T)⟩ − ⟨β(X, T), β(Z, Y)⟩ = 0, for all X, Y, Z, T ∈ V.

Assume 1 ≤ p ≤ 6 and dim N(β) ≤ n − p − q − 1. Then there is an orthogonal decomposition W^{p,q} = W_{1}^{ℓ,ℓ} ⊕ W_{2}^{p−ℓ,q−ℓ}, 1 ≤ ℓ ≤ p, such that the W_{j}-components β_{j} of β satisfy

i) β_{1} ≠ 0 and S(β_{1}) is null.

ii) β_{2} is flat and dim N(β_{2}) ≥ n − dim W_{2}.

For the proof of the above result we need several lemmas. The first one is a consequence of Theorem 3.8 in [Ar].

Lemma 4. Given a degenerate subspace U ⊂ W^{p,q}, set E = U ∩ U^{⊥} and let S ⊂ U be a subspace such that E ⊕ S = U. Then, there is a (not necessarily unique) subspace Ê ⊂ W with dim Ê = dim E so that E ⊕ Ê is nondegenerate and E ⊕ Ê ⊂ S^{⊥}.

The following result is due to Moore (see [Mo] or [Da]).

Lemma 5. Let β: V ×V → W^{p,q} be a flat bilinear form. Then,

β(ker B_{X}, V) ⊂ B_{X}(V) ∩ B_{X}(V)^{⊥}

for any X ∈ RE(β).

Finally, we recall an elementary fact from [CD] which is proved using that the subset of not asymptotic regular elements

RE^{∗}(β) = {X ∈ RE(β): β(X, X) ≠ 0},

is still open and dense in V.

Lemma 6. Let β: V × V → W^{k} be a symmetric bilinear form such that S(β) = W^{k}. Given X ∈ RE^{∗}(β), take X = X_{1}, …, X_{r} ∈ V such that B_{X}(V) = span{B_{X}(X_{j}), 1 ≤ j ≤ r}, where r = dim B_{X}(V). Then,

S(β) = span{β(X_{i}, X_{j}), 1 ≤ i ≤ j ≤ r}.

In particular, r(r + 1) ≥ 2k.

Proof of Theorem 3 : First suppose that S(β) is degenerate. By Lemma 4, there is a decomposition W^{p,q} = E ⊕ Ê ⊕ V such that S(β) ⊂ E ⊕ V, where E = S(β)∩S(β)^{⊥} ≠ 0 and V^{⊥} = E ⊕Ê. Accordingly, there is a decomposition β = β_{1} + β_{2}, where S(β_{1}) = E and S(β_{2}) ⊂ V. Hence, β_{1} ≠ 0 is null and β_{2} = β − β_{1} is flat. Moreover, S(β_{2}) is nondegenerate. Otherwise, there is 0 ≠ η = ∑_{j} β_{2}(X_{j}, Y_{j}) ⊂ V so that ⟨η, S(β_{2})⟩ = 0, i.e.,

0 = ⟨∑ β_{2}(X_{j}, Y_{j}), β_{2}(Z, T)⟩ = ⟨∑ β(X_{j}, Y_{j}), β(Z, T)⟩, for all Z, T ∈ V.

Hence, ∑_{j} β(X_{j}, Y_{j}) ∈ E. Thus η = 0, which is a contradiction.

To complete the proof of the theorem it suffices to assume that S(β) is nondegenerate and conclude that dim N(β) ≥ n − p − q. We claim that U(X) = B_{X}(V) ∩ B_{X}(V)^{⊥} satisfies U = U(X) ≠ 0 for any X ∈ RE(β). Otherwise, N ⊂ N(β) from Lemma 5, where N = N (X) = ker B_{X}. Since N(β) ⊂ N, we conclude that dim N(β) = dim N ≥ n − p − q, which is a contradiction and proves the claim.

Set τ = min{dim U(X) : X ∈ RE(β)}. The subset

R(β) = {X ∈ RE(β) : dim U(X) = τ}

is open and dense in V; see [DR] or [D]. We fix an element X ∈ R(β) for the remaining of the proof. Lemma 4 yields a decomposition

W^{p,q} = U ⊕ Û ⊕ V,  (3)

where V^{⊥} = U ⊕ Û and B_{X}(V) ⊂ U ⊕ V. Let β̂: V^{n} × V^{n} → Û be the Û-component of β according to (3). Set B̂_{X} = β̂(X, ·) and κ = dim B̂_{Y} (V) for Y ∈ RE(β̂). Being S(β) nondegenerate, for any vector ξ ∈ U there are vectors Y, Z ∈ V such that 0 ≠ ⟨ξ, β(Y, Z)⟩ = ⟨ξ, β̂(Y, Z)⟩. It follows that

S(β̂) = Û.  (4)

Lemma 7. Given Y ∈ R(β) ∩ RE(β̂), we define ρ ≥ 0 by

2ρ = rank (B_{Y} (V) ∩ U) ⊕ B̂_{Y} (V).

Then, ρ ≤ p − τ and dim B_{Y} (N) ≤ p − κ.

Proof: Set V^{n} = L ⊕ L̃, where L̃ = ker B̂_{Y}. Thus, B_{Y} (L̃) ⊂ U ⊕ V and B_{Y} (L) ∩ (U ⊕ V) = 0. Hence, dim B_{Y} (L) = κ. The matrix of inner products of the elements of a basis of B_{Y} (V) associated to the decomposition

B_{Y} (V) = B_{Y}^{0} ⊕ B_{Y} (L) ⊕ B_{Y}^{1} ,

where B_{Y}^{0} = B_{Y} (L̃) ∩ U = B_{Y} (V) ∩ U and B_{Y} (L̃) = B_{Y}^{0} ⊕ B_{Y}^{1} , has the form

[ \begin{matrix}0 & A & 0 \\ A^{t} & B & C \\ 0 & C^{t} & D \end{matrix} ] .

It follows that rank B_{Y} (V) ≥ 2ρ+rank B_{Y}^{1} since ρ = rank A. We obtain from B_{Y}^{1} ⊂ U ⊕ V and B_{Y}^{1} ∩ U = 0 that rank B_{Y}^{1} ≥ dim B_{Y}^{1} − p + τ. Therefore, dim B_{Y} (V) − τ = rank B_{Y} (V) ≥ 2ρ + dim B_{Y}^{1} − p + τ. We conclude that

2ρ ≤ dim B_{Y} (V) ∩ U + κ + p − 2τ.  (5)

Clearly, ρ ≥ dim B_{Y} (V) ∩ U + κ − τ. We get using (5) that

dim B_{Y} (V) ∩ U ≤ p − κ.  (6)

Then, the first statement follows from (5) and (6) whereas the second one from Lemma 5 and (6).

Fix Y_{1} ∈ R(β) ∩ RE^{∗}(β̂). Lemma 6 and (4) yield κ(κ + 1) ≥ 2τ and

Û(X) = span{β̂(Y_{i}, Y_{j}) : 1 ≤ i ≤ j ≤ κ},  (7)

where B̂_{Y_{1}}(V) = span{B̂_{Y_{1}}(Y_{j})}: 1 ≤ j ≤ κ}. Given any N ∈ N, we have from Lemma 5 that β(N, Z) ∈ U for all Z ∈ V. It follows from (7) that

β(N, Z) = 0 if and only if ⟨β(N, Z), β̂(Y_{i}, Y_{j})⟩ = 0, 1 ≤ i, j ≤ κ. (8)

We conclude the proof arguing for the most difficult case p = 6, being the cases p ≤ 5 similar. Suppose first that 4 ≤ τ ≤ 6, and assume that κ = 3 which is its lowest possible value. Thus, there are vectors Y_{1}, Y_{2}, Y_{3} ∈ R(β) ∩ RE^{∗}(β̂) such that

Û(X) = span{β̂(Y_{i}, Y_{j}) : 1 ≤ i ≤ j ≤ 3}.  (9)

Using Lemma 6 again, we choose Y_{2} so that in (9) we may drop the element corresponding to (i, j) = (3, 3) when τ = 5, and when τ = 4 the ones for (i, j) = (2, 3), (3, 3). Hence, Û = B̂_{Y_{1}}(V) + B̂_{Y_{2}}(V) for τ = 4, 5. Consider the linear map B_{1} = B_{Y_{1}}|_{N}: N → B_{Y_{1}}(N). Then, dim B_{Y_{i}}(N) ≤ 3 by Lemma 7. Hence, N_{1} = ker B_{1} satisfies

dim N_{1} ≥ dim N − 3.  (10)

Flatness gives ⟨β(N_{1}, V), B̂_{Y_{1}}(V)⟩ = 0. In particular,

rank B_{Y_{2}}(N_{1}) ⊕ B̂_{Y_{1}}(V) = 0.  (11)

We use Lemma 7 again. If τ = 6, then ρ = 0 and

rank B_{Y_{2}}(N_{1}) ⊕ (B̂_{Y_{1}}(V) + B̂_{Y_{2}}(V)) = 0.  (12)

If τ = 5, then ρ ≤ 1. Thus,

rank B_{Y_{2}}(N_{1}) ⊕ Û ≤ 2.  (13)

In fact, it follows from (11) that (13) holds for τ = 4. We get from (12) and (13) that dim B_{Y_{2}}(N_{1}) ≤ 1 for 4 ≤ τ ≤ 6. Set B_{2} = B_{Y_{2}}|_{N_{1}}: N_{1} → B_{Y_{2}}(N_{1}). It follows using (10) and dim N ≥ n − 6 − q + τ that N_{2} = ker B_{2} satisfies

dim N_{2} ≥ dim N_{1} − dim B_{Y_{2}}(N_{1}) ≥ dim N − 4 ≥ n − q − 6.

By a similar argument as above, B_{Y_{3}}(N_{2}) = 0 when τ = 6. It follows from (8) that N_{2} ⊂ N(β). In particular, dim N(β) ≥ n − q − 6 as we wished. Finally, one can easily check that the estimate for dim N_{2} is even larger if κ > 3, and this concludes the proof for 4 ≤ τ ≤ 6. The argument for the remaining cases is similar and easier.

§2 The proofs.

It is easy to see that the following result is equivalent to Theorem 5 in [DT] when f is an embedding.

Proposition 8. Given an isometric immersion f: M^{n} → ℝ^{n+p}, suppose that the second fundamental form of an isometric immersion g: M^{n} → ℝ^{n+p+q}, q ≥ 1, decomposes regularly. Assume that τ is parallel for the induced connection on L. If

W = span{( ˜∇_{X}ξ)_{T M⊕L}: X ∈ T M and ξ ∈ L^{⊥}}

satisfies W ∩ L = 0, then g is a composition.

Proof: Observe that W ⊂ N(γ)^{⊥} ⊕ L, where T M = N(γ) ⊕ N(γ)^{⊥}. From W ∩ L = 0, we easily see that dim W = dim N(γ)^{⊥}. Being N(γ) smooth, it follows that the subspaces Γ = (N(γ)^{⊥} ⊕ L) ∩ W^{⊥} form a rank-p vector bundle. Moreover, we have that Γ∩T M = 0. Let F: Γ → ℝ^{n+p} be defined as

F (ξ_{x}) = f(x) + ξ_{x}, ξ_{x} ∈ Γ(x).

Thus, there is an open neighborhood N_{0}^{n+p} of the 0-section f′: M^{n} → N_{0}^{n+p} of Γ such that F |_{N_{0}^{n+p}} is a local diffeomorphism onto the open subset F (N_{0}^{n+p}) ⊂ ℝ^{n+p}. We consider in N_{0}^{n+p} the metric induced by F and identify (Id⊕τ)^{−1}Γ with Γ. Let h: N_{0}^{n+p} → ℝ^{n+p+q} be the immersion h(ξ_{x}) = g(x) + ξ_{x}. A straightforward computation shows that F and h are isometric; see the proof of Theorem 5 in [DT]. Hence, h is an isometric immersion and g = h ∘ f′.

Proof of Theorem 1: Let β: T M × T M → T_{f}^{⊥}M ⊕ T_{g}^{⊥}M be defined as β = α_{f} ⊕ α_{g}. Endow T_{f}^{⊥}M ⊕ T_{g}^{⊥}M with the indefinite inner product of type (p, p + q) given by

⟨⟨ , ⟩⟩_{T_{f}^{⊥}M⊕T_{g}^{⊥}M} = ⟨ , ⟩_{T_{f}^{⊥}M} − ⟨ , ⟩_{T_{g}^{⊥}M}.

Then β is a symmetric bilinear form which is flat from the Gauss equation for f and g. Theorem 3 applies to β(x) at each x ∈ M^{n}. With the notations there, suppose that ℓ < p. It follows that

ν_{p−ℓ}^{f}(x) ≥ N(π_{T_{f}^{⊥}M} ∘ β_{2}(x)) ≥ N(β_{2}(x)) ≥ n − q − 2(p − ℓ).

This is a contradiction to our hypothesis on ν_{s}^{f} for s = p − ℓ, and implies that ℓ = p. We conclude that at each point α_{g} decomposes as in (2) and Ω = N(γ) satisfies

dim Ω ≥ n − q.  (14)

Assume now that the subspaces S(γ) have constant dimension. Clearly, the same holds for the subspaces S(α_{g}) since ν_{1}^{f} < n everywhere. We claim that the decomposition is smooth in the sense that the subspaces L form a vector subbundle and that τ: T_{f}^{⊥}M → L ⊂ T_{g}^{⊥}M is a bundle isometry. To prove the claim, observe that S(β) ∩ S(β)^{⊥} ⊂ T_{f}^{⊥}M ⊕ T_{g}^{⊥}M is a smooth subbundle of rank p and that L = S(α_{f}) ⊂ T_{g}^{⊥}M is its orthogonal projection onto T_{g}^{⊥}M.

We identify T_{f}^{⊥}M and L by means of τ. Let K: T M → End(L) be the linear map into the skew-symmetric endomorphisms of L defined by

K(X)η = ∇_{X}^{⊥}η − ( ˆ∇_{X}^{⊥}η)_{L},

where ∇^{⊥} and ˆ∇^{⊥} denote the normal connection for f and g, respectively, and writing a linear space as subscript indicates taking the orthogonal projection of the vector onto that subspace. We need the following result.

Lemma 9. The tensor K satisfies K(Z) = 0 and

K(X)α_{f}(Y, Z) = K(Y)α_{f}(X, Z),

for all Z ∈ Ω and X, Y ∈ T M.

Proof: Since ν_{1}^{f} < n − q by assumption, we get using (14) that

L = span{α_{f}(Ω, X): X ∈ T M}.  (15)

It follows easily from the Codazzi equation for f and g that

K(Z_{1})α_{f}(Z_{2}, Z_{3}) = K(Z_{2})α_{f}(Z_{1}, Z_{3})

if either Z_{1}, Z_{2} ∈ Ω or Z_{3} ∈ Ω. Denote

(X_{1}, X_{2}, X_{3}, X_{4}, X_{5}) = ⟨K(X_{1})α_{f}(X_{2}, X_{3}), α_{f}(X_{4}, X_{5})⟩,

and take Z_{1}, Z_{2}, Z_{3} ∈ Ω. We have,

(Y, Z_{1}, Z_{2}, Z_{3}, X) = −(Y, Z_{3}, X, Z_{1}, Z_{2}) = −(X, Z_{3}, Y, Z_{1}, Z_{2})
= (X, Z_{1}, Z_{2}, Z_{3}, Y) = (Z_{2}, Z_{1}, X, Z_{3}, Y) = −(Z_{2}, Z_{3}, Y, Z_{1}, X)
= −(Z_{3}, Z_{2}, Y, Z_{1}, X) = (Z_{3}, Z_{1}, X, Z_{2}, Y) = (Z_{1}, Z_{3}, X, Z_{2}, Y)
= −(Z_{1}, Z_{2}, Y, Z_{3}, X) = −(Y, Z_{1}, Z_{2}, Z_{3}, X) = 0.

Hence, ⟨K(Z_{1})α_{f}(Y, Z_{2}), α_{f}(Z_{3}, X)⟩ = 0, and this concludes the proof.

Proceeding with the proof of the theorem, we claim that K = 0, that is, τ is parallel. Let us assume otherwise. Define ϕ: Ω × T M → S by ϕ = α_{S}|_{Ω×T M}, where

S = span{K(X)L: X ∈ T M}

and α_{S} = π_{S} ∘ α_{f}. It follows from (15) that

S = span{ϕ(Ω, X): X ∈ T M}.  (16)

Take Y ∈ T M so that K(Y) has maximal rank. Set Ω_{Y} = ker ϕ( · , Y). Then,

dim Ω_{Y} ≥ n − q − k  (17)

from (14), where k = dim ϕ(Ω, Y). By Lemma 9, we have

K(Y)ϕ(Ω_{Y} , X) = K(X)ϕ(Ω_{Y} , Y) = 0, for all X ∈ T M.

Being K(Y) skew-symmetric, we easily obtain that

α_{K(Y)L}(Ω_{Y} , X) = 0, for all X ∈ T M.

Set dim K(Y)L = r ≥ 2. From our assumption on ν_{r}^{f} and (17), we get

n − q − 2r − 1 ≥ ν_{r}^{f} ≥ dim Ω_{Y} ≥ n − q − k.

Hence, 5 ≤ 2r + 1 ≤ k ≤ p ≤ 6. Thus, r = 2, k = 5, 6 and dim S = 5, 6. From (16), there is Y_{1} = Y, Y_{2} ∈ T M such that S = ϕ(Ω, Y_{1}) + ϕ(Ω, Y_{2}). By Lemma 9, we have

S = span{K(X)S: X ∈ T M}
= span{K(X)(ϕ(Ω, Y_{1})+ϕ(Ω, Y_{2})): X ∈ T M}
⊆ K(Y_{1})L + K(Y_{2})L,

which implies that dim S ≤ 4, a contradiction. This proves the claim.

To reduce the proof of the theorem to Proposition 8, all it remains to show is that the condition W ∩ L = 0 is satisfied. This is divided in two cases.

Case 1. Assume dim Ω > n − q. We claim that L is parallel along Ω. The subspaces D = S(γ) have constant dimension by assumption, hence, there is a smooth orthogonal splitting L^{⊥} = D ⊕ D^{⊥}. Taking the difference between the Codazzi equations for f and g and using that τ is parallel gives for the second fundamental form of g that

A_{( ˆ∇_{X}^{⊥}ξ)_{D}}Y = A_{( ˆ∇_{Y}^{⊥}ξ)_{D}}X, for all X, Y ∈ T M and ξ ∈ L.  (18)

Taking X ∈ Ω in (18), we get

ˆ∇_{X}^{⊥}ξ ⊂ D^{⊥}, for all X ∈ Ω and ξ ∈ L.  (19)

The Codazzi equation also yields

A_{( ˆ∇_{X}^{⊥}η)_{L}}Y = A_{( ˆ∇_{Y}^{⊥}η)_{L}}X, for all X, Y ∈ Ω and η ∈ D^{⊥}.  (20)

For η ∈ D^{⊥}, consider the linear map ϕ_{η}: Ω → L defined as ϕ_{η}(X) = ( ˆ∇_{X}^{⊥}η)_{L}, and set r = dim Im ϕ_{η}. It follows from (20) that

⟨α(ker ϕ_{η}, T M), Im ϕ_{η}⟩ = 0.

Thus, ν_{r}^{f} ≥ dim Ω − r. This is not possible by (14) and the hypothesis on ν_{r}^{f} unless r = 0, that is,

ˆ∇_{X}^{⊥}ξ ⊂ D, for all X ∈ Ω and ξ ∈ L,  (21)

and the claim follows from (19) and (21).

From the Codazzi equation and the claim, we have that

⟨∇_{Z}Y, A_{ξ}X⟩ = ⟨α(Z, Y), ˆ∇_{X}^{⊥}ξ⟩,

for all Z, Y ∈ Ω and ξ ∈ L^{⊥}, or equivalently,

˜∇_{Z}Y ⊥ W, for all Z, Y ∈ Ω.  (22)

Assume that there is a normal vector 0 ≠ ξ ∈ W ∩ L. Then (22) implies that

⟨A_{ξ}Y, Z⟩ = 0, for all Y, Z ∈ Ω.

It follows easily that ν_{1}^{f} ≥ 2 dim Ω − n ≥ n − 2q + 2, which is in contradiction with our assumption on ν_{1}^{f}.

Case 2. Assume dim Ω = n − q. We follow closely the argument in [DT]. It is shown there that γ smoothly decomposes as the orthogonal sum of one-dimensional orthogonal forms of rank one, namely,

γ = γ_{1} ⊕ ⋯ ⊕ γ_{q},

with correspondent linearly independent unit eigenvectors Z_{1}, …, Z_{q} and nonzero eigenvalues λ_{1}, … λ_{q}. Therefore, there exists an orthonormal basis {ξ_{1}, …, ξ_{q}} of L^{⊥} = D such that

γ_{j}(X, Y) = λ_{j}⟨X, Z_{j}⟩⟨Y, Z_{j}⟩ξ_{j}, 1 ≤ j ≤ q.

In fact, the existence of such a decomposition goes back to Cartan ([Ca]). It follows from (18) that

∑_{k=1}^{q} λ_{k} (⟨ ˆ∇_{Z_{i}}^{⊥}η, ξ_{k}⟩⟨Z_{j}, Z_{k}⟩ − ⟨ ˆ∇_{Z_{j}}^{⊥}η, ξ_{k}⟩⟨Z_{i}, Z_{k}⟩) Z_{k} = 0,

for any η ∈ L. We easily get that ( ˆ∇_{Y}^{⊥}ξ_{k})_{L} = 0, for any Y ⊥ Z_{k}. We conclude that W = span{λ_{j}Z_{j} − ( ˆ∇_{Z_{j}}^{⊥}ξ_{j})_{L}, 1 ≤ j ≤ q}, and thus, W ∩ L = 0.

When p = 1 and q ≥ 5, the additional assumption in [DT] is that M^{n} is not (n − q + 1)–ruled. Our additional assumption on ν_{1}^{f} is used in Case 1 to make sure that Γ has dimension p. Thus, when p = 1 it just assures that Γ ≠ 0. But if Γ = 0, then Ω is clearly totally geodesic in the ambient space, that is, the submanifold is ruled by the leaves of Ω. Hence, both assumptions are equivalent.

Proof of Theorem 2: Let U ⊂ M^{n} be an open connected neighborhood of x_{0} where ν_{s}^{f} ≤ n − q − 2s − 1 for 1 ≤ s ≤ p. By Theorem 3, we have that α_{g} decomposes as in (2) along U. Moreover. since γ is flat and traceless, an elementary argument shows that γ = 0. From Theorem A in [DR] (or Theorem 4.5 in [D]), it follows that g reduces codimension to p. Thus, g is congruent to f along U, and the result follows using that minimal euclidean submanifolds are real-analytic.

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IMPA – Estrada Dona Castorina, 110 22460-320 — Rio de Janeiro — Brazil marcos@impa.br —— luis@impa.br