Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 34 (1993), No. 2, 209-215.

The Semiring of Immersions of Manifolds

Dedicated to the 50th birthday of Prof. Bang-Yen Chen

F. Decruyenaere^{(1)}, F. Dillen^{(2)}, L. Verstraelen, L. Vrancken^{(2)}

Katholieke Universiteit Leuven, Departement Wiskunde, Celestijnenlaan 200B, B - 3001 Leuven

Abstract. In [2], [3], [4], [5] B.-Y. Chen introduced the tensor product immersion of a given Riemannian manifold. In this paper we study the tensor product of two immersions of differentiable manifolds. Although inspired by Chen’s definition, our tensor product is a somewhat different concept, since it realizes an immersion of the product manifold. Basically, we will obtain a commutative semiring structure on the set of transversal immersions of manifolds in Euclidean spaces with operations direct sum ⊕ and tensor product ⊗. As a geometrical application of our notion of tensor products, we will make a remark on a conjecture of H. Hopf concerning Riemannian metrics on S^{2} × S^{2}. 1991 Mathematics Subject Classification. 53C40, 53B25, 58G25

1. Introduction

We review some elementary facts about the tensor product of Euclidean spaces. Let V and W be two finite–dimensional real vector spaces of dimensions m resp. n, equipped with metrics g_{V} resp. g_{W}. Then the direct sum V ⊕ W is a (m + n)-dimensional vector space with metric g_{V ⊕W} defined by :

g_{V ⊕W} ((v, w), (v′, w′)) = g_{V} (v, v′) + g_{W} (w, w′),

for all v, v′ ∈ V and w, w′ ∈ W. The tensor product V ⊗ W is a mn-dimensional vector space with metric g_{V ⊗W} defined by :

g_{V ⊗W} (v ⊗ w, v′ ⊗ w′) = g_{V} (v, v′).g_{W} (w, w′),

(1) Supported by a research fellowship of the Research Council of the Katholieke Universiteit Leuven.

(2) Senior Research Assistant of the National Fund for Scientific Research.

0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix}1993 Heldermann Verlag, Berlin

for all v, v′ ∈ V and w, w′ ∈ W, and extended linearly.

Denote by E^{n} the n-dimensional Euclidean space with the Euclidean metric g_{n}. Define

i_{p,q} : E^{p} → E^{q} : (x_{1},…, x_{p}) ↦ (x_{1},…, x_{p}, 0,…, 0),

for each p ≤ q.

Lemma 1. E^{p} ⊕ E^{q} ≅ E^{p+q}; (ii) E^{p} ⊗ E^{q} ≅ E^{pq}.

Proof. The first part is obvious. For (ii), the dimensions do agree. We still check the metrics. If X = (x_{1},…, x_{p}), X′ = (x′_{1},…, x′_{p}) ∈ E^{p} and if Y = (y_{1},…, y_{q}), Y′ = (y′_{1},…, y′_{q}) ∈ E^{q}, then

g_{E^{p}⊗E^{q}}(X ⊗ Y, X′ ⊗ Y′) = g_{p}(X, X′)g_{q}(Y, Y′) = (∑_{i=1}^{p} x_{i}x′_{i})(∑_{j=1}^{q} y_{j}y′_{j})

= ∑_{i=1}^{p} ∑_{j=1}^{q} x_{i}x′_{i}y_{j}y′_{j} = ∑_{i=1}^{p} ∑_{j=1}^{q}(x_{i}y_{j})(x′_{i}y′_{j})

= g_{pq}(X ⊗ Y, X′ ⊗ Y′).

2. The set of transversal immersions in Euclidean spaces.

Definition. Consider two differentiable manifolds M and N of dimensions r resp. s and assume that

f : M → E^{m} : p ↦ (f_{1}(p),…, f_{m}(p))

and

h : N → E^{n} : q ↦ (h_{1}(q),…, h_{n}(q))

are two immersions. Then we define the direct sum map

\begin{matrix}f ⊕ h : M × N → E^{m+n} : \\ (p, q) ↦ (f(p), h(q)) = (f_{1}(p),…, f_{m}(p), h_{1}(q),…, h_{n}(q)) \end{matrix}

and the tensor product map

\begin{matrix}f ⊗ h : M × N → E^{mn} : \\ (p, q) ↦ f(p) ⊗ h(q) = (f_{1}(p)h_{1}(q),…, f_{1}(p)h_{n}(q),…, f_{m}(p)h_{n}(q)). \end{matrix}

Lemma 2. Assume M and N are differentiable manifolds and f : M → E^{m} and h : N → E^{n} are two immersions. Then

(1) (f ⊕ h)_{∗}T_{(p,q)}(M × N) ≅ f_{∗}T_{p}M ⊕ h_{∗}T_{q}N;

(2) (f ⊗ h)_{∗}T_{(p,q)}(M × N) ≅ f_{∗}(T_{p}M) ⊗ h(q) + f(p) ⊗ h_{∗}(T_{q}N).

Proof. Assume X_{p} ∈ T_{p}M and Y_{q} ∈ T_{q}N. It is well known that (f ⊕ h)_{∗}(X_{p}, Y_{q}) = (f_{∗}(X_{p}), h_{∗}(Y_{q})) and thus (1) follows. Assume now α : I ⊆ R → M and β : J ⊆ R → N are curves such that p = α(0), X_{p} = α′(0) and q = β(0), Y_{q} = β′(0). Then, the (ij)- component of (f ⊗ h)_{∗}(X_{p}, Y_{q}) equals

[(f_{i} ∘ α).(h_{j} ∘ β)]′(0, 0) = (f_{i} ∘ α)′(0).h_{j}(q) + f_{i}(p).(h_{j} ∘ β)′(0)

and therefore

(f ⊗ h)_{∗}(X_{p}, Y_{q}) = f_{∗}(X_{p}) ⊗ h(q) + f(p) ⊗ h_{∗}(Y_{q}).

Definition. Let f : M → E^{m} be an immersion of a differentiable manifold in a Euclidean space. Then f is said to be transversal in a point p ∈ M if and only if the position vector f(p) is not tangent to M at p, i.e. f(p) ∉ f_{∗}(T_{p}M). If f is transversal in every point of M, then f shortly is called transversal .

Proposition 3. Consider two differentiable manifolds M and N of dimensions r resp. s and assume that f : M → E^{m} and h : N → E^{n} are two transversal immersions. Then f ⊕ h and f ⊗ h are transversal immersions of M × N into E^{m+n}, resp. E^{mn}.

Proof. Assume X_{p} ∈ T_{p}M, Y_{q} ∈ T_{q}N. Then, as f and h are immersions, it follows that (f ⊕ h)_{∗}(X_{p}, Y_{q}) = (f_{∗}(X_{p}), h_{∗}(Y_{q})) = 0 if and only if X_{p} = Y_{q} = 0. Thus f ⊕ h is an immersion. Moreover, as f and h are transversal, so is f ⊕ h. Indeed, assume (f(p), h(q)) ∈ (f ⊕ h)_{∗}T_{(p,q)}(M × N), then it immediately follows from Lemma 2 that f(p) ∈ f_{∗}T_{p}(M) and h(q) ∈ h_{∗}T_{q}N, a contradiction.

We now prove that f ⊗ h is an immersion. Suppose (f ⊗ h)_{∗}(X_{p}, Y_{q}) = 0. Then, by Lemma 2, f(p) ⊗ h_{∗}(Y_{q}) = −f_{∗}(X_{p}) ⊗ h(q) and so there exists λ ∈ R_{0} such that f(p) = −λf_{∗}(X_{p}) and h_{∗}(Y_{q}) = (1/λ)h(q); a contradiction. Moreover, f ⊗ h is transversal. For assume that f ⊗ h is not transversal. Then there exist p ∈ M, q ∈ N such that (f ⊗ h)(p, q) ∈ T_{(p,q)}(f ⊗ h)(M × N). Thus f(p) ⊗ h(q) = f(p) ⊗ h_{∗}(w_{q}) + f_{∗}(v_{p}) ⊗ h(q) for v_{p} ∈ T_{p}M and w_{q} ∈ T_{q}N. But then f(p) ⊗ (h(q) − h_{∗}(w_{q})) = f_{∗}(v_{p}) ⊗ h(q) and there exists μ ∈ R_{0} such that

\{ \begin{matrix}f(p) & = μf_{∗}(v_{p}) \\ h(q) − h_{∗}(w_{q}) & = 1/μh(q), \end{matrix}

a contradiction.

Remark. Note that the transversality condition implies that the images of the immersions do not contain the origins of the Euclidean spaces. But to assure that the tensor product of two immersions is an immersion again, the latter condition is not sufficient, as the following example shows. Put f : E^{1} → E^{2} : s ↦ (s, (s − 1)^{2}) and h : R_{>0} → E^{2} : t ↦ (t, t). Then these are two immersions, yielding a parabola resp. a halfline, but the tensor product f ⊗h : E^{1} ×R_{>0} → E^{4} : (s, t) ↦ (st, st, (s − 1)^{2}t, (s − 1)^{2}t) is not an immersion, as its Jacobian

( \begin{matrix}t & s \\ 2(s − 1)t & (s − 1)^{2} \\ t & s \\ 2(s − 1)t & (s − 1)^{2} \end{matrix} )

has rank one in all points (1, t), t > 0. Indeed, f is not transversal in 1 and h is nowhere transversal.

It directly follows from its proof that Proposition 3 can also be stated in the following way (in the above terminology) :

Proposition 3’. The tensor product f ⊗ h is an immersion if and only if f and h are immersions that satisfy the three following conditions :

(1) either f or h is transversal;

(2) 0 ∉ Im(f);

(3) 0 ∉ Im(h). If both f and h are transversal immersions (and then conditions (2) and (3) are automatically satisfied), then the tensor product f ⊗ h is a transversal immersion too.

Consequently, for example, the tensor product of the above parabola immersion f with h′ : E^{1} → E^{2} : t ↦ (t, t − 1) is indeed an immersion again (as h is transversal).

We first look how these operations behave under orthogonal transformations :

Proposition 4. Assume f : M → E^{m} and h : N → E^{n} are two transversal immersions. Let A : E^{m} → E^{m} and B : E^{n} → E^{n} be two orthogonal transformations of E^{m}, resp. E^{n}. Let d_{1} : M′ → M and d_{2} : N′ → N be two diffeomorphisms. Then

(i) Af ⊕ Bh and f ⊕ h are orthogonally equivalent transversal immersions in E^{m+n}; (ii) Af ⊗ Bh and f ⊗ h are orthogonally equivalent transversal immersions in E^{mn}; (iii) (i_{m,N} ∘ f) ⊕ (i_{n,N} ∘ h) and i_{m+n,2N} ∘ (f ⊕ h) are orthogonally equivalent transversal

immersions for all N ≥ m + n; (iv) (i_{m,N} ∘ f) ⊗ (i_{n,N} ∘ h) and i_{mn,N^{2}} ∘ (f ⊗ h) are orthogonally equivalent transversal

immersions for all N ≥ mn;

(v) (f ∘ d_{1}) ⊕ (h ∘ d_{2}) = (f ⊕ h) ∘ (d_{1}, d_{2}), where (d_{1}, d_{2}) : M′ × N′ → M × N : (p′, q′) ↦ (d_{1}(p′), d_{2}(q′)) is also a diffeomorphism;

(vi) (f ∘ d_{1}) ⊗ (h ∘ d_{2}) = (f ⊗ h) ∘ (d_{1}, d_{2}).

Proof. First remark that Af and Bh are still transversal immersions and therefore it follows from Theorem 4 that Af ⊕ Bh and Af ⊗ Bh are still transversal immersions in E^{m+n} resp. E^{mn}. Analogously, (i_{m,N} ∘ f) ⊕ (i_{n,N} ∘ h) and (i_{m,N} ∘ f) ⊗ (i_{n,N} ∘ h) are also transversal immersions.

(i) It is clear that Af ⊕ Bh = (A ⊕ B)(f ⊕ h) and obviously, A ⊕ B is an orthogonal transformation of E^{m+n}. (ii) Put A = (a_{i,j})_{1≤i,j≤m} and B = (b_{i,j})_{1≤i,j≤n}. Then, the (i, j)-component in E^{mn} (1 ≤ i ≤ m, 1 ≤ j ≤ n) of (Af ⊗ Bh)(α, β) is

∑_{k=1}^{m} ∑_{l=1}^{n}(a_{ik})(b_{jl})f_{k}h_{l}.

So Af ⊗ Bh = C(f ⊗ h), where C is a (mn × mn)-matrix with entries C_{i,j;k,l} = a_{i,k}b_{j,l} for 1 ≤ i, k ≤ m and 1 ≤ j, l ≤ n. Therefore C = A ⊗ B. Since the tensor product of two orthogonal matrices is orthogonal again, this proves the second point of the Theorem.

(iii) until (vi) are easy.

Remark. It is easily seen from the Remark after Proposition 3 that the tensor product is not invariant under translations. Consider also the following example. Put

f : E^{1} → E^{2} : s ↦ (s, s + 1).

Then the tensor product f ⊗ f is the intersection of the hyperquadric xu = yz with the hyperplane x−y−z+u = 1 in E^{4} (with coordinates (x, y, z, u)). Let A be the translation by (0, 1). Then Af ⊗ Af is the intersection of the hyperquadric xu = yz with the hyperplane x − y − z + u = 4 in E^{4}. Clearly these two surfaces are not isometrically equivalent.

Define Z to be the set of all transversal immersions of all differentiable manifolds in Euclidean spaces. We define a symmetric relation ∼ on Z as follows :

(1) if f : M → E^{m} ∈ Z, then f ∼ i_{m,n} ∘ f for all n ≥ m;

(2) if f : M → E^{m}, h : N → E^{n} ∈ Z and there exists an orthogonal transformation of E^{N}, N ≥ max(m, n), that carries i_{m,N} ∘ f into i_{n,N} ∘ h, then f ∼ h;

(3) if f : M → E^{m} ∈ Z and d : N → M is a diffeomorphism, then f ∼ f ∘ d. Obviously, ∼ is an equivalence relation on Z and we put I = Z/ ∼. Then we obtain

Theorem 5. I, ⊕, ⊗ is a commutative semiring, with e : {0} → E^{1} : 0 ↦ 1 as identity element.

Proof. It directly follows from Propositions 3 and 4 and the definition of I that ⊕ and ⊗ are well defined operations on I. Associativity and distributivity are readily checked. Moreover it is clear that, if f, h ∈ I, then f ⊗h and h⊗f are equivalent in Z, since M ×N and N × M are diffeomorphic and the respective images are orthogonally equivalent (by a change of coordinates). Thus they are equal in I. Now take f : M → E^{m} ∈ Z, then

f ⊗ e : M × {0} → E^{m} : (p, 0) ↦ f(p) ⊗ 1 = f(p).

Thus f ⊗ e is diffeomorphic to f and thus f ⊗ e ∼ f in Z.

3. An application.

As one example of a geometrical application, we consider the following well known

Conjecture of H. Hopf ([1], [6]). On S^{2} × S^{2} there exists no Riemannian metric with strictly positive sectional curvature.

We consider the tensor product of the immersions of two spheres in E^{3} each being immersed as an ellipsoid with mass center the origin. Then it follows from Theorem 3 that, eventually after applying an orthogonal transformation of E^{3}, we may assume the immersions are of the form

\begin{matrix}f_{1} : S^{2} × E^{3} : & p = (x_{1}, y_{1}, z_{1}) ↦ (a_{1}x_{1}, b_{1}y_{1}, c_{1}z_{1}), \\ f_{2} : S^{2} × E^{3} : & p′ = (x_{2}, y_{2}, z_{2}) ↦ (a_{2}x_{2}, b_{2}y_{2}, c_{2}z_{2}), \end{matrix}

where a_{1}, b_{1}, c_{1}, a_{2}, b_{2}, c_{2} > 0. Then the tensor product f : S^{2} × S^{2} → E^{9} is given by

\begin{matrix}f(p, p′) = \\ (a_{1}a_{2}x_{1}x_{2}, a_{1}b_{2}x_{1}y_{2}, a_{1}c_{2}x_{1}z_{2}, b_{1}a_{2}y_{1}x_{2}, b_{1}b_{2}y_{1}y_{2}, b_{1}c_{2}y_{1}z_{2}, c_{1}a_{2}z_{1}x_{2}, c_{1}b_{2}z_{1}y_{2}, c_{1}c_{2}z_{1}z_{2}). \end{matrix}

We show that, according to the conjecture of H. Hopf, the sectional curvature of this immersed surface can not be strictly positive everywhere. A parametrization around ((1, 0, 0), (1, 0, 0)) is given by

x_{1} = cos θ_{1} cos ϕ_{1}
y_{1} = cos θ_{1} sin ϕ_{1}
z_{1} = sin θ_{1},
x_{2} = cos θ_{2} cos ϕ_{2}
y_{2} = cos θ_{2} sin ϕ_{2}
z_{2} = sin θ_{2}.

Then a basis of the tangent space at this point is given by

\begin{aligned}f_{θ_{1}}(0, 0, 0, 0) & = (0, 0, 0, 0, 0, 0, c_{1}a_{2}, 0, 0), \\ f_{θ_{2}}(0, 0, 0, 0) & = (0, 0, a_{1}c_{2}, 0, 0, 0, 0, 0, 0), \\ f_{ϕ_{1}}(0, 0, 0, 0) & = (0, 0, 0, b_{1}a_{2}, 0, 0, 0, 0, 0), \\ f_{ϕ_{2}}(0, 0, 0, 0) & = (0, a_{1}b_{2}, 0, 0, 0, 0, 0, 0, 0). \end{aligned}

The normal space is spanned by

\begin{aligned}f_{θ_{1}θ_{1}}(0, 0, 0, 0) & = (−a_{1}a_{2}, 0, 0, 0, 0, 0, 0, 0, 0) = h(f_{θ_{1}}, f_{θ_{1}}), \\ f_{θ_{1}θ_{2}}(0, 0, 0, 0) & = (0, 0, 0, 0, 0, 0, 0, 0, c_{1}c_{2}) = h(f_{θ_{1}}, f_{θ_{2}}), \\ f_{θ_{1}ϕ_{2}}(0, 0, 0, 0) & = (0, 0, 0, 0, 0, 0, 0, c_{1}b_{2}, 0) = h(f_{θ_{1}}, f_{ϕ_{2}}), \\ f_{θ_{2}ϕ_{1}}(0, 0, 0, 0) & = (0, 0, 0, 0, 0, b_{1}c_{2}, 0, 0, 0) = h(f_{θ_{2}}, f_{ϕ_{1}}), \\ f_{ϕ_{1}ϕ_{2}}(0, 0, 0, 0) & = (0, 0, 0, 0, b_{1}b_{2}, 0, 0, 0, 0) = h(f_{ϕ_{1}}, f_{ϕ_{2}}), \end{aligned}

where h denotes the second fundamental form of f(S^{2} × S^{2}) in E^{9}.

If all sectional curvatures were to be positive, then

\begin{aligned}g(R(f_{θ_{1}}, f_{θ_{2}})f_{θ_{2}}, f_{θ_{1}}) & = g(h(f_{θ_{1}}, f_{θ_{1}}), h(f_{θ_{2}}, f_{θ_{2}})) − g(h(f_{θ_{1}}, f_{θ_{2}}), h(f_{θ_{1}}, f_{θ_{2}})) \\ & = a_{1}^{2}a_{2}^{2} − c_{1}^{2}c_{2}^{2} > 0. \end{aligned}

But similarly, by working with a parametrization around ((0,0,1),(0,0,1)), we obtain c_{1}^{2}c_{2}^{2} − a_{1}^{2}a_{2}^{2} > 0, a contradiction. Therefore we obtain

Proposition 6. If two spheres are immersed as ellipsoids in E^{3}, with mass center the origin, then the tensor product immersion of their product manifold confirms the Conjecture of H. Hopf.

References

[1] J.-P. Bourguignon: La conjecture de Hopf sur S^{2} × S^{2}. Géométrie Riemannienne en dimension 4, Séminaire A. Besse 3 (1981), 347-355.

[2] B.-Y. Chen: Differential Geometry of semiring of immersions. I: General Theory. Bull. Inst. Math. Acad. Sinica 21 (1993), 1-34.

[3] B.-Y. Chen: Classification of tensor product immersions which are of 1-type. Glasgow J. Math. (to appear).

[4] B.-Y. Chen: Two theorems on tensor product immersions. Rend. Sem. Math. Messina, Série II, Tomo XIV, Vol I, 69–83 (1991).

[5] B.-Y. Chen: Differential geometry of tensor product immersions. Ann. Global Anal. Geom. (to appear).

[6] S.S. Chern: Differential Geometry : its past and its future. Actes du Congrès International des Math. 1970 Nice 1 (1970), 41-53.