Embeddings of S^p× S^q× S^rin S^p+q+r+1

Abstract

Let f : S p × S q × S r → S p+q+r+1 , 2 ≤ p ≤ q ≤ r, be a smooth embedding. In this paper we show that the closure of one of the two components ofprovided that p + q = r or p + q = r with r even. We also show that when p + q = r with r odd, there exist infinitely many embeddings which do not satisfy the above property. We also define standard embeddings of S p × S q × S r into S p+q+r+1 and, using the above result, we prove that if C 1 has the homology of S p × S q , then f is standard, provided that q < r.

The Fox property for codimension one embeddings of products of three spheres into spheres

“…by an argument similar to that in the proof of [11,Lemma 5.2], we see that for some diffeomorphism b  W S 1 S 1 ! S 1 S 1 , the composite…”

“…By the same argument as in [11,Section 4], we see that the natural inclusion S 1 S 1 f g ! C 1 induces a homology equivalence.…”

“…S pCqCr C1 is standard 1 if its image coincides with the boundary of a tubular neighborhood of a product of two spheres standardly embedded in S pCqCr C1 , or more precisely, if its image is isotopic to @N p 0 ;q 0 ;r 0 C D @N p 0 ;q 0 ;r 0 .Š S p 0 S q 0 S r 0 / for some permutation .p 0 ; q 0 ; r 0 / of .p; q; r /. In [11], the authors have shown that if 2 Ä p Ä q Ä r and either p C q ¤ r or p C q D r is even, then every smooth embedding f W S p S q S r ! S pCqCr C1 has the Alexander property.…”

“…In this paper we will show that in fact, they do not have the Fox property, provided n ¤ 0; 1 (see Theorem 3.2 in Section 3). This means that the embeddings f n , n ¤ 0; 1, constructed in [11] are highly knotted. This answers the question posed at the end of [12].…”

“…Case 1 1 < q < r and r ¤ q C 1 By Alexander duality, we can show that either C 1 or C 2 has the same homology as S 1 S q , S 1 S r or S q S r (for details, see the argument in [11,Section 2]). In the following, we assume that C 1 has such a homology.…”