Let n be any integer greater than two. We prove that there exists a projection P having the following properties. (1) P is not the projection of any unknotted knot. (2) The singular point set of P consists of double points. (3)P is the projection of an n-knot which is diffeomorphic to the standard sphere.We prove there exists an immersed n-sphere ( ⊂ R n+1 × {0} ) which is not the projection of any n-knot (n > 2). Note that the second theorem is different from the first one. §1. Introduction and Main resultsThe study of n-dimensional knots and links has a long history. The research was connected with surgery theory. (See [9], [20] etc. for the history. ) There are many fruitful results. There are many outstanding problems. For example, n-links have not been classified. (This open problem is not discussed explicitly in this paper but it is one of motivations of this paper.)When we study n-knots and n-links (n ≥ 2), we sometimes see similarities and differences between the theory of 1-links and that of n-links. In this paper we point out one difference between them, associated with the projections of knots. (Theorem 1.3.). When one studies classical knots (in R 3 ), it is important to consider the projections of knots into R 2 . See [1], [3], [8], [12], [16], [23], [25], etc. For 2dimensional knots in R 4 , one considers the projections of 2-knots into R 3 . See [2], [4], [5], [7], [10], etc.In order to state our problems (Problem 1.1 and 1.2) and our main theorem (Theorem 1.3.), we prepare some definitions on n-knots and on their projections.We work in the smooth category. An (oriented) n-(dimensional) knot K is a smooth oriented submanifold of R n+1 × R which is PL homeomorphic to the standard n-sphere. We say that n-knots K 1 and K 2 are equivalent if there exists an orientation preserving Mathematics Subject Classification (1991): 57M25, 57Q45We continue the proof of Claim 3.3. If Lemma 3.5 below is true, then P (n) satisfies condition (1) of Theorem 1.3.Lemma 3.5. Let K be an n-knot whose projection is P (n) . Then K is not unknotted. (n ≧ 3.)Proof. The n = 3 case holds by §2. We prove n > 3. By Lemma 3.4, there are a 3-knot Z (3) , a 4-knot Z (4) ,..., and an n-knot Z (n) such that Z (q+1) is the 8 spun knot of Z (q) (q = 3, ..., n − 1), that Z (n) = K, and that the projection of Z (r) is P (r) (r = 3, ..., n).Recall that the following facts hold by Theorem 4,1 of [11] or by using the Mayer-Vietoris exact sequence. See, e.g., §14 of [19] for the Alexander polynomials. See, e.g., p.160 of [21] and [18] for the Alexander invariant. Let X K denote the canonical infinite cyclic covering of the complement of the knot K.