Two more notes on embeddings. I believe that the second one is handled by the change from considering the embedding to be in S^{2n-1} rather than R^{2n-1}.........DMD _______________________________ Date: Tue, 23 Jan 2001 16:48:34 -0500 From: Frank Connolly Subject: More on embeddings EVERY COMPACT ORIENTABLE n-MANIFOLD EMBEDS IN S^{2n-1}. As Mark Mahowald explained earlier, Massey-Peterson (1964) proved, based on work of Haefliger-Hirsch, that if n is different from 4, an orientable n-manifold embeds in R^{2n-1}. But it turns out that if n=4, every smooth compact oriented 4 manifold embeds in R^7. To see this, note that the paper of Boechat-Haefliger in the volume dedicated to deRham (Essays on Topology and Related Topics, Springer, 1970; p.165), proves that a connected oriented compact 4-manifold M smoothly embeds in R^7 if and only if there is a CHARACTERISTIC class V in H^2(M;Z) (i.e. one whose reduction mod 2 is w_2) whose square, V^2, in H^4(M;Z)=Z, is equal to the index of M. In 1970, they could not settle whether such a V could always be found. However, when the intersection form of M is indefinite, it is well known that this form contains a hyperbolic plane. Quite easy algebra using this hyperbolic plane then shows that any characteristic class V can be modified to get another such that V^2 = \tau. When the intersection form is definite, the answer depends on Donaldson's 1987 theorem that the intersection form must be diagonal. If M is oriented to be positive definite, and x_1, x_2, ... x_n is an orthonormal basis, then V= x_1 + x_2 + ... + x_n is the class which Boechat-Haefliger require. This is discussed briefly in a paper by Fang Fuquan in Topology 1994. _______________________________________ Date: Mon, 22 Jan 2001 21:23:34 -0500 (EST) From: Joseph Roitberg Subject: Re: more on embeddings I'm puzzled by remark (1) in Mark's reply to Frank. Wouldn't that imply that the circle embeds in the real line? Joe Roitberg