THEOREM II. Suppose n = 4k + 1, but k is not a power of 2. Then Pa can be imbedded in (2n 2)-space. In each case the result would be false for k a power of 2. Pa cannot be imbedded in (2n 1)-space for n = 2r, by a result of Stiefel or Thom, [11, p. 157, III. 16]. Pa cannot be imbedded in (2n 2)-space for n = 2r + 1, by a result of Levine [5] and Mahowald. There is a tentative conjecture, due to Atiyah, that Pn can be imbedded in (2n a(n) + 1)-space but not in (2n a(n))-space. Here a(n) is the number of non-zero terms in the dyadic expansion of n. Our results agree with the first part of this conjecture for n = 2r, 2r + 1, 2r + 28, 2r+1 + 28+1 + 1(r > s > 0).