Some New Immersion Results for Complex Projective Space

Abstract

Abstract. We prove the following two new optimal immersion results for complex projective space. First, if n ≡ 3 mod 8 but n ≡ 3 mod 64, and α(n) = 7, then CP n can be immersed in R 4n−14 . Second, if n is even and α(n) = 3, then CP n can be immersed in R 4n−4 . Here α(n) denotes the number of 1's in the binary expansion of n. The first contradicts a result of Crabb, which said that such an immersion does not exist, apparently due to an arithmetic mistake. We combine Crabb's method with that developed by the author and Mahowald. Main theoremsWe prove the following two new optimal immersion results for 2n-dimensional complex projective space CP n . Theorem 1.1. If n ≡ 3 mod 8 and n ≡ 3 mod 64, and α(n) = 7, then CP n can be immersed in R 4n−14 . Theorem 1.2. If n is even and α(n) = 3, then CP n can be immersed in R 4n−4 .Here and throughout, α(n) denotes the number of 1's in the binary expansion of n. Theorem 1.1 contradicts a result of Crabb ([2]). In Section 2, we prove Theorem 1.1 by an adaptation of Crabb's argument, and point out what we believe to be his mistake, apparently arithmetic. We prove Theorem 1.2 in Section 3.We now summarize what we feel to be the status of the immersion question for CP n . In addition to incorporating the two new immersion results above, we list as unresolved one immersion result claimed by Crabb. We will discuss our reason Date: February 21, 2006.

An algebraic characterization of injectivity in phase retrieval

“…Therefore a different approach would be necessary to prove the conjecture for other M. Remark 5.6. Proposition 5.4 is similar to, but does not seem to follow from, previous results [7,10]. Heinosaari, Mazzarella and Wolf use embedding results from topology to show that when N ≤ 4M − 2s 2 (M − 1) − 4, the map A Φ is never injective [10].…”

“…(M + i) − s 2 (M + i − 2) − M −(i + 2) − s 2 (i) . Now if M = 2 k + 1 and 0 ≤ n ≤ M − 2 then s 2 (M − 1 + n) = s 2 (n) + 1,so most of the terms in the two summations of (6) cancel leaving just four terms(7) (s2 (M) − s 2 (M − 2)) − (s 2 (M − 1) − s 2 (M − 3)) . When M = 2 k + 1, s 2 (M) = 2, s 2 (M − 1) = 1, s 2 (M − 2) = k − 1 and s 2 (M − 3) = k − 2.…”