# Differential geometric methods in mathematical physics. by H.-D. Doebner, S. I. Andersson, H. R. Petry By H.-D. Doebner, S. I. Andersson, H. R. Petry

Read or Download Differential geometric methods in mathematical physics. Proc. conf. Clausthal, 1980 PDF

Similar mathematics books

The Mathematics of Paul Erdos II (Algorithms and Combinatorics 14)

This can be the main accomplished survey of the mathematical lifetime of the mythical Paul Erd? s, probably the most flexible and prolific mathematicians of our time. For the 1st time, all of the major components of Erd? s' examine are lined in one venture. as a result of overwhelming reaction from the mathematical neighborhood, the venture now occupies over 900 pages, prepared into volumes.

Extra info for Differential geometric methods in mathematical physics. Proc. conf. Clausthal, 1980

Sample text

Then the set coincides with the segment [0,2]. Proof. We use the standard geometric argument presented, for example, in . Namely, let us introduce a mapping This mapping can be described in another way. Denoting we see that 4 is identical with the projection whose direction is determined by the straight line I . Now, from the geometric viewpoint it is almost evident that C x C = n{z, :n < w), where {Zn : n < w) is some decreasing (with respect to inclusion) sequence of compact subsets of the unit square [O,l] x [0,1] and, in addition, the equality ~ r l ( z n= ) [O, 21 38 CHAPTER 3 holds for any natural number n.

0 Observe also that m 2 2 since in view of the definition of H. Notice, in addition to this, that no set En possesses the Steinhaus property. ,ezn) C H and denoting e = e o + e l ... +e2,, + we have e # 0 and (En + qe) n En = 0 for any nonzero rational number q. Since q can be arbitrarily small, we claim that the Steinhaus property does not hold for En. It immediately follows from this fact that all the sets are nonmeasurable in the Lebesgue sense. To finish the proof, let us consider two possible cases.

Indeed, suppose otherwise. Then at least one of these sets is Lebesgue measurable and, in view of the relation -A = B , we claim that both these sets must be Lebesgue measurable. Since we derive that X(A) = X(B) > 0. On the other hand, the metrical transitivity of the Lebesgue measure (see Exercise 7 from Chapter 1) implies which leads to a contradiction. We thus conclude that each of the sets A and B is nonmeasurable in the Lebesgue sense. Moreover, an easy argument based on the same property of metrical transitivity of X shows that both these sets are A-thick in R; in other words, we have However, the last relation enables us to consider the sets A and B as measurable ones with respect to some measure on R which extends X and is 44 CHAPTER 3 invariant under the group of all motions (isometric transformations) of R .

Download PDF sample

Rated 4.31 of 5 – based on 32 votes