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**Sample text**

We prove that ker p 1 7/i for some l < m. Indeed, let us choose l linearly independent vectors {c1, ... , cl } in the kernel ker p 1 , and consider the subspace El they generate. Then Hl = El n ker p 1 is a closed subgroup in E,. 2 3 (a), Hl RP + 7Lq , p + q < l . Since the kernel ker p 1 does not contain subgroups topologically isomorphic to R and Hl contains l linearly independent vectors, it follows that Hl 7/i and p1 (E1) ee%zzoo T 1 . Hence l < m . , kerp1 7L , 1 < M. Denote by l2 the subspace in R°° consisting of t = (t1, ...

20). 11 and formula (4) that p(v)(d) = exp{-(Af(d), f(d))}. 10 (b) that u = p (v) . In the general case X Rn +K one may assume Y = Rn +D and the proof is similar. A continuous monomorphism f : Y - RI is determined in the following way: f((s; d)) = f((s1 , ... , sn ; d)) = (s1 , ... , sn , k1 /k , ... , km/k), l = n+m , where s = (Si ... , SO E Rn , d E D , and k , k1 , ... , km are cho- sen as in the case X = K. One can easily see that the matrix A = () =1 1*
*

16(i). Then µ has an indecomposable divisor. PROOF. , e(kmG) = (1 - a)E0 + amG , a = 1 - e-k. (1) Therefore, it is sufficient to prove that the distribution e (k mG) has an indecomposable divisor. Choose nonzero elements xl , x2 , and x3 in G such that x3 = xl + x2 (x, and x2 being not necessarily different). This can be done since G ;;6 Z(2). Let U and UX3 be neighborhoods of the elements x1 and x3 such that UX = x2 + UX , UX n UX = 0, and 0, x2 ¢ UX . Then the open set V = G\Ux3 satisfies the condition (2)V = G.