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Then the time–space projection of the Killing operator acting on Ka is [3,4]. ... aAa, andthe third term is also zero because Ka is a Killing vector.

Since K H+ {0},we have θHa θ. Ka and p. Ha p. ... Ka , and therefore pg p. Ha by Theorem 5. We pose two questions.

p(m)qkd (kA,kB) = Pr(KA = kA,KB = kB|M = m). (1). ... KA, kA, KB, kB: output keys for Alice and Bob.

harmonic with a period of 41 ka but it bears periods as long as 1,200 ka. ... solutions at t = 0 when started from a grid of initial conditions at tI = 700 ka.

ùEû=,?:7A9 @46D'Dû @lZE4D@l=" Ká Ço$#$$B>Ñ.H/2«K W Q û@«ù»=,ZKZK@rB[¢=&9 _)7:9ý[ Z=)_)ú ù»D«[ ... A@ =,Z4B[û@rZ7FùT4,?ùT=,?:?A=&7:_7FB ka«$è:L O á5ú$øCù»G$2û7FùEûa7:5[?

Limitations on Quantum Key Repeaters. Stefan Bäuml,1, 2, Matthias Christandl,3, † Karol Horodecki,4, 5, ‡ and Andreas Winter6, 2, 1,. 1Department of Mathematics, University of Bristol, Bristol BS8 1TW, UK. 2F́ısica Teòrica: Informació i

The vector ω is an interior point (since a‖ω‖ωTω = a1 < 0), and thus Ka issolid. ... Consider x Ka and y Kb, which we take to have unit norm without lossof generality.

ma,ry,eo}. = {ka,en,at,te}. c c c. c. c. c. cc. ... re,eo}. = {ka,ar,re,en,at,te}. = {ka,ar,re,en,at,te}. = {ma,ar,ry,re,eo}.

4). At the quantum level the theory has the symmetryxa xa+ 2πika2naQ , ka Z coming from the periodicity oft, obtaining T 2 as the dual target space.

that KA is not separable). Suppose for the contradiction that TA/Fin =n<ω Cn and each Cn is centered. ... For each A V we fix a pair (kA,UA) such that kA ω{0}.