I think there is a minor correction to add p.249 in the proof of Theorem 6.2. It says that “there are only elements of larger than ” but with the ordering, I think it should be . Then for all , and the following sum. Of course the result is the same though.

*[Correction added, thanks – T.]*

I have a question:

– page 163, exercise 4.3.4 : I understand how to prove it if the group is cyclic.

– does the fact, that the group has bilinear form that indues a isomorpism means, that this is always the case ( because the circle group is cyclic)?

all the best

Tomer

– you wrote that that P(A_1)***P(A_n) is the quantity one would

expect if random selection is involved conditioning on x =a_1 + .. + a_n,

restricted to these random sets.

– essentially meaning, that the representation function at a fixed point

is expected to be P(A_1)***P(A_n)*(|Z|^(n-1)) restricted to these sets.

the first part of P(A_1)***P(A_n) is understood, but the last part (|Z|^(n-1))

would describe the representation function at a fixes point where solutions

are unconditioned. I find it hard to believe and follow this claim.

please clarify, it is important for me to understand.

T

]]>*[Uncountable sets can still have zero measure. For instance, the set has zero two-dimensional Lebesgue measure for any given number . -T.]*