@QiaochuYuan Edited the typo, thank you.

The reason I said that was this: Any representation gives rise to a Herz-Schur multiplier $\varphi(g)=<g\xi | \eta>$ which, as you said, gives the kernel $K(g, h)=\varphi(g^{-1}h)$. In my mind the question was rather about analogy between Schur multipliers as $H_2$ and Herz-Schur multiplier, but I decided to not put the Herz word, for a better impact on the reader. I really wanted both to really have the same name :)

I have a questions (5 years after the post): Say $B(m,n)$ is generated by $a_1,\ldots, a_m$. When you say "six-power-free word (without subwords of the form $u^6$) in the generators of the Burnside group ..." do you mean that $u$ should contain only positive powers of the generators $a_i$ or you can also use $a_i^{-1}$ for your statement to be true?

A paper where you can see who the Kernel is: download.springer.com/static/pdf/512/…*~hmac=178bfde5b6b4481‌​665e0b3ecf50d793193a‌​203b2c749fca0ea6247a‌​1e09c54e2

I want to see in what conditions the non-abelian tensor product of a group G, acting by conjugation on itself, $G\otimes G$ is a reduced free group. If not him, at least something similar to him. He is the quotient of $[G,\overline{G}]\subset G*\overline{G}$ where $\overline{G}$ is just a copy of $G$. So all I need to do is prove is that the kernel of this projection is verbal. But I doubt it is true in general.

I agree, but I got stuck at: why is $H_2(G)$ a p-group? I didn't find any references anywhere for the infinite case, and I couldn't prove this by myself. To be honest, that's what I actually wanted from the very beginning, but I decided to write the post in terms of $G\otimes G$ rather than $H_2(G)$.