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In Nica's dissertation Group actions on median spaces, we can read the following assertion:

Braid groups do not contain infinite subgroups satisfying Kazhdan's property (T).

This is used in order to motivate the question (which is still open up to my knowledge) of whether braid groups satisfy the Haagerup property. Does anyone have a reference and/or a quick justification of the above claim?

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    $\begingroup$ How about Exercise 1.8.14 of Khazdan's Property (T) by B. Bekka, P. de la Harpe and A. Valette? $\endgroup$
    – Will Sawin
    Commented Apr 6, 2021 at 14:09
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    $\begingroup$ As was already mentioned at this question, pure braid groups are residually (torsion-free nilpotent). Hence for no $n$ there exists any nontrivial subgroup $H\le B_n$ such that $(B_n,H)$ has relative Property T. In particular there's no infinite subgroup with Property T (but the latter is quite immediate since pure braid groups are iterated extensions of free groups, as mentioned in this answer). $\endgroup$
    – YCor
    Commented Apr 6, 2021 at 14:58

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It is enough to see that the pure braid group $P_n$ does not contain subgroups which have Kazhdan property (T). There is an exact sequence $$1 \rightarrow F_{n-1} \rightarrow P_n \rightarrow P_{n-1} \rightarrow 1 $$

which reduces, by induction, to showing that $F_n$ does not contain subgroups with Kazhdan property (T). The latter is easy.

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  • $\begingroup$ $F_n$ is the free group on $n$ generators $\endgroup$
    – Venkataramana
    Commented Apr 6, 2021 at 15:38
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As suggested by Will Sawin in the comments, I took a look at Bekka, de la Harpe and Valette's book. The claim is a straightforward consequence of the following statement:

Theorem. (Artin) For every $k \geq 2$, the morphism between pure braid groups $P_k \to P_{k-1}$ that forgets the last strand has a free kernel.

Indeed, let $k \geq 1$ be an integer and let $H \leq B_k$ be a subgroup having (T). We want to prove that $H$ is trivial. Because $P_k$ has finite index in $B_k$, we can assume that $H$ lies in $P_k$ up to replacing $H$ with $H \cap P_k$. Now, we argue by induction over $k$. If $k=1$ then the conclusion is clear. If $k \geq 2$, then the image of $H$ under $P_k \to P_{k-1}$ must be trivial as a consequence of our induction hypothesis. In other words, $H$ must lie in the kernel of $P_k \to P_{k-1}$, which is free according to Artin's theorem. This implies that $H$ is trivial, as desired.

Actually, the argument applies to any group property $(P)$ satisfying the following axioms:

  • If $G$ has $(P)$, then so does the image of $G$ under any homomorphism.
  • If $G$ has $(P)$, then so do the finite-index subgroups of $G$.
  • Free groups of rank $\geq 1$ do not have $(P)$.

This includes many properties weaker than (T) such that the hereditary property (FA) or the property (FW).

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    $\begingroup$ Actually this shows that $P_n$ is "locally indicable" in the sense that every nontrivial finitely generated subgroup surjects onto $\mathbf{Z}$, as this property is satisfied by free groups and stable under taking extension. [It even has the property that every nontrivial subgroup surjects $\mathbf{Z}$ for the same reason, but I'm not aware of a name for this. For $B_n$, this implies that every nontrivial subgroup has an infinite virtually abelian finitely generated quotient.] $\endgroup$
    – YCor
    Commented Apr 6, 2021 at 17:10

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