Let $G$ be a simple (i.e. *every proper normal subgroup is discrete*) simply connected compact Lie group. Define **the degree of $k$-nilpotence** of $G$ to be the Haar measure of the set
$$\{(x\_1,\dotsc,x\_{k+1}): [x\_1,\dotsc,x\_{k+1}]=1\}.$$
($[x,y]=x^{-1}y^{-1}xy$ and $[x\_1,\dotsc,x\_{k+1}]:=[[x\_1,\dotsc,x\_k],x\_{k+1}]$.)
The following question is raised in [Martino, Tointon, Valiunas, and Ventura - Probabilistic nilpotence in infinite groups](https://doi.org/10.1007/s11856-021-2168-3):



> 
> Does a compact group with positive degree of $k$-nilpotence have an
> open $k$-step nilpotent subgroup?
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> 
> 


Is it easy to answer this question when $G$ is a simple simply connected compact Lie group?



Can the degree of $k$-nilpotence of a simple simply connected compact Lie group be in $(0,1)$?
Here's a positive answer of the question for arbitrary compact Lie groups. For a group $G$, denote $W\_k(G)=\{x\in G^k:[x\_1,\dots,x\_k]=1\}$.



> 
> Let $G$ be a compact Lie group. Then $W\_k(G)$ has nonzero Haar measure for some $k\ge 1$ if and only if $G^0$ is a torus.
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> 
> 


One direction is obvious: if $G^0$ is a torus, then $(G^0)^k\subset W\_k^G$ has positive measure (for every $k\ge 2$).


Conversely, we assume $W\_k$ has nonzero Haar measure. If $k=1$, it follows that $G$ is finite and we are done. Assuming otherwise, $k\ge 2$.


The subset $W\_k(G)\subset G^k$ is Zariski-closed. Since it has nonzero Haar measure, it therefore contains $\prod\_{i=1}^k C\_i$ for cosets $C\_i$ of $G^0$. [If $G$ is connected, this is $G^k$, so $G$ is nilpotent, hence is a torus and the proof is finished.]


I assume the conventions $[x,y]=xyx^{-1}y^{-1}$ and that the iterated commutator is defined by $[x]=x$ and $[x\_1,\dots,x\_k]=[x\_1,[x\_2,\dots,x\_k]]$ for $k\ge 2$. Also, by $[X\_1,\dots,X\_k]$ I mean the set $\{[x\_1,\dots,x\_k]:x\_i\in X\_i\}$ (rather than the group it generates).


Now assume that $G^0$ is semisimple and nontrivial. I need an inductive argument. Namely, to reach a contradiction, choose $k\ge 2$ minimal for the property that there are cosets $C\_1,\dots,C\_k$ of $G^0$ such that $[C\_1,\dots,C\_k]$ is a singleton, say $w$. Necessarily $k\ge 2$.


Define $Y=\{[x\_2,\dots,x\_k]:x\_i\in C\_i,\forall i\ge 2\}$. Then $[x,y]=w$ for all $x\in C\_1$ and $y\in Y$. Then for all $s\in G^0$, $x\in C\_1,y\in Y$, we have $$w=[x,y]=[xs,y]=x[s,y]x^{-1}[x,y]= x[s,y]x^{-1}w,$$
and hence $[s,y]=1$. Thus, $Y$ is contained in the centralizer of $G^0$. Since $G^0$ is semisimple, its centralizer is finite. Hence $Y=[C\_2,\dots,C\_k]$ is finite, hence a singleton by connectedness. This contradicts the minimality of $k$.


Now the result is proved when $G^0$ is semisimple (i.e., when $G^0$ is semisimple, if $W\_k(G)$ has nonempty interior implies $G^0=1$). In general, let $Z$ be the center of $G^0$. If $W\_k(G)$ has nonempty interior, since the projection $G\to G/Z$ is open and so is $G^k\to (G/Z)^k$, the subset $W\_k(G/Z)$ (which contains the projection of $W\_k(G)$) has nonempty interior. So, by the semisimple case $(G/Z)^0$ is trivial. That is, $Z=G^0$.


[Edit: I fixed my previous argument]


