I'm having trouble understanding a definition of a quiver Hecke algebra.
Suppose $k$ is a commutative ring, and $\Omega$ a finite set. We build a quiver $Q_{\Omega,n}$ with vertex set $\Omega^n$. Then $S_n$ acts on $\Omega^n$, so define the arrow set to be $$ s_i=s_{i,v}\colon v\to s_i(v) $$ for $1\leq i\leq n-1$, $v\in\Omega^n$ and $$ x_i=x_{i,v}\colon v\to v $$ for $1\leq i\leq n$ and $v\in \Omega^n$.
The define the quiver algebra $A$ with the quiver above and relations $$ s_i^2=1, s_is_j=s_js_i,\ |i-j|>1,\ s_is_{i+1}s_i=s_{i+1}s_is_{i+1} $$ $$ x_ix_j=x_jx_i,\ s_ix_j=x_js_i,\ j\neq i,i+1,\ s_ix_i=x_{i+1}s_i. $$
Supposedly this implies $A=k[x]^\Omega\wr S_n$, the wreath product. I don't understand what it means that the quiver algebra is a wreath product. I only know how to make sense of a wreath product of groups, so even viewing $k[x]$ (I assume this is shorthand for $k[x_1,\dots,x_n]$) as a group, it seems like $k[x]^\Omega\wr S_n$ only carries a group structure, so how can it be the quiver algebra? (Also, I suspect should $k[x]$ be indexed by $\Omega^n$, not $\Omega$? I can't think of a sensible way $S_n$ would act on $\Omega$.)