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Let $X:=\mathbb{C}^n$, and let the symmetric group $S_n$ act by permutation of coordinates in the obvious way; let $X_n:=X/S_n$ be the quotient by the group action. Now, $X_n\simeq \mathbb{C}^n$, so we get a map $\pi: X\to X_n$ with finite fibers. Away from the discriminant $\Delta:=\prod_{i\lt j}(x_i-x_j)$, $\pi$ is a covering map $X\setminus\Delta\to X_n\setminus\pi(\Delta)$, and this map has monodromy group $S_n$. In similar fashion, we can let the subgroup $S_k$ act on the first $k$ coordinates, and let $X_k$ be the resulting quotient; again away from the discriminant the map $X\to X_k$ is a covering map and has monodromy group $S_k$. (We could of course generalise to $S_\lambda$ for a partition $\lambda$ of $n$, but let's keep it simple.)

Now, $\pi:X\to X_n$ factors through any of the $X_k$, so we have maps $X_k\to X_n$, and in particular $X_n\to X_{n+1}$, and away from the branching locus they are also covering maps.

Question: What are the monodromy groups of these covering maps?

This seems like the sort of thing someone might have worked out long ago, is anything known about this?

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1 Answer 1

up vote 7 down vote accepted

For every $k=1,\dots,n-1$, the monodromy group of $q_k:X_k\to X_n$ is isomorphic to the symmetric group $S_n$; one argument uses the fundamental theorem of Galois theory (or its analogue for topological covering spaces, if you prefer to work topologically). First of all $q_1:X_1\to X_n$ equals $\pi$, which is a Galois cover with group of deck transformations $S_n$ by construction. Next, for $k=1,\dots,n$, the morphism $q_k$ factors $\pi$, with the corresponding fixed subgroup being isomorphic to the copy of $S_k$ in $S_n$ that you indicated. Hence also the Galois closure of $q_k$ factors $\pi$, with the corresponding fixed subgroup $N$ being the intersection of all conjugates of $S_k$, i.e., the largest normal subgroup of $S_n$ contained in $S_k$. For $k=n$, $N$ equals $S_n$, so assume $k < n$. For $n\geq 5$, of course, $S_n$ has only the normal subgroups $\{1\}$, $A_n$ and $S_n$, so $N$ equals $\{1\}$. Also for $n=2,3,4$, it is straightforward to compute that $N$ equals $\{1\}$. Thus $\pi$ is the Galois closure of $q_k$, and the Galois group $S_n/N$ equals $S_n$.

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Beautiful, thank you! – Ketil Tveiten Jul 5 '12 at 13:39

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