Let $X$, $Y$ be smooth, connected, compact manifolds (for instance, projective varieties) and $f \colon X \longrightarrow Y$ be a finite, branched cover of degree $n$, with branch locus $B \subset Y$. We can then associate to $f$ its monodromy representation $$\theta_f \, \colon \pi_1(Y-B) \longrightarrow S_n,$$ so that $\mathrm{im} \, \theta_f$ is a transitive subgroup of $S_n$. Conversely, isomorphism classes of connected covers of $Y$, branched over $B$, are in bijection to monodromy representations of the type above, up to conjugacy in $S_n$.

We now assume that $f$ is simple, that is that for all $b \in B$ the fibre $f^{-1}(b)$ consists of exactly $n-1$ points, and we call $f$ an elementary cover if it is not possible to factor it as $$X \stackrel{g}{\longrightarrow} Z \stackrel{h}{\longrightarrow} Y,$$ where $g\colon X \longrightarrow Z$ is a branched cover and $h \colon Z \longrightarrow Y$ is an ordinary (i.e, unramified) cover.

Q. What is the characterization of simple, elementary covers $f \colon X \longrightarrow Y$, branched over $B$, in terms of their monodromy representation $\theta_f$?

  • $\begingroup$ The cover is elementary if and only if the image subgroup of the symmetric group preserves no nontrivial partition of $\{1,\dots,n\}$. The cover is simple if and only if the loops around $B$ map to transpositions in the symmetric group. $\endgroup$ Jun 8, 2017 at 14:52
  • $\begingroup$ @JasonStarr: thanks. Then, is it true that simple+elementary means that the monodromy representation is surjective? $\endgroup$ Jun 8, 2017 at 15:38
  • $\begingroup$ In fact, I'm looking for a statement of this type: every transitive subgroup of $S_n$ which is not $S_n$ itself and that contains at least one transposition must preserve a non-trivial partition of $\{1, \ldots, n\}$. Is this true? Any reference? $\endgroup$ Jun 8, 2017 at 16:06
  • $\begingroup$ For a subgroup $H$ of the symmetric group, define a relation on elements of $\{1,\dots,n\}$ by $a \sim b$ if either $a$ equals $b$ or if the transposition $(a,b)$ is in $H$. This is symmetric and reflexive. For pairwise distinct elements $a,b,c$ with $(a,b)$ and $(b,c)$ in $H$, then $(a,c) = (a,b)\circ (b,c)\circ (a,b)$ is in $H$. Thus it is also reflexive. So if $H$ contains at least one transposition, then either it contains every transposition (and thus $H$ equals the entire symmetric group) or it preserves a partition of $\{1,\dots,n\}$. $\endgroup$ Jun 8, 2017 at 18:00
  • $\begingroup$ . . . Therefore, a simple, elementary cover with nonempty branch locus has monodromy group equal to the entire symmetric group. I believe that a version of this is in Joe Harris's article, "Galois groups of enumerative problems". It may also be in Fulton's paper (pre-Deligne-Mumford) on the problem of irreducibility of $\mathcal{M}_g$ in positive characteristic via Hurwitz spaces (it has been some time since I looked at that paper). $\endgroup$ Jun 8, 2017 at 18:03

1 Answer 1


I am just writing my comments as an answer. For every subgroup $H$ of the symmetric group $\mathfrak{S}_n$, define a relation on $\{1,\dots,n\}$ by $a\sim b$ if either $a$ equals $b$ or if the transposition $(a,b)$ is contained in $H$. By definition, this is symmetric and reflexive. Let $a,b,c$ be pairwise distinct elements of $\{1,\dots,n\}$ such that $a\sim b$ and $b\sim c$. Since $H$ contains $(a,b)$ and $(b,c)$, and since $H$ is a subgroup, also $H$ contains $$(a,c) = (a,b)\circ (b,c) \circ (a,b).$$ Thus, also $a\sim c$. So this is an equivalence relation. The associated partition of $\{1,\dots,n\}$ is preserved by $H$.

Therefore, every subgroup of $H$ that contains at least one transposition and that preserves no nontrivial partition must equal all of $\mathfrak{S}_n$. Geometrically, this says that a simple, elementary cover with nonempty branch locus has monodromy group equal to the full symmetric group. If the branch locus is empty, this can fail, e.g., it fails for unbranched cyclic covers.

  • $\begingroup$ Thank you very much for the clear explanation and the useful references provided in the comments. $\endgroup$ Jun 9, 2017 at 6:13

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