15
$\begingroup$

A covering map $f:X\rightarrow Y$ is called Galois if for each $y\in Y$ and each pair of lifts $x, x^{'}$, there is a covering transformation taking $x$ to $x^{'}$. What is a good way to understand this definition? It seems to me that $f$ is Galois if and only if $Y$ is obtained from $X$ as a quotient of some group.

$\endgroup$
1
  • 6
    $\begingroup$ Indeed. In fact $Y$ will be the quotient of $X$ by the group of deck transformations of $f$. (This is exactly like the case of a Galois extension of fields.) $\endgroup$ – Zhen Lin Nov 29 '12 at 20:15
9
$\begingroup$

In the setting of (complex) algebraic geometry, the covering is Galois if and only if the function field $K(X)$ is a Galois extension of the function field $K(Y)$. Moreover, if $f$ is Galois, then the Galois group of the extension is exactly the deck transformation group $G$. As you've already noticed. If $f$ is Galois, then $Y$ is isomorphic to $X/G$, where $G$ is the Galois group.

$\endgroup$
3
  • $\begingroup$ Any book references for this? $\endgroup$ – Hodge-Tate Oct 1 '18 at 18:07
  • $\begingroup$ yes there is a very nice book concerning this subject @Hodge-Tate: Galois Groups and Fundamental Groups by Szamuely $\endgroup$ – Simonsays Nov 11 '18 at 11:02
  • $\begingroup$ where in this reference (Szamuely) is this correspondence Galois extension and Galois covering stated? My assumptions are $X$ and $Y$ Are Noetherian, separated, integral schemes over $\mathbb C$ (or at least varieties with base field $\mathbb C$) $\endgroup$ – quantum Sep 7 '20 at 7:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.