I've come across a double cover of $\mathbb P_1(\mathbb C)$, ramified at $[1:1]$ and $[-1:1]$ in homogeneous coordinates, given as the quotient by the natural $\mathbb Z/2\mathbb Z$-action generated by $[x:y]\mapsto[y:x]$. This space is totally natural and must have a name! For instance, it can be obtained as the space of all ratios $\lambda_1/\lambda_2$ of eigenvalues of non-nilpotent $2\times2$ complex matrices (which is the case I'm interested in).

In the litterature the notion of "reduced symmetric product" sometimes arises as $M\times M\setminus\Delta$, where $\Delta$ is the diagonal. This is somehow related, but in the case I'm confronted to I want to keep the branch-point (corresponding to $\Delta\setminus\{(0,0)\}$ in the case $M:=\mathbb C$, before projectivization). Also I feel the context is not completely the same.

My question: how should I call/write $\mathbb P_1(\mathbb C)~/~\mathbb Z/2\mathbb Z$? Is there a standard denomination? Notation?

Edit: as Noam D. Elkies very rightly mentions in a comment, the quotient is conformally equivalent to $\mathbb P_1(\mathbb C)$, via the explicit parameterization $[x^2+y^2:xy]$. It doesn't settle the question of whether this particular perception of the projective line has a particular name, though. But maybe this is not interesting.

  • 7
    $\begingroup$ You could just call it ${\bf P}^1({\bf C})$ because it's a genus-zero Riemann surface; an explicit coordinate is $(x^2+y^2 : xy)$. $\endgroup$ – Noam D. Elkies May 25 '15 at 21:50
  • $\begingroup$ Hi @Balarka, is that true? For example, you can exhibit any Riemann surface as a branched double cover of $\mathbb{CP}^1$ (to exhibit this geometrically, put a skewer symmetrically through the surface and let $\mathbb{Z}/2$ act as rotation by 180 degrees about the skewer). $\endgroup$ – Amitesh Datta May 25 '15 at 22:12
  • 1
    $\begingroup$ @AmiteshDatta, indeed, hyperelliptic curves $y^2=f(x)$ with polynomial $f$ are branched double covers, and can have any genus, so, as you hint, it is certainly not true that any branched double cover of $\mathbb P^1$ is again $\mathbb P^1$. :) $\endgroup$ – paul garrett May 25 '15 at 23:35
  • $\begingroup$ Dear @paul, thank you for your comment! :) I guess in that case one can also state that the genus of the surface will be the ceiling of $\frac{\text{deg}(f)-2}{2}$. $\endgroup$ – Amitesh Datta May 26 '15 at 1:20
  • $\begingroup$ Note too that Loïc Teyssier's double cover has two branch points (as expected by Riemann-Hurwitz): the involution $(x:y) \leftrightarrow (y:x)$ fixes not just $(1:1)$ but also $(-1:1)$. $\endgroup$ – Noam D. Elkies May 26 '15 at 3:26

Your Answer

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

Browse other questions tagged or ask your own question.