3
$\begingroup$

I apologize if this question is sophomoric as my knowledge of projective geometry is rather elementary. But I'm curious if there exists a good intuitive geometric explanation for why the curve $y-x^3=0$ in $\mathbb{P}^2(\mathbb{R})$ has a singularity at infinity.

Thanks

$\endgroup$
  • 3
    $\begingroup$ Just a (non-geometric) thought, don't know if this helps or you already know this: if you mean $y-x^3$ in $\mathbb{P}^2$ you really mean the curve $yz^2-x^3$. The curve $y-x^3$ is then the "affine part" of $yz^2-x^3$ when looked at in the affine chart $z=1$ in $\mathbb{P}^2$. However, you can also look at $yz^2-x^3$ using other affine charts, for example the chart $y=1$ and then $yz^2-x^3$ becomes $z^2-x^3$, from which it is very apparent that the cubic $yz^2-x^3$ has a singularity (an ordinary cusp, to be more precise) at the point $[0:1:0]$. That's the way I like to think about it. $\endgroup$ – Nils Matthes Dec 2 '12 at 9:32
  • $\begingroup$ @Nils Matthes: Yes looking at it in the $y=1$ chart was how I discovered its singularity at infinity. My understanding was that looking at it in a different affine chart was just a tool to gain a 'fuller' understanding of the geometry of $y-x^3$. But it appears what you're implying is that the its projective version is in some sense the 'true' curve and it just so happens that its $z=1$ portion happens to miss the singularity. Nevertheless, I can't help but feel that there's probably something interesting to say which relates the way in which $y-x^3$ diverges to its singularity at infinity. $\endgroup$ – Thoth Dec 2 '12 at 19:57

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.