I'm stuck in trying to understand the graphical part of the following problem.

Let $\mathcal C = \{ [x:y:w:z] \in \mathbb P ^3: x^2 +xy +yw -w^2 = 0 \}$. Graph the intersection $\mathcal C \cap \mathbb A ^3$.

What I have tried is the following. I found the canonical projective form of the quadric which if I'm not mistaken is: $\mathcal C(x^2 - y^2) $. The notation means the quadric generated by that form. Now I know that $\mathbb A ^3$ are the points where $x \neq 0$. So if everything is correct I'm left with the following points in $\mathbb P ^3 \cap \mathbb A ^3$.

$$[x,x,w,z], x\neq 0 \hspace0.75cm \text{and} \hspace0.75cm [x,-x,w,z], x\neq 0 .$$

Are these plane (two dimensionals as $z$ and $w$ are free) where the $y$ value is fixed at either $1$ or $-1$? I'm following the idea that $\mathbb A ^3$ is an hyperplane in $\mathbb A^4$ where the first coordinate is fixed at 1 as it's different from 0.

Sorry if it's not clear enough as I'm still wrapping my head around these concepts.


In what way did you arrive at that canonical form? It sounds to me as though you had essentially diagonalized the matrix of the quadratic form. You can do that, but doing so corresponds to a projective transformation so it will change the relationship between the affine space and the quadric. Therefore I'd suggest not doing this here.

Instead of merely going for $x\neq 0$ you might as well pick one specific representative and assume $x=1$. So you are looking for points $(y,w,z)\in\mathbb A^3$ with $y+yw-w^2=0$. This is independent of $z$, so you can plot this as a planar curve and then extrude it along the $z$ axis. To graph the planar curve, a hyperbola, you can compute asymptotes and probably a few values, then judge a reasonable interpolation in between. Unless you have learned different techniques.

  • $\begingroup$ Sorry for not including the details of the canonical form. Yes, I applied several projective transformations to get to that. So, the intersection in this case actually corresponds to the points $y + yw - w^2 = 0$ so that means you are left with a curve in the $(y,w)$ plane and as $z$ is free, you are left with a surface in $\mathbb A ^3$. Is this correct? $\endgroup$ – Leo Sep 23 '18 at 22:16
  • $\begingroup$ @LeoLerena: That's the way I understand it. Personally I'm more used to the last coordinate being used for homogenization, not the first, so all of this depends a bit on context. $\endgroup$ – MvG Sep 24 '18 at 19:18
  • $\begingroup$ @LeoLerena $\mathcal C$ factorizes into $(x + w) (x + y - w)$. $\endgroup$ – Maxim Nov 7 '18 at 20:52

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