6
$\begingroup$

Consider the family: $$ \begin{matrix} X & = & \textbf{Proj}\left(\frac{\mathbb{C}[s,t,u][x,y,z]}{s(x^4 -y^2z^2) + t(x^2y^2 - z^4) - u(x^4 + y^4 + z^4)}\right) & \\ \downarrow & & \downarrow\\ B & = & \textbf{Proj}(\mathbb{C}[s,t,u]) \end{matrix} $$ How can I find the hyperplane $H_x \subset B$ of points whose fibers contain a point $x \in X$? For example, let $x = [0:1:1:1:(-1)^{(1/3)}:(-1)^{(1/3)}]$. For reference, I am looking at Nicolaescu's introduction to morse theory (page 252).


For reference, here is what I have tried: if I have the point $[s:t:u:a:b:c] \in X$, then I think the hyperplane will be spanned by the set of points $$ \{s,t,u \in \mathbb{C}: s(a^4 - b^2c^2) + t(a^2b^2 - c^4) + u(a^4 + b^4 + c^4)\} $$ I am guessing this because of the statement afterwards $$ H_x = \{ P \in U : P(x) = 0 \} $$ where $U$ is the projective plane parametrizing this family, but I am not sure why this is true.

$\endgroup$
  • $\begingroup$ If someone has enough reputation, can they add a linear system tag? $\endgroup$ – 54321user Aug 4 '16 at 23:51
  • $\begingroup$ There is only a nonlinear system tag. $\endgroup$ – 关一骏 Aug 15 '16 at 0:47
  • $\begingroup$ If that is the right one, will do. Actually, I think it has already been done by someone else. $\endgroup$ – 关一骏 Aug 15 '16 at 0:47
  • 1
    $\begingroup$ Nonlinear systems are a far distance from algebraic geometry... $\endgroup$ – 54321user Aug 15 '16 at 5:21
  • $\begingroup$ In fact, the more appropriate tag is divisors. I'll add it now. $\endgroup$ – paf Aug 15 '16 at 17:46
3
+50
$\begingroup$

If the fiber over $[s:t:u]\in B$ contains $x=[s':t':u':a:b:c]\in X$ then this means that, denoting by $P$ the equation of $X$ you wrote :

$$[s:t:u]=[s':t':u']\text{ and }P(s',t',u',a,b,c)=0.$$

In fact, by definition of $X\to B$, the fiber over $[s_0:t_0:u_0]\in B$ is $$\{[x:y:z]\in \Bbb P^2(\Bbb C) : P(s_0,t_0,u_0,x,y,z)=0\}\subseteq \mathrm{Proj}(\Bbb C[s_0,t_0,u_0][x,y,z])\subseteq \mathrm{Proj}(\Bbb C[s,t,u][x,y,z]).$$

So, what you wrote is almost correct : I would write $H_x=\{ [s:t:u]\in \Bbb P^2(\Bbb C) : s(a^4 - b^2c^2) + t(a^2b^2 - c^4) + u(a^4 + b^4 + c^4)\}$ instead of $\{ s,t,u\in\Bbb C : \dots\}$.

Edit: Now, denoting by $\hat{U}$ the dual projective space of $U$, if we want to describe explicitly the map $X\to \hat{U}$ defined by $x\mapsto H_x$, we can write it in coordinates. For this purpose, note that if $(As+Bt+Cu=0)$ is the equation of a hyperplane $H$ of $\text{Proj}(\Bbb C[s,t,u])$, then the coordinates of $H$ in $\hat{U}$ are $[A:B:C]$.

Hence, in your example, the map $x\mapsto H_x$ is defined by $$[s:t:u:a:b:c]\mapsto [a^4 - b^2c^2:a^2b^2 - c^4:a^4 + b^4 + c^4].$$

$\endgroup$
  • $\begingroup$ This doesn't entirely answer the part of the question I raised in the bounty message. $\endgroup$ – 54321user Aug 13 '16 at 19:15
  • $\begingroup$ In fact, I don't understand very well what you wrote here. What is "the vanishing locus of the hyperplanes"? (i.e. what does mean "vanishing" for a hyperplane?) Moreover, I don't know the definition of the "modification" of a linear system (and I don't have Nicolaescu's notes). $\endgroup$ – paf Aug 13 '16 at 19:32
  • 1
    $\begingroup$ This is on page 212 of www3.nd.edu/~lnicolae/Morse2nd.pdf $\endgroup$ – 54321user Aug 13 '16 at 19:44
  • $\begingroup$ Ok, thanks, and what's your problem with the definition of "modification"? $\endgroup$ – paf Aug 13 '16 at 20:06
  • $\begingroup$ There is a lot of jumping around with information. I think the computation I was doing above was computing a correspondence between $X$ and $U$ not $\hat{U}$. I'm not sure how to figure out this morphism to the dual variety. $\endgroup$ – 54321user Aug 13 '16 at 20:28

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.