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Let $X$ be an irreducible hypersurface defined by a polynomial $f$ of degree $5$ in $\mathbb{P}^3$. Let the homogeneous co-ordinates is given by $[x, y, z, w]$ and let $H$ be a hyperplane given by $w= 0$. Assume that the intersection $X \cap H$, a curve of degree $5$ in the plane $H$, is given by an equation of the form $hg^2=0$, where $h$ is a linear form and $g$ is of degree $2$. Further assume that $X$ is singular along the plane curve $g= 0$. In this situation one can write $f$ as $f = hg^2 + wgf_2 + w^2f_3$, where $f_2$ is a polynomial of degree $2$ in $x, y, z$. My question is the following: Can $X$ have isolated singularities? If yes then is it possible to give an upper bound on the number of isolated singularity ?

Thanks in advance.

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  • $\begingroup$ You say $X$ is singular along the plane curve and ask whether $X$ can have isolated singularities. Do you mean "isolated singularities away from the plane curve"? Also it is better to correct a typo in the third sentence. $\endgroup$
    – Sasha
    Commented Dec 3, 2018 at 8:12
  • $\begingroup$ Yes. I meant to say isolated singularity away from the plane curve $\endgroup$
    – user130022
    Commented Dec 3, 2018 at 9:06
  • $\begingroup$ It can certainly have some isolated singularities outside $H$: just take a point $p$ outside $H$, assume that $g,f_2,f_3$ vanish at $p$, with $\operatorname{mult}_p( f_3)\geq 2$. A rough upper bound will be given by the second Betti number of the normalization of a general such $X$. $\endgroup$
    – abx
    Commented Dec 3, 2018 at 9:12
  • $\begingroup$ Is there any way to calculate the second Betti number of the normalization of a general such $X$ ? $\endgroup$
    – user130022
    Commented Dec 3, 2018 at 9:28
  • $\begingroup$ in the previous example why $p$ is isolated ? it could be the case that $p$ lies in another plane curve on another plane $H_2$ and $X$ is singular along that curve also. For example if $f$ is of the form: $xg^2 + yz^2w^2$, then $f_3 =yz^2$ and such a point $p$ lies in the plane curve $g$ in the plane given by $z= 0$ and $X$ is also singular along that curve. $\endgroup$
    – user130022
    Commented Dec 3, 2018 at 12:11

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Consider the blowup $BL_C\mathbb{P}^3$ of $\mathbb{P}^3$ along the conic $$ C = \{ w = g = 0 \} \subset \mathbb{P}^3. $$ Let $H$ be the pullback of the class of a hyperplane in $\mathbb{P}^3$ and $E \subset Bl_C\mathbb{P}^3$ is the exceptional divisor. Then the linear system of quintics in $\mathbb{P}^3$ singular along $C$ identifies with the linear system $|5H - 2E|$ on the blowup. Furthermore, a general divisor in $|5H - 2E|$ is the blowup of such a quintic along $C$, i.e., its normalization. Thus, your question is about Betti numbers of a general divisor in $Bl_C\mathbb{P}^3$ from the linear system $|5H - 2E|$.

Next, note that $$ Bl_C\mathbb{P}^3 \cong Bl_PQ^3, $$ the blowup of a 3-dimensional quadric at a point. Furthermore, if $H'$ and $E'$ are the preimage of the hyperplane class of $Q^3$, and the exceptional divisor of the econd vblowup, then $$ H = H' - E',\qquad E = H' - 2E'. $$ Therefore, $$ 5H - 2E = 5(H' - E') - 2(H' - 2E') = 3H' - E', $$ hence the required surface is the strict transform of a general cubic hypersurface through $P$. This means it is isomorphic to the blowup at point of $Y$, the intersection of a quadric and cubic in $\mathbb{P}^4$. The Betti numbers of $Y$ are well known, these are $$ b_0(Y) = b_2(Y) = b_4(Y) = b_6(Y) = 1,\qquad b_3(Y) = 40, $$ hence the Betti numbers are $$ b_0 = b_6 = 1,\qquad b_2 = b_4 = 2, \qquad b_3 = 40. $$

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  • $\begingroup$ Thank you very much for the answers. $\endgroup$
    – user130022
    Commented Dec 3, 2018 at 12:22

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