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I am considering the quintic:

$$ x_1^5 +x_2^5 +x_3^5 +x_4^5 +x_5^5 - 5 x_1x_2x_3x_4x_5=0$$

and it is said (Candelas et al.) that

only one singular point which we may as well take to be the point $(1, 1, 1, 1, 1)$. The singularity is a node, that is a point for which $p$ and $dp$ both vanish but the matrix of second derivatives is nonsingular.

But I obtain that the previous statement is false, since the determinant of the Hessian is

$$ -12500 (-255 x_1^3 x_2^3 x_3^3 x_4^3 x_5^3 + 3 x_1^2 x_2^2 x_3^2 x_4^2 x_5^2 (x_1^5 + x_2^5 + x_3^5 + x_4^5 + x_5^5) + 8 x_1 x_2 x_3 x_4 x_5 (x_3^5 x_4^5 + x_3^5 x_5^5 + x_4^5 x_5^5 + x_2^5 (x_3^5 + x_4^5 + x_5^5) + x_1^5 (x_2^5 + x_3^5 + x_4^5 + x_5^5)) + 16 (x_3^5 x_4^5 x_5^5 + x_2^5 (x_4^5 x_5^5 + x_3^5 (x_4^5 + x_5^5)) + x_1^5 (x_4^5 x_5^5 + x_3^5 (x_4^5 + x_5^5) + x_2^5 (x_3^5 + x_4^5 + x_5^5))))$$

which vanishes at the point $(1,1,1,1,1)$.

This does not seem an erratum since it is mentioned in other places. The only solution I can imagine is that we are in fact considering a parameter $\psi$ as a point of the singularity, such that

$$ x_1^5 +x_2^5 +x_3^5 +x_4^5 +x_5^5 - 5 \psi x_1x_2x_3x_4x_5=0$$

and we evaluate it at $\psi=1$, in that case, the Hessian (form deriving wrt. $\psi$ too) is non-singular.

jinawee
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  • I punched this into a CAS and I can confirm that the determinant of the Hessian vanishes at every singularity of that hypersurface. If you vary the constant $5$ by some small amount, the projective hypersurface becomes nonsingular. This doesn't help you understand the quote from that paper, but what you say is correct. I have no idea what they mean. – Jesko Hüttenhain Jun 06 '17 at 07:58
  • The matrix of second derivatives should be computed in local AFFINE coordinates (in particular, it is a $4\times 4$ matrix). – Sasha Jun 07 '17 at 20:29
  • @Sasha Could you elaborate a bit the procedure or give a reference? I have never worked with affine spaces. In the paper they do define a symplectic basis where you have four coordinates only, but I guess it's something different. – jinawee Jun 07 '17 at 21:17
  • Consider the affine chart $x_5 = 1$. Then $x_1,x_2,x_3,x_4$ are affine coordinates. In this chart the equation looks $x_1^5 + x_2^5 + x_3^5 + x_4^5 + 1 - 5x_1x_2x_3x_4 = 0$. Its Hessian (at point $(1,1,1,1)$) is nondegenerate. – Sasha Jun 07 '17 at 21:20

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