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Let $(L, 0) \subset \mathbb C^4$ be a germ of Lagrangian surface with isolated singularity at $0$ ($\mathbb C^4$ equipped with the standard symplectic form $\omega = \sum dx_i \wedge dy_i$). What are the known examples in which $L$ has a normal singularity at $0$?

It is stated by Claus Hertling, Frobenius manifolds and moduli spaces for singularities, Cambridge Tracts in Mathematics, 151, Cambridge: Cambridge University Press, pp. ix+270 (2002), MR1924259, Zbl 1023.14018, that such a singularity exists (chapter 14.2), but no explicit example is presented there. I know only one example: we can take $L$ to be an affine cone over a suitably chosen twisted cubic $C$ in $\mathbb P^3$.

Edit: as was suggested, one could possibly construct other examples of normal Lagrangian affine cones. To achieve this, however, one needs a projectively normal Legendrian curve in $\mathbb P^3$ that is not a twisted cubic. The question of whether the twisted cubic is the only linearly normal Legendrian curve in $\mathbb P^3$ was raised by Wahl J. Wahl, "Introduction to Gaussian maps on an algebraic curve", in: Complex Projective Geometry: Selected Papers, London Math. Soc. Lecture Note Ser., vol. 179, Cambridge Univ. Press, 1992, pp. 304–323, §5.

Let $C \subset \mathbb P^3$ be a linearly normal Legendrian curve of degree $d$ and genus $g$, distinct from a twisted cubic.

Several constraints are known:

  • Wahl proved that $C$ cannot lie on a smooth quadric or a smooth cubic surface.
  • Degree of $C$ must satisfy $d \geq 7$. This follows from the classification results in Quo-Shin Chi and Xiaokang Mo, "The moduli space of branched superminimal surfaces...", Osaka J. Math. 33 (1996), no. 3, 669–696. Specifically:
    1. By Theorem 1 in §4, there are no Legendrian curves with $g \geq 1$ and $d \leq 5$.
    2. By Theorem 2 in §4, any Legendrian curve with $g \geq 1$ and $d = 6$ must be hyperelliptic.
    3. Theorem 3 in §4 implies that for a degree $6$ map $u\colon C' \to \mathbb P^3$ with $g(C') \geq 3$ and $u(C')$ Legendrian, the image $u(C')$ is a cubic curve; hence $u$ cannot be an embedding.
    4. The remaining cases with $1 \leq g \leq 2$ and $d = 6$ are excluded by Riemann--Roch.
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  • $\begingroup$ There should be many more cones, as in your construction. The symplectic form induces a foliation in projective $3$- space of rank $2$ (corank $1$). Every integral curve will give such a cone. $\endgroup$ Commented Dec 11 at 13:17
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    $\begingroup$ Yes, but such a cone will not, in general, be normal. If $D\subset \mathbb P^n$ is a variety, then $Cone(D)$ is normal iff $D$ is projectively normal. My question is motivated by the following conjecture: if $D \subset \mathbb P^3$ is a smooth linearly normal Legendrian curve, then $D$ must be a twisted cubic. The foliation corresponding to the symplectic form defines a contact structure on $\mathbb P^3$. Then $D\subset \mathbb P^3$ is Legendrian iff $Cone(D)$ is Lagrangian. So to get a new example, one would need a projectively normal Legendrian curve which is not a twisted cubic. $\endgroup$ Commented Dec 11 at 14:01
  • $\begingroup$ You can deform multiple covers of your twisted cubic. A general deformation of a double cover of a twisted cubic by a genus $3$ curve is a sextic curve of genus $3$. A general such curve in $\mathbb{P}^3$ is projectively normal. By my computation, the local deformation space for integral "stable maps" is unobstructed of dimension $1$ greater than the space of double covers of integral twisted cubics. So a generic deformation as an integral curves is an embedded curves of degree $6$ and genus $3$ (I have not yet checked whether these are contained in quadric surfaces). $\endgroup$ Commented Dec 11 at 17:24

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I am posting my comments as one answer.

Edit. The poster is correct, I was wrong. I was studying infinitesimal deformations in a leaf of a foliation, but the poster is asking about deformations that remain tangent, not deformations in a leaf of a foliation. Of course that main difference is that the foliation need not be involutive, so the infinitesimal deformation theory in a leaf is the wrong notion in that setting. Sorry for the mistake, and thanks to the poster for (patiently) explaining what he wants. I am leaving the original (wrong) answer below.

Original post. A smooth, projective curve of genus $3$ and degree $6$ in $\mathbb{P}^3$ is linearly nondegenerate. By Halphen's theorem, the curve is projectively normal unless it is contained in a quadric surface as a curve of bidegree $(4,2)$, in which case it is hyperelliptic. So to prove that there are projectively normal curves of genus $3$ and degree $6$ that are integral to the foliation, it suffices to prove that for double covers of integral twisted cubic curves by genus $3$ curves, a sufficiently general deformation as an integral curve is an embedded, non-hyperelliptic curve.

For an integral twisted cubic curve, $$\nu:\mathbb{P}^1\hookrightarrow \mathbb{P}^3,$$ the pullback $\nu^*\mathcal{O}(1)$ is isomorphic to $\mathcal{O}(3)$. The pullback $\nu^*\mathcal{F}$ of the foliation is isomorphic to $\mathcal{O}(3)\oplus \mathcal{O}(3)$. For a double cover by a genus $3$ curve, $$f:C\to \mathbb{P}^1,$$ the pushforward $f_*\mathcal{O}_C$ is isomorphic to $\mathcal{O}\oplus \mathcal{O}(-4)$. Thus, $f_*\mathcal{O}_C \otimes_{\mathcal{O}_{\mathbb{P}^1}} \nu^*\mathcal{F}$ is isomorphic to $\mathcal{O}(3)^{\oplus 2} \oplus \mathcal{O}(-1)^{\oplus 2}$. Notice that this has vanishing $h^1$. This implies smoothness of the $1$-morphism to the moduli space of genus $3$ curves from the stack of stable maps to $\mathbb{P}^3$ that have genus $3$ domain, that have degree $6$ and that are integral to the foliation. In particular, a sufficiently general deformation of $f$ as an integral curve is non-hyperelliptic. Thus, it is also an embedded curve (else it would be contained in a quadric surface, hence hyperelliptic).

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  • $\begingroup$ It seems that for a map $u\colon C \to \mathbb P^3$ which is integral w.r.t. $\mathcal F$ the space $H^0(C, u^* \mathcal F)$ does not parametrise local deformations of $u$ as an integral map. For example, $h^0(\mathbb P^1, \nu^*\mathcal F) = 8$, but the space of degree $3$ integral maps from $\mathbb P^1$ to $\mathbb P^3$ has dimension $10$ (the dimension of $\mathrm{Sp}(4, \mathbb C)$). Apologies for the delayed response. I've been thinking about your answer and have sent you an email with a follow-up question. $\endgroup$ Commented Dec 16 at 21:25
  • $\begingroup$ I believe you computed the dimension incorrectly. $\endgroup$ Commented 2 days ago
  • $\begingroup$ The group $\mathrm{Sp}(4, \mathbb C)$ acts free and transitively on the space of integral degree $3$ maps from $\mathbb P^1$ to $\mathbb P^3$ since the group $\mathrm{PGL}(4, \mathbb C)$ acts free and transitively on the space of all degree $3$ maps from $\mathbb P^1$ to $\mathbb P^3$. To be sure, the dimension of the space of integral degree $d$ maps from $\mathbb P^1$ to $\mathbb P^3$ is proven to be equal to $2d+4$ (see, e.g., Bonaventure Loo, "The Space of Harmonic Maps of $S^2$ into $S^4$", Trans. Amer. Math. Soc. 313 (1989), 81–102.) $\endgroup$ Commented 2 days ago
  • $\begingroup$ I believe there might be an inaccuracy. It is proved in Quo-Shin Chi and Xiaokang Mo, "The moduli space of branched superminimal surfaces...", Osaka J. Math. 33 (1996), no. 3, 669–696, §4, Prop. 5. that a curve $C$ of genus $3$ admits an integral degree $6$ map whose image is not a line if and only if $C$ is hyperelliptic. $\endgroup$ Commented 2 days ago
  • $\begingroup$ I think the problem is that a first-order deformation that is tangent to $\mathcal F$ is not the same thing as a first-order deformation of an integral map $u\colon C \to \mathbb P^3$ within the class of integral maps. One-parameter subgroup of $\mathrm{Sp}(4, \mathbb C)$ can produce a section of $H^0(C,u^*T_{\mathbb P^3})$ that is not tangent to $\mathcal F$. The space of integral degree $3$ maps from $\mathbb P^1$ to $\mathbb P^3$ has dimension $10$ as it is a torsor over $\mathrm{Sp}(4, \mathbb C)$. $\endgroup$ Commented 2 days ago

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