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Let $g\ge 0$, $d\ge 1$ be integers. We consider the Moduli space $H_{g,d}$ parametrising smooth irreducible closed curves $C\subset \mathbb{P}^3$ of degree $d$ and genus $g$. Let us denote by $U_{g,d}$ the subset consisting of curves $C$ such that $\{f\in \mathrm{Aut}(\mathbb{P}^3)\mid f(C)=C\}$ is trivial.

For which $g,d$ is $U_{g,d}$ a dense open subset of $H_{g,d}$ ?

I have the impression that $U_{g,d}$ is always open in $H_{g,d}$. I would for this consider $X=\{(f,C)\mid f\in \mathrm{Aut}(\mathbb{P}^3, C\in H_{g,d}\mid f(C)=C\}$, and look at the projection to $H_{g,d}$. The fibres with at least two elements should be closed I guess (unless the ones with one element in the fibre come with multiplicity...).

For $g=0$ and $d\in \{1,2,3\}$, then $U_{g,d}$ is simply empty, as all automorphisms of lines, conics and twisted cubics extend to $\mathbb{P}^3$. Similarly, $U_{1,3}$ is also empty as plane cubics always have some automorphisms extending (finitely many). Is $U_{g,d}$ a dense open subset of $H_{g,d}$ for the others $(g,d)$?

Edit: As pointed by Robert Bryant, all curves contained in a plane gives $U_{g,d}=\varnothing$. One should then forget the cases where $g=\frac{(d-1)(d-2)}{2}$, so $(0,1),(0,2),(1,3),(3,4), (6,5), (10,6),\ldots$

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    $\begingroup$ Well, you'll also have $U_{3,4}$ is empty because Castelnuovo's bound implies that an irreducible space curve of degree $4$ and genus $3$ must lie in a $\mathbb{P}^2\subset\mathbb{P}^3$ (and be smooth), so there will always be nontrivial automorphisms of $\mathbb{P}^3$ that fix the $\mathbb{P}^2$ pointwise. I think you need to at least restrict to non-degenerate curves. $\endgroup$ Nov 8, 2021 at 10:38
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    $\begingroup$ $U_{1,4}$ also is empty: quartic elliptic curves have nontrivial automorphisms. $\endgroup$
    – abx
    Nov 8, 2021 at 10:44
  • $\begingroup$ Thanks for the comments. @abx where is the automorphism coming from? $\endgroup$ Nov 8, 2021 at 11:09
  • $\begingroup$ I think that a "rigorous" argument for the openness could proceed as follows (although it's a bit overkill): Let $X$ be the quotient stack $\mathcal{X}:=[H_{g,d}/ \mathrm{PGL}_4]$. To show that $U_{g,d}$ is open it suffices to show that it is the pull-back of an open $U$ in $\mathcal{X}$. By (include reference to stacks project), the substack $U$ of $\mathcal{X}$ of points with trivial inertia group is open in $\mathcal{X}$. (Short argument: It is the locus over which the inertia stack is an isomorphism.) Note that $U_{g,d}$ is the pull-back of $U$. $\endgroup$ Nov 8, 2021 at 12:36
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    $\begingroup$ I think what @abx is alluding to is that a space curve of genus 1 and degree 4 is the intersection of two quadrics that can be simultaneously diagonalized, i.e., $${X_0}^2+{X_1}^2+{X_2}^2+{X_3}^2 =a_0 {X_0}^2+a_1{X_1}^2+a_2{X_2}^2+a_3{X_3}^2=0,$$ and the maps $(X_i)\mapsto(\epsilon_iX_i)$ with $\epsilon_i = \pm1$ (and not all equal) are then nontrivial automorphisms of the curve, since they fix the quadrics. $\endgroup$ Nov 8, 2021 at 14:27

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