# Deformations of the Fermat curve

We work over an algebraically closed field $$k$$, say of characteristic $$0$$, just in case, and we let $$C$$ be a smooth curve over $$k$$. First-order deformations of $$C$$ (or of any smooth variety for that matter) are captured by the cohomology group $$H^1(C,\mathcal{T}_C)$$, where $$\mathcal{T}_C$$ is the tangent sheaf of $$C$$. Via Serre duality, $$\dim H^1(C,\mathcal{T}_C) = \dim H^0(C,\omega_C^{\otimes 2})$$. Now $$\omega_C^{\otimes 2}$$ has degree $$4g - 4$$, which, if $$g \geq 2$$, is large enough to deduce that $$\dim H^0(C,\omega_C^{\otimes 2}) = 3g - 3$$.

I am interested in the case where $$C$$ is the Fermat curve $$C_n$$, cut out by the equation $$x^n + y^n = z^n$$. Through e.g. Riemann--Hurwitz or the theory of Hilbert polynomials, one deduces that this curve has genus $$(n-1)(n-2) / 2$$; thus, if $$n \geq 4$$, we find that $$C_n$$ has a deformation space of dimension $$3n(n-3) /2$$. For instance, the equation $$x^4 + y^4 = z^4$$ must have a $$6$$-dimensional space of deformations.

Can these deformations be seen explicitly?

• Do you want to see explicitly all smooth curves of genus $3$? They are all canonically embedded plane quartics, unless they are hyperelliptic. Commented Nov 18, 2020 at 11:09
• If you are asking whether a general curve of genus $g=\frac{1}{2}(n-1)(n-2)$ can be described explicitely, the answer is no for, say, $g>10$. Note that MO is for questions at research level, please consider using MSE instead.
– abx
Commented Nov 18, 2020 at 11:19
• @abx That's not what I'm asking, I'm asking if there's a good way of describing the deformations of the Fermat curve.
– Jim
Commented Nov 18, 2020 at 13:14
• A general curve of genus $g=\frac{1}{2} (𝑛−1)(𝑛−2)$ is a deformation of the Fermat curve.
– abx
Commented Nov 18, 2020 at 13:28

It might be helpful for you to distinguish embedded deformations of the plane curve $$C\subset{{\mathbb P}_k^2}$$ from general deformations of the projective variety $$C$$. The embedded deformations are all of the form $$\{x^n+y^n-z^n+f(x,y,z)=0\}\subset {{\mathbb P}_k^2}$$ for a general homogeneous polynomial $$f$$ of degree $$n$$ whose coefficients we think of as "small". These are the deformations that are easy to see explicitly. Once you are off this locus, you are essentially asking for the description of the general curve of some genus, and as noted in the comments, this is hopeless to see "explicitly".
The embedded deformations deform the curve $$C$$ in the family of all plane curves of degree $$n$$. This deformation space has dimension $${n+2\choose 2} - 9$$, given by the dimension of the space of polynomials of degree $$n$$ in three variables, less the dimension of GL(3) (or take one off each term if you wish). For $$n=4$$, this number is equal to $$6=3g-3$$, reflecting the fact that every genus $$3$$ curve near the Fermat curve is a plane quartic. So every deformation of the quartic $$C$$ is one of these plane curves. For $$n>4$$, this number gets much smaller than $$3g-3 = 3 {n-1\choose 2} -3$$.
There is one remaining question of some interest to which I don't know the answer, and that is whether there is a particularly natural $$6$$-dimensional family of quartic polynomials that describes all deformations of the Fermat quartic curve. This is asking for an explicit slice of the action of $$GL(3)$$ on the space of quartic polynomials near the Fermat quartic. I don't know if such a family has been written down anywhere in the literature.
• All Fermat curves have very ample canonical bundle I believe, so you have the canonical embedding $C\hookrightarrow {\mathbb P}(H^0(C,\omega_C)^\vee)\cong{\mathbb P}^{g-1}$. All deformations can also be embedded in this same space. But this is not at all "explicit". Commented Nov 18, 2020 at 13:45