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?