de Rham cohomology of $x^3+y^3 +z^3+c \,xyz= 0$ find representatives

Question

The equation $x^3+y^3 +z^3+c\, xyz= 0$ defines a non-singular elliptic curve $X$ in $\mathbb{C}P^2$ projective space.

In fact, how do we prove it has genus 1 both as an algebraic curve and as a Riemann surface?

Since $H^1(X,\mathbb{C})\simeq \mathbb{C}^2$ what are the explicit representatives of the de Rham cohomology? There should be two of them.

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This question is (verbatim) on Math.Stackexchange unanswered even with a bounty.

• despite a change of variables, I am guessing it might be best to maintain the symmetry between $x,y,z$ if possible.

• I am asking for explicit sections in a manner that might have been done in the 19th century (but could be done in many situations today). they should correspond to classical functions in some way.

• there's an underlying question of whether the "cohomology" of this curve as a topological space matches up with the sheaf cohomology or de Rham cohomology. Especially when the curve becomes singular.

• I'm also looking for examples where they disagree

Answer 1

A possible method for getting explicit representatives involves Griffiths's Residues. More in general, let $X \subset \mathbb{P}^{n+1}$ a smooth hypersurface of degree $d$. Denote by $H^n_{\textrm{prim}}(X)$ the primitive subspace of the n-th cohomology of $X$, that is $$ H^n_{\textrm{prim}}(X)= \textrm{Ker} ( \lambda: H^n(X) \to H^{n+2}(X) ),$$ $$\lambda(\alpha)= \alpha \cup c_1 (\mathcal{O}_X(1)).$$ The (restriction on) the Hodge filtration $F^p H^n_{\textrm{prim}}(X)$ can be identified with $$H^0(\Omega^{n+1}_{\mathbb{P}}(n-p+1)(X)) / d H^0((\Omega^{n}_{\mathbb{P}}(n-p)(X)).$$

Now, if $$\omega= \sum_i (-1)^i x_i dx_0 \wedge \ldots d\hat{x_i} \wedge \ldots dx_{n+1}$$ write $$H^0(\Omega^{n+1}_{\mathbb{P}}(k))=\big\{ \frac{g \omega}{f^k}\ | \ g \in \mathbb{C}[x_0, \ldots, x_{n+1}]_{kd-n-2} \big \} $$ and use this identification in the quotient above.

One has that if $R_f:= \mathbb{C}[x_0, \ldots, x_{n+1}]/J_f$ with $J_f$ being the ideal generated by the partial derivatives of $f$ then $$H^{n-p,p}_{\textrm{prim}}(X):= H^n_{\textrm{prim}}(X) \cap H^{n-p,p}(X) \cong (R_f)_{(p+1)d-n-2},$$ where the maps sends the class of $g\omega/f^k$ to the class of $g$.

Now this is all a very nice bit of theory, but in this case the situation is very simple. For a curve we have $H^1(C, \mathbb{C}) \cong H^{1,0}(C) \oplus H^{0,1}(C)$ and moreover $H^1_{\textrm{prim}}(C) \cong H^1(C, \mathbb{C})$ (and the same for the subspaces of course). The degree of the equation is three, therefore one has $H^0(K_C) \cong (R_f)_0$ and $H^1(\mathcal{O}_C) \cong (R_f)_3$. The former is generated by 1 (it is canonically identified with $\mathbb{C}$). For the latter, your equation for $C$ implies that you can pick as generator for $(R_f)_3 \cong \mathbb{C}\cong \langle z^3 \rangle$. Note that $R_3$ is the socle of the Jacobian ring, that is $(R_f)_t=0$ for $t>3$. Moreover (in the non weighted case) if $\rho$ is the degree corresponding to the socle of this (artinian, Gorenstein) ring one has $R_a \cong (R_{\rho-a})^*$, and this is in turn just another expression of Serre Duality.

Plugging these polynomials in the quotient above, one gets an explicit expression as desired.

Answer 2

Let me try to answer a few of your questions.

As long as $X$ is smooth (it seems like this is the case you are most interested in anyway), the degree-genus formula $g = (d-1)(d-2)/2$ tells us that $g = 1$, since the degree $d$ is 3. For smooth projective curves, GAGA applies, so $X$ has genus 1 in both the algebraic and analytic sense.

By Hodge theory, $H^1(X,\mathbb{C}) = \mathbb{C}\omega \oplus \mathbb{C} \overline{\omega}$, where $\omega$ is a non-trivial holomorphic differential on $X$. Enrico's answer gives a nice way of finding $\omega$. Alternatively, it is always possible to transform $X$ so that it has affine equation $y^2 = h(x)$ for a polynomial of degree 3 or 4 with distinct roots. In these coordinates, $\omega$ has the expression $\displaystyle \frac{dx}{y}$. Finally, $X$ can be parametrized by $z\rightarrow (\mathcal{p}_{\tau}, \mathcal{p}_{\tau}'(z))$ where $\mathcal{p}_{\tau}(z)$ is the Weierstrass $p$-function and $\tau$ is an element of the upper half-plane determined by the periods of $X$. In these coordinates, $\omega = dz$.

On manifolds, de Rham cohomology agrees with Cech cohomology because the de Rham complex gives a resolution of the constant sheaf. Sheaf cohomology $H^j(X,\mathcal{O}_X)$ doesn't typically agree with $H^j(X,\mathbb{C})$, however. In fact, on a compact Kähler manifold $X$, we have that $H^j(X,\mathcal{O}_X)$ is a direct summand of $H^j(X,\mathbb{C})$ (again by Hodge theory).