# Direct image of the structure sheaf by an endomorphism of $\mathbb{P}^2$

Let us take 3 quadratic forms on $$\mathbb{P}^2$$ with no common zero; they define a map $$\pi : \mathbb{P}^2\rightarrow \mathbb{P}^2$$ of degree 4. It is not difficult to see that $$\pi _*\mathscr{O}_{\mathbb{P}^2}\cong \mathscr{O}_{\mathbb{P}^2}\oplus \mathscr{O}_{\mathbb{P}^2}(-1)^3$$. Does anyone know how to write down the algebra stucture of $$\pi _*\mathscr{O}_{\mathbb{P}^2}$$ in terms of this decomposition?

Here's a partial answer for a particular case. I suspect you probably already knew this. (So it is really just a comment, but it is too long for the comment box.)

If the polynomials are $$x_0^2, x_1^2, x_2^2$$, and the characteristic is not 2, then in affine coordinates the cover $$\pi$$ is $$y_1=x_1^2, y_2=x_2^2$$. So it is Galois, with Galois group $$G=(\mathbb{Z}/2)^2$$. You can then break up $$\pi_*\mathcal{O}_{\mathbb{P}^2}$$ using 4 characters of $$G$$, which I'll label $$1, \sigma_1,\sigma_2,\sigma_1\sigma_2$$. The last 3 corresponds to the involutions $$(x_1\mapsto -x_1)$$, $$( x_2\mapsto -x_2)$$ and, I'm guessing $$(x_0\mapsto -x_0)$$. The invariant part of $$\pi_*\mathcal{O}$$ is just $$\mathcal{O}$$, and the remaining isotypic factors are isomorphic to $$\mathcal{O}(-1)$$. Under the identification $$\Gamma((\pi_*\mathcal{O})(1))\cong \Gamma(\mathcal{O}(2))$$ coming from the projection formula, the sections coming from the $$\mathcal{O}(-1)$$ factors correspond to $$x_1^2,x_2^2, x_0^2$$ (modulo the guess above). Let me label the factors accordingly by $$1,2,0$$. You presumably want the multiplication table. I'm guessing, for instance, $$\mathcal{O}(-1)_{i}\otimes \mathcal{O}(-1)_{i}\to \mathcal{O}$$ is $$\mathcal{O}(-1)_{i}\otimes \mathcal{O}(-1)_{i}\cong \mathcal{O}(-2)\stackrel{x_i^2}{\to} \mathcal{O}$$ Etc.

• Thanks, Donu. Yes, I was more or less aware of this particular Galois case. I am intrigued by the general case... – abx Jun 20 at 16:46
• I imagine you can do something similar if only one of the quadratic forms is a square. – Piotr Achinger Jun 22 at 17:47

In general, if $$f \colon X \to Y$$ is a quadruple cover of algebraic varieties, the structure of the algebra $$f_* \mathcal{O}_X$$ is studied in the paper

D. Hahn, R. Miranda: Quadruple Covers in Algebraic Geometry, Journal of Algebraic Geometry 8 (1999).

You can download the paper from the webpage of the second author.

• Thank you @Francesco Polizzi, but I am not sure this helps for my problem. According to this paper, my $\pi$ corresponds to a certain section of $\bigwedge^2 \mathsf{S}^2 \mathscr{0}_{\mathbb{P}^2}(1)^3\otimes \mathscr{0}_{\mathbb{P}^2}(-3)$, that is, $\mathscr{0}_{\mathbb{P}^2}(1)^{15}$. How do I find such a section starting from 3 quadratic forms $P,Q,R$ in 3 variables? – abx Jun 24 at 6:01