I have difficulties understanding a part of a paper. I think it should be kind of easy, but I'm just a lowly physicist trying to understand all of this stuff... be gentle ;)

Let $T_3$ be the blow-up of $\mathbb C \mathbb P^3$ in a point. This is a toric variety which can be described using coordinates $(z_0, z_1, z_2, z_3, v)$ with scaling $$ (z_0, z_1, z_2, z_3, v) \sim (\lambda z_0, \lambda z_1, \lambda z_2, \lambda\mu z_3, \mu v) $$ and a Stanley-Reisner Ideal generated by $z_0 z_1 z_2$ and $z_3 v$. Let $S$ be a hypersurface given as the zero set of a polynomial $P_{3,2}$ of class $$ [ P_{3,2} ] = 3 [z_0] + 2[v] \;, $$ $S$ is in fact the blow-up of a cubic in $\mathbb C \mathbb P^3$.

If $P_{3,2}$ has the form $$ P_{3,2} = z_0 z_1 v\, F_1(z_0 v, z_1 v, z_2 v, z_3) + z_2 z_3\, \tilde F_1(z_0 v, z_1 v, z_2 v, z_3)$$ (where $F_1$ and $\tilde F_1$ are polynomials of degree $1$), then the curves $C_{02} = \{z_0 = 0 = z_2\}$, $C_{v2} = \{v = 0 = z_2\}$ and $C_{13} = \{z_1 = 0 = z_3\}$ are linearly independent algebraic curves (in the homology of $S$, not the one of $T_3$!).
The self-intersection numbers are $C_{02}^2 = -2$, $C_{v2}^2 = -1$, $C_{13}^2 = -1$.

According to the paper, this is obvious because of the analogy to the following (which I understand). A quadric $z_0 z_3 - z_1 z_2$ in $\mathbb C \mathbb P^3$ is a Segre variety, isomorphic to $\mathbb C \mathbb P^1 \times \mathbb C \mathbb P^1$. Explicitly: $$ ( (s_0 : s_1), (t_0 : t_1) ) \mapsto (s_0 t_0 : s_0 t_1 : s_1 t_0 : s_1 t_1) \;. $$ The curves $C_{01} = \{ z_0 = 0 = z_1 \}$ and $C_{02} = \{ z_0 = 0 = z_2 \}$ are linearly independent because they correspond to the (independent) classes $[s_0]$ and $[t_0]$ in $\mathbb C \mathbb P^1 \times \mathbb C \mathbb P^1$.

This totally makes sense to me, I still fail to see why the claim above should be true though.


1 Answer 1


I have found an answer myself, I'll post it for readers from the future.

It is key to calculate the intersection products (in $S$) of the three curves. The easy part is $$ C_{02} \cdot C_{13} = 0, C_{02} \cdot C_{v2} = 1, C_{13} \cdot C_{v2} = 0 $$ (by simply counting points).

For calculating the self-intersection product of a curve $C$ in $S$, we will use the genus formula (see e.g. Hartshorne V.1.5) $$ 2g - 2 = C \cdot (C + K_S) $$ where $g$ is the genus of the curve and $K_S = -c_1(S)$ the canonical class on $S$, by adjunction $K_S = -[z_0] |_S$. As mentioned, $S$ is the blow-up of a cubic in $\mathbb C \mathbb P^3$ - therefore $S$ is a $dP_7$ surface, the blow-up of $\mathbb C \mathbb P^2$ in $7$ generic points. This shows that all $2$-cycles in $S$ are $\mathbb C \mathbb P^1$s with genus $g = 0$.

The intersection product $C \cdot K_S$ can be calculated as $\int_S [C]_S \wedge K_S$ (where $[ \cdot ]_S$ is the Poincaré dual with respect to $S$). Using the definition of Poincaré duality twice, $$ C^2 = -2 + \int_S [C]_S \wedge [z_0] |_S = -2 + \int_C [z_0] |_C = -2 + \int_{T_3} [C] \wedge [z_0]. $$

One can easily show that $\int_{T_3} [z_0]^3 = 0$ and $\int_{T_3} [z_0]^2 [v] = 1$. Using this, we get $$ C_{02}^2 = -2, C_{13}^2 = -1, C_{v2}^2 = -1. $$

This is all we wanted. For example, we can directly read off $C = C_{02} + C_{v2} - C_{13} = 0$ in the homology of $T_3$, but $C$ is non-trivial in the homology of $S$ since $C \cdot C \neq 0$.


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