# Faulty calculation in Schubert calculus: the number of lines on the intersection of two planes.

As a easy test of the correctness of the results of Schubert calculus I tried to solve the following problem with the symbolic method.

Q: How many lines are there lying on the intersection of two planes in $$\mathbb{P}^3$$?

The correct answer is one, the intersection of two planes is a line. However in the following calculation I seem to find that there are none. Note that the lines on a plane $$H$$ are represented by $$\Omega(L_i, H)\subseteq G_{1,3}$$ for any line $$L_i \subseteq H$$. Therefore we must calculate the degree of $$\Omega(1,2)^2$$. Giambelli's formula yields that \begin{align*}\Omega(1,2) &= \det \begin{pmatrix} \sigma(1) & \sigma(0)\\ \sigma(2) & \sigma(1) \end{pmatrix} \\ &= \sigma(1)^2 - \sigma(0)\sigma(2) \end{align*} Hence, by Pierri's formula \begin{align*} \Omega(1,2)^2 &= \Omega(1,2)\sigma(1)^2 - \Omega(1,2)\sigma(0)\sigma(2)\\ &= \Omega(0,2)\sigma(1) - \Omega(0,1)\sigma(2)\\ &= \Omega(0,1) - \Omega(0,1)\\ &= 0 \end{align*} Suggesting that $$\deg \Omega(1,2)^2 = 0$$.

What am I doing wrong here?

Note: I am using the notation of Kleiman and Laksov's paper on the subject.

The faulty step is using $$\Omega(1,2)\sigma(0)=\Omega(0,1).$$ Pieri's formula gives $$\Omega(1,2)\sigma(0)=\sum\Omega(b_0,b_1)$$ where the sum is over integers $$b_0,b_1$$ such that $$0\leq b_0\leq 1 There are no such choices of $$b_0,b_1,$$ so $$\Omega(1,2)\sigma(0)=0.$$