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 $$0\leq b_0\leq 1<b_1\leq 2,\qquad b_0+b_1=1$$ There are no such choices of $b_0,b_1,$ so $$\Omega(1,2)\sigma(0)=0.$$


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