I am unfortunately unable to solve the following exercise:
Let $\Lambda_1,\ \Lambda_2$ be two-planes in $\mathbb{P}^4$ meeting in one point $p$, and let $C_i\subset \Lambda_i$ be conic curves. Concretely: \begin{equation*} \Lambda_1: W_0=W_1=0,\ \ \Lambda_2: W_3=W_4=0\end{equation*}
- Let $C_1$ and $C_2$ be disjoint given by the equations $C_1: W_2^2=W_3W_4,\ C_2: W_0W_1=W_2^2$. Show that in this case the join $J(C_1,C_2)$ is a quartic hypersurface.
- Now look at $C_1: W_4^2=W_2W_3,\ C_2: W_1W_2=W_0^2$ i.e. non-disjoint conic curves. Show that now the join $J(C_1,C_2)$ becomes a cubic hypersurface.
Here is a link to Harris' book for the definitions of joins etc. http://userpage.fu-berlin.de/aconstant/Alg2/Bib/Harris_AlgebraicGeometry.pdf, p. 88,89.
Here is my approach: We have defined the join to be the image of $j: X\times Y -->\mathbb{G}(1,n),\ ([v],[w])\mapsto [v\wedge w]$. In this notation, I have implicitly used the Pluecker Embedding. My idea is to essentially look at what happens with the Pluecker Relations: \begin{equation*} W_{12}W_{34}-W_{13}W_{24}+ W_{14}W_{23}=0 \end{equation*} However, when I plug in $v_1=v_2=1,\ w_4=w_5=0$ for the two rows in the matrix, I simply get 0=0 which is definitely not a quartic hypersurface. How do I get this hypersurface?
Any help would be very greatly appreciated :).