I started to learn algebraic geometry by myself not long ago. This question arised from the notes by Andreas Gathmann on Algebraic geometry. I do not understand why we are working in projective space.
In chapter 11, the author introduced the 27 lines on Fermat cubic. But a Fermat cubic is of the form $V(x^3+y^3+z^3-1)$ in $\mathbb{C}^3$. In the notes, the author did not explain the situation in $\mathbb{C}^3$ but in $\mathbb{C}P^3$ instead.
Note that in $\mathbb{C}P^3$ a line corresponds to a 2-dimensional subspace of $\mathbb{C}^4$.
So I think there should be a bijection between the set of lines on $V(x^3+y^3+z^3-1)$ and the set of lines on $V_+(x^3+y^3+z^3+w^3)$. I tried to prove the bijectivity in the following way
Given a line $l$ in $V_+(x^3+y^3+z^3+w^3)$, according to the notes, it can be written in the form \begin{bmatrix} 1 & 0 & 0 & -\omega^j\\ 0 & 1 & -\omega^k & 0 \\ \end{bmatrix} for $\omega$ a primitive 3rd root of unity and $j,k=0,1,2$. We take $j=k=1$ and now the line has this expression $$(s:t:-\omega t:-\omega s), (s:t)\in\mathbb{C}P^1$$ If we let the last coordinate be $-1$, then $-\omega s=-1$ means $s=\omega^2$. The line is $$(\omega^2:t:-\omega t:-1)$$ It satisfy $(\omega^2) ^3+t^3+(-\omega^3 t^3)+(-1)=0$, which is $1+t^3-t^3-1=0$. If we consider the line $$(\omega^2,t,-\omega t)\subseteq\mathbb{C}^3$$ we do have a line on $V(x^3+y^3+z^3-1)$.
On the other hand, if we have a line $l$ on $V(x^3+y^3+z^3-1)$, I do not know how to get a line in $\mathbb{C}P^3$. But I think since we want a 2-dimensional subspace in $\mathbb{C}^4$, so when we think of the set $$\{(x_l,y_l,z_l,w)|w\in\mathbb{C},(x_l,y_l,z_l)\in l\}$$ we should have a plane but it is not a 2-dimensional subspace because it does not pass origin. Then how do we prove the bijection?
Thank you really for helping and reading.



You must log in to answer this question.