Skip to main content
7 events
when toggle format what by license comment
Feb 24, 2019 at 19:40 comment added IMeasy Yes, thanks, I looked at that and it is very well explained. Actually the invariants are much simpler than I expected.
Feb 18, 2019 at 21:38 comment added Jason Starr I found a reference in Igor Dolgachev's book, "Lectures on Invariant Theory". The computation is precisely Section 11.3, "Lines in $\mathbb{P}^3$". Dolgachev proves that the geometric quotient is isomorphic to the surface $\mathbb{P}^2$. He writes the three generators of the ring of invariants explicitly. Each of these $3$ generators corresponds to a partition of the set of $4$ lines into two subsets of size $2$.
Feb 18, 2019 at 15:17 comment added Jason Starr My parameter count is $2$. Please note, there is a $1$-dimensional stabilizer of a generic $4$-tuple of lines. If we write our $4$ dimensional vector space $V=H^0(\mathbb{P}^3,\mathcal{O}(1))^\vee$ as $V=W\oplus W$, then we can consider the lines that are $\mathbb{P}(W\oplus\{0\})$, $\mathbb{P}(\{0\}\oplus W)$, $\mathbb{P}\Delta(W)$, and $\mathbb{P}\Gamma_g(W)$ where $\Delta$ and $\Gamma_g$ are the respective graphs of $\text{Id}_W:W\to W$ and a linear automorphism $g:W\to W$. For every $h$ in the centralizer of $g$ in $\textbf{PGL}_2$, also $h\times h:V\to V$ fixes all $4$ lines.
Feb 18, 2019 at 14:00 comment added IMeasy @Jason, Yes sorry the lines are ordered as usual. But by naive dimension count the quotient should have dimension 1, am I wrong?
Feb 18, 2019 at 12:48 comment added Jason Starr In my previous comment I forgot about the centralizer of a regular element of $\mathbf{PGL}_2$. So the correct quotient is the quotient of $\widehat{\mathbf{PGL}}_2$ by this centralizer (I need to think about the minimal generators for the ring of invariants).
Feb 18, 2019 at 12:24 comment added Jason Starr Are the $4$ lines ordered? If so, it appears to me that the geometric quotient is simply $\widehat{\textbf{PGL}}_2 \cong \mathbb{P}^3$. I realize that you are asking for the formulas for the $4$ explicit generators of the homogeneous coordinate ring in terms of Pluecker coordinates on the $4$-fold product of the Grassmannian. However, it helps to know that there are $4$ generators (and not some other number).
Feb 18, 2019 at 10:06 history asked IMeasy CC BY-SA 4.0