2021年11月16日 · Complex plane pairs quadrics have eigenvalues with the same sign like diag([1,1,0,0]) which is just the z-axis for real points. The solution to this polynomial system may be simpler if the quadrics are already rank deficient (e.g. cones), so an initial discovery of cone quadrics in the pencil using simple eigenvalue decomposition could simplify ...

For the real cases (1, 2, 4) I'm fairly convinced of my result: given two quadrics from the same equivalence class, I could always find an affine transformation to transform one into the other. For the complex cases (3, 5) I was not as sure, since intuition fails me. So in those cases you might want a more reliable proof.

2023年1月18日 · $\begingroup$ @JosephHarrison Also, note that i’m studying computer science. From what I’ve seen, in maths you (obviously) study geometry much more in depth and you study quadrics as surfaces with a lot of geometry, so I can’t really understand much of the stuff that I find online on quadrics because they are really “geometric-imprinted”.

2016年8月6日 · First, you should decide whether or not your quadrics are projective. Also, you should be asking which field you are working over. For example $$ \textbf{Proj}\left( \frac{\mathbb{R}[x,y,z,w]}{x^2 + y^2 + z^2 + w^2} \right) $$ has no solutions, hence it's just the empty variety which cannot intersect any other quadrics in $\mathbb{P}^3(\mathbb ...

How to understand (maybe, informally) why the intersection of two quadrics in general position in $\mathbb{CP}^3$ is an elliptic curve? It is obvious that it is a compact 2-manifold, i.e. a sphere with handles, but how to find its genus? I suppose one can apply Mayer-Vietoris sequence, but the union of two quadrics seems to be hard to realise.

2020年5月31日 · quadrics; Share. Cite. Follow edited May 31, 2020 at 23:46. altugkarakurt. 749 4 4 silver badges 11 11 ...

2024年7月4日 · My question is, given two quadrics of this form, how do I reverse this map, i.e. how can I show that every intersection of this form corresponds to a unique (up to hometheticity) lattice, and subsequently an ellitpic curve over the complex numbers?

2016年6月17日 · And since a projective transformation (which is an invertible linear transformation of the containing space) maps quadrics to quadrics, I see no benefit in adding the orthogonality requirement. Particularly since normalization to length $1$ fits in poorly with the arbitrary choice of representatives you have with homogeneous coordinates.

2015年6月26日 · quadrics; Share. Cite. Follow asked Jun 25, 2015 at 17:50. user46348 user46348. 935 4 4 silver badges 14 ...

2015年4月23日 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.