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I am working on the following question from an old exam:

Give the equation of a conic in the real projective plane with the homogeneous coordinates $\{x, y, z\}$ such that it is singular in $[1, i, 1]$ and $[i, -1, i]$. Let $L \subset \mathbb{R}[x, y, z]_2$ be the vector space of the conics with this property. What is its dimension?

My solution: $[1, i, 1]$ and $[i, -1, i]$ are the same point because $(i, -1, i) = i \cdot (1, i, 1)$. The following double line is singular in all its points and particularly in $[1, i, 1]$: $(x - z)^2 = 0$.

Now let $\mathcal C : Ax^2 + By^2 + Cz^2 + Dxy + Exz + Fyz = 0$ be a general conic in the real projective plane. Then $$ \mathcal C_x = 2Ax + Dy + Ez = 0 \\ \mathcal C_y = 2Bx + Dx + Fz = 0 \\ \mathcal C_z = 2Cz + Ex + Fy = 0 $$ are three linearly independent restrictions on the parameters $A, \dots, F$, so $\dim L = 6 - 3 = 3$.

Am I correct with this? Is there another (maybe more intuitive) way to see this?

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You are correct in that the space of possible parameter combinations has dimension $3$.

Personally I'd not say that the space of conics has dimension $3$, though. That's because multiples of a given equation represent the same conic. In general (i.e. barring some specific exceptions) you can normalize your equation somehow, e.g. make the sum of all coefficients equal to $1$ (unless it is equal to $0$) without changing the conic. Thus one of your degrees of freedom is without geometric meaning.

I'm actually no sure which of these answers your question would want, since the notation $\mathbb R[x,y,z]_2$ doesn't imply an identification of scalar multiples. So relying on that notation, your answer of $3$ would be correct even though I find this viewpoint highly ungeometric.

If you know how to classify conics, you may know that a degenerate conic factors into a pair of lines, and that the point of intersection is the singular point your question mentions. So intuitively speaking you are looking for the space of all pairs of lines through a given point. Now the space of lines through a given point is a topological circle, and as such one-dimensional. Combining two of these leads to a two-dimensional space of possible solutions.

You can use this to parametrize that space. Pick any two distinct lines through the given point, e.g. $x-z=0$ which you represent as $[1:0:-1]$ and $x+iy=0$ which you represent as $[1:i:0]$. Then the pencil of lines through that point would consist of all linear combinations of these two:

$$l_{[a:b]}=a[1:0:-1]+b[1:i:0]$$

Then you could write the conic as

$$\begin{bmatrix}2A&D&E\\D&2B&F\\E&F&2C\end{bmatrix}= l_{[a:b]}^{\phantom T}\cdot l_{[c:d]}^T + l_{[c:d]}^{\phantom T}\cdot l_{[a:b]}^T$$

Here $l_{[a:b]}\cdot l_{[c:d]}^T$ means you take two elements of the family of lines, use one as a row vector and the other as a column vector and multiply them to obtain a rank $1$ matrix. You then add the transpose of that to obtain a symmetric matrix which in general will have rank $2$. You see four real parameters describing this family of conics, but they form two vectors of homogeneous coordinates so the whole thing has merely two dimensions.

In case you were wondering how these four real parameters relate to the dimension $3$ you found: If you multiply $a$ and $b$ by some $\lambda\neq 0$ and at the same time divide $c$ and $d$ by that $\lambda$, you get the same matrix. So there is a one-dimensional space of parameters which will result in the same equation, just like there is a one-dimensional space of equations which describe the same conic.

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  • $\begingroup$ Sorry for the late reply and thank you very much. This is a very clear and helpful answer! $\endgroup$ – bbrot Feb 11 '17 at 12:34

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