When does a conic represents a pair of planes?

Given a conic equation $$H:= ax^2+by^2+cz^2+2fyz+2gzx+2hyx=0,$$

What is (are) the conditions on the coefficients that it represents a pair of planes?

I know that if this is true, then $H$ can be written as $$H=(px+qy+rz)(sx+ty+uz)$$ for some constants $p, q, r, s,t, u$. Expanding gives

$$ps x^2 + qt y^2 + ruz^2 + (pt+qs)xy+ (pu+ rs) xz+ (qu+rt) yz = H,$$

which gives us a system of $6$ equations \begin{align} ps &=a \\ qt &= b \\ ru &= c \\ pt+qs &= 2f \\ pu + rs &= 2g \\ qu+ rt &= 2h. \end{align}

The algebra gets a little bit too messy for me and I do not know how to move on. In the solution, it is given that $$f^2 \geq bc, \ \ g^2 \geq ac, \ \ h^2 \geq ab$$

and the determinant of some minor of \begin{pmatrix} a & h & g \\ h & b & f \\ g & f & c \end{pmatrix} are zero.

• Any thoughts of your own?
– amd
Dec 20 '17 at 0:01
• Actually, I am ambiguous. In a text, it was given that some determinant is zero and $f^2\geq bc$, $g^2\geq ac$ and $h^2\geq ab$. But I think these conditions $f^2\geq bc$, $g^2\geq ac$ and $h^2\geq ab$. are obvious. Dec 20 '17 at 0:05
• Here's a small hint: Why don't you start with the equation of a pair of planes and see what that will look like? How do you get a pair of planes? Start with an easy case, like the $xy$-plane and the $xz$-plane. Dec 20 '17 at 0:06
• I know the solution part, but are the conditions $f^2\geq bc$, $g^2\geq ac$ and $h^2\geq ab$ necessary? Dec 20 '17 at 0:08

We can confirm that

$$f^2 \geq bc, \ \ g^2 \geq ac, \ \ h^2 \geq ab$$

and the determinant of a certain matrix equal to $0$ are necessary conditions.

Here is how. The quadratic form can be written under a matrix form as follows :

$$\tag{1}\begin{bmatrix} x&y&z \end{bmatrix}\underbrace{\begin{bmatrix} a&h&g\\h&b&f\\g&f&c \end{bmatrix}}_Q\begin{bmatrix} x\\y\\z\\ \end{bmatrix}$$

If the quadratic form can be expressed as the product of two linear forms (when we will equal it to zero, we will get the equations of the two planes passing through the origin):

$$2(px+qy+rz)(sx+ty+uz).$$

The equivalent matrix form of this expression is (many thanks to @Will Jagy) :

$$\begin{bmatrix} x&y&z \end{bmatrix}\begin{bmatrix} p\\q\\r\\ \end{bmatrix}\begin{bmatrix} s&t&u \end{bmatrix}\begin{bmatrix} x\\y\\z\\ \end{bmatrix}+\begin{bmatrix} x&y&z \end{bmatrix}\begin{bmatrix} s\\t\\u\\ \end{bmatrix}\begin{bmatrix} p&q&r \end{bmatrix}\begin{bmatrix} x\\y\\z\\ \end{bmatrix}=$$

$$\tag{2}\begin{bmatrix} x&y&z \end{bmatrix}\underbrace{\begin{bmatrix} 2ps&(pt+qs)&(pu+rs)\\(pt+qs)&2qt&(qu+rt)\\(pu+rs)&(qu+rt)&2ru\end{bmatrix}}_Q \begin{bmatrix} x\\y\\z\\ \end{bmatrix}$$

And yes, comparing (1) and (2), we check that necessarily, for example the "diagonal minor":

$$\begin{vmatrix} b&f\\f&c \end{vmatrix}=bc-f^2=4qurt-(qu+rt)^2=-(qu-rt)^2 \leq 0$$

the same for the other expressions...

Remark : As we can write :

$$Q=\begin{bmatrix} p&s\\q&t\\r&u \end{bmatrix} \begin{bmatrix} s&t&u\\p&q&r \end{bmatrix},$$

the rank of $Q$ is at most 2.

Thus, $det(Q)=0$.

The rank of $Q$ falls to $1$ iff $\begin{bmatrix} p\\q\\r \end{bmatrix}$ and $\begin{bmatrix} s\\t\\u \end{bmatrix}$ are proportional (the case of a double plane).

• Jean, I got the symmetric matrix form of the product of two linears was $P S^T + S P^T,$ where $P$ and $S$ are yor column vectors of coefficients. In particular, symmetric matrix was rank two when $P,S$ not parallel. Dec 20 '17 at 2:05
• @Will Jagy I must leave now, but I will think a little more about your expression $PS^T+SP^T$... Dec 20 '17 at 2:27
• @Will Jagy One thing is sure, we haven't to consider the (degenerate) case of parallel planes because all planes here must contain the origin. Dec 20 '17 at 2:37
• Looks good to me. I upvoted a few minutes ago. It must be pretty late where you live. Dec 20 '17 at 3:36
• Very interesting. I scratch my head : this case of "two planes degeneracy" allows to interpret the matrix of our quadratic form as a certain Hessian with the form $PS^T+SP^T$, which, besides, has eigenvalues $(0, \lambda_1 \geq 0, \lambda_2 \leq 0)$ (I haven't a real proof, it is only based on extensive simulations). Does this constitutes a characterization of this case ? Hum, hum... Dec 20 '17 at 23:09

This is a quadratic form, which is associated with the symmetric matrix $$Q_H=\begin{bmatrix} a&h&g\\h&b&f\\g&f&c \end{bmatrix}$$ and the quadric is degenerate if and only if the discriminant of $H$, i.e. $\det Q_H$ is $0$.

• (ctd...) or example taking $a=h=1, b=2$ and all other coefficients set to $0$, you have $(x+y)^2+y^2$. Setting this expression to zero, you get a degenerate case which is a line (with equations $y=-x=0$), but not a pair of planes. Dec 20 '17 at 1:52
• I am not the downvoter ! I am fully aware that in this kind of situations, there are subtle differences and degeneracies of one kind that aren't degeneracies of another kind... Dec 20 '17 at 2:02
• I was completly wrong about rank=1 !!! (I have erased it because it would mislead a future reader, but I recognize my error). Thanks to a remark of @Will Jagy I have written a new version. I was right only on one thing: the remark about subtleties of this kind of situations... Dec 20 '17 at 3:27
• @Jean Marie: I focused only on the degeneracy criterion, not the type of degeneracy. Dec 20 '17 at 9:40
• That's right. And there are other degeneracy cases than those that have been considered. Regards. Dec 20 '17 at 21:07