# Quadric preserved by the $SO(3)$ action on $\mathbb{P}^6$

Disclaimer: this question is a -more preicse- version of this one Orbits of $SO(3)$, where thanks to the help of a user I realize I din't write things down precisely. This is a more serious attempt.

We work over the field of complex numbers. We define $$SO(3)$$ as $$SO(3)=\{A\in Gl(3,\mathbb{C}\mid A^t Q A=Q,\text{ }\det(A)=1\},$$ where $$Q$$ is the $$3\times 3$$-matrix $$\begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\\ \end{pmatrix}$$ that is the bilinear form associated to the quadric $$C$$ define as $$x_0x_2+x_1^2=0$$.

Consider the action of $$SO(3)$$ on $$\mathbb{P}^2(\mathbb{R})$$, with homogeneous coordinates $$x_0,x_1,x_2$$, of the form $$SO(3)\times\mathbb{P}^2\to \mathbb{P}^2$$ $$(A,p)\mapsto Ap$$ I have to prove that, given a point $$p\in C$$, the orbit of $$p$$ is $$C$$, that is $$SO(3)p\simeq C$$. In order to do so, since $$SO(3)/SO(3)_p\simeq SO(3)p$$, I have to prove that the quotient $$SO(3)/SO(3)_p\simeq C$$ For doing so I considered for simplicity the point $$p=(1:0:0)$$, and I've found that $$SO(3)_p=\{A\in SO(3)\mid \text{the first column of A is equal to p}\}.$$ In order to conclude, I should show that given a point $$y\in C$$, there exists a matrix $$B\in SO(3)$$ such that $$Bp=y$$, i.e. the first column of $$B$$ is equal to $$y$$. Unfortunately now I'm stuck, because I don't know how to create a matrix of determinant $$1$$ out of simply a column $$y$$.

Finally, I've a very (dumb) question: in order to do so, I considered a point $$p$$ belonging to the quadric; if I choose $$p\not\in C$$, I can consider the orbit $$SO(3)p$$: in that case $$SO(3)p=\mathbb{P}^2\setminus C$$? Thanks in advance.

• Your title mentions $\Bbb P^6$; is that exponent an error? – John Hughes Aug 29 '20 at 13:26

Answer in development -- not yet complete

Ah. This makes much more sense, although I find it a little odd to call that particular thing $$SO(3)$$, which is a little like deciding that from now on, we're going to use the symbol $$8$$ to denote the successor of the integer we usually denote by $$22$$. Anyhow, letting that go, almost exactly the same argument as before works. I want to define a new symbol, $$\odot$$, by saying that for vectors $$v$$ and $$w$$ in 3-space, $$v \odot w = v^t Q w.$$

Now noting that $$Qw$$ is just $$w$$ with its first and third entries swapped, it's pretty easy to write this down explicitly: $$\pmatrix{a\\b\\c} \odot \pmatrix{u\\v\\w} = aw + bv + cu.$$ This gives me a result that I'll use over and over: if vectors $$\alpha$$ and $$\beta$$ are orthogonal (i.e., $$\alpha \cdot \beta = 0$$), then $$\alpha$$ and $$\beta'$$, where $$\beta'$$ is just $$\beta$$ with its first and third entries swapped, are in fact $$\odot$$-orthogonal, i.e., $$\alpha \odot \beta' = 0.$$

Let's suppose that $$A = \pmatrix{a\\b\\c}$$ is a point of your curve $$C$$, so that $$A \odot A = 0$$. We'd like to find a matrix $$M \in SO(3)$$ with $$A$$ (or some scalar multiple of $$A$$) as its first column.

Letting $$U = \pmatrix{u\\v\\w}$$, and $$R = \pmatrix{r\\s\\t}$$ denote the second and third columns, that means finding numbers $$u,v,w,r,s,t$$ such that \begin{align} A \odot A &= 0 & A \odot U &= 0 & A \odot R &= 1 \\ & & U \odot U &= 1 & U \odot R &= 0 \\ & & & & R \odot R &= 0 \\ \end{align} where I've left out the other three products because of symmetry. The good news is that we have six equalities to satisfy, and six free variables. Actually, we have a seventh: we can multiply $$A$$ by any constant and still have the same point of the curve $$C$$, so for the first row, for instance, doing so won't change $$A \odot U = 0$$, but it can be used to adjust $$A \odot R$$ from "some nonzero number" to $$1$$.

Now let's specialize a little bit: I'm going to assume that $$b \ne 0$$. Then the equation of $$C$$, namely $$xz + y^2 = 0$$ tells us that both $$a$$ and $$c$$ are nonzero. The remaining cases, where $$b = 0$$, are $$\pmatrix{0\\0\\1}, \pmatrix{1\\0\\0}.$$ These can be solved by hand, which I leave to you as an instructive exercise. I'll call those "exceptional" points, and the other points of $$C$$ (those with $$b \ne 0$$) the "good" points, just to have a name.

Having restricted to $$b \ne 0$$, we can write all possible good points in the form $$\pmatrix{a\\b\\-b^2/a}$$, or equivalently (up to scale) in the form $$\pmatrix{a^2 \\ ab \\ -b^2}.$$

I want to address finding $$R$$ first, because it seems to be harder. We need $$R \odot R = 0$$, so $$R$$ must be a good point, and $$A \odot R = 1$$, a linear constraint on $$R$$. Now for $$R$$ to be a good point, some multiple of it must have the form $$\pmatrix{u^2 \\ uv \\ -v^2},$$ and then $$A \odot R = 1$$ becomes \begin{align} 1 &= -a^2v^2 + abuv -b^2 u^2\\ -1 &= (av)^2 - (av)(bu) + (bu)^2\\ -1 &= (av - bu)^2 + (av)(bu)\\ \end{align}

Abandoned for now

SCRATCH WORK follows.

Now pick $$U_0 = \pmatrix{a\\0\\-c}$$

The other observation is that if we're working sequentially, there's not a lot of constraint on $$U$$ initially --- it has to be $$\odot$$-orthogonal to $$A$$, and have $$\odot$$-squared-length $$1$$. So we can just pick ANYTHING that's $$\odot$$-orthogonal, and then adjust its length.

Anyhow, let's get moving. The vector $$A$$ is nonzero, so we can pick some unit vector $$\alpha$$ such that $$A \cdot \alpha = 0$$. (My answer to your prior question gives one method, using a gram-schmidt-like technique.) A typical method might be to take any two entries of $$A$$, at least one nonzero, swap them and negate one, and set the third entry to $$0$$, and call that new vector $$\beta$$; then you observe that $$A \cdot \beta = 0$$. And then you let $$\alpha = \beta / \| \beta \|$$ to get yourself a unit vector in that direction. Anyhow, ANY unit vector $$\alpha$$ perpendicular to $$A$$ will suffice. Now let $$U_0 = Q \alpha,$$ i.e., let $$U$$ be $$\alpha$$ with its first and third entries swapped. At this point, we have $$A \cdot \alpha = 0$$, so we also know that $$A \odot U_0 = 0$$. We've fixed up that $$(1,2)$$ entry in our system of equations.

What about $$U_0 \odot U_0 = 1$$? That might be true, or it might not. The case $$U_0 \odot U_0 = 0$$ is a special one; let's assume that's not true (i.e., that we picked $$U_0$$ wisely, or got lucky, or something. In that case, let $$U_0 \odot U_0 = d \ne 0$$, and picking either square root, let $$U = \frac1{\sqrt{d}} U_0.$$ Then by bilinearity of $$\odot$$, we have $$A \odot U = 0$$ (i.e., our success with the $$(1,2)$$ entry is unchanged), but now we also know that $$U \odot U = 1$$ i.e., we've got the $$(2,2)$$ entry in our system of equations satisfied.

Now we need to find a vector $$R$$ for which $$A \odot R = 1, U \odot R = 0, R \odot R = 0$$.}