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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.

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  • $\begingroup$ Your title mentions $\Bbb P^6$; is that exponent an error? $\endgroup$ – John Hughes Aug 29 '20 at 13:26
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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 $A \odot R = 1, U \odot R = 0, R \odot R = 0$.}

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