# The algebra associated to SO(1,1)

The special orthogonal group $SO(2)$ sits inside the space of $2\times 2$ real matrices, and by taking the closure under scalar multiplication we obtain the set of all matrices of the form

$$\begin{array}{r}\left[\begin{array}{cc}a& -b\\ b& a\end{array}\right].\end{array}$$ This set of matrices forms an $\mathbb{R}$-algebra, and it is isomorphic to $\mathbb{C}$ by having the above matrix correspond to $a+bi$.In Math 54, we are covering quadratic forms right now, and I wondered what this sort of construction would give for $SO(1,1)$. Let’s set $q(x)={x}_{1}^{2}-{x}_{2}^{2}$, so then a $2\times 2$ matrix $A$ is in $SO(1,1)$ if $detA=1$ and $q(Ax)=q(x)$ for all $x\in {\mathbb{R}}^{2}$. Let’s just solve for what form $A$ must have, since there’s a certain diagram I want to draw for later. Suppose

$$\begin{array}{r}A=\left[\begin{array}{cc}a& c\\ b& d\end{array}\right].\end{array}$$ Plugging in ${e}_{1}$ to $q(Ax)=q(x)$ we get ${a}^{2}-{b}^{2}=1$, and plugging in ${e}_{2}$ we get ${c}^{2}-{d}^{2}=-1$. The quadratic form has a corresponding bilinear form $\u27e8x,y\u27e9={x}_{1}{y}_{1}-{x}_{2}{y}_{2}$ satisfying $\u27e8x,x\u27e9=q(x)$, and one can show $\u27e8Ax,Ay\u27e9=\u27e8x,y\u27e9$. Hence, $\u27e8A{e}_{1},A{e}_{2}\u27e9=ac-bd$ and $\u27e8{e}_{1},{e}_{2}\u27e9=0$, giving an additional equation. Geometrically speaking, this is that the columns of $A$ are orthogonal (via the dot product) after reflecting the second vector over the ${e}_{1}$-axis. The following plot shows the hyperbolae ${x}^{2}-{y}^{2}=1$ and ${x}^{2}-{y}^{2}=-1$ which respectively contain the columns of $A$ along with the two points on the second hyperbola that are orthogonal to $A{e}_{1}$.Now let’s consider the corresponding algebra, which is from taking all scalar multiples of these matrices. The effect of this is to take all matrices of the form

$$\begin{array}{r}\left[\begin{array}{cc}a& b\\ b& a\end{array}\right]\end{array}$$ with no constraints on $a$ and $b$. (If I hadn’t done the above analysis, I would have been surprised to learn that these matrices form a commutative subalgebra!)The only two-dimensional algebras over $\mathbb{R}$ are $\mathbb{R}\oplus \mathbb{R}$, $\mathbb{R}[x]/({x}^{2})$, and $\mathbb{C}$, and it seems like it should be the first. (It can’t have nilpotent elements since the square of the above matrix has ${a}^{2}+{b}^{2}$ in the $(1,1)$ entry, and for a similar reason it can’t be $\mathbb{C}$ since it lacks periodic elements.) How do we decompose it, then?

It’s easy enough to solve for a set of minimal orthogonal idempotents. They turn out to be the orthogonal projections onto $y=x$ and $y=-x$:

$$\begin{array}{rlrl}{\pi}_{1}& ={\textstyle \frac{1}{2}}\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]& {\pi}_{2}& ={\textstyle \frac{1}{2}}\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right]\end{array}$$ These certainly have the property that ${\pi}_{1}+{\pi}_{2}=\mathrm{id}$ and ${\pi}_{1}{\pi}_{2}=0$ (while also being well outside of $SO(1,1)$ proper!). Thus, we have a decomposition $$\begin{array}{rl}\left[\begin{array}{cc}a& b\\ b& a\end{array}\right]& ={\pi}_{1}\left[\begin{array}{cc}a& b\\ b& a\end{array}\right]+{\pi}_{2}\left[\begin{array}{cc}a& b\\ b& a\end{array}\right]=(a+b){\pi}_{1}+(a-b){\pi}_{2}\end{array}$$ and furthermore an isomorphism to $\mathbb{R}\oplus \mathbb{R}$ by sending such a matrix to $(a+b,a-b)$.A consequence to this is that we get a sort of geometric rule for multiplication of matrices in $SO(1,1)$.
We regard a matrix in $SO(1,1)$ as being a point on the hyperbola ${x}^{2}-{y}^{2}=1$.
**Step 1:** take each point and project them onto the lines $y=x$ and $y=-x$.
**Step 2:** think of these two lines as being $\mathbb{R}$ (with $({\textstyle \frac{1}{2}},{\textstyle \frac{1}{2}})$ and $({\textstyle \frac{1}{2}},-{\textstyle \frac{1}{2}})$ as the units) and multiply the correspond numbers.
**Step 3:** “un-project” the resulting two numbers to form a new point on the hyperbola.
(**Or said more simply**, rotate the plane by $\pi /4$ while scaling by $2$, do component-wise multiplication of the points, then rotate and scale back. The fact that $SO(1,1)$ is closed under multiplication from this point of view is that if you have pairs $(c,d)$ and $(e,f)$ with $cd=1$ and $ef=1$, then $(ce,df)$ has $cedf=1$.)

For example, since points of the positive sheet of the hyperbola have the parameterization $(\mathrm{cosh}t,\mathrm{sinh}t)$, then under the assumption that this is supposed to be a homomorphism $\mathbb{R}\to SO(1,1)$ with $t$ behaving like an angle, we can recover some identities. Set $a=(\mathrm{cosh}t+\mathrm{sinh}t)(\mathrm{cosh}u+\mathrm{sinh}u)$ and $b=(\mathrm{cosh}t-\mathrm{sinh}t)(\mathrm{cosh}u-\mathrm{sinh}u)$. Then $\mathrm{cosh}(t+u)={\textstyle \frac{1}{2}}(a+b)$ and $\mathrm{sinh}(t+u)={\textstyle \frac{1}{2}}(a-b)$. That is,

$$\begin{array}{rl}\mathrm{cosh}(t+u)& =\mathrm{cosh}t\mathrm{cosh}u+\mathrm{sinh}t\mathrm{sinh}u\\ \mathrm{sinh}(t+u)& =\mathrm{cosh}t\mathrm{sinh}u+\mathrm{sinh}t\mathrm{cosh}u\end{array}$$Something to notice for the positive hyperbola is that the projections onto the two lines, as numbers relative to $({\textstyle \frac{1}{2}},{\textstyle \frac{1}{2}})$ and $({\textstyle \frac{1}{2}},-{\textstyle \frac{1}{2}})$, are always greater than or equal to $1$. By taking natural logarithms, one can add instead of multiplying for the composition law. This corresponds to $(\mathrm{cosh}t,\mathrm{sinh}t)=({\textstyle \frac{1}{2}}({e}^{t}+{e}^{-t}),{\textstyle \frac{1}{2}}({e}^{t}-{e}^{-t}))$ (Note: the base of the exponentials are not fixed by this observation.)

## As a Clifford algebra

Another way to write $\mathbb{R}\oplus \mathbb{R}$ is as $\mathbb{R}[x]/({x}^{2}-1)$, where the isomorphism from $\mathbb{R}[x]({x}^{2}-1)$ is from $[p]\mapsto (p(1),p(-1))$ with $p\in \mathbb{R}[x]$. The corresponding isomorphism from the above algebra to $\mathbb{R}[x]/({x}^{2}-1)$ is given by

$$\begin{array}{rl}\left[\begin{array}{cc}a& b\\ b& a\end{array}\right]& \mapsto a+bx\end{array}$$ so $x$ corresponds to the permutation matrix of order $2$. Hence, the algebra is isomorphic to the Clifford algebra on a 1-dimensional space with a positive-definite quadratic form. (These are also known as the split-complex numbers.)## SO(2,1)

Although it seemed unlikely, I wondered whether $SO(2,1)$ might give an interesting subalgebra in the space of $3\times 3$ matrices. It turns out the space is spanned by $SO(2,1)$, so the answer is no.

It wasn’t necessary for this, but through a similar process I worked out a parameterization of the component of $SO(2,1)$ containing the identity:

$$\begin{array}{r}\left[\begin{array}{ccc}\mathrm{cos}\alpha \mathrm{cos}\theta \mathrm{cosh}t-\mathrm{sin}\alpha \mathrm{sin}\theta & -\mathrm{sin}\alpha \mathrm{cos}\theta -\mathrm{cos}\alpha \mathrm{sin}\theta \mathrm{cosh}t& \mathrm{cos}\alpha \mathrm{sinh}t\\ \mathrm{sin}\alpha \mathrm{cos}\theta \mathrm{cosh}t+\mathrm{cos}\alpha \mathrm{sin}\theta & \mathrm{cos}\alpha \mathrm{cos}\theta -\mathrm{sin}\alpha \mathrm{sin}\theta \mathrm{cosh}t& \mathrm{sin}\alpha \mathrm{sinh}t\\ \mathrm{cos}\theta \mathrm{sinh}t& -\mathrm{sin}\theta \mathrm{sinh}t& \mathrm{cosh}t\end{array}\right].\end{array}$$ This factorizes as $$\begin{array}{r}\left[\begin{array}{ccc}\mathrm{cos}\alpha & -\mathrm{sin}\alpha & 0\\ \mathrm{sin}\alpha & \mathrm{cos}\alpha & 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}\mathrm{cosh}t& 0& \mathrm{sinh}t\\ 0& 1& 0\\ \mathrm{sinh}t& 0& \mathrm{cosh}t\end{array}\right]\left[\begin{array}{ccc}\mathrm{cos}\theta & -\mathrm{sin}\theta & 0\\ \mathrm{sin}\theta & \mathrm{cos}\theta & 0\\ 0& 0& 1\end{array}\right].\end{array}$$