The algebra associated to SO(1,1)

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

[abba]. This set of matrices forms an R-algebra, and it is isomorphic to 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)=x12x22, so then a 2×2 matrix A is in SO(1,1) if detA=1 and q(Ax)=q(x) for all xR2. Let’s just solve for what form A must have, since there’s a certain diagram I want to draw for later. Suppose

A=[acbd]. Plugging in e1 to q(Ax)=q(x) we get a2b2=1, and plugging in e2 we get c2d2=1. The quadratic form has a corresponding bilinear form x,y=x1y1x2y2 satisfying x,x=q(x), and one can show Ax,Ay=x,y. Hence, Ae1,Ae2=acbd and e1,e2=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 e1-axis. The following plot shows the hyperbolae x2y2=1 and x2y2=1 which respectively contain the columns of A along with the two points on the second hyperbola that are orthogonal to Ae1.
Both options for Ae2 can be obtained from reflections of Ae1 over the lines y=x and y=x. Hence, A must be of the form [a±bb±a] for a2b2=1. One of the options has determinant 1, and the other (the corresponding to b and a) has determinant 1. Hence, SO(1,1) corresponds to symmetric matrices of determinant 1 that are also symmetric across the minor axis.

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

[abba] 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 R are RR, R[x]/(x2), and C, and it seems like it should be the first. (It can’t have nilpotent elements since the square of the above matrix has a2+b2 in the (1,1) entry, and for a similar reason it can’t be 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:

π1=12[1111]π2=12[1111] These certainly have the property that π1+π2=id and π1π2=0 (while also being well outside of SO(1,1) proper!). Thus, we have a decomposition [abba]=π1[abba]+π2[abba]=(a+b)π1+(ab)π2 and furthermore an isomorphism to RR by sending such a matrix to (a+b,ab).

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 x2y2=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 R (with (12,12) and (12,12) 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 π/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 (cosht,sinht), then under the assumption that this is supposed to be a homomorphism RSO(1,1) with t behaving like an angle, we can recover some identities. Set a=(cosht+sinht)(coshu+sinhu) and b=(coshtsinht)(coshusinhu). Then cosh(t+u)=12(a+b) and sinh(t+u)=12(ab). That is,


Something to notice for the positive hyperbola is that the projections onto the two lines, as numbers relative to (12,12) and (12,12), 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 (cosht,sinht)=(12(et+et),12(etet)) (Note: the base of the exponentials are not fixed by this observation.)

As a Clifford algebra

Another way to write RR is as R[x]/(x21), where the isomorphism from R[x](x21) is from [p](p(1),p(1)) with pR[x]. The corresponding isomorphism from the above algebra to R[x]/(x21) is given by

[abba]a+bx 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.)


Although it seemed unlikely, I wondered whether SO(2,1) might give an interesting subalgebra in the space of 3×3 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:

[cosαcosθcoshtsinαsinθsinαcosθcosαsinθcoshtcosαsinhtsinαcosθcosht+cosαsinθcosαcosθsinαsinθcoshtsinαsinhtcosθsinhtsinθsinhtcosht]. This factorizes as [cosαsinα0sinαcosα0001][cosht0sinht010sinht0cosht][cosθsinθ0sinθcosθ0001].