Orthogonal group, indefinite orthogonal group, orthochronous stuff

This post appears in the Linear Algebra and Special Relativity courses.

There are several ways to see that the matrices satisfying $A^*A=I$ are related to rotations in some way, other than just expanding out the components like a dumb pygmy chimp -- no, we are the normal chimp:
  1. Write it as $A^TIA=I$ -- i.e. the set of matrices that preserve the identity quadratic form. The identity quadratic form corresponds to the $n$-sphere (e.g. a circle), so we're looking for transformations that preserve the $n$-sphere. A clearer way to see this is that preserving the quadratic form $I$ is equivalent to preserving the valuation $x^TIy$ for all $x, y$, i.e. $(Ax)^TI(Ay)=x^Ty$, so it preserves the value of each contour.
  2. With the same logic as above, $(Ax)^T(Ay)=x^Ty$, i.e. the preservation of the Euclidean dot product means that all lengths and angles are preserved. These are called "rigid rotations", and are basically the kind of stuff we can do to a sheet of paper without compressing or stretching it in any way -- i.e. if we nudge a vector by a certain angle, every other vector should also be nudged by the same angle.
What kind of transformations preserve the unit sphere? 

The reason this is a good way of understanding things is that there are plenty of other such "dot products" you can define in mathematics, corresponding to different geometries -- each can be based on the bilinear form it preserves, see this later linear algebra article for more details, relating to isomorphisms of such geometries etc.

As for discriminating between rotations and reflections, suppose we define rotations in a completely geometric way -- for a matrix to be a rotation, all its eigenvalues are either 1 or in pairs of unit complex conjugates.

What do the eigenvalues of orthogonal matrices look like? For each eigenvalue, you need $\overline{\lambda}\lambda=1$, i.e. all the eigenvalues are unit complex numbers. If a complex eigenvalue isn't paired with a corresponding conjugate, you will not get a real-valued transformation on $\mathbb{R}^n$. Meanwhile if an eigenvalue of -1 isn't paired with another -1 -- i.e. if there are an odd number of reflections -- you get a reflection. In this sense, the "conjugate eigenvalues" property of rotations can be seen as a generalisation of the "$s_1s_2=r$" property which you may have learned from plane geometry or dihedral groups. The orthogonal (or rather unitary) transformations that do not behave this way are precisely the rotations.

The similarity between unpaired unit complex eigenvalues and unpaired -1's is interesting, by the way -- when thinking about reflections, you might have gotten the idea that reflections are $\pi$-angle rotations in a higher-dimensional space -- like the vector was rotated through a higher-dimensional space and then landed on its reflection -- like it was a discrete snapshot of a process as smooth as any rotation.

Well, now you know what this higher-dimensional space is -- precisely $\mathbb{C}^n$. And the determinant of a unitary matrix also takes a continuous spectrum -- the entire unit circle. In this sense (among other senses) complex linear algebra is more "complete" than real linear algebra. In fact, you will see in Lie theory that the group $SO(n)$ is connected but $O(n)$ is not, while $SU(n)$ and $U(n)$ are both connected. Can you see why?

(original version of above originally posted to math stackexchange)

Well, here, we benefited from the fact that the product of two reflections is a rotation -- so we could just enforce the "even number of flips", i.e. that $\det A=1$, to specify rotations. But what if we're dealing with one of the "generalised geometries" we discussed? What if instead of preserving $I$, we wanted the group $O(m\mid n)$, i.e. that preserves some $\mathrm{diag}(m\mid n)$ with $m$ 1's and $n$ -1's along the diagonal?

Well, then we don't have rotations between the "1"-labeled (spatial) axes and the "-1"-labeled (temporal) axes, only boosts. But compositions between such reflections form rotations! So simply restricting that $\det A = 1$ will -- while still forming a group $SO(m \mid n)$ -- retain all these rotations which can only be understood as compositions of reflections.

So how do we extract the transformations we want? (What transformations do we want? The ones that correspond to changes of reference frame, in special relativity language -- well, in the sense of Lie theory, this means we're looking for the "component connected to the identity" -- do you see why?)

Let's think about this more clearly. Start by noting that not all reflections in spacetime preserve the Minkowski metric $\mathrm{diag}(m\mid n)$ -- only those that preserve the invariant hyperboloids. In the case of 3+1-spacetime, this means infinite spatial reflections and one time-reversal -- in the case of a general $m+n$-spacetime, this means infinite spatial reflections and infinite temporal reflections (in any $m+n-1$-plane whose normal vector is temporal, not to be confused with time-like). When you multiply an odd temporal reflection with an odd spatial reflection, you get an even time-space rotation, which is in $SO(3\mid 1)$.

$$A = \left[ {\begin{array}{*{20}{c}}{{A_t}}&B^T\\C&{{A_s}}\end{array}} \right]$$
(Note on notation: we'll use ${A_T} = \left[ {\begin{array}{*{20}{c}}{{A_t}}&0\\0&I\end{array}} \right]$ and analogously ${A_S} = \left[ {\begin{array}{*{20}{c}}I&0\\0&{{A_s}}\end{array}} \right]$, where $A_t$ and $A_T$ are "basically the same thing", and analogously for $A_s$ and $A_S$ -- in particular $\det A_t=\det A_T$ and $\det A_s=\det A_S$.)

We see the problem: instead of just mandating $\det A=1$, we must mandate that the temporal minor and the spatial minor of the matrix both have determinant 1, $\det A_t=\det A_s = 1$. But this isn't right -- if you have a boost, i.e. some mixing between the space and time co-ordinates, then $A\ne A_TA_S$ and the component determinants are multiplied by a Lorentz factor (even though still $\det A = 1$). So we mandate instead that $\det A_t>0$, $\det A_s>0$ (equivalently $\ge 1$). Such transformations are called the proper orthochronous Lorentz transformations, because in the context of special relativity they are proper Lorentz transformations that do not flip time:

$$SO^{+}(3 \mid 1)=\{A\in O(3 \mid 1) \mid \det A_t >0, \det A_s >0\}$$
OK, how do we show $SO^{+}(m\mid n)$ is a subgroup? You might get the notion that because of the "two sheets hyperbola" topology of the group, the sheet connected to the identity must be a subgroup (and the other sheet a coset) because moving about on the sheet keeps you on the sheet (and that's what group multiplication is -- moving about on the sheet). The formal way to say this is to say that the map $A\mapsto \mathrm{sgn}(\det A_t )$ is a group homomorphism to the cyclic group $\{1,-1\}$, so its kernel is necessarily a normal subgroup (do you see how these are the same thing?).

So the key is to prove that for two matrices satisfying $\mathrm{sgn}(\det A_t )>0$, their product does too. A proof of the $SO^+(m\mid 1)$ case (relevant for relativity) can be found here -- I'm not sure how that proof can be appropriately generalised to $SO^+(m\mid n)$. I've written out the first few steps here:
  1. Multiply the two matrices $A$ and $\tilde{A}$ to show $(A\tilde{A})_t=A_t\tilde{A}_t+B^T\tilde{C}$. We want to show the determinant of this is positive.
  2. From multiplying out $A^T\eta A=\eta$ and $A\eta A^T=\eta$, we see that $A_t^2-C^TC=A_t^2-B^TB=I$ and analogous for $\tilde{A}$.
  3. So $\det((A\tilde{A})_t-A_t\tilde{A}_t)=\det(B^T\tilde{C})=\sqrt{\det(A_t^2-I)\det(\tilde{A}_t^2-I)}$
  4. Well, I'm not sure how to proceed at this point. Does $\det(X-PQ)=\det((P^2-I)(Q^2-I))^{1/2}$ imply that $\det P\ge1\land\det Q\ge1\Rightarrow \det X>0$?
Well, I can't think of a way to continue -- and certainly one can think of a much wider category of problems like this, where we have a much simpler topological picture in our heads than rubbish algebra like the above would betray. So we need a topological way of looking at Lie groups.

You might think of just considering something like the orbit of a vector -- e.g. the unit time vector -- under the group for the topology, but this does not fully describe the topology of the group. As an illustration, in the above example, for $n>1$, the orbit of the time vector under $O^+(m\mid n)$ is actually connected (prove this -- you need to count the number of sheets a general hyperbola has), while the entire topology of the group is actually disconnected, as we will see. A simple way to see that these are two different topologies is that spatial rotations/reflections leave the unit time vector unchanged and therefore all correspond to a single point on the orbit.

This will be our starting point to motivate the study of the topology of a Lie group in the Lie theory articles.

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