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This is a linear algebra question, but it's about an article about Minkowski spacetime, so I think it's appropriate to post it here. The article is The rich structure of Minkowski space by Domenico Giulini. The detail I'm asking about is at the top of page 16.
The article is describing the 3+1-dimensional version of the "nothing but relativity" argument that's been discussed here in a few threads recently. (The idea is to prove that the group of functions that make a coordinate change from one global inertial coordinate system to another, is either the group of Galilean boosts or the Lorentz group. So at the start, we do not assume that spacetime is Minkowski spacetime. We just assume that spacetime is some structure with underlying set ℝ4).
The article assumes that the group has two subgroups, one corresponding to rotations, and one corresponding to boosts. The rotations are 4×4 matrices
$$R(D)=\begin{pmatrix}1 & 0\\ 0 & D\end{pmatrix},$$ where the zeroes are a 3×1 matrix and a 1×3 matrix, and D is a member of SO(3). The boosts can be expressed as a function of velocity, and the relationship between boosts and rotations is assumed to be
$$B(Dv)=R(D)B(v)R(D^{-1}).$$ Now the author claims that by chosing v to be a multiple of e1, and D to be an arbitrary rotation around the 1 axis, we can see that
$$B(v)=\begin{pmatrix}A & 0\\ 0 & \alpha I\end{pmatrix},$$ where A is a 2×2 matrix, I is the 2×2 identity matrix, and ##\alpha## is a real number. This result looks wrong to me. I want to know if I'm missing something. So here's my argument:
First write
$$B(v)=\begin{pmatrix}K & L\\ M & N\end{pmatrix}.$$ We have
$$R(D)=\begin{pmatrix}I & 0\\ 0 & D'\end{pmatrix},$$ where I is the 2×2 identity matrix and D is a member of SO(2). So $$R(D^{-1})=\begin{pmatrix}I & 0\\ 0 & D'^{-1}\end{pmatrix}$$ and
\begin{align}\begin{pmatrix}K & L\\ M & N\end{pmatrix} &=B(v)=B(Dv)=R(D)B(v)R(D^{-1})\\
&=\begin{pmatrix}I & 0\\ 0 & D'\end{pmatrix}\begin{pmatrix}K & L\\ M & N\end{pmatrix} \begin{pmatrix}I & 0\\ 0 & D'^{-1}\end{pmatrix} =\begin{pmatrix}K & LD'^{-1}\\ D'M & D'ND'^{-1}\end{pmatrix}
\end{align} Now it's easy to see that L=M=0. For example, we have M=D'M for all D' in SO(2), and if we e.g. choose D' to be a rotation by ##\pi/2##, we can easily see that M=0.
However, the same choice of D in the equation for N yields that N is of the form
$$N=\begin{pmatrix}a & b\\ -b & a\end{pmatrix},$$ and this is a number times a member of SO(2), but that member doesn't have to be the identity. And I don't think that there's a way to get b=0 by choosing another D', because ##D'ND'^{-1}## will just be (a number times) a product of 3 rotations by angles θ,λ,-θ that add up to λ, and this turns the equation ##N=D'ND'^{-1}## into N=N, which tells us nothing.
The article is describing the 3+1-dimensional version of the "nothing but relativity" argument that's been discussed here in a few threads recently. (The idea is to prove that the group of functions that make a coordinate change from one global inertial coordinate system to another, is either the group of Galilean boosts or the Lorentz group. So at the start, we do not assume that spacetime is Minkowski spacetime. We just assume that spacetime is some structure with underlying set ℝ4).
The article assumes that the group has two subgroups, one corresponding to rotations, and one corresponding to boosts. The rotations are 4×4 matrices
$$R(D)=\begin{pmatrix}1 & 0\\ 0 & D\end{pmatrix},$$ where the zeroes are a 3×1 matrix and a 1×3 matrix, and D is a member of SO(3). The boosts can be expressed as a function of velocity, and the relationship between boosts and rotations is assumed to be
$$B(Dv)=R(D)B(v)R(D^{-1}).$$ Now the author claims that by chosing v to be a multiple of e1, and D to be an arbitrary rotation around the 1 axis, we can see that
$$B(v)=\begin{pmatrix}A & 0\\ 0 & \alpha I\end{pmatrix},$$ where A is a 2×2 matrix, I is the 2×2 identity matrix, and ##\alpha## is a real number. This result looks wrong to me. I want to know if I'm missing something. So here's my argument:
First write
$$B(v)=\begin{pmatrix}K & L\\ M & N\end{pmatrix}.$$ We have
$$R(D)=\begin{pmatrix}I & 0\\ 0 & D'\end{pmatrix},$$ where I is the 2×2 identity matrix and D is a member of SO(2). So $$R(D^{-1})=\begin{pmatrix}I & 0\\ 0 & D'^{-1}\end{pmatrix}$$ and
\begin{align}\begin{pmatrix}K & L\\ M & N\end{pmatrix} &=B(v)=B(Dv)=R(D)B(v)R(D^{-1})\\
&=\begin{pmatrix}I & 0\\ 0 & D'\end{pmatrix}\begin{pmatrix}K & L\\ M & N\end{pmatrix} \begin{pmatrix}I & 0\\ 0 & D'^{-1}\end{pmatrix} =\begin{pmatrix}K & LD'^{-1}\\ D'M & D'ND'^{-1}\end{pmatrix}
\end{align} Now it's easy to see that L=M=0. For example, we have M=D'M for all D' in SO(2), and if we e.g. choose D' to be a rotation by ##\pi/2##, we can easily see that M=0.
However, the same choice of D in the equation for N yields that N is of the form
$$N=\begin{pmatrix}a & b\\ -b & a\end{pmatrix},$$ and this is a number times a member of SO(2), but that member doesn't have to be the identity. And I don't think that there's a way to get b=0 by choosing another D', because ##D'ND'^{-1}## will just be (a number times) a product of 3 rotations by angles θ,λ,-θ that add up to λ, and this turns the equation ##N=D'ND'^{-1}## into N=N, which tells us nothing.