- #71

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If X and Y are members of the subgroup that consists of all the matrices of the form

$$\begin{pmatrix}A & B\\ 0 & C\end{pmatrix}$$ with det A>0, then there's an R in SO(3) such that ##XY^{-1}=U(R)##, where the right-hand side is defined by

$$U(R)=\begin{pmatrix}I & 0\\ 0 & R\end{pmatrix}.$$ This implies that if

$$X=\begin{pmatrix}A & B\\ 0 & C\end{pmatrix},\quad Y=\begin{pmatrix}D & E\\ 0 & F\end{pmatrix},\quad V(X)=V(Y),$$ we have

$$X=U(R)Y=\begin{pmatrix}D & E\\ 0 & RF\end{pmatrix}.$$ So two members of this subgroup with the same velocity differ only in the lower right, and there they differ only by multiplication of a member of SO(2).

Since for all R, ##V(U(R)X)=V(X)=V(XU(R))##, this implies that the following statements are true:

$$\begin{pmatrix}a & \pm b\\ b & \pm a\end{pmatrix}$$ and it turns out that three of the four possible sign combinations can be ruled out by the observation that an SO(2) matrix acting from the left on a 2×2 matrix doesn't change the inner product of the columns. So the final result is that there exist numbers a,b such that

$$C=\begin{pmatrix}a & -b\\ b & a\end{pmatrix}.$$ The columns (and the rows) are orthogonal and have the same norm. So if we define ##k=\sqrt{a^2+b^2}##, there's an R in SO(2) such that C=kR. This means that the X that we started with is of the form

$$\begin{pmatrix}A & 0\\ 0 & kR\end{pmatrix}$$ where k is a real number and R is a member of SO(2). I don't see a way to prove that k=0 right now, but I have only just started to think about it.

It seems impossible to me to prove that the group contains a member with velocity v for each v with |v|<c. If I'm right, it's a pretty big flaw, and the theorem would have be repaired by adding an assumption like my "0 is an interior point of V(G)". Then we would have to go through the same sort of stuff I did in my pdf for the 1+1-dimensional case.

$$\begin{pmatrix}A & B\\ 0 & C\end{pmatrix}$$ with det A>0, then there's an R in SO(3) such that ##XY^{-1}=U(R)##, where the right-hand side is defined by

$$U(R)=\begin{pmatrix}I & 0\\ 0 & R\end{pmatrix}.$$ This implies that if

$$X=\begin{pmatrix}A & B\\ 0 & C\end{pmatrix},\quad Y=\begin{pmatrix}D & E\\ 0 & F\end{pmatrix},\quad V(X)=V(Y),$$ we have

$$X=U(R)Y=\begin{pmatrix}D & E\\ 0 & RF\end{pmatrix}.$$ So two members of this subgroup with the same velocity differ only in the lower right, and there they differ only by multiplication of a member of SO(2).

Since for all R, ##V(U(R)X)=V(X)=V(XU(R))##, this implies that the following statements are true:

- For all R in SO(2), we have BR=B.
- For all R in SO(2), there's an R' in SO(2) such that ##RC=CR'##.
- For all R' in SO(2), there's an R in SO(2) such that ##RC=CR'##.

^{2}). An SO(2) matrix acting on a 2×2 matrix from the right doesn't change the norm of the rows. These observations and the results 2-3 above imply that ##a^2+c^2=b^2+d^2## and ##a^2+b^2=c^2+d^2##. These results imply that ##a^2=d^2## and ##b^2=c^2##. So C is of the form$$\begin{pmatrix}a & \pm b\\ b & \pm a\end{pmatrix}$$ and it turns out that three of the four possible sign combinations can be ruled out by the observation that an SO(2) matrix acting from the left on a 2×2 matrix doesn't change the inner product of the columns. So the final result is that there exist numbers a,b such that

$$C=\begin{pmatrix}a & -b\\ b & a\end{pmatrix}.$$ The columns (and the rows) are orthogonal and have the same norm. So if we define ##k=\sqrt{a^2+b^2}##, there's an R in SO(2) such that C=kR. This means that the X that we started with is of the form

$$\begin{pmatrix}A & 0\\ 0 & kR\end{pmatrix}$$ where k is a real number and R is a member of SO(2). I don't see a way to prove that k=0 right now, but I have only just started to think about it.

It seems impossible to me to prove that the group contains a member with velocity v for each v with |v|<c. If I'm right, it's a pretty big flaw, and the theorem would have be repaired by adding an assumption like my "0 is an interior point of V(G)". Then we would have to go through the same sort of stuff I did in my pdf for the 1+1-dimensional case.

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