Please explain this statement about Lorentz transformations

Question

I'm reading Sternberg's Group Theory and Physics. I have a question about chapter 1.2 Homeomorphisms.

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Background:

A Lorentz Metric is defined as $||{\bf x}||^2=x_0^2-x_1^2-x_2^2-x_3^2$

And a Lorentz Transformation $B$ as one which satisfies $||B{\bf x}||^2=||{\bf x}||^2$

Identify ${\bf x}$ with $x=x_0I+x_1\sigma_1+x_2\sigma_2+x_3\sigma_3$ where $\sigma_i$ are the Pauli matrices.

We have $x=x^*$ ($x$ is self-adjoint) and $\det x=||{\bf x}||^2$

Let $A$ be any 2x2 matrix and define the action of $A$ on $x$ as $x \to AxA^*$ and denote the corresponding action on the vector ${\bf x}$ as $\phi(A){\bf x}$

$AxA^*$ is also self-adjoint and $\det AxA^*=(\det A)^2\det x$

Therefore if $A \in SL(2,\mathbb{C})$ then $\phi(A)$ is a Lorentz transformation and also a homeomorphism.

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Ok so far, here begins my question.

The next bit of the text says: suppose $A \in SU(2)$, then $AIA^*=I$

so if ${\bf e}_0 = \left( \begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \end{array} \right)$ then $\phi(A){\bf e}_0 = {\bf e}_0$

I see why this is so. Now here's the bit I don't understand.

If a Lorentz Transformation $C$ satisfies $C{\bf e}_0 = {\bf e}_0$ then $C$ also carries the space of vectors $\left( \begin{array}{c} 0 \\ x_1 \\ x_2 \\ x_3 \end{array} \right)$ onto itself.

I don't see why this is so. Can someone please explain the logical steps which I'm missing? Thanks.

Answer 1

The space of vectors $(0,x_1,x_2,x_3)^T$ may also be defined as the space of 4-vectors $\vec v$ that satisfy $\vec v\cdot \vec e_0 = 0$. The inner product is Lorentzian. If all vector $\vec u$ are transformed to $C\vec u$, then this space of vectors is – by definition – transformed to the space of vectors that obey $$ (C\vec v)\cdot \vec e_0=0 $$ for any allowed transformation $C$. However, the condition above is equivalent to $$ \vec v\cdot (\tilde C\vec e_0) =0$$ because the inner product may also be written as $\vec u\cdot \vec v = u^T \cdot \eta\cdot v$ where $\eta$ is the usual metric tensor. I used $(CD)^T = D^TC^T$ etc. Here, $\tilde C = \eta C^T \eta^{-1}$. However, $\tilde C \vec e_0$ may be easily seen to be $\vec e_0$ as well, so the condition in the displayed formula is equivalent to $$\vec v \cdot \vec e_0$$ so the whole space (as a set) is invariant under the transformation by $C$.

The text above is an unnecessary formalism that may be reproduced by self-evident words. The space of vectors given by the three coordinates is the space of all vectors normal to $\vec e_0$, and because both $\vec e_0$ and the inner product is conserved by the transformations given by $C$ and the definition of the 3D space depends on these invariant things only, this 3D space as a set is conserved as well. $C$ must be just a rotation of the spatial coordinates into each other.

Answer 2

If $C$ doesn't change the time component then it is just a spatial rotation!

To show this mathematically we can use $\boldsymbol{e}_{0}$ to construct a projection operator that projects down to the space orthogonal to $\boldsymbol{e}_{0}$:

$$ \Pi = \mathbb{I} - \boldsymbol{e}_{0} \otimes \boldsymbol{e}_{0} $$

Then the statement is that

$$ C \boldsymbol{x} = \Pi C \boldsymbol{x}$$

for spatial $\boldsymbol{x}$, that is $\boldsymbol{x}$ obeying

$$ \boldsymbol{x} = \Pi \boldsymbol{x} $$

Applying this above we get

$$ C \Pi \boldsymbol{x} = \Pi C \boldsymbol{x}$$

So if $ C\Pi - \Pi C = 0$ then the result holds. Calculating this:

$$ C \Pi = C (\mathbb{I} - \boldsymbol{e}_{0} \otimes \boldsymbol{e}_{0}) = C - (C\boldsymbol{e}_{0})\otimes \boldsymbol{e}_{0} = C - \boldsymbol{e}_{0} \otimes \boldsymbol{e}_{0} $$

and similarly for $\Pi C$ after using orthogonality of $C$. Subtracting gives $C\Pi - \Pi C = 0$.

Answer 3

Hints:

1. Note that the time coordinate $x^0=\frac{1}{2} {\rm tr}(x)$ is given by the trace.

2. Note that the trace ${\rm tr}(x)$ is invariant under the $SU(2)$ action.

3. The subspace with $x^0=0$ is invariant under $SU(2)$.

4. Note that the stabilizer/isotropy subgroup of $e_0$ is $SU(2)$.