What exactly means that a certain mathematical statement is invariant under a group? For Ex:$$O(1,3)$$ for $$x^2_0-x^2_1-x^2_2-x^2_3$$

and how do you check it?

  • $\begingroup$ The point which is important is called group action (en.wikipedia.org/wiki/Group_action) an element or a set is invariant under a group $G$ if it is invariant under the group action. For example if $g(x^2) = x^2$ $\forall g \in G$ in your example. $\endgroup$ – Alpha001 Feb 20 '17 at 15:14

Take the 4-vector $(x^0,x^1,x^2,x^3)$. Its length is given by $x^\mu x^{\nu}g_{\mu\nu}$. This length is invariant under $SO(3,1)$ because, if you take any transformation $\Lambda \in SO(3,1)$ taking, $$(x^0,x^1,x^2,x^3)^T\to (\bar x^0,\bar x^1,\bar x^2,\bar x^3)^T :=\Lambda (x^0,x^1,x^2,x^3)^T,$$ then the length $x^\mu x^{\nu}g_{\mu\nu}= \bar x^\mu \bar x^{\nu}g_{\mu\nu}$ is preserved, or invariant. A similar concept applies when saying the Cartesian length of a vector is invariant under rotation. Invariance under a group means the numerical value of the quantity is unchanged under group transformation.

In the specific case of your question, you need to write the most general group transformation that would act on the components of your 4-vector, i.e. write $\Lambda$ as a general $4\times 4$ matrix in $SO(3,1)$, work out $(\bar x^0,\bar x^1,\bar x^2,\bar x^3)^T$ and then verify that the result is the same as that for $(x^0,x^1,x^2,x^3)^T$.

Note that the condition equivalent to the requirement that $\Lambda^T\cdot g\cdot \Lambda= g$ if you think of $g$ as a matrix.


ZeroTheHero's argument is the correct one, but just as a tack-on to see why it is $SO(3,1)$, specifically, the easiest way to see is by transforming variables:

$$\begin{align} x_{1} &= r \cos \phi \sin \theta\\ x_{2} &= r \sin \phi \sin \theta\\ x_{3} &= r \cos \theta \end{align}$$

then, $x_{0}^2 - x_{1}^2 - x_2^2 - x_3^2$ becomes $x_{0}^{2} - r^{2}$

Now, do a transformation:

$$\begin{align} x_{0} &= \rho \cosh \psi\\ r &= \rho \sinh \psi \end{align}$$

Applying this transformation, $x_{0}^{2} - r^{2}$ becomes $\rho^2$. So, you can see that the interval does not depend on $(\psi, \theta, \phi)$. Since your original coordinates are periodic in $\theta, \phi$, and the transformation is that of ordinary three-dimensional space into spherical coordinates, they describe rotations in 3-d space, so variations in $\theta \phi$ are equivalent to transformations in $SO(3)$. Meanwhile, there is no periodicity in $\psi$, increasing it just will always just create a different transformation, so your overall group is $SO(3,1)$

  • $\begingroup$ Absolutely! Nicely done too. $\endgroup$ – ZeroTheHero Feb 20 '17 at 15:24

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