# Conditions to determine if a matrix is a Lorentz matrix and if it is a Galilean matrix

Consider a generic 4x4 matrix $$\Lambda$$ $${\begin{bmatrix}{\Lambda ^{0}}_{0}&{\Lambda ^{0}}_{1}&{\Lambda ^{0}}_{2}&{\Lambda ^{0}}_{3}\\{\Lambda ^{1}}_{0}&{\Lambda ^{1}}_{1}&{\Lambda ^{1}}_{2}&{\Lambda ^{1}}_{3}\\{\Lambda ^{2}}_{0}&{\Lambda ^{2}}_{1}&{\Lambda ^{2}}_{2}&{\Lambda ^{2}}_{3}\\{\Lambda ^{3}}_{0}&{\Lambda ^{3}}_{1}&{\Lambda ^{3}}_{2}&{\Lambda ^{3}}_{3}\\\end{bmatrix}}$$

(and use the metric signature $$-+++$$).

What is the right way to determine if $$\Lambda$$ is a matrix belonging to Lorentz group, i.e. is a Lorentz matrix? What are the conditions to use to test if a matrix is Lorentz matrix?

Also, how do I determine if $$\Lambda$$ is a matrix belonging to Galilean group? Is a Galilean matrix necessarily a Lorentz matrix? In this case, what are the addititional conditions for which a Lorentz matrix is also a Galilean matrix?

My attempt:

• Lorentz matrices: the condition

$$\Lambda^T \eta \Lambda =\eta$$

with $$\eta$$ metric tensor

must hold for a Lorentz matrix. Nevertheless I don't understand if a Lorentz matrix necessarily belongs to $$SO(3)$$, therefore is special ortogonal, which means that another condition is

$$\Lambda^{-1}=\Lambda^T$$ and $$det{\Lambda}=1$$

Therefore is it correct to say the following? $$\Lambda \,\,\, is \,\,\,Lorentz \,\, \iff \,\,\, \begin{cases} \Lambda^{-1}=\Lambda^T \\ det{\Lambda}=1 \\ \Lambda^T \eta \Lambda =\eta \end{cases}$$

• Galilean matrices: is necessarily Lorentz, but it must also have the first row $$(1,0,0,0)$$ and also the submatrix $$A={\begin{bmatrix}{\Lambda ^{1}}_{1}&{\Lambda ^{1}}_{2}&{\Lambda ^{1}}_{3}\\{\Lambda ^{2}}_{1}&{\Lambda ^{2}}_{2}&{\Lambda ^{2}}_{3}\\{\Lambda ^{3}}_{1}&{\Lambda ^{3}}_{2}&{\Lambda ^{3}}_{3}\\\end{bmatrix}}$$ is $$SO(3)$$

Therefore is it correct to say the following? $$\Lambda \,\,\, is \,\,\,Galilean \,\, \iff \,\,\, \begin{cases} \Lambda \,\,\, is \,\,\,Lorentz \\ the \,\,\,first\,\,\, row\,\,\, is\,\,\, (1,0,0,0) \\ the \,\,\, submatrix \,\,\,\, {\begin{bmatrix}{\Lambda ^{1}}_{1}&{\Lambda ^{1}}_{2}&{\Lambda ^{1}}_{3}\\{\Lambda ^{2}}_{1}&{\Lambda ^{2}}_{2}&{\Lambda ^{2}}_{3}\\{\Lambda ^{3}}_{1}&{\Lambda ^{3}}_{2}&{\Lambda ^{3}}_{3}\\\end{bmatrix}} \,\,\,is \,\,\,SO(3) \end{cases}$$