# How to prove that all Minkowski spacetime isometries (transformations in Poincare group) are compositions of translation and Lorentz Transformations?

It is said in wikipedia that Minkowski spacetime isometries, i.e. the transformation that preserves $$(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2-(t_1-t_2)^2$$ between points, can be represented as $$\mathbb{R}^{1,3}\rtimes O(1,3)$$, meaning that it consists of transformations with the form $$x \mapsto \Lambda x+b, \Lambda \in O(1,3) (\text{Lorentz group}), b\in \mathbb{R}^{1,3}.$$ How can we prove this?

It seems that the techniques that is available for proving the form of Euclidean group is unavailable here since the "distance" is not positive definite.

• One way would be to use the fact that a four dimensional spacetime has at most 10 Killing fields, and all of those are represented in the Poincaré group, but I'm not sure whether this is what you're going for Jun 29 at 1:39
• To expand a bit on that, a space of dimension $n$ can have at most $\frac12n(n+1)$ independent Killing vector fields. A space that has precisely this number of Killing vectors is known as "maximally symmetric". The 4-dimensional spacetime thus can have at most 10 continuous symmetries, and the Poincaré group has dimension 10, so there are no more. The structure of the Lie algebra of the Poincaré group can be worked out explicitly, by looking at commutation relations of the relevant differential operators. Jun 30 at 2:17

Let $$\left(\mathbb{R}^4, \mathcal{O}, \mathcal{A}^{\uparrow}, \nabla^{L.C.}, g, T\right)$$ be a four dimensional Minkowski spacetime.

We denote $$g$$ as the Minkowski metric and $$\nabla^{L.C.}$$ the metric induced Levi-Civita connection, which satisfies: $$\nabla^{L.C.} g=0$$ (metric compatible) and $$T=0$$ (torsion free). The spacetime being flat immediately implies the curvature to vanish everywhere on the manifold.

An isometry is defined with the help of the Lie-derivative: $$$$\mathcal{L}_{\xi} g=0.$$$$ Suppose we have another isometry: $$$$\mathcal{L}_{\Psi} g=0.$$$$ Then by definition of the symmetry, which is a Lie-subalgebra,we get, that: $$$$\mathcal{L}_{[\xi, \Psi]} g=0.$$$$ Because we know that the commutator can be expressed in terms of linear combination of the vector fields with the help of the structure constants (this follows from the definition of the Lie-subalgebra): $$$$[V_i,V_j]=C^k_{ij} V_k.$$$$ Thus the coordinate-free expression of the Poincaré algebra is given by the span of vector fields satisfying the Killing equation for a Minkowski metric together with their Lie-Bracket: $$$$\mathcal{L}_{V_i} g=0.$$$$ We will expand all of this in charts,then show it is chart independent. Expanding we get: $$$$(\mathcal{L}_{\xi} g)^{ab}=\xi^{i}\frac{\partial}{\partial x^i} g^{ab}-\frac{\partial \xi ^a}{\partial x^s} g^{bs}-\frac{\partial \xi^b}{\partial x^l} g^{al}=0.$$$$ Remember the definition of covariant derivative: $$$$(\nabla_{\xi}^{L.C.} g)^{ab}=\xi^{i} \frac{\partial}{\partial x^i} g^{ab}+\Gamma^{a}_{sm} \xi^m g^{sb}+\Gamma^{b}_{sm} \xi^m g^{as}.$$$$ We know that in flat spacetime all the connection coefficient functions vanish. This leads to: $$$$(\nabla_{\xi}^{L.C.} g)^{ab}=\xi^{i} \frac{\partial}{\partial x^i} g^{ab}.$$$$ Since the metric is connection-compatible, $$\nabla^{L.C} g$$=0. Introducing: $$\frac{\partial}{\partial x^{s}}=\partial_{s}$$ and $$\frac{\partial}{\partial x^{l}}=\partial_l$$ gives: $$$$(\mathcal{L}_{\xi} g)^{ab}=-g^{bs} \partial_{s} \xi^{a}-g^{al} \partial_{l} \xi^{b}=0.$$$$ Simplifying further leads to: $$$$(\mathcal{L}_{\xi} g)^{ab}=-\partial^{b} \xi^{a}-\partial^{a} \xi^{b}=0.$$$$ This can be also read as: $$$$\tag{Lie condition}\label{equation} \partial^{b} \xi^{a} + \partial^{a} \xi^{b}=0.$$$$ Taking a derivative: $$$$\label{1} \tag{1} \partial^{c}\partial^{b} \xi^{a}+\partial^{c}\partial^{a}\xi^{b}=0.$$$$ Using cyclic permutation on the indices: $$$$\label{2} \tag{2} \partial^{a}\partial^{c} \xi^{b}+\partial^{b}\partial^{c} \xi^{a}=0.$$$$ $$$$\label{3} \tag{3} \partial^{b}\partial^{a} \xi^{c}+\partial^{a}\partial^{b} \xi^{c}=0.$$$$ Substracting (\ref{1}) and (\ref{2}) from (\ref{3}) yields: $$$$\partial^{b}\partial^{a} \xi^{c}+\partial^{a}\partial^{b} \xi^{c}-\partial^{a}\partial^{c} \xi^{b}-\partial^{b}\partial^{c} \xi^{a}- \partial^{c}\partial^{b} \xi^{a}-\partial^{c}\partial^{a}\xi^{b}=2\partial^{b} \partial^{a} \xi^{c}=0.$$$$ This means the second derivatives vanish, therefore it can only be linear: $$$$\label{solution} \tag{solution} \boxed{ \xi^{c}=a^{c}+b^{c k}x_{k}. }$$$$

An interesting property of the coefficients $$b^{ck}$$ is that they are antisymmetric. this can be shown substituting \eqref{solution} into \eqref{equation}: $$$$\frac{\partial}{\partial x_b} \left(a^a +b^{ak} x_{k} \right)+\frac{\partial}{\partial x_a} \left(a^b +b^{bk}x_k \right)=0.$$$$ The partial derivative gives $$0$$ when acting on $$a^a$$, respectively $$a^b$$, since they are constant and $$b^{ak}$$, $$b^{bk}$$ can be pulled out, since they are expansion functions: $$$$b^{ak} \frac{\partial x_k}{\partial x_b}+b^{bk} \frac{\partial x_k}{\partial x_a}=0 \implies \boxed{ b^{ab}+b^{ba}=0. }$$$$ The whole vector field is given by: $$$$\xi^{c} \frac{\partial}{\partial x^c}=a^{c}\frac{\partial}{\partial x^c}+b^{c k}x_{k}\frac{\partial}{\partial x^c}.$$$$ Now we just name the objects we got that correspond to the literature. Namely, the translation generator is: $$$$\label{translation} \tag{translation generators} \boxed{ P_{c}=\frac{\partial}{\partial x^c} .}$$$$ Using the antisymmetry of the expansion coefficients, we can write: $$$$b^{ck} x_k \frac{\partial}{\partial x^c}=\frac{1}{2} b^{ck} \left(x_k \frac{\partial}{\partial x^c}-x_c \frac{\partial}{\partial x^k} \right).$$$$ We introduce the Lorentz generator: $$$$\boxed { M_{ck}=\left(x_k \frac{\partial}{\partial x^c}-x_c \frac{\partial}{\partial x^k} \right) .}$$$$ APPENDIX

Here we show that the thus introduced generators indeed satisfy the Poincare algebra: $$$$\label{poincare1} \tag{Poincare1} {\left[P_{\mu}, P_{\nu}\right]=0}$$$$ $$$$\label{poincare2} \tag{Poincare2} \left[M_{\mu \nu}, P_{\rho}\right]=\eta_{\mu \rho} P_{\nu}-\eta_{\nu \rho} P_{\mu} .$$$$ $$$$\label{poincare3} \tag{Poincare3} \left[M_{\mu \nu}, M_{\rho \sigma}\right]=\eta_{\mu \rho} M_{\nu \sigma}-\eta_{\mu \sigma} M_{\nu \rho}-\eta_{\nu \rho} M_{\mu \sigma}+\eta_{\nu \sigma} M_{\mu \rho}.$$$$ The first property \eqref{poincare1} immediately follows from the Schwartz rule in a chart: $$$$\label{property1}\tag{Poincare 1} \boxed{[P_{\mu},P_{\nu}]=\left[\frac{\partial}{\partial x^{\mu}},\frac{\partial}{\partial x^{\nu}} \right]=0 .}$$$$ Using the definition of the Lorentz and translation generators, one obtains: $$$$\left[M_{\mu \nu}, P_{\rho}\right]=\left[\left( x_{\nu} \frac{\partial}{\partial x^{\mu}}-x_{\mu}\frac{\partial}{\partial x^{\nu}} \right),\frac{\partial}{\partial x^{\rho}}\right].$$$$ Using the property of the commutator:$$[A-B,C]=[A,C]-[B,C]$$ gives: $$$$\left[M_{\mu \nu}, P_{\rho}\right]=\left[ x_{\nu} \frac{\partial}{\partial x^{\mu}},\frac{\partial}{\partial x^{\rho}} \right] - \left[x_{\mu}\frac{\partial}{\partial x^{\nu}},\frac{\partial}{\partial x^{\rho}} \right].$$$$ Using $$[AB,C]=A[B,C]+[A,C]B$$ leads to: $$$$\left[M_{\mu \nu}, P_{\rho}\right]= x_{\nu} \left[\frac{\partial}{\partial x^{\mu}},\frac{\partial}{\partial x^{\rho}} \right]+\left[x_{\nu},\frac{\partial}{\partial x^{\rho}} \right] \frac{\partial}{\partial x^{\mu}}-x_{\mu} \left[\frac{\partial}{\partial x^{\nu}},\frac{\partial}{\partial x^{\rho}} \right]-\left[x_{\mu},\frac{\partial}{\partial x^{\rho}} \right] \frac{\partial}{\partial x^{\nu}}.$$$$ Using \eqref{property1} or equivalently Schwartz rule leads to the first and third term dropping out, which leaves: $$$$\left[M_{\mu \nu}, P_{\rho}\right]=x_{\nu} \frac{\partial}{\partial x^{\rho}}\frac{\partial}{\partial x^{\mu}}-\frac{\partial x_\nu}{\partial x^{\rho}} \frac{\partial}{\partial x^{\mu}} - x_{\mu} \frac{\partial}{\partial x^{\rho}} \frac{\partial}{\partial x^{\nu}} +\frac{\partial x_{\mu}}{\partial x^{\rho}} \frac{\partial}{\partial x^{\nu}}.$$$$ With the help of the metric one concludes $$\frac{\partial x_\mu}{\partial x^{\rho}}=\eta_{\mu \rho}$$, respectively $$\frac{\partial x_{\nu}}{\partial x^{\rho}}=\eta_{\nu \rho}$$. We can use the Schwartz rule for the first and third term: $$$$\left[M_{\mu \nu}, P_{\rho}\right]=x_{\nu} \frac{\partial}{\partial x^{\mu}} \frac{\partial}{\partial x^{\rho}}- \eta_{\nu \rho} \frac{\partial}{\partial x^{\mu}} -x_{\mu}\frac{\partial}{\partial x^{\nu}} \frac{\partial}{\partial x^{\rho}} + \eta_{\mu \rho} \frac{\partial}{\partial x^{\nu}}.$$$$ Regroupping the terms: $$$$\left[M_{\mu \nu}, P_{\rho}\right]=\left(x_{\nu} \frac{\partial}{\partial x^{\mu}}-x_{\mu}\frac{\partial}{\partial x^{\nu}} \right) \frac{\partial}{\partial x^{\rho}}-\eta_{\nu \rho} \frac{\partial}{\partial x^{\mu}}+ \eta_{\mu \rho} \frac{\partial}{\partial x^{\nu}}.$$$$ By renaming the indices, the first term cancels out. Plugging in the definition of \eqref{translation} leads to the desired result: $$$$\boxed{ \left[M_{\mu \nu}, P_{\rho}\right]=\eta_{\mu \rho} P_{\nu} -\eta_{\nu \rho} P_{\mu} }.$$$$ The last thing to show is \eqref{poincare3}. Using the definition: $$$$\left[M_{\mu \nu},M_{\rho \sigma} \right]= \left[\left(x_{\nu} \frac{\partial}{\partial x^{\mu}}-x_{\mu}\frac{\partial}{\partial x^{\nu}}\right), \left(x_{\sigma} \frac{\partial}{\partial x^{\rho}}-x_{\rho}\frac{\partial}{\partial x^{\sigma}} \right) \right].$$$$ Using $$[A-B,C]=[A,C]-[B,C]$$ leads to: $$$$\left[M_{\mu \nu},M_{\rho \sigma} \right]=\left[x_{\nu} \frac{\partial}{\partial x^{\mu}},\left(x_{\sigma} \frac{\partial}{\partial x^{\rho}}-x_{\rho}\frac{\partial}{\partial x^{\sigma}} \right) \right]- \left[x_{\mu}\frac{\partial}{\partial x^{\nu}}, \left(x_{\sigma} \frac{\partial}{\partial x^{\rho}}-x_{\rho}\frac{\partial}{\partial x^{\sigma}} \right) \right].$$$$ Using $$[A,B-C]=[A,B]-[A,C]$$ one obtains: \begin{align} \left[M_{\mu \nu},M_{\rho \sigma} \right]=\left[ x_{\nu}\frac{\partial}{\partial x^{\mu}},x_{\sigma} \frac{\partial}{\partial x^{\rho}}\right]- \left[x_{\nu} \frac{\partial}{\partial x^{\mu}},x_{\rho}\frac{\partial}{\partial x^{\sigma}} \right] \\ - \left[ x_{\mu}\frac{\partial}{\partial x^{\nu}},x_{\sigma} \frac{\partial}{\partial x^{\rho}}\right]+\left[ x_{\mu}\frac{\partial}{\partial x^{\nu}}, x_{\rho}\frac{\partial}{\partial x^{\sigma}}\right]. \end{align} As we can see, all the terms are similar. Thus we have to compute only one. Using the identity $$[A B, C D]=A[B, C] D+[A, C] B D+C A[B, D]+C[A, D] B$$ for the first term gives: \begin{align} \left[ x_{\nu}\frac{\partial}{\partial x^{\mu}},x_{\sigma} \frac{\partial}{\partial x^{\rho}}\right]=x_{\nu} \left[\frac{\partial}{\partial x^{\mu}},x_{\sigma} \right] \frac{\partial}{\partial x^{\rho}}+\left[x_{\nu}, x_{\sigma} \right]\frac{\partial}{\partial x^{\mu}} \frac{\partial}{\partial x^{\rho}}\\ +x_{\sigma} x_{\nu} \left[\frac{\partial}{\partial x^{\mu}},\frac{\partial}{\partial x^{\rho}} \right] \frac{\partial}{\partial x^{\mu}}+x_{\sigma} \left[x_{\nu},\frac{\partial}{\partial x^{\rho}} \right]\frac{\partial}{\partial x^{\mu}}. \end{align} We immediately observe, that the second and third term drop out, leaving us with: $$$$\left[ x_{\nu}\frac{\partial}{\partial x^{\mu}},x_{\sigma} \frac{\partial}{\partial x^{\rho}}\right]=x_{\nu} \left[\frac{\partial}{\partial x^{\mu}},x_{\sigma} \right] \frac{\partial}{\partial x^{\rho}}+x_{\sigma} \left[x_{\nu},\frac{\partial}{\partial x^{\rho}} \right]\frac{\partial}{\partial x^{\mu}}.$$$$ Expanding the commutators: \begin{align} \left[ x_{\nu}\frac{\partial}{\partial x^{\mu}},x_{\sigma} \frac{\partial}{\partial x^{\rho}}\right]=x_{\nu} \left(\frac{\partial x_{\sigma}}{\partial x^{\mu}} -x_{\sigma}\frac{\partial}{\partial x^{\mu}} \right) \frac{\partial}{\partial x^{\rho}} +x_{\sigma} \left(x_{\nu} \frac{\partial}{\partial x^{\rho}}-\frac{\partial x_{\nu}}{\partial x^{\rho}} \right)\frac{\partial}{\partial x^{\mu}}. \end{align} With the help of the metric we can conclude that $$\frac{\partial x_{\sigma}}{\partial x^{\mu}}=\eta_{\sigma \mu}$$, respectively $$\frac{\partial x_{\nu}}{\partial x^{\rho}}=\eta_{\nu \rho}$$. Thus: $$$$\left[ x_{\nu}\frac{\partial}{\partial x^{\mu}},x_{\sigma} \frac{\partial}{\partial x^{\rho}}\right]= x_{\nu} \eta_{\sigma \mu} \frac{\partial}{\partial x^{\rho}}-x_{\nu} x_{\sigma} \frac{\partial}{\partial x^{\mu}}\frac{\partial}{\partial x^{\rho}}+x_{\sigma} x_{\nu}\frac{\partial}{\partial x^{\rho}} \frac{\partial}{\partial x^{\mu}} - x_{\sigma} \eta_{\nu \rho} \frac{\partial}{\partial x^{\mu}}.$$$$ Because of the Schwartz rule the second term drops out with the third, leaving us with: $$$$\left[ x_{\nu}\frac{\partial}{\partial x^{\mu}},x_{\sigma} \frac{\partial}{\partial x^{\rho}}\right]=x_{\nu} \eta_{\sigma \mu} \frac{\partial}{\partial x^{\rho}}-x_{\sigma} \eta_{\nu \rho} \frac{\partial}{\partial x^{\mu}}.$$$$ The other three commutators give similar result: $$$$\left[x_{\nu} \frac{\partial}{\partial x^{\mu}},x_{\rho}\frac{\partial}{\partial x^{\sigma}} \right]= x_{\nu} \eta_{\rho \mu} \frac{\partial}{\partial x^\sigma }-x_{\rho} \eta_{\nu \sigma} \frac{\partial}{\partial x^{\mu}}.$$$$ $$$$\left[ x_{\mu}\frac{\partial}{\partial x^{\nu}},x_{\sigma} \frac{\partial}{\partial x^{\rho}}\right]=x_{\mu} \eta_{\sigma \nu } \frac{\partial}{\partial x^\rho}-x_{\sigma} \eta_{\mu \rho} \frac{\partial}{\partial x^\nu}.$$$$ $$$$\left[ x_{\mu}\frac{\partial}{\partial x^{\nu}}, x_{\rho}\frac{\partial}{\partial x^{\sigma}}\right]=x_{\mu} \eta_{\rho \nu} \frac{\partial}{\partial x^\sigma}-x_{\rho} \eta_{\mu \sigma}\frac{\partial}{\partial x^\nu}.$$$$ Putting them together yields: \begin{align} \left[M_{\mu \nu},M_{\rho \sigma} \right]= x_{\nu} \eta_{\sigma \mu} \frac{\partial}{\partial x^{\rho}}-x_{\sigma} \eta_{\nu \rho} \frac{\partial}{\partial x^{\mu}}+x_{\mu} \eta_{\rho \nu} \frac{\partial}{\partial x^\sigma}-x_{\rho} \eta_{\mu \sigma}\frac{\partial}{\partial x^\nu} \\ -\left( x_{\mu} \eta_{\sigma \nu } \frac{\partial}{\partial x^\rho}-x_{\sigma} \eta_{\mu \rho} \frac{\partial}{\partial x^\nu} + x_{\nu} \eta_{\rho \mu} \frac{\partial}{\partial x^\sigma }-x_{\rho} \eta_{\nu \sigma} \frac{\partial}{\partial x^{\mu}}\right). \end{align} Regrouping the terms by the diagonalized metric components: \begin{align} \left[M_{\mu \nu},M_{\rho \sigma} \right]= \eta_{\sigma \mu} \left(x_{\nu} \frac{\partial}{\partial x^\rho}-x_{\rho}\frac{\partial}{\partial x^\nu} \right)+ \eta_{\rho \nu} \left(x_{\mu}\frac{\partial}{\partial x^\sigma}-x_{\sigma} \frac{\partial}{\partial x^\mu} \right) \\ + \eta_{\mu \rho} \left(x_{\sigma} \frac{\partial}{\partial x^\nu} -x_{\nu} \frac{\partial}{\partial x^\sigma} \right) +\eta_{\nu \sigma} \left(x_{\rho} \frac{\partial}{\partial x^\mu}-x_{\mu}\frac{\partial}{\partial x^\rho} \right). \end{align} Using the definition of the Lorentz generator $$M_{ck}=\left(x_k \frac{\partial}{\partial x^c}-x_c \frac{\partial}{\partial x^k} \right)$$ the expression simplifies: $$$$\left[M_{\mu \nu},M_{\rho \sigma} \right]=\eta_{\sigma \mu} M_{\rho \nu}+\eta_{\rho \nu} M_{\sigma \mu}+\eta_{\mu \rho} M_{\nu \sigma}+\eta_{\nu \sigma} M_{\mu \rho}.$$$$ We also know that the Lorentz generators are antisymmetric: $$$$M_{\rho \nu}=-M_{\nu \rho}.$$$$ $$$$M_{\sigma \mu}=-M_{\mu \sigma}.$$$$ This leads to the final result: $$$$\boxed{ \left[M_{\mu \nu},M_{\rho \sigma} \right]=\eta_{\mu \rho} M_{\nu \sigma}+\eta_{\nu \sigma} M_{\mu \rho} -\eta_{\sigma \mu} M_{\nu \rho} - \eta_{\rho \nu} M_{\mu \sigma} .}$$$$ We conclude: The exponential map from the Lie algebra given above maps into the Proper,Ortochronous Poincare group, which is connected to the identity.

• I do not know how to make the numbering of the equations and referencing to them work. If someone can help, let me know. Jun 29 at 3:57
• Thank you! I'll edit the equations. What is your concern about the Lie derivative? Do you mean I should write it like the metric evaluated in the push forward of the vector fields is the same, i.e. doesn't change? And relate this to the Lie derivative? or? Jun 29 at 17:49
• there are other transformations which preserve (x(2)-x(1))²+.....of the following form: X'= X-kVt , t'=kt (capital letters ~ vector), k= {bcos(n)+ [(1-b²sin²(n)]^(1/2)}^(-1), with b=v/c
– The Tiler
Jun 30 at 15:18
• I have no idea if this argument works but the proof is much more complicated than necessary. A much better argument is to note that a semi Riemannian isometry of a connected manifold which fixes a point and has identity derivative at that points has to be the identity map. Jul 3 at 3:11
• Thanks so much! Jul 3 at 17:17

$$\newcommand\R{\mathbb{R}}$$In fact, the proof that a Euclidean isometry is a rigid motion works here, too. Here is the proof for the Minkowski case. No differential geometry is needed.

Given points $$p_1 = (x_1,y_1,z_1,t_1)$$ and $$p_2 = (x_2,y_2,z_2,t_2)$$, let $$Q(p_1,p_2) = x_1x_2 + y_1y_2 + z_1z_2 - t_1t_2.$$ You should view $$Q$$ as the Minkowski analogue of the dot product. The key facts used below are that it is bilinear and nondegenerate.

A map $$F: \R^{1,3} \rightarrow \R^{1,3}$$ is an isometry, if for any points $$p_1, p_2 \in \R^{1,3}$$, $$Q(F(p_2)-F(p_1),F(p_2)-F(p_1)) = Q(p_2-p_1,p_2-p_1).$$
Observe that we make no other assumptions about the map $$F$$, not even that it is continuous.

Let $$(e_1,e_2,e_3,e_4)$$ be the standard basis. It satisfies \begin{align*} Q(e_i,e_j) &= \begin{cases} 0 &\text{ if }i\ne j\\ 1 &\text{ if }i=j < 4\\ -1 &\text{ if }i = j = 4 \end{cases}. \end{align*} In particular, if $$p = (x,y,z,t) = xe_1 + ye_2 + ze_3 + te_4$$, then \begin{align*} x &= Q(p,e_1)\\ y &= Q(p,e_2)\\ z &= Q(p,e_3)\\ t &= - Q(p,e_4). \end{align*} More generally, we call $$(f_1,f_2,f_3,f_4)$$ an orthonormal basis if \begin{align*} Q(f_i,f_j) &= \begin{cases} 0 &\text{ if }i\ne j\\ 1 &\text{ if }i=j < 4\\ -1 &\text{ if }i = j = 4 \end{cases}. \end{align*} Moreover, if $$p = af_1 + bf_2 + cf_3 + df_4$$, then \begin{align*} a &= Q(p,f_1)\\ b &= Q(p,f_2)\\ c &= Q(p,f_3)\\ d &= - Q(p,f_4). \end{align*}

Now let $$F$$ be an isometry. Since translation by a vector $$v$$ is an isometry, the map $$\widetilde{F}(p) = F(p) - F(0)$$ is an isometry. So we assume $$F(0) = 0$$. This implies that \begin{align*} Q(F(p),F(p)) &= Q(F(p)-F(0), F(p)-F(0))\\ &= Q(p-0,p-0)\\ &= Q(p,p). \end{align*} This in turn implies that \begin{align*} Q(F(p),F(q)) &= -\frac{1}{2}(Q(F(p)-F(q),F(p)-F(q)) - Q(F(p),F(p)) - Q(F(q),F(q))\\ &= -\frac{1}{2}(Q(p-q,p-q) - Q(p,p) - Q(q,q))\\ &= Q(p,q). \end{align*}

Now let $$f_k = F(e_k)$$, for $$k = 1, 2, 3, 4$$. Then, for any $$1 \le j,k \le 4$$, \begin{align*} Q(f_j,f_k) &= Q(F(e_j),F(e_k))\\ &= Q(e_j,e_k). \end{align*} This implies that $$(f_1,f_2,f_3,f_4)$$ is an orthnormal basis. Therefore, if $$p = xe_1+ye_2+ze_3 + te_4$$ and $$F(p) = af_1+ bf_2 + cf_3 + df_4,$$ then \begin{align*} a &= Q(f_1, F(p)) = Q(F(e_1),F(p)) = Q(e_1,p) = x\\ b &= Q(f_2, F(p)) = Q(F(e_2),F(p)) = Q(e_2,p) = y\\ c &= Q(f_3, F(p)) = Q(F(e_3),F(p)) = Q(e_3,p) = z\\ d &= -Q(f_4, F(p)) = -Q(F(e_4),F(p)) = -Q(e_4,p) = t. \end{align*} This implies that $$F(xe_1+ye_2+ze_3+te_4) = xF(e_1)+yF(e_2) + zF(e_3) + tF(e_4).$$ In other words, $$F$$ is a linear map.

Finally, if $$F$$ is linear, then $$F(p) = Ap$$ for a matrix $$A$$. Moreover, for any $$p, q \in \R^{1,3}$$, $$Q(Ap,Aq) = Q(p,q).$$ By definition, this means $$A \in O(1,3)$$.

Putting this all together we have shown that if $$F$$ is a Minkowski isometry, then it is of the form $$F(p) = Ax + b,$$ where $$A \in O(1,3)$$.

• Thanks so much! Jul 3 at 17:17