I'm studying Minkowski spacetime $\Bbb{M}$, and I would like to make the following statement about its symmetry transformations. Since $\Bbb{M}$ is the product manifold of time and space, it inherits their symmetries (i.e. translations, inversions, rotations); since it is the product manifold of $\Bbb{R}^{2}$ with the metric

$$(\eta_{\mu\nu})=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$$

and $\Bbb{R}^{2}$ with the metric

$$(-g_{ij})=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$$

it inherits the symmetries of $(\Bbb{R}^{2},\eta)$ too; since it doesn't matter which dimension is taken to be the minus-one signed one in $(\Bbb{R}^{2},\eta)$, this last product induces a total of three independent symmetries, i.e. the symmetries by boost transformations in the three directions. The statement I would like to make is: no symmetry transformation other than the former (and their compositions) is allowed (of course, as we know, this is the correct answer to the problem of the symmetries of Minkowski spacetime). How can I justify this statement without solving the equation

$$\Lambda^*\,\eta=\eta$$ ?

  • As I think about the isometries of $\eta$, aka Lorentz transformations, I see first the orthogonal transformations which just fix time. Then, if I understand your post, you obtain $x,y,z$ boosts just from looking at the product of space time with space space. Now, the space space also picks up fixed time orthogonal transformations in which the remaining space is also fixed, that is O(2) embedded. Those are already included in the orthogonal transformations O(3) embedded to fix time. So, I suppose it makes sense, each of the products you consider add just one boost and we know those three... – James S. Cook Jun 25 '15 at 4:07
  • boosts together with orthogonal transformations on space space space (time fixed) are the Lorentz group. All of this said, I'm still not entirely sure I understand the question... – James S. Cook Jun 25 '15 at 4:08
  • Yes. Well, my question is, as written above, how can I state in a rigorous way that those are the ONLY symmetries, without resorting to the differential equations for the pullback metric, i.e. state that by only taking into account the manifolds that make up $\Bbb{M}$ and their products? (If you want to see why I don't want to turn to the PDEs, visit my other question, math.stackexchange.com/questions/1336870/…) – Giorgio Comitini Jun 25 '15 at 8:41
  • Ok, maybe I don't understand what you mean by Minkowski spacetime. Could you clarify, is this just $\mathbb{R}^4$ or is it a $4$-dimensional manifold paired with a particular Lorentzian structure ? (or, you pick the description). If the Minkowski spacetime is $\mathbb{R}^4$ then I think we can settle the question with linear algebraic techniques. – James S. Cook Jun 25 '15 at 14:31
  • It is the four dimensional manifold diffeomorphic to $\Bbb{R}^{4}$ endowed with the Minkowski metric $(\eta_{\mu\nu})=\begin{pmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{pmatrix}$. So it is not the euclidean $\Bbb{R}^4$. Nonetheless, if you have algebraic tecniques for $\Bbb{R}^{4}$, they could have an application to Minkowski spacetime too, as long as they don't depend on the signature of the metric. – Giorgio Comitini Jun 25 '15 at 14:52

We call Minkowski spacetime and denote with $\Bbb{M}$ the smooth manifold $\Bbb{R}^{4}$ endowed with the constant, symmetric, non-degenerate, covariant $2$-tensor field $\eta$ of coefficients

$$ (\eta_{\mu\nu})=\begin{pmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{pmatrix} $$ in the standard coordinate representation, i.e. $p\in \Bbb{R}^{4}\Rightarrow p=(x^{0},x^{1},x^{2},x^{3})$, then $$ \eta=(dx^{0})^{2}-(dx^{1})^{2}-(dx^{2})^{2}-(dx^{3})^{2} $$ Minkowski spacetime is thus a four-dimensional pseudo-riemannian manifold with metric $\eta$. The latter is called a Minkowski metric. The restriction of $\eta$ to each point of $\Bbb{M}$ is, by definition, a symmetric bilinear form that assigns to pairs of tangent vectors in $T_{p}\,\Bbb{M}$ a real number: $$ \eta: T_{p}\,\Bbb{M}\times T_{p}\,\Bbb{M}\to \Bbb{R} $$ Notice that, as $\eta$ is a non-degenerate $2$-tensor field, the following applies: $$ \eta(v,w)=0\quad\forall\ w\in T_{p}\Bbb{M}\quad\Longrightarrow\quad v=0 $$ As the underlying manifold of $\Bbb{M}$ is $\Bbb{R}^{4}$, the tangent spaces $T_{p}\,\Bbb{M}$ are canonically identified with the space $\Bbb{R}_{p}^{4}$ of geometric four-dimensional vectors based at $p$. In particular, $\eta$ defines the Minkowski length of such vectors to be $$ |v|=\sqrt{\eta(v,v)} $$ Thus, the length of any vector $v$ can be an imaginary number as well as a real number. As the difference $q-p$ between two points $p,q\in\Bbb{M}$ defines a geometric four-dimensional vector based at $p$, we define the Minkowski distance between two points $p,q\in\Bbb{M}$ to be $$ d_{\gamma}(p,q)=\sqrt{\eta(q-p,q-p)} $$ Again, the Minkowski distance (henceforth, the distance) between two points in Minkowski spacetime can be imaginary as well as real. An isometry of Minkowski spacetime is a smooth function $F:\Bbb{R}^{4}\to \Bbb{R}^{4}$ whose pullback preserves the Minkowski metric, i.e. it must satisfy the pullback equation $$ F^{*}\,\eta=\eta $$

$$\\$$ $$\\$$

Lemma : A smooth map from $\Bbb{M}$ to itself is an isometry if and only if it preserves the Minkowski distance: $$ d_{\eta}(F(p),F(q))=d_{\eta}(p,q) $$

Proof : If $F$ is an isometry, it preserves the metric and, as the distance is defined with respect to the metric, it also preserves the distances. Now suppose that $F$ preserves the distance, and let $v,w$ be vector fields on $\Bbb{M}$. First of all, we have $$ \eta(v-w,v-w)=\eta(v,v)+\eta(w,w)-2\eta(v,w) $$ and, as smooth maps pull back symmetric tensor fields to symmetric tensor fields, $$ F^{*}\eta(v-w,v-w)=F^{*}\eta(v,v)+F^{*}\eta(w,w)-2F^{*}\eta(v,w) $$ Then we observe that, pointwise, $(v-w)|_{p}$ is the tangent vector that defines the distance between the points $p+v|_{p}$ and $p+w|_{p}$, so as $F$ preserves the distances, it pointwise preserves the length of the vector $(v-w)|_{p}$: $|F_{*}(v-w)||_{p}=|v-w|_{p}\ \forall\ p\in\Bbb{M}$. This implies that between the above equations holds equality, and so we have $$ F^{*}\eta(v,v)+F^{*}\eta(w,w)-2F^{*}\eta(v,w)=\eta(v,v)+\eta(w,w)-2\eta(v,w) $$ As a linear combination of symmetric tensor fields is still a symmetric tensor field, the latter translates into $$ (F^{*}\eta-\eta)(v-w,v-w)=0 $$ and this is true for every vector field $v-w$. Thus $F^{*}\eta-\eta$ either is antisymmetric, or it is the zero covariant $2$-tensor field. Since $F^{*}\eta-\eta$ can't be both symmetric and antisymmetric at the same time, it follows that it is zero, from which $$ F^{*}\eta=\eta $$ follows.

$$\\$$

$$\\$$

We would like to make the statement that all the isometries of Minkowski spacetime are those inherited from the euclidean spaces $\Bbb{R}$ and $\Bbb{R}^{3}$ and from the pseudo-riemannian manifold $(\Bbb{R}^{2},\,^{2}\eta)$, where $^{2}\eta$ is the Minkowski metric expressed in standard coordinates as $$ ^{2}\eta=\begin{pmatrix}1&0\\0&-1\end{pmatrix} $$ Identifying $\Bbb{R}$ with the subspace spanned by the coordinate $x^{0}$ in $\Bbb{M}$, and $\Bbb{R}^{3}$ with the subspace spanned by the coordinates $(x^{i}),\ i=1,2,3$ in $\Bbb{M}$, the isometries that $\Bbb{M}$ inherits from $\Bbb{R}$ are the inversion $x^{0}\mapsto -x^{0}$ and the translation $x^{0}\mapsto x^{0}+c^{0}$, while the isometries that it inherits from $\Bbb{R}^{3}$ are the inversions $x^{i}\mapsto -x^{i}$, the translations $x^{i}\mapsto x^{i}+c^{i}$ and the three-dimensional rotations. All these isometries are inherited by $\Bbb{M}$ as the latter is the product manifold of $\Bbb{R}$ and $\Bbb{R}^{3}$. (Notice that the minus signs of the metric $\eta$ do not prevent $\Bbb{M}$ from inheriting the isometries of $\Bbb{R}^{3}$, as the pullback equation for any map that does not act on the $x^{0}$-subspace is unchanged by simultaneous multiplication of both sides by a constant). As for the isometries of $(\Bbb{R}^{2},\,^{2}\eta)$, upon identifying $\Bbb{R}^{2}$ with the subspace spanned by the coordinates $(x^{0},x^{i})$ for any $i=1,2,3$, those are computed by setting $F:\Bbb{R}^{2}\to \Bbb{R}^{2},\ F(x^{0},x^{i})=(F^{0},F^{i})$ and solving the system of partial differential equations: \begin{align*} \left(\frac{\partial F^{0}}{\partial x^{0}}\right)^{2}-\left(\frac{\partial F^{i}}{\partial x^{0}}\right)^{2}&=1\\ \\ \left(\frac{\partial F^{0}}{\partial x^{i}}\right)^{2}-\left(\frac{\partial F^{i}}{\partial x^{i}}\right)^{2}&=-1\\ \\ \frac{\partial F^{0}}{\partial x^{0}}\frac{\partial F^{0}}{\partial x^{i}}-\frac{\partial F^{i}}{\partial x^{0}}\frac{\partial F^{i}}{\partial x^{i}}&=0 \end{align*}

obtained by the explicitly writing the equation for the pullback $$ F^{*}\ ^{2}\eta=\,^{2}\eta $$ An easy computation shows that the only non-traslative solution to the system given above is $$ F(x^{0},x^{i})=(x^{0}\cosh\theta+x^{i}\sinh\theta,x^{0}\sinh\theta+x^{i}\cosh\theta) $$ where $\theta$ is a real constant. Thus $F$ is linear, and to $F$ corresponds the matrix $$ F\equiv\begin{pmatrix}\cosh\theta&\sinh\theta\\\sinh\theta&\cosh\theta\end{pmatrix} $$ Setting $\beta=-\tanh\theta$ and $\gamma=\cosh\theta$, the former translates into $$ F\equiv\begin{pmatrix}\gamma&-\gamma\beta\\-\gamma\beta&\gamma\end{pmatrix} $$ As $\gamma$ is the function of $\beta$ $$ \gamma=\frac{1}{\sqrt{1-\beta^{2}}} $$ this is a one-parameter set of isometries. We call these isometries boosts in the $i$-th direction. Because the subspace of $\Bbb{M}$ that is to be taken as the second dimension in $(\Bbb{R}^{2},\,^{2}\eta)$ along with that spanned by $x^{0}$ is arbitrary, $\Bbb{M}$ inherits from $(\Bbb{R}^{2},\,^{2}\eta)$ three one-parameter sets of isometries. By composing $F$ with the rotations of $\text{Span}(x^{1},x^{2},x^{3})$, we find that the general non-translative isometry that $\Bbb{M}$ inherits from $(\Bbb{R}^{2},\,^{2}\eta)$ (the symmetry of the expression having been obtained only thanks to the the intervention of the rotations) is linear and associated to the matrix $$ \Lambda_{\boldsymbol{\beta}}= \begin{pmatrix} \gamma&-\gamma\boldsymbol{\beta^{\text{T}}}\\ -\gamma\boldsymbol{\beta}&\boldsymbol{I}+\frac{(\gamma-1)}{\beta^{2}}\,\boldsymbol{\beta\beta^{\text{T}}} \end{pmatrix} $$ where $\gamma=[1-(\beta_{1}^{2}+\beta_{2}^{2}+\beta_{3}^{2})]^{1/2}$ and $$ \boldsymbol{\beta}=\begin{pmatrix}\beta_{1}\\ \beta_{2}\\ \beta_{3}\end{pmatrix} $$ is the column matrix containing the parameters of the isometry.

$$\\$$ Given the linearity of the isometries that we derived so far, and given the identity $$ \eta(q-p,q-p)=\eta\left(Fq-Fp,Fq-Fp\right) $$ We find, taking $p$ to be zero, that any isometry inherited by $\Bbb{M}$ as a product manifold obeys the matrix equation $$ F^{T}\,\eta\,F=\eta $$ We call any non-translative isometry that obeys the above equation a Lorentz transformation, and denote it with the symbol $\Lambda$.

$$\\$$

Theorem : Every isometry of Minkowski spacetime is a linear transformation. Moreover, it can be expressed in a unique way as the composition of a Lorentz transformation and a translation.

Proof : Let $F$ be a isometry of Minkowski spacetime such that $F(0)=0$, and let $p\in\Bbb{M}$. Then, under the identification of points in $\Bbb{M}$ with tangent vectors in zero, $$ p^{T}\,\eta\, p=\eta(p,p)=d^{2}_{\eta}(p,0)=d^{2}_{\eta}(F(p),0)=\eta(F(p),F(p))=F(p)^{T}\,\eta\, F(p) $$ Moreover, we have \begin{align*} d_{\eta}^{2}(p,q)&=\eta(p-q,p-q)=\\ &=q^{T}\,\eta\, q-2\,q^{T}\,\eta\,p+p^{T}\,\eta\, p \end{align*} and similarly \begin{align*} d_{\eta}^{2}(F(p),F(q))&=\eta(F(p)-F(q),F(p)-F(q))=\\ &=F(q)^{T}\,\eta\, F(q)-2\,F(q)^{T}\,\eta\,F(p)+F(p)^{T}\,\eta\, F(p)=\\ &=q^{T}\,\eta\,q-2\,F(q)^{T}\,\eta\,F(p)+p^{T}\,\eta\,p \end{align*} so that $d_{\eta}^{2}(F(p),F(q))=d_{\eta}^{2}(p,q)$ implies $$ F(q)^{T}\,\eta\,F(p)=q^{T}\,\eta\,p\quad\Longrightarrow\quad\eta(F(q),F(p))=\eta(q,p) $$ Then, for each and every $p,q,s\in\Bbb{M}$ and $a\in\Bbb{R}$, we have \begin{align*} \eta(F(ap+q),F(s))&=\eta(ap+q,s)=a\eta(p,s)+\eta(q,s)=\\ &=a\eta(F(p),F(s))+\eta(F(q),F(s))=\\ &=\eta(aF(p)+F(s),F(s)) \end{align*} which implies, for the non-degeneracy of $\eta$ and the arbitrariness of $F(s)$, that $$ \eta(F(ap+q)-aF(p)-F(q),F(s))=0\quad\Longleftrightarrow\quad F(ap+q)=aF(p)+F(q) $$ Thus any $F$ that fixes $0$ is a linear transformation. If $F$ doesn't fix $0$, then the transformation $\Lambda$ such that $\Lambda(p)=F(p)-F(0)$ does, and by the preceding result we can associate to $\Lambda$ a matrix $\Lambda$, so that $$ F(p)=\Lambda p+F(0)=\Lambda p+c $$ where $c$ is a constant four-dimensional vector. Thus, we found that every isometry of Minkowski spacetime is a linear transformation. A trivial calculation shows that $\Lambda$ indeed is a Lorentz transformation: $$ p^{T}\eta\,p=(\Lambda p)^{T}\eta\,(\Lambda p)=p^{T}\Lambda^{T}\eta\,\Lambda\,p\quad\Longleftrightarrow\quad\Lambda^{T}\eta\,\Lambda=\eta $$ The fact that $F(p)=\Lambda p+c$ holds for every $p\in\Bbb{M}$ makes $\Lambda$ uniquely determined.

  • I recently read the McInerney's First Steps in Differential Geometry to get some extra perspective as I'm teaching based on O'neill. In his Section 5.6 on isometries he also defines an isometry as a map which preserves the metric tensor through the pull-back operation. He even states the PDEs you mention, but in Riemannian context, and examines what happens if the isometry is linear. BUT, he does not explain why the isometry must be linear in the case of euclidean space. It seems to me that is the sticky thing. As I think about it more, it is only O'neill who defines isometry for points... – James S. Cook Jul 4 '15 at 4:47
  • So, because I've been immersed in O'neill lately I naturally took that as the definition of isometry (despite you telling me otherwise) Anyway, I am happy to see a proof of the equivalence of the pull-back definition and the global distance definition in this case... but, I'm not sure I follow: "pointwise preserves the difference" where you have just $F_*(v-w)=v-w$. I think there is a distance function missing in that step. Also, I like that you did without the absolute value bars, I put it there because I once did some homework on this thing where I had to prove the "reverse triangle... – James S. Cook Jul 4 '15 at 4:51
  • inequality" (which is not a general truth, just an example to exhibit that this "distance" function is not really a distance function). Anyway, nice work! One other thing, in McInerney, he uses Killing Vectors to derive isometries in the Euclidean case. I realize these are local, but, I wonder if that might be another approach to this study... – James S. Cook Jul 4 '15 at 4:53
  • You're right, I think I'll have to make that statement more explicitly. As for the Killing fields, I've only heard about them. The only book I've read so far on differential geometry is Lee's "Introduction to Smooth Manifolds", and he makes mention of them only once throughout the book. Nevertheless, I was able to prove what I needed computing the Lie algebra of the Lorentz group. The proof is exquisitely elegant, but of course it requires some basic Lie Theory, and moreover it renders completely useless nearly three quarters of the contents of our answers, so I think this is not the... – Giorgio Comitini Jul 4 '15 at 7:44
  • ... right context in which to post it. – Giorgio Comitini Jul 4 '15 at 7:45

So, $\mathbb{R}^4$ paired with $\eta = \text{Diag}(1,-1,-1,-1)$ is an example of a Minkowski space. Thus, if we can analyze the question of isometries for this space then I expect you can lift the result via the diffeomorphism to the abstract case.

My idea is to follow the usual derivation of isometries in Euclidean three dimensional space. First, we show every isometry is the composition of an orthogonal transformation and translation (I expect this translates to Lorentz transformation [meaning $\Lambda$ such that $\Lambda^T \eta \Lambda = \eta$] and space time translation). Second, for vectors the push-forward kills the translational piece and leaves us with a Lorentz matrix for the Jacobian matrix driving the push-forward. So, the question of what isometries are given for $\eta$ acting on the tangent space to an event will boil down to analyzing what the set of Lorentz matrices looks like. That, in my view, is just linear algebra. This is a sketch. I'll fill in some of the details now and leave the hard part for later (unless this is not what you want in which case I will delete this)

The psuedo metric is defined by $d_{\eta}(p,q) = \sqrt{|\eta(q-p,q-p)|}$. An isometry of $\mathbb{R}^4$ is defined to be function $F: \mathbb{R}^4 \rightarrow \mathbb{R}^4$ for which $d_{\eta}(F(p),F(q)) = d_{\eta}(p,q)$. We define a translation of space time by $T(p)=p+a$ where $p,a \in \mathbb{R}^4$. Clearly $T$ serves as an isometry since $T(q)-T(p)=q+a-(p+a)=q-p$. Likewise, we define a Lorentz Transformation to be $F(p) = \Lambda p$ where $\Lambda^T \eta \Lambda = \eta$. Here I abuse notation slightly in that $\eta(v,w) = v^T\eta w$ where $\eta$ on the LHS is a bilinear mapping yet $\eta$ on the RHS is a matrix $\text{Diag}(1,-1,-1,-1)$. Notice, $$ \eta(\Lambda v, \Lambda w) = (\Lambda v)^T\eta (\Lambda w) = v^T\Lambda^T\eta \Lambda w = v^T\eta w = \eta(v,w). $$ Hence, as a Lorentz transformation is linear $F(q)-F(p)=F(q-p)$ we have $$d_{\eta}(F(p),F(q)) = \sqrt{ | \eta(F(q-p),F(q-p))|} = \sqrt{ | \eta(q-p,q-p)|} = d_{\eta}(p,q). $$ It is easy to prove the composition of isometries is once again an isometry. What is harder is to show that any isometry can be written as a composition of a spacetime translation and a Lorentz transformation. My intuition is that I can mimic the standard Euclidean argument found in Chapter 3 of Barret O'neill's Elementary Differential Geometry to complete the story for isometries of points. Then the case of vectors will also be dealt with similarly to the story told in O'neill for the Euclidean case.

Lemma: if $F$ is an isometry of $\mathbb{R}^4$ with the Minkowski product and $F(0)=0$ then there is a unique $\Lambda \in \mathbb{R}^{ 4 \times 4}$ such that $\Lambda^T \eta \Lambda = \eta$ for which $F(x) = \Lambda x$. In other words, every isometry of $(\mathbb{R}^4, \eta)$ which fixes $0$ is a Lorentz transformation.

Proof: Let $p \in \mathbb{R}^4$ and consider: $$ d_{\eta}(p,0) = d_{\eta}(F(p),F(0)) = d_{\eta}(F(p),0) \ \ \Rightarrow \ \ p^T\eta p = F(p)^T \eta F(p). $$ Where I squared both sides to obtain the implication above. Likewise, for $p,q \in \mathbb{R}^4$ consider: $$ d_{\eta}(p,q)^2 = (q-p)^T\eta (q-p) = q^T\eta q - 2q^T\eta p + p^T\eta p $$ where we noted $\eta^T=\eta$ hence $v^T\eta w = w^T \eta v$ to collect the terms with $p$ and $q$ into one summand. Likewise, $$ d_{\eta}(F(p),F(q))^2 = F(q)^T\eta F(q) - 2F(q)^T\eta F(p) + F(p)^T\eta F(p). $$ However, as $F$ is an isometry we have $d_{\eta}(p,q)^2 = d_{\eta}(F(p),F(q))^2$ and as $F(0)=0$ we also have $p^T\eta p = F(p)^T \eta F(p)$ and $q^T\eta q = F(q)^T \eta F(q)$ thus cancelling terms we deduce: $$ F(q)^T\eta F(p) = q^T\eta p $$ for all $p,q \in \mathbb{R}^4$. Notice that $\eta$ defines a bilinear form $\langle x,y \rangle = x^T \eta y$ for all $x,y \in \mathbb{R}^4$. It is easy to prove: $$ \langle cx+y, z \rangle = c\langle x,z \rangle+\langle y,z \rangle$$ Moreover, the standard basis is almost orthogonal, I suppose some would call the standard basis psuedo-orthogonal and you can prove: for $(e_i)_j = \delta_{ij}$ for $i,j =0,1,2,3$ $$ x = \langle x, e_0 \rangle e_0-\langle x, e_1 \rangle e_1-\langle x, e_2 \rangle e_2-\langle x, e_3 \rangle e_3$$ as $\langle e_0, e_0 \rangle = 1$ and $\langle e_i, e_i \rangle = -1$ for $i=1,2,3$. In particular, $v=w$ if and only if $\langle v, e_i \rangle =\langle w, e_i \rangle$ for $i=0,1,2,3$. Yet, as $\langle F(e_i), F(e_j) \rangle =\langle e_i, e_j \rangle$ we can likewise prove psuedo-orthogonality of the $F(e_0),F(e_1),F(e_2),F(e_3)$ basis and obtain the result that $v=w$ if and only if $\langle v, F(e_i) \rangle = \langle w, F(e_i) \rangle$ for $i=0,1,2,3$. With this in mind, consider: $$ \langle F(cx+y),F(z) \rangle = \langle cx+y,z \rangle = c\langle x,z \rangle +\langle y,z \rangle = c\langle F(x),F(z) \rangle +\langle F(y),F(z) \rangle$$ where I used linearity of $\langle, \rangle$ in the middle. Next, use linearity of the Minkowski product once more: $$ \langle F(cx+y),F(z) \rangle = \langle cF(x)+F(y),F(z) \rangle $$ Finally, set $z=e_o,e_1,e_2,e_3$ to obtain that $F(cx+y)=cF(x)+F(y)$ for all $x,y \in \mathbb{R}^4$. Thus there exists $\Lambda \in \mathbb{R}^{4 \times 4}$ for which $F(x) = \Lambda x$. But, $F(x)^T\eta F(y) = x^T\eta y$ for all $x,y$ hence $$ (\Lambda x)^T \eta (\Lambda y) = x^T\Lambda^T \eta \Lambda y = x^T\eta y$$ for all $x,y$ and it follows $\Lambda^T \eta \Lambda = \eta$. end of proof of Lemma. $\Box$

Theorem: every isometry $F$ of $\mathbb{R}^4$ with psuedo distance function $d_{\eta}$ can be written $F(x) = \Lambda x +a$ for a unique $\Lambda \in \mathbb{R}^{4 \times 4}$ such that $\Lambda^T\eta \Lambda = \eta$ and $a \in \mathbb{R}^4$. In words, every isometry of $\mathbb{R}^4$ with the Minkowski metric is formed by a unique composition of a Lorentz transformation and a spacetime translation.

Proof: Let $F$ be an isometry and define $G(x) = F(x)-F(0)$ notice $G(0)=F(0)-F(0)=0$ and $G$ is formed by the composition of an isometry and a translation hence is again an isometry. By the Lemma, there exists a Lorentz matrix $\Lambda$ for which $G(x) = \Lambda x$ for all $x \in \mathbb{R}^4$. Thus $F(x)=\Lambda x+F(0)$. The question of uniqueness remains. Suppose $F(x) = \Lambda_1 x+a_1$ and $F(x) = \Lambda_2 x+a_2$ note $F(0)=a_1=a_2$. Thus, $\Lambda_1 x = \Lambda_2 x$ for all $x \in \mathbb{R}^4$ hence $\Lambda_1 = \Lambda_2$ and this concludes the proof of the theorem. $\Box$

  • Thank you for the answer. What I really need is this last part. I only need an argument to show that any isometry can be written as a composition of spacetime translations and Lorentz transformations. – Giorgio Comitini Jun 26 '15 at 7:42
  • Ok, I'll work on that soon, the proof in O'neill procedes in two parts (1.) to show any isometry which fixes $0$ is orthogonal and (2.) the composition result. – James S. Cook Jun 26 '15 at 14:23
  • Ok great, thank you. I'll have a look at the proof too, later – Giorgio Comitini Jun 26 '15 at 16:01
  • Hey, it'll be a bit until I have the details fleshed out, just so you know where I'm going in more detail, see page 60 of supermath.info/DifferentialGeometry2015.pdf Basically I'm trying to adapt my argument (really O'neill's argument) there to this context. I'm worried a little about the $\pm$ signs which clutter things. I may have to break into spacelike and timelike cases... – James S. Cook Jun 26 '15 at 16:05
  • No problem, as long as it works out! Even if you don't manage to get the correct answer, any attempt to solve the problem would at least be a step forward in its understanding, so I'd be glad to hear even an incomplete proof. Anyway, thank you for your efforts!! – Giorgio Comitini Jun 26 '15 at 16:10

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