I am quite confused with the groups $\operatorname{Diff}(M)$ and $\operatorname{GL}(4,\mathbb{R})$ in the context of general relativity. I understand that the symmetries of GR are the transformations that leave the equations invariant under under continuous smooth coordinate transformations. That is, automorphisms $f\colon M \to M$ form an infinite dimensional Lie group, the group $\operatorname{Diff}(M)$.

Now, I had the impression that the isometry group of GR is $\operatorname{GL}(4,\mathbb{R})$ and something like that is stated at the Wiki page of general covariance. In specific it says that

A more modern interpretation of the physical content of the original principle of general covariance is that the Lie group $\operatorname{GL}(4,R)$ is a fundamental "external" symmetry of the world.

I am struggling to realise if $\operatorname{Diff}(M)$ is isomorphic to $\operatorname{GL}(4,\mathbb{R})$ and if in this context talking about diffeomorphism group of GR is the same as talking about $\operatorname{GL}(4,\mathbb{R})$ or if this is a subgroup of $\operatorname{Diff}(M)$. What exactly is their relation?


3 Answers 3

  1. Let there be given a 4-dimensional real manifold$^1$ $M$. As OP says, the set ${\rm Diff}(M)$ is the group of globally defined $C^{\infty}$-diffeomorphisms $f:M\to M$. The set ${\rm Diff}(M)$ is an infinite-dimensional Lie group. (To actually explain mathematically what the previous sentence means, one would have to define what an infinite-dimensional manifold is, which is beyond the scope of this answer.)

  2. There is also the groupoid ${\rm LocDiff}(M) \supseteq {\rm Diff}(M)$ of locally defined $C^{\infty}$-diffeomorphisms $f:U\to V$ (i.e. the invertible morphisms in the category). Here $U,V\subseteq M$ are open neighborhoods (i.e. objects in the category).

  3. The above is part of the active picture. Conversely, in the passive picture, there is the groupoid $LCT(M)$ of local coordinate transformations $f:U\to V$, where $U,V\subseteq\mathbb{R}^4$. Heuristically, due to the dual active & passive formulations, the two groupoids ${\rm LocDiff}(M)$ and $LCT(M)$ must be closely related at "the microscopic level". (We leave it to the reader to try to make the previous sentence precise.)

  4. Next let us consider the frame bundle $F(TM)$ of the tangent bundle $TM$. It is a principal bundle with structure group $GL(4,\mathbb{R})$, which is a 16-dimensional Lie group.

  5. Given two locally defined sections $$(e_0, e_1, e_2, e_3), (e^{\prime}_0, e^{\prime}_1, e^{\prime}_2, e^{\prime}_3)~\in~ \Gamma(F(TM_{|W})), \tag{1}$$ in some neighborhood $W\subseteq M$, then there is a locally defined $GL(4,\mathbb{R})$-valued section $$\Lambda~\in~\Gamma(GL(4,\mathbb{R})\to W), \tag{2}$$ such than the two sections (1) are related via$^2$ $$e^{\prime}_{b}~=~ \sum_{a=0}^3 e_{a} \Lambda^{a}{}_{b}, \qquad b~\in~\{0,1,2,3\} \tag{3}. $$

  6. Conversely, given only one of the sections in eq. (1), we can use an arbitrary $GL(4,\mathbb{R})$-valued section (2) to define the other frame via eq. (3).

  7. Let us now return to the groupoid $LCT(M)$ and see how the structure group $GL(4,\mathbb{R})$ comes in. In details, let there be given two local coordinate charts $U,U^{\prime}\subseteq M$, with non-empty overlap $U\cap U^{\prime}\neq \emptyset$, and with local coordinates $(x^0,x^1,x^2,x^3)$ and $(x^{\prime 0},x^{\prime 1},x^{\prime 2},x^{\prime 3})$, respectively. Then we have two locally defined sections $$\left(\frac{\partial}{\partial x^0}, \frac{\partial}{\partial x^1}, \frac{\partial}{\partial x^2},\frac{\partial}{\partial x^3}\right)~\in~ \Gamma(F(TM_{|U})), $$ $$\left(\frac{\partial}{\partial x^{\prime 0}}, \frac{\partial}{\partial x^{\prime 0}}, \frac{\partial}{\partial x^{\prime 0}},\frac{\partial}{\partial x^{\prime 0}}\right)~\in~ \Gamma(F(TM_{|U^{\prime}})), \tag{4}$$ in the frame bundle. The analogue of the $GL(4,\mathbb{R})$-valued section (2) is given by the (inverse) Jacobian matrix$^1$ $$ \Lambda^{\mu}{}_{\nu} ~=~\frac{\partial x^{\mu}}{\partial x^{\prime \nu}},\qquad \mu,\nu~\in~\{0,1,2,3\}, \tag{5} $$ cf. the chain rule.

  8. Conversely, note that not all $GL(4,\mathbb{R})$-valued sections (2) are of the form of a Jacobian matrix (5). Given one local coordinate system $(x^0,x^1,x^2,x^3)$ and given a $GL(4,\mathbb{R})$-valued section (2), these two inputs do not necessarily define another local coordinate system $(x^{\prime 0},x^{\prime 1},x^{\prime 2},x^{\prime 3})$. The $GL(4,\mathbb{R})$-valued section (2) in that case evidently needs to satisfy the following integrability condition $$ \frac{\partial (\Lambda^{-1})^{\nu}{}_{\mu}}{\partial x^{\lambda}} ~=~ (\mu \leftrightarrow \lambda). \tag{6}$$

  9. So far we have just discussed an arbitrary $4$-manifold $M$ without any structure. For the rest of this answer let us consider GR, namely we should equip $M$ with a metric $g_{\mu\nu}\mathrm{d}x^{\mu}\odot \mathrm{d}x^{\nu}$ of signature $(3,1)$.

  10. Similarly, we introduce a Minkowski metric $\eta_{ab}$ in the standard copy $\mathbb{R}^4$ used in the bundle $GL(4,\mathbb{R})\to W$. We now restrict to orthonormal frames (aka. as (inverse) tetrads/vierbeins) $$(e_0, e_1, e_2, e_3)~\in~ \Gamma(F(TM_{|W})), \tag{7}$$ i.e. they should satisfy the orthonormal condition $$ e_a\cdot e_b~=~\eta_{ab}. \tag{8} $$

  11. Correspondingly, the structure group $GL(4,\mathbb{R})$ of the frame bundle $F(TM)$ is replaced by the proper Lorentz group $SO(3,1;\mathbb{R})$, which is a $6$-dimensional Lie group. In particular the sections (2) are replaced by locally defined $SO(3,1;\mathbb{R})$-valued sections $$\Lambda~\in~\Gamma(SO(3,1;\mathbb{R})\to W). \tag{9}$$ This restriction is needed in order to ensure the existence of finite-dimensional spinorial representations, which in turn is needed in order to describe fermionic matter in curved space. See also e.g. this Phys.SE post and this MO.SE post.

  12. Consider a covariant/geometric action functional $$S[g, \ldots; V]~=~\int_V \! d^4~{\cal L} \tag{10}$$ over a spacetime region $V\subseteq M$, i.e. $S[g, \ldots; V]$ is independent of local coordinates, i.e. invariant under the groupoid $LCT(M)$. The action $$S[g, \ldots; V]~=~S[f^{\ast}g, \ldots; f^{-1}(V)] \tag{11}$$ is then also invariant under pullback with locally defined diffeomorphisms $f\in{\rm LocDiff}(M)$.

  13. In summary, the symmetries of GR are:

    • Pullbacks by the group ${\rm Diff}(M)$ of globally defined diffeomorhisms.
    • Pullbacks by the groupoid ${\rm LocDiff}(M)$ of locally defined diffeomorphisms.
    • The groupoid $LCT(M)$ of local coordinate transformations, and
    • The local $SO(3,1;\mathbb{R})$ Lorentz transformations (9) of the tetrads/vierbeins.


$^1$ In most of this answer, we shall use the language of a differential geometer where e.g. a point/spacetime-event $p\in M$ or, say, a worldline have an absolute geometric meaning. However, the reader should keep in mind that a relativist would say that two physical situations which differ by an active global diffeomorphism are physically equivalent/indistinguishable, and hence a point/spacetime-event $p\in M$ does only have an relative geometric meaning.

$^2$ Conventions: Greek indices $\mu,\nu,\lambda, \ldots,$ are so-called curved indices, while Roman indices $a,b,c, \ldots,$ are so-called flat indices.

  • 4
    $\begingroup$ $LCT(M)$ isn't an actual symmetry, is it? Every single physical theory is invariant under coordinate transformations. GR is special because it's invariant under active diffeomorphisms (if $g$ is a solution, so is $\phi^* g$, $\phi\in Diff(M)$). Although the effect is the same as in a coordinate transf. (at least locally), the interpretation is quite different. The $SO(3,1)$ is also just some sort of change of coordinates on the tangent bundle, so it's not a proper symmetry as well. $\endgroup$
    – user74106
    Commented Oct 11, 2016 at 21:14
  • 1
    $\begingroup$ $\uparrow$ I agree. $\endgroup$
    – Qmechanic
    Commented Jan 26, 2018 at 12:50
  • $\begingroup$ Correction to the answer (v6): In footnote 1 the word an should be a. $\endgroup$
    – Qmechanic
    Commented Jan 27, 2018 at 11:33
  • $\begingroup$ Great answer! I Believe points 10, 11, and part of 13 are relevant to a question I posted here:physics.stackexchange.com/q/445738 I was specifically thinking of a finite spinor representation when I asked that question. $\endgroup$
    – R. Rankin
    Commented Dec 13, 2018 at 11:31

A very good answer to the above question has already been provided here, where Marek summarises the differences between the symmetry group of a theory and the groups of coordinates transformations leaving the equations invariant.

In a nutshell (but it is more complex) let $f\colon U\to V$ be any coordinates transformation on charts of a manifold $U,V\subset\mathcal{M}$ (i. e. a change of coordinates). Under such transformation fields $\phi(x)$ are sent into $\phi'(f(x)) = S(x)\phi(x)$.

In order the equations of motion to be satisfied, one must require certain appropriate conditions on the factor $S(x)$ (in particular one can see that these could be related to the representations of the underlying group of transformations $f$). The set of all allowed operators $S(x)$ defines the symmetry group of the theory for the general mapping $f$ as defined above. In the case of general relativity $f$ are the diffeomorphisms and $S(x)$ span the general linear group (up to isomorphisms and cartesian products).

  • $\begingroup$ I cannot really understand what this transformation you have written is. I see a scalar field that you call $\phi'(f(x))$ and you ask that this $S(x)$ satisfies some properties. Any tensorial object, like $\phi$ should have the Jacobian of the coordinate transformations in front of it and this is related to the Diff($M$). I do not see how the $GL$ group comes into the picture. $\endgroup$
    – Marion
    Commented Dec 23, 2015 at 9:56
  • $\begingroup$ There is no requirement on $\phi(x)$ to be a scalar field, nor to be a tensor. It is any type of field that transforms in any possible way, this being determined by its nature and the equation of motion. For example if you take the Dirac equation the field $\phi$ is a spinor field and $S(x)$ are representations of the Clifford algebra. $\endgroup$
    – gented
    Commented Dec 23, 2015 at 10:36
  • $\begingroup$ Moreover, in the special case you have mentioned, the set of all Jacobian matrices is exactly a subset of $\textrm{GL}(n, \mathbb{R})$. $\endgroup$
    – gented
    Commented Dec 23, 2015 at 10:39
  • 1
    $\begingroup$ The representations of the (quotient of the) diffeomorphisms group can be (in some cases) described by $\textrm{GL}(n)$ (up to cartesian products, isomorphisms and quotients). For the general case, I must say I'm not an expert but that can be looked at investigating standard representations of the diffeomorphisms group. $\endgroup$
    – gented
    Commented Dec 23, 2015 at 13:16
  • 1
    $\begingroup$ @GennaroTedesco could you cite a reference for all of this? The fact the group $GL(n,\mathbb{R})$ appears in GR is because it's the structure group of the frame bundle. I'm just having a hard time understanding why the representation of Diff happens to be the same as the structure group. BTW, it's weird that the infinite dimensional group Diff has a finite dimensional representation on $GL(n,\mathbb{R})$. Why is $GL(n,\mathbb{R})$ the gauge group of GR and not Diff? $\endgroup$
    – user74106
    Commented Aug 2, 2016 at 10:41

It is helpful to remember that the fields of general relativity are maps

\begin{equation} M \rightarrow \Sigma \,, \end{equation}

where $M$ is spacetime and you can think of $\Sigma$ either as the space of metrics or, better yet, as the space of frame fields at a point. In either case, the field is a four by four invertible matrix, thus an element of $\mathrm{GL}(4,\mathbb{R})$. Of course it is not just any element of this group: the metric is symmetric, and the frame field is upper trianglular. In either case, what you really have is a field valued in $\Sigma=\mathrm{GL}(4,\mathbb{R})/\mathrm{SO}(1,3)$.

Now returning to the question: $\mathrm{Diff}(M)$ acts on $M$ and $\mathrm{GL}(4,\mathbb{R})$ acts on $\Sigma$. So the $\mathrm{Diff}(M)$ symmetries are spacetime symmetries and the $\mathrm{GL}(4,\mathbb{R})$ symmetries are internal symmetries.


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