The only numerical tensors that are invariant under some relevant symmetry group (the Euclidean group in Newtonian mechanics, the Poincare group in special relativity, and the diffeomorphism group in general relativity) are the metric $g_{\mu \nu}$, the inverse metric $g^{\mu \nu}$, the Kronecker delta $\delta^\mu_\nu$, and the Levi-Civita tensor $\sqrt{|\det g_{\mu \nu}|} \epsilon_{\mu \nu \dots}$. (Note that $\delta^\mu_\nu = g^{\mu \rho} g_{\rho \nu}$, so only two of the first three invariant tensors are independent). This result is extremely useful for constructing all possible scalar invariants of a given set of tensor fields, but I've never actually seen it proven. How does one prove that there are no other invariant tensors?

  • $\begingroup$ I did actually come across a proof of this, but I forget where. $\endgroup$ – Mozibur Ullah Aug 3 '17 at 23:34
  • $\begingroup$ Are you considering sums as "not independent"? Or is this more restrictive than I think it is? $\endgroup$ – CR Drost Aug 3 '17 at 23:47
  • $\begingroup$ @CRDrost Could you clarify your question? The four listed tensors all have different index structures, and cannot be added together. $\endgroup$ – tparker Aug 4 '17 at 0:19
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    $\begingroup$ This is the topic of study in Invariant theory. They think in terms of polynomials rather than tensors, so there is often some translating work to do. $\endgroup$ – Oscar Cunningham Aug 4 '17 at 5:47
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    $\begingroup$ @AlexNelson The Riemann tensor (and its contractions, the Ricci tensor and scalar) can be calculated from the metric and the inverse metric, so it isn't independent of the listed tensors. $\endgroup$ – tparker Aug 9 '17 at 16:18

For a proof that the only ${\rm SO}(N)$ invariant tensors are products of $\delta_{ab}$'s and Levi-Civita symbols see M. Spivak, A Comprehensive Introduction to Differential Geometry (second edition) Vol. V, pp. 466-481. The number of pages required for the argument shows that it is not trivial.

I've just looked up the third edition of Spivak vol V. What is needed is theorem 35 on page 327. This is the section entitled "A Smattering of Classical Invariant Theory." He writes in terms of scalar invariants, but of course an invariant tensor becomes a scalar invariant when contracted with enough vectors, and any such scalar invariant arises from an invariant tensor.

  • $\begingroup$ Do you know the page numbers for the (current) third edition? $\endgroup$ – tparker Aug 11 '17 at 19:23
  • $\begingroup$ If not, could you give more context for the section, so that one could find the corresponding material in the third edition? $\endgroup$ – tparker Aug 11 '17 at 19:29
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    $\begingroup$ I;ve amended my answer to give the third edition refernce. $\endgroup$ – mike stone Aug 13 '17 at 14:45

The rotation group ($O(3)$ for Euclidean space, $O(1,3)$ for Minkowski signature) is defined as the group of all linear transformations that preserve the metric tensor. If there were other independent tensors, this would give additional restrictions, so it would necessarily define a subgroup of the rotation group.

Demanding that the volume form (~ the Levi-Civita) is also preserved indeed restricts to the special rotation group.

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    $\begingroup$ It's in principle possible that there are other 'accidentally' invariant tensors that don't place additional constraints on the group in question, no? $\endgroup$ – gj255 Aug 4 '17 at 0:15
  • $\begingroup$ That explains why the metric tensor $g_{\mu \nu}$ is invariant, but then why is the Kronecker delta $\delta^\mu_\nu$ (or equivalently, the inverse metric $g^{\mu \nu}$) also invariant? And how can we prove that the volume form is the only additional tensor invariant under $SO(n, m)$ transformations? $\endgroup$ – tparker Aug 4 '17 at 1:00
  • $\begingroup$ The delta is the identity tensor, so it is invariant under any transformations. This is not equivalent to the inverse metric tensor being invariant. $\endgroup$ – Robin Ekman Aug 4 '17 at 1:00
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    $\begingroup$ I believe that it is equivalent given that the metric is invariant. $\endgroup$ – tparker Aug 4 '17 at 4:33
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    $\begingroup$ Depending on your background and motivation, you might find the "birdtracks monograph" by Cvitanović interesting (available [free][1] from the author online). Although it's largely about the author's diagrammatic approach to group theory (and the sort of group theory that arises in perturbative QFT, in particular), it contains many interesting, idiosyncratic, and useful observations about group theory more generally. The book spells out how to "define" Lie groups / algebras in terms of their invariant tensors along the lines that @Robin_Ekman mentions. [1]: birdtracks.eu $\endgroup$ – wijay Aug 6 '17 at 20:49

First of all, Kronecker $\delta$ is $\textit{Diff}$-invariant since it is defined as the dual pairing $<.,.>:V^*\otimes V\to k.$ Its components are ${\delta^\mu}_\nu=\delta(\mathrm d x^\mu,\partial_\nu)=<\mathrm d x^\mu,\partial_\nu>=\delta^\mu_\nu$, where the last "$\delta^\mu_\nu$" is simply the Kronecker symbol, equal to 1 or 0 the usual way. It is the same in any coordinate system. Feel free to flesh out the details yourself.

Now, I feel the need to clarify something regarding covariance, invariance and the diffeomorphism group. You list the Galilean and Poincare groups along with diffeomorphisms. This is wrong, and many textbooks on GR gloss over it in a hurry to introduce the Einstein equations and discuss solutions. Great, but it gave me many unnecessary headaches, and those are arguably the worst kinds of headaches.

Let's consider $(\mathcal M,g,\nabla)$ where $\nabla$ is Levi-Civita (ie metric compatible) connection (meaning $\nabla g =0$). We also note that any vector field $\xi$ generates a 1-parameter diffeomorphsim via the pullback by its flow ${\phi_t}:\mathbb R\times \mathcal M\to\mathcal M$, which satisifes $\phi_{t+s}=\phi_t\circ \phi_s$ and $\phi_0=1$. Locally, $\phi_t = \exp{t\xi}$. Forgive my glossing over details. It just means that if things are the same in the direction a vector points, they will remain the same if you follow the vector along the whole path, using it like a compass.

Now, it can be shown that diffeomorphism invariance ${\phi_t}^*g =g$ implies $\mathcal L_\xi g =0.$ However, $$\mathcal L_\xi g_{ab} = 2 \nabla_{(a}\xi_{b)}\neq 0$$ in general, except for Killing fields. The metric is always covariant, but it's not in general invariant under diffeomorphisms. Minkowski space is has the maximal possible number of Killing vectors: it has 10 vectors which correspond to the 10 generators of the Poincare group. Besides Minkowski space, only deSitter and AntideSitter spaces have this property. This is not something we can expect from GR. The Kerr metric describing a rotating black hole only has a 2 dimensional subgroup of the Poincare group: time translations and rotations around only one axis.

The full diffeomorphism group isn't a symmetry but a gauge symmetry. It represents our freedom to describe nature in a different coordinate system, but not a symmetry of the physical system itself. For this, we need Killing fields which generate what we usually mean by a "symmetry". To illustrate my claim about it being a gauge symmetry, in linearized gravity we have the gauge transformation $$\gamma_{ab}\to\gamma_{ab}'=\gamma_{ab}-\nabla_a\xi_b-\nabla_b\xi_a$$ which corresponds to $(\mathcal M,g)$ and $(\mathcal M,{\phi_t}^*g)$ representing the same spacetime. We see that $\gamma_{ab}'=\gamma_{ab}$ iff this is a "proper symmetry". Moreover, we can define conserved charges only using the Killing symmetries. In the Kerr black hole case, these correspond to the mass and to the total angular momentum.

Any clearer? To recap: anything we write as $T_{a_1a_2,...}^{b_1 b_2,...}$ is covariant. This just means that gets all those partial derivative factors under a coordinate change (ie diffeomorphism). As an example, take the Minkowski metric $$\eta=-\mathrm dt^2 + \mathrm d x^2+ \mathrm d y^2+ \mathrm d z^2$$ If we switch to spherical coordinates, we get $$\eta=-\mathrm dt^2 + \mathrm d r^2+ r^2\mathrm d \theta^2 + r^2sin^2\theta\mathrm d\phi^2$$ It is clearly not the same, but it is covariant. A real symmetry transformation would be to use the Poincare group, and then the metric would look exactly the same, so it would be invariant.

There are many invariant tensors if we have a Killing field. Given a Killing field $\xi_a$, we can construct a new invariant tensor, $\xi_a\xi_b$. For instance, $$h_{ab}\equiv g_{ab} - (\xi^a\xi_a)^{-1}\xi_a\xi_b$$ is the metric of a spacelike or timelike hyperplane orthogonal to the orbit of $\xi_a$, and is invariant. These may not be invariant under the action of another Killing field, though. Only under special circumstances. Using tensors made up from the metric is a safe bet, since symmetries are explicitly given by invariance of the metric.


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