What a metric should necessarily do is it should give me a way to associate a frame invariant number with a given pair of spacetime events. Now, if I use a higher rank tensor field (say, for example, a tensor field of rank 3: $g_{\mu \nu \rho}$) then also I can certainly produce a frame invariant scalar out of a given displacement $\vec{A}$ in this trivial way: $ I := g_{\mu \nu \rho} A^{\mu}A^{\nu}A^{\rho}$. Since the components of the higher rank tensor and that of the displacement vector are going to transform in a covariant and a contravariant manner respectively, the cooked up quantity is certainly a scalar.

Another crucial property, which I think has more direct physical content than the previous one, is that between two inertial frames, at least one such transformation should exist that leaves at least one metric invariant. i.e., There should exist at least one combination of transformation matrix $\displaystyle\frac{\partial{x^{\alpha}}}{\partial{x^{\mu '}}}$ and metric $g_{\alpha \beta \gamma}$ that satisfies the following equation:

$\displaystyle\frac{\partial{x^{\alpha}}}{\partial{x^{\mu '}}}\displaystyle\frac{\partial{x^{\beta}}}{\partial{x^{\nu '}}}\displaystyle\frac{\partial{x^{\gamma}}}{\partial{x^{\rho '}}} g_{\alpha \beta \gamma} - \delta_{\mu '}^{\alpha}\delta_{\nu '}^{\beta}\delta_{\rho '}^{\gamma} g_{\alpha \beta \gamma} =0$

The last thing I can think of that can put a restriction on the choice of a tensor as a metric is the existence of a possibility of finding a metric compatible symmetric connection field. Following the usual procedure of finding the expression for a metric compatible symmetric connection field, I reached following condition (unlike the case of the usual two rank metric where we get a full-fledged expression) for the connection in the terms of the metric:

$g_{\nu \rho k_1} \Gamma^{k_1}_{\mu \lambda} - g_{\lambda \mu k_2} \Gamma^{k_2}_{\rho \nu} = \displaystyle\frac{1}{2} (\partial_{\nu}g_{\rho \lambda \mu} + \partial_{\rho}g_{\lambda \mu \nu} - \partial_{\lambda}g_{\mu \nu \rho} - \partial_{\mu}g_{\nu \rho \lambda}) $

My question is that if it is possible to satisfy the two highlighted conditions then can we use such a 3 rank (or even higher rank tensors with similarly produced conditions) tensor fields as metric fields?

PS: This is NOT a proposal for a new home-production theory of gravity (or that of anything for that matter) but rather it is just that I am trying to understand why a two rank tensor is used in General Relativity as the metric. Thank you.

  • 1
    $\begingroup$ you can certainly consider higher rank tensors, but why would you want to call them metric? a metric is, by definition, $\langle\cdot,\cdot\rangle$, i.e., it is an inner product. You can define objects like $\langle\cdot,\cdot,\cdot\rangle$, but these have nothing to to with the standard definition of a metric... $\endgroup$ Nov 28, 2016 at 22:26
  • $\begingroup$ Metric is not (by its own definition) what defines the inner product. Metric is (by definition) what measures. The point is should I always represent my measure as the inner product of a displacement with itself or can I do a triple inner product of the displacement with itself to represent the measure and formulate the theory in those terms? Is there something wrong with the Physics in doing so or not? I mean I can go out in Nature and determine what my metric is. Now it certainly matters whether it has 10 independent bits of information or more. $\endgroup$
    – user87745
    Nov 28, 2016 at 23:03
  • $\begingroup$ Seems to me that this is more of a maths question than a physics one $\endgroup$
    – Kyle Kanos
    Nov 28, 2016 at 23:05
  • $\begingroup$ Would it help if I say I am trying to understand with the method of contradiction? i.e., prove the other tensors can't serve and thus, two rank tensor must be the only option. $\endgroup$
    – user87745
    Nov 29, 2016 at 4:06
  • 1
    $\begingroup$ OP, the metric is just a way of defining the length of a vector $\sqrt{g(x,x)}$ and the angle between two vectors $g(x,y)=\cos\theta\sqrt{g(x,x)}\sqrt{g(y,y)}$ on a manifold. It has two arguments, hence is a 2-tensor, and the usual rules of inner products give it symmetry. It is not clear what a symmetric trilinear product would give you mathematically. Whether it is physically useful is another question. But it would certainly not be a "metric." $\endgroup$
    – Ryan Unger
    Nov 29, 2016 at 17:01

3 Answers 3


I) OP is interested in totally symmetric covariant $(0,r)$ tensor fields $$g\in\Gamma\left({\rm Sym}^r(T^{\ast}M)\right) \tag{A}$$ on an $n$-dimensional manifold $M$. The number of totally symmetric tensor components are $$\begin{pmatrix} n+r-1 \cr r\end{pmatrix} .\tag{B}$$

II) If the manifold $M$ is paracompact, we can use the partition of unity to prove

  1. that there exist globally defined positive definite$^{\dagger}$ tensor fields (A).

  2. that there exist globally defined torsion-free tangent-bundle connections $\nabla$.

(To see point 2, use point 1 for the case $r=2$ to deduce the existence of a globally defined positive definite metric tensor field, and hence a globally defined Levi-Civita connection.)

III) Next we extract the interesting part of OP's question as follows:

Can we choose a torsionfree tangent bundle connection $\nabla$ that is compatible $$\nabla g~=~0 \tag{C}$$ with a given tensor field (A)?

Generically the answer is No, not even locally, if the rank $r\geq 3$. This is because the number $$n\begin{pmatrix} n+r-1 \cr r\end{pmatrix} \tag{D}$$ of compatibility conditions (C) is greater than the number $$n \begin{pmatrix} n+1 \cr 2\end{pmatrix} \tag{E}$$ of Christoffel symbols, if $r\geq 3$. So the equations are overconstrained.

IV) We leave it to the reader to generalize the above to (not necessarily totally symmetric) higher-rank $(s,r)$ tensor fields. Higher-rank tensor fields appear e.g. in string theory, AKSZ sigma models & higher spin theories. For generalizations of Riemannian geometry, see also Finsler geometry.


$^{\dagger}$ We define that the tensor field $g$ is positive definite if $$\forall p\in M ~\forall X_p\in T_pM\backslash\{0\}: \quad g_p(\underbrace{X_p,\ldots, X_p}_{r\text{ entries}}) ~>~ 0. \tag{F}$$

  • $\begingroup$ Thanks for resolving my doubt! I found this article on arXiv (which also has some similarities to your answer in its arguments concerning the connection) discussing the introduction of higher rank symmetric tensors along with a metric and torsion-free connection field in the construction of a modified Einstein-Hilbert action. Would you comment on the relevance of such proposals to the mainstream Theoretical Physics? arxiv.org/abs/1409.6757 $\endgroup$
    – user87745
    Nov 29, 2016 at 16:04
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Nov 29, 2016 at 18:46
  • $\begingroup$ @Qmechanic, wow this answer is very knowledge-dense. Would you happen to know of a book where I could read about this topics and get to understand your answer? For the record, I am currently reading chapter 2 of Fecko's Differential geometry and Lie groups for physicists. $\endgroup$
    – user137661
    Feb 27, 2020 at 18:26
  • $\begingroup$ Thanks for the feedback. I don't recall any relevant references just now. $\endgroup$
    – Qmechanic
    Feb 27, 2020 at 19:50

Given a vector space $V$, a metric is defined as a map that, given any two elements $v,u\in V$, associates a real positive number $d(v,u) \geq 0$, namely the distance between the two. Since a distance can be derived from a scalar product $\sigma$ by means of $d(v,u) = \sqrt{\sigma(v-u, v-u)}$ assigning a metric is equivalent to assigning a scalar product.

A 2-rank tensor is, by definition, a multilinear map $\tau\colon V\times V\to\mathbb{C}$ and therefore exactly a scalar product. Given $\mathcal{M}$ as space-time manifold with charts $U_i$ and tangent spaces $T_m\mathcal{M}$ in each point $m\in\mathcal{M}$, it is natural to define scalar products in each point as the action of a $(2,0)$-type tensor onto the vectors, evaluated at each point: namely $$ \sigma(X_m, Y_m) = g(m)(X_m, Y_m) $$ induces a positive defined distance in each $T_m\mathcal{M}$ provided $g$ be positive-definite.

This seems to be the most natural choice and although one might use higher rank tensor one would need to evaluate the remaining components onto fixed bases in each $T_m\mathcal{M}$ (to exhaust the remaining entries), which reduces the action to exactly a 2-rank tensor again.


You could definitely use higher order metrics, so that the length element would be $(ds^2)^n$, but that would be non-Riemannian geometry, i.e not GR, plus you have to confront the possibility of $c_n$ light-like characteristics, general Lorentz-invariance failure etc.

  • 2
    $\begingroup$ The metric is not an 2-form, it's a symmetric tensor. $\endgroup$
    – Slereah
    Nov 29, 2016 at 8:19
  • $\begingroup$ @Slereah Thanks for pointing out the correction. What is the reason of using a 2-rank symmetric covariant tensor as the metric in GR if I can also get "a metric compatible symmetric connection as well as the transformation that, in a flat space, leave my metric components invariant" via using a general n-rank symmetric covariant tensor as the metric? $\endgroup$
    – user87745
    Nov 29, 2016 at 8:24
  • $\begingroup$ @Slereah yes, you are right, but in the scope of vector spaces and mappings I think 2-forms and bilinear forms (metric) are seen as the same, apart from the fact that the metric tensor must by definition be symmetric and positive definite, when in Riemmanian geometry. Isn't that right? $\endgroup$ Nov 29, 2016 at 10:54
  • $\begingroup$ No, they are not seen as the same thing. A $k$-form on a vector space is a totally antisymmetric $k$-tensor, this is standard terminology. $\endgroup$
    – Ryan Unger
    Nov 29, 2016 at 16:55
  • $\begingroup$ @0celo7 you're absolutely right. Got a bit confused here. Thank you! $\endgroup$ Nov 29, 2016 at 17:15

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