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Why is it that all problems I encountered until now have metrics that when represented in a matrix form turn out to be symmetric? Aren't there asymmetric matrices representing some metrics?

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    $\begingroup$ In some string theories/supergravities there is an antisymmetric tensor (the NS-NS or Kalb-Ramond 2-form) that, in some contexts, is natural to add to the metric to make a sort of generalized `non-symmetric metric'. The geometrical interpretation is far from clear though. $\endgroup$ – Holographer Jan 31 '15 at 22:19
  • $\begingroup$ As far as I'm aware generalized metric is still symmetric, even though it involves $B$. The sacrifice you have to make is to generalize the tangent bundle to include the cotangent bundle also. This is the idea of generalized geometry, where you include gauge transformations as well as diffeomorphisms when defining the spacetime manifold. See for example Nigel Hitchin's excellent notes. $\endgroup$ – Edward Hughes Apr 24 '15 at 13:16
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A metric on a manifold $M$ is, by definition, a symmetric 2-tensor field $g$ with the property that $g_x$ is positive-definite for every $x\in M$ (plus some smoothness/continuity requirements if $M$ is smooth/topological). This ensures that the norm of a vector in a fibre of the tangent bundle to $M$ is a non-negative number, and that the angle between vectors doesn't depend on the order in which you choose them, i.e. $$\frac{g(u,v)}{\sqrt{g(u,u)g(v,v)}} = \frac{g(v,u)}{\sqrt{g(u,u)g(v,v)}}.$$

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    $\begingroup$ Perhaps it's a good idea to mention the relaxation of positive-definiteness in the physically relevant case of Semi-Riemannian manifolds. $\endgroup$ – Danu Jan 31 '15 at 23:01
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    $\begingroup$ +1 (fake edit: cosine of the) angle being independent of order is a reasonable and intuitive physical requirement. $\endgroup$ – Stan Liou Feb 1 '15 at 1:02
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    $\begingroup$ This sounds impressive but doesn't really answer the question. As dicsussed in some of the other answers, Einstein and others did consider non-symmetric metrics as possible ways of constructing unified field theories. So we need a physical argument, not just an assertion that certain mathematical properties are desirable. $\endgroup$ – Ben Crowell Jan 15 '19 at 14:42
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    $\begingroup$ There is no such thing as non-symmetric metric. Appealing to the definition of the word doesn't constitute an answer. Mathematicians also reserve the word "metric" for positive-definite metrics, but physicists use it more inclusively in the semi-Riemannian case. The question is a physics question that requires a physical answer, and it needs to be an answer that includes the semi-Riemannian case. $\endgroup$ – Ben Crowell Jan 15 '19 at 14:52
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    $\begingroup$ Note that symmetric doesn't imply positive definite. I didn't say that it did. The point, which was originally made in Danu's comment, is that your answer only discusses the Riemannian case, whereas what we care about here is the semi-Riemannian case. $\endgroup$ – Ben Crowell Jan 15 '19 at 15:05
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Technically, yes (for loose enough definition of "metric"), but there's very little point to it.

Some attempts at unification of gravity and electromagnetism, including several attempts by Einstein and various co-authors, basically amount to some variation of trying to interpret the antisymmetric part $g_{[ab]}$ as the electromagnetic Faraday tensor $F_{ab}$.

These tend to be unworkable because in general relativity, the metric acts like a potential for the gravitational field, so the antisymmetric part should work like some kind of potential too. But the electromagnetic potential is a four-vector instead. However, as mentioned in the comments, a Kalb–Ramond field generalizes electromagnetic potential would be of the right type.

A bigger obstacle to that to some kind of asymmetric metric, however, is that it's just not very useful. You could always decompose it into symmetric and anti-symmetric parts, so it's effectively "usual sort of metric plus an extra field" anyway. Since a symmetric metric has a much clear geometrical interpretation and rich mathematical theory, you won't gain anything if you forcibly jam an antisymmetric potential into a metric.

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  • $\begingroup$ it's effectively "usual sort of metric plus an extra field" anyway. Einstein actually addressed that criticism: symmetry of the tensors is non-essential for the formalism; instead, he demands 'transposition symmetry' of the laws and links it to charge conjugation $\endgroup$ – Christoph Feb 1 '15 at 2:45
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In general, having an asymmetric matrix for a metric won't really help, because only its symmetric part will contribute to the norm of any given vector.

Take some finite-dimensional real vector space $V$ with an inner product $\langle\cdot, \cdot\rangle$ represented by some matrix $g_{ij}$ in a given basis $\beta$. Then for any vector $v$ with components $v_i$ in $\beta$, you can rewrite its norm as $$ \langle v,v\rangle= v_i g_{ij}v_j= v_j g_{ji}v_i $$ by changing the indices. This is equivalent to taking the transpose of the matrix equation $$\langle v,v \rangle =v^Tgv$$ to get $$\langle v,v \rangle =v^Tgv=v^T g^T v.$$

If you now add both sides of the equations and divide by two, you get $$ \langle v,v \rangle=v^T\frac{g+g^T}{2}v=v_i\frac{g_{ij}+g_{ji}}{2}v_j. $$ In essence, you can safely replace $g$ by its symmetric part $g_S=\tfrac12 (g+g^T)$, because its antisymmetric part $g_A=\tfrac12 (g-g^T)=g-g^T$ does not contribute to the norm of any element (since $v^T g_A v=(v^T g_A g)^T=-v^Tg_A g$ and therefore $v^T g_A v=0$).

Now, of course, the antisymmetric part does play a role in the calculation of any general inner product $\langle v,w\rangle$. However, it is only norms which are physically measurable, so the consequences of an antisymmetric metric would not be measurable.

Such consequences, however, would go deeply against the mathematics, since one of the basic axioms of inner product spaces is that they be symmetric, $$ \langle v,w\rangle=\langle w,v\rangle, $$ which in turn requires a symmetric matrix. This means that it's OK to have an antisymmetric matrix for the norm, because it doesn't actually change the norm of any vector, but you will be unable to use this matrix for an inner product, and you will therefore miss out on the inner-product structures on which all of GR theory rests.

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  • $\begingroup$ This is interesting, but I don't think I buy the claim that only norms are observable. For example, it's certainly possible to carry out a series of measurements that verifies that two spacetime vectors are orthogonal; this is essentially what we do in Einstein synchronization. $\endgroup$ – Ben Crowell Jan 15 '19 at 14:40
  • $\begingroup$ @BenCrowell "orthogonal" doesn't make sense (by itself) in an asymmetric metric - for two given vectors $v$ and $w$, you might have $\langle v,w\rangle = 0$ but $\langle w,v\rangle \neq 0$; this almost certainly translates into an asymmetric synchronization procedure, which doesn't sound right to me. It's possible that what you indicate can be developed into a full procedure, but without seeing a concrete and fully-worked-out example, I'm extremely skeptical. $\endgroup$ – Emilio Pisanty Jan 15 '19 at 14:46
  • $\begingroup$ @EmilioPisanty in principle, such a procedure would not be able to simply assume that the two-way light travel time is half the one-way time (as Einstein assumed)? $\endgroup$ – N. Steinle Jan 15 '19 at 15:13
  • $\begingroup$ @N.Steinle You'd have to ask Ben. $\endgroup$ – Emilio Pisanty Jan 15 '19 at 16:42
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Asymmetric tensors have been considered in the quest for a unified classical field theory.

Einstein in particular went through a whole series of candidate theories. His last paper on the subject - co-authored by Bruria Kaufman, submitted 3 months before and published 3 months after his death - is about a field of this type; the theory actually was referred to as 'the theory of the non-symmetric field'.

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  • $\begingroup$ How did he reconcile that with @danu's answer? Did he think distance $A\to B \neq B\to A$? $\endgroup$ – innisfree Jan 31 '15 at 22:42
  • $\begingroup$ @innisfree: the field in question is no longer the metric tensor representing gravity, but the field, from which you need to derive all other objects of the theory; I'd have to look up how you get to the metric (there were approaches where the metric is just the symmetric part, but I don't remember if this particular theory is one of them) $\endgroup$ – Christoph Jan 31 '15 at 22:54
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    $\begingroup$ @innisfree there's nothing to reconcile because asymmetry in the metric tensor doesn't imply that distance between points is asymmetrical. $\endgroup$ – Stan Liou Feb 1 '15 at 0:12
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In general relativity we describe spacetime as a manifold with a semi-Riemannian metric. However, this metric needs to be locally compatible with the flat-spacetime metric of special relativity, and the SR metric in turn needs to be compatible with the spatial metric of Euclidean geometry, on a 3-surface of simultaneity.

The spatial metric of Euclidean geometry is symmetric. For example, Euclidean geometry has a dot product, which can be used to measure the angle between two vectors. Angles are symmetric. Say we notate the angle between vector A and vector B as $\theta(A,B)$. Then we have an assumption built in to Euclidean geometry that $\theta(A,B)=\theta(B,A)$. Euclidean geometry is verified experimentally on small distance scales, so we need to have this property.

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