Can a metric in General Relativity, Supergravity, String Theory, etc., be asymmetric?

Question

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?

Answer 1

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. $$\frac{g(u,v)}{\sqrt{g(u,u)g(v,v)}} = \frac{g(v,u)}{\sqrt{g(u,u)g(v,v)}}.$$

Answer 2

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.

Answer 3

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.

Answer 4

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'.

Answer 5

In order for a non-symmetric metric to have any use or relevance, the two indices of the metric should have different interpretations and play different roles. That, right there, is a strong hint. There is a geometry that most naturally suits a non-symmetric metric, as you'll soon see; and in it, the two indices refer to different types of objects.

Spoiler Alert: a non-symmetric metric may be considered as the components of a second set of "pseudo-translation" generators that are added to the $GA(4)$ structure of the Cartan Structure Equations. The resulting Lie algebra associated with this inclusion may be considered as a contraction of $SL(5)$.

So, start by considering a geometry that includes a symmetric non-degenerate metric $g = g_{μν} dx^μ ⊗ dx^ν$, where I will be using the Einstein summation convention here and below has world indices ($μ,ν,…$) ranging over $0,1,…,n-1$, where $n$ is the dimension of the underlying geometry.

It reduces, locally, has a singular value decomposition into a constant frame metric $g = η_{ab} θ^a ⊗ θ^b$, involving a field of co-frames $\left(θ^a = h^a_μ dx^μ: a = 0,1,…,n-1\right)$, where $g_{μν} = η_{ab} h^a_μ h^b_ν$. The frame is non-singular, since the metric is, and has an inverse $(e^μ_a) = (h^a_μ)^{-1}$ and corresponding dual frame $e_a ≡ e^μ_a ∂/∂x^μ$. The frame indices ($a,b,…$) therefore also range over $0,1,…,n-1$.

If the decomposition can be done globally, the metric can effectively be replaced by the frame field. This does not actually get rid of the metric, completely: it's still there (as $η_{ab}$), but normalizes it. So, the geometry contains, as its basic ingredients: a frame field, and a frame metric.

In addition, we also add in a connection which may or may not be the Levi-Civita connection given by the metric. The connection, in turn, may be treated as a linear functional: $$\left(\begin{align} α &= α^μ ∂_μ\\X &= X^ν dx^ν\\Y &= Y^ρ dx^ρ \end{align}\right) ↦ Γ^α_{XY} ≡ \left(∇_X Y\right) ˩ α$$ which, when applied to the coordinate frame $∂_μ ≡ ∂/∂x^μ$ and co-frame $dx^μ$ yields the usual components $$Γ^ρ_{μν} ≡ Γ^{dx^ρ}_{∂_μ ∂_ν} = ∇_{∂_μ} ∂_ν ˩ dx^ρ.$$ This linear function is tensorial with respect to the $α$ and $X$ indices: $$Γ^{fα}_{(gX)Y} = fg Γ^α_{XY}$$ but only affine with respect to the $Y$ index: $$Γ^α_{X(hY)} = (Xh) Y ˩ α + h Γ^α_{XY}$$ so it is not quite a tensor. So, its decomposition has an affine part: $$Γ^α_{XY} = α_ρ \left(X^μ Y^ν Γ^ρ_{μν} + X^μ ∂_μ Y^ρ\right),$$ and, when expressed in terms of frame indices, becomes $$Γ^c_{ab} ≡ Γ^{θ^c}_{e_ae_b} = h^c_ρ \left(e^μ_a e^ν_b Γ^ρ_{μν} + e^μ_a ∂_μ e^ρ_b\right).$$

(The chief example of this kind of geometry are the Riemann-Cartan geometries, where the frame components $Γ^c_{ab}$ are often written as $ω^c_{ab}$ and treated as an separate object than $Γ^ρ_{μν}$; when in reality it is just the non-tensorial object $Γ$, itself, referred to frame indices. So, I prefer to use $Γ$ uniformly throughout.)

The connection comprises a field of $1$-forms, $ω^c_b ≡ Γ^c_{μb} dx^μ = Γ^c_{ab} θ^a$ that is rank $(1,1)$ with respect to the frame indices. So, all together, the geometry has $3$ basic objects: the connection $Γ$, the frame fields $θ$ and the metric $g$, which is still there as the frame metric $η_{ab}$ ... and still symmetric.

Other objects can be formed from these, by way of the Cartan Structure Equations $$\begin{align} Θ^a &≡ dθ^a + ω^a_c ∧ θ^c,\\ Ω^a_b &≡ dω^a_b + ω^a_c ∧ ω^c_b, \end{align}$$ that give us the torsion $2$-form $Θ^a = ½ T^a_{μν} dx^μ ∧ dx^ν$ and curvature $2$-form $Ω^a_b = ½ R^a_{bμν} dx^μ ∧ dx^ν$, from which we read off the components $T^a_{μν}$ of the torsion tensor and $R^a_{bμν}$ of the Riemann curvature tensor.

The connection $ω$ is associated with the symmetry group $GL(n)$, which may expressed as a Lie algebra with generators $\left(E^a_b\right)$ satisfying the Lie bracket relations $$[E_a^b, E_c^d] = E_a^b E_c^d - E_c^d E_a^b = δ^b_c E_a^d - δ^d_a E_c^b.$$ In particular, when we treat the connection $ω$ as a Lie-valued $1$-form $ω ≡ ω^a_b E_a^b$, then its structure equation reduces to: $$Ω = dω + ω^2$$ for the Lie-valued curvature $2$-form $Ω = Ω^a_b E_a^b$, where we expand $ω^2$ as $$ω^2 = \left(ω^a_c E_a^c\right) \left(ω^d_b E_d^b\right) = ω^a_c ∧ ω^d_b (E_a^c E_d^b).$$ By anti-symmetry $ω^a_c ∧ ω^d_b = -ω^d_b ∧ ω^a_c$, so we can equally-well write this (after a little index-renaming) as: $$ω^2 = ½ ω^a_c ∧ ω^d_b (E_a^c E_d^b - E_a^c E_d^b) = ½ ω^a_c ∧ ω^d_b [E_a^c, E_d^b],$$ provided we treat the generators $E_a^b$ as residing in their covering algebra. Applying the Lie bracket relations and contracting the Kronecker deltas, we get $ω^2 = ω^a_c ∧ ω^c_b$. So, this works out.

The frame field $θ$ can be thrown into this by treating $(ω,θ)$ as a connection for the general affine group GA(n), which has an extra set of generators $E_b$ satisfying and Lie bracket relations: $$[E_a^b, E_c] = δ^b_c E_a, \quad [E_a, E_c] = 0.$$ Calling the combined object $\hat ω ≡ ω^a_b E_a^b + θ^a E_a$, the two Cartan structure equations now combine to yield: $$\hat Ω = d\hat ω + {\hat ω}^2,$$ where the $2$-form $\hat Ω ≡ Ω^a_b E_a^b + Θ^a E_a$ now includes the torsion and curvature $2$-forms in it.

But, we still have the metric dangling out in the cold. Where does it fit in? The answer to that question also answers your question and reveals the geometry that lies under all of this in disguise.

So, now, treat the metric $g_{μν}$ (or $η_{ab}$) as non-symmetric and treat its indices differently; keeping all the infrastructure already laid out above intact. The second index is actually a world index (or equivalently, a co-frame index) for the $1$-form $θ_b ≡ g_{bμ} dx^μ = η_{ba} θ^a$. These $1$-forms together comprise a second co-frame, and the index $a$ is actually an index for it. We can write down another Cartan structure equation for it $$Θ_b ≡ dθ_b - ω^c_b ∧ θ_c$$ providing us with a second "torsion" $2$-form $Θ_b ≡ ½ T_{bμν} dx^μ ∧ dx^ν$. The second torsion $2$-form is related to the first co-frame and its torsion form by: $$Θ_b ≡ Q_{ba} ∧ θ^a + η_{ba} Θ^a,$$ where $$Q_{ba} ≡ dη_{ba} - ω^c_b η_{ca} - ω^c_a η_{bc}$$ comprise the components for the covariant derivative of the metric. For Riemann-Cartan geometries, it is assumed to be $0$. Connections which satisfy this property are called metrical, which may be characterized here by the condition: $$Θ_b = η_{ba} Θ^b.$$

The metric, itself, may be expressed in terms of the two co-frames as $$g = η_{ba} θ^a ⊗ θ^b = θ_b ⊗ θ^b.$$ The requirement that it be symmetric now becomes the condition $$θ_b ∧ θ^b = η_{ba} θ^a ∧ θ^b = ½ \left(η_{ba} - η_{ab}\right) θ^a ∧ θ^b = 0.$$

The third Cartan structure equation can be thrown in with the other two by treating the second co-frame as the components of a connection for a symmetry group containing a extra set of generators $E^b$ satisfying the Lie bracket relations $$[E^b, E_c^d] = δ_b^c E^d, \quad [E^b, E_c] = δ_b^c E, \quad [E^b, E^d] = 0.$$ To complete this picture, we can add in another $1$-form $h$ and $2$-form $H$, to go with $E$, an extra set of Lie bracket relations $$[E_a^b,E] = 0, \quad [E^b,E] = 0, \quad [E_a,E] = 0$$ and a fourth Cartan equation $$dh + θ_c ∧ θ^c = H.$$ This is where "non-symmetricity" $2$-form $θ_c ∧ θ^c$ can be fit in.

Defining the connection by $$\tilde ω ≡ ω^a_b E_a^b + θ^a E_a + θ_b E^b + h E,$$ the result is that all $4$ structure equations now combine to give you $$\tilde Ω = d\tilde ω + {\tilde ω}^2,$$ where $\tilde Ω = Ω^a_b E_a^b + Θ^a E_a + Θ_b E^b + H E$ is the combined $2$-form.

In the general case of $n$ dimensions, it is also possible to set up the geometry to work with $SL(n+1)$ instead, but the structure equations will be slightly different. In that case, the two co-frames fit in with the $GL(n)$ connection as two sets of pseudo-translation generators that make up an $SL(n+1)$ connection, which is $SL(5)$, when $n = 4$. Either geometry provides a way to removing the frame metric from the background, resulting in a truly background-free gauge theory ... in which the metric itself emerges from the gauge field one-form.

Why $SL(5)$? Because the symmetry group of the Law Of Inertia - when stated in this form $$\frac{d^2y^a}{ds^2} = 0 \quad (a = 0,1,2,3),$$ is $SL(5)$. So, an $SL(5)$ foundation for a theory of gravity is one which gauges the symmetry group of the Law Of Inertia.

Answer 6

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.