Torsion & rotation of basis

Question

Recently I find someone declared that: ‘Even if we write $g_{\mu\nu}=\delta_{\mu\nu }$ everywhere in some patch, we can still find a non-zero Riemann tensor if our basis vectors don't commute’ From this I find an interesting example:

If we parallel a basis $e_\mu(x_0)$ on a manifold $M$ which equipped with a connection field $\Gamma^\rho_{\mu\nu}$. Then we get: $$D_\mu e_\nu(x)=\partial_\mu e_\nu(x)-\Gamma^\rho_{\mu\nu}(x)e_\rho(x)=0$$ And $\Gamma^\rho_{\mu\nu}(x)\delta x^\mu \in SO(n)$. Then parallel transport $e_\mu(x_0)(e_\mu(x_0) \cdot e_\nu(x_0)= \delta_{\mu\nu })$ to another point $x$ along a particular path, we can get $$ e_\mu(x)= e_\mu(x_0)R(x)= e_\mu(x_0)\prod \Gamma^\rho_{\mu\nu }dx^\mu $$

Obviously $ R(x)\in SO(n)$, and the corresponding metric is $g_{\mu\nu}=\delta_{\mu\nu }$ . If $R(x) $ is a function on the manifold, then $ e_\mu(x)$ is a non-coordinate basis field. If the curvature of $ \Gamma^\rho_{\mu\nu }$ is not zero. Then when basis $e_\mu(x)$ moves along an infinitesimal closed path, it will rotate an infinitesimal angle contrast to the initial basis $e_\mu(x)$, So on the manifold we can’t define a global basis field, but we can define a global metric $g_{\mu\nu}=\delta_{\mu\nu }$..

The connection $\Gamma^\rho_{\mu\nu}$ is not always symmetric in the indices $\mu,\nu$.But when we calculate the curvature from the metric $\delta_{\mu\nu }$, the curvature we get is zero, this obviously contradict that the curvature of connection $\Gamma^\rho_{\mu\nu}$ is not zero. the reason is when we calculate the connection from metric, we require $\Gamma^\rho_{\mu\nu}$ must symmetric in indices $\mu,\nu$(torsion free). So

(1) Can we conclude that the torsion should be relevant to the rotation of the basis when parallel transporting along a path?

(2) Because the definition of curvature is when a vector parallel transport along a closed path, the change of vector is not zero, the effect of torsion can rotate the vector when it parallel transport a closed path, then can we say the torsion is only part of curvature?

(3) If we do not neglect the torsion part of the connection, then when we parallel transport a vector on the manifold either we will get a vector field whose curl isn’t zero or we can’t get a vector field at all. So why can we neglect the torsion?

• About ‘if the frame is holonomic if and only if the torsion vanishes’，It is unquestionably true, if there is a frame $ e_\mu (x)$, then the connection coefficients defined as: $\Gamma^\rho_{\mu\nu}=e_\mu^a \partial_a e^b_\nu e_b^\rho$, if $ e_\mu^a $ can be equivalently expressed with a global coordinate transformation $ e_\mu^a =\frac{\partial x^a}{\partial x^\mu}$, then $ e_\mu $ is a coordinate basis, then $[ e_\mu, e_\nu]=0$, The connection $\Gamma^\rho_{\mu\nu}$ is torsion free, and it can be transformed to 0 using a coordinate transformation.

  If $ e_\mu $ is a non- coordinate basis, then the corresponding connection $\Gamma^\rho_{\mu\nu}$ is not symmetric in $\mu,\nu$, it contains the part of torsion, and obviously it can’t be transformed to 0 under a coordinate transformation. Further more, to some $\Gamma^\rho_{\mu\nu}$ fields, they can’t be described using a global basis fields.

Answer 1

I do not have time to post a super detailed answer right now, but OP appears to have some things confused.

Let $\Gamma_{\kappa\ \ \nu}^{\ \mu}$ denote the connection coefficients in a holonomic frame and $\Gamma_{\mu\ \ b}^{\ a}$ in an orthonormal frame. The corresponding connection form is $\Gamma^a_{\ b}=\Gamma_{\mu\ \ b}^{\ a}\mathrm dx^\mu$. I will assume the connection is metric, but not torsionful. Some comments are in order:

• Suppose that there is a point $x_0\in M$ ($M$ is the manifold we are working on) such that $\Gamma^a_{\ b}(x_0)=0$. Then at that point $$ \Theta^a(x_0)=\mathrm d\vartheta^a(x_0)+\Gamma^a_{\ b}(x_0)\wedge\vartheta^b(x_0)=\mathrm d\vartheta^a(x_0)=-\frac{1}{2}C^a_{\ bc}(x_0)\vartheta^b(x_0)\wedge\vartheta^c(x_0). $$ Here $\Theta^a$ is the torsion form, $\vartheta^a$ is the covielbein and $C^c_{ab}$ are the frame commutators $[e_b,e_c]=C^a_{bc}e_a$. This shows that at a point $x_0$ where the connection form vanishes $$ T^c_{\ ab}=-C^c_{ab}, $$ so a parallel frame is holonomic if and only if the torsion vanishes (in a pointwise manner).
• Torsion doesn't affect the integrability of the parallel transport whatsoever. Any infinitesimal holonomy formula one writes involves the curvature tensor only. The only exception is if special parallel displacements are considered that are nonlinear but affine, like the so-called Cartan displacements. If a connection is used in a linear manner then those are never considered though.
• When torsionful connections are allowed (and basically whenever any connection other than the Levi-Civita connection of a metric is allowed), one must be careful about the definition of flatness one uses. $$ $$ A connection is flat if its parallel transport is locally integrable, and this condition is equivalent to the vanishing of the curvature tensor of that connection. $$ $$ A metric tensor is flat if it can be locally transformed into its canonical form by a coordinate transformation (can always be done by anholonomic frame transformation). This condition is equivalent to the vanishing of the curvature tensor of its Levi-Civita connection.
• The situation is further complicated by the fact that if a metric compatible but torsionful connection is given, that connection does not only differ by torsion from the LC (Levi-Civita) connection. The geodesics and curvature of a metric compatible connection absolutely doesn't have to agree with that of the LC connection.