# Torsion & rotation of basis

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.

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).
• 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.