Torsion and Non-metricity Tensor on a Surface

Question

In differential geometry of surfaces, how can one define a non-zero Torsion tensor? It seems that the connection you provide has always to be symmetric since, by definition, $$\Gamma^{\gamma}_{\alpha\beta}\equiv\mathbf{a}^{\gamma}\cdot\mathbf{a}_{\alpha,\beta}=\mathbf{a}^{\gamma}\cdot\mathbf{r}_{,\alpha\beta}=\mathbf{a}^{\gamma}\cdot\mathbf{r}_{,\beta\alpha}=\Gamma^{\gamma}_{\beta\alpha},$$ where $\mathbf{r}:U\to\mathbb{R}^3$, $U\subset\mathbb{R}^2$, is an embedded $C^3$ surface with parametrization $(\theta^1,\theta^2)\in U$, $\mathbf{a}_\alpha\equiv\mathbf{r}_{,\alpha}$ are the tangent vectors to the coordinate curves $\theta^\alpha$, $\alpha=\{1,2\}$, and $\mathbf{a}^\gamma$ is the covector of $\mathbf{a}_\alpha$.

This definition also implies that the connection is metric compatible: $$\Gamma^{\gamma}_{\alpha\beta}=\frac{1}{2}a^{\gamma\lambda}(a_{\beta\lambda,\alpha}+a_{\gamma\alpha,\beta}-a_{\alpha\beta,\lambda}).$$ So there is no non-zero Non-metricity Tensor either. ($a_{\alpha\beta}\equiv\mathbf{a}_\alpha\cdot\mathbf{a}_\beta$,$a^{\alpha\beta}\equiv\mathbf{a}^\alpha\cdot\mathbf{a}^\beta$.)

Existence of non-zero Torsion tensor and Non-metricity tensor is important in studies of defects in two-dimensional crystals because in continuum model, they represent certain defect densities.

Answer 1

I think that the OP is asking a more specific question than whether or not a surface has a connection that is not metric or not torsion free. It seems that the OP is assuming that the surface $M$ comes equipped with an immersion $\mathbf{r}:M\to\mathbb{E}^3$ into (oriented) Euclidean $3$-space and is asking whether, using the data of the immersion $\mathbf{r}$, it is possible to define, in a canonical way, a connection that has torsion and/or is not metric compatible.

His question includes the argument that the usual induced connection associated to a given $\mathbf{r}$ discussed in all curves-and-surfaces books is both compatible with the induced metric and is torsion-free.

Now, it's true that the only canonical connection induced by $\mathbf{r}$ that uses at most second-order information from $\mathbf{x}$ at a point is the Levi-Civita connection. However, there are other canonical connections definable using $\mathbf{r}$ that use higher order information, and these need be neither torsion-free nor compatible with any metric (let alone the induced metric), at least for the general immersion. (Obviously, any canonical formula using higher order information will just produce the Levi-Civita connection when applied to an immersion whose image is either a plane or a sphere.)

Example: Given an immersion $\mathbf{x}:M\to\mathbb{E}^3$, there is an associated mean curvature function $H$ that, unfortunately, depends on a choice of orientation of the surface $M$; it switches sign if one reverses the orientation of $M$ (always, assuming, of course, that the target space $\mathbb{E}^3$ is oriented). However, the $1$-form $\eta = \ast dH$ is independent of a choice of orientation of the surface, since both $H$ and $\ast$ reverse sign when one reverses orientation. Let $\nabla$ be the Levi-Civita connection on $M$ associated to the metric induced on $M$ by the immersion $\mathbf{x}$, and define a second connection $\tilde\nabla$ on $M$ by the formula $$ \tilde\nabla_XY = \nabla_XY + \eta(X)Y $$ Then $\tilde\nabla$ is a connection canonically associated to $\mathbf{x}$ (whose local formula depends on third order derivatives of $\mathbf{x}$). One computes (using the fact that the torsion of $\nabla$ vanishes) that $$ T^{\tilde\nabla}(X,Y) = \tilde\nabla_XY - \tilde\nabla_YX - [X,Y] = \eta(X)Y - \eta(Y)X, $$ so the torsion of $\tilde\nabla$ vanishes if and only if $\eta=0$, i.e., $H$ is locally constant.

Meanwhile, it is easy to compute that the curvatures of the two connections are related by $$ R^{\tilde\nabla}(X,Y)Z = R^{\nabla}(X,Y)Z + d\eta(X,Y)\ Z, $$ so $\tilde\nabla$ does not even have a parallel $2$-form, let alone a parallel metric, unless $d\eta=0$, i.e., unless $H$ is (locally) a harmonic function on the surface.

Thus, in general, $\tilde\nabla$ is neither torsion-free nor metric compatible.

Answer 2

Levi-Civita means metric compatible and torsion free. Adding a skew symmetric $\binom{1}{2}$ tensor field (= your favorite torsion) to a covariant derivative does not change metric compatibility.