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.



Torsion and Non-metricity Tensor on a Surface
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.


