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.

  • 2
    $\begingroup$ @AyanVivek: It's certainly true that the only connection canonically induced on a surface in Euclidean $3$-space is metric and torsion-free, but there are many other ways to define connections on surfaces (based on other given data than an immersion into Euclidean $3$-space), and these do not always yield a torsion-free connection, or even a metric connection. On any surface, there will always exist connections that have nonzero torsion and/or are not compatible with any Riemannian metric. $\endgroup$ Jun 11, 2013 at 12:09
  • $\begingroup$ Sir, can you give an example of such a connection, say on a plane $\mathbf{r}=\theta^{\alpha}\mathbf{e}_\alpha$? $\endgroup$
    – Ayan
    Jun 11, 2013 at 12:25
  • 1
    $\begingroup$ @AyanVivek: Well, obviously, there exist many such connections, just choose arbitrary functions $\Gamma^\alpha_{\beta\gamma}(\theta^1,\theta^2)$ and define a connection by $$ \nabla_{\mathbf{e}_\beta}\mathbf{e}_\gamma = \Gamma^\alpha_{\beta\gamma}\ \mathbf{e}_\alpha\ . $$ However, I grant you that this method makes arbitrary choices that are not forced by the given immersion. If you want to make it depend canonically on the given immersion, that's another matter. See my answer below. $\endgroup$ Jun 11, 2013 at 12:42
  • $\begingroup$ Thank you sir. As per you answer below, the second order information could only provide a torsion-free metric connection. And only higher order informations, e.g. using the third order partial derivatives in the definition of the connection, can provide a fully non-Riemannian surface. So if we do not put the usual induced metric on a cylinder, say, and also provide it with such a non-cannonical connection, then in can have a non-zero Riemann curvature and a torsion too? $\endgroup$
    – Ayan
    Jun 11, 2013 at 13:00

2 Answers 2


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.

  • $\begingroup$ "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." Why is it so Sir? $\endgroup$
    – Ayan
    Jun 11, 2013 at 13:48
  • $\begingroup$ @AyanVivek: A precise explanation will hinge on a precise definition of 'canonical' in this context, so, when I have a little more time, I'll edit my answer above to address this point. In the meantime, the point is that if you had such a higher order formula, it would construct a connection on the sphere, say, that was invariant under the isometry group of the sphere. Looking at the difference between this connection and the Levi-Civita connection, you'd get a tensor of type (1,2) that was invariant under the isometry group of the sphere, but the only such invariant tensor is the zero tensor. $\endgroup$ Jun 11, 2013 at 15:45

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.