# A When can a nonvanishing torsion occur?

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1. Dec 6, 2016

### EnigmaticField

Since the first day I learnt the torsion I keep having the question how a nonvanishing torsion is likely to occur because based on the definition formula of the torsion, it looks like the torsion always vanishes. I have come back to think about this question a couple of times after my first encounter with it, but always feel the same and am puzzled. I know I shall be wrong because if the torsion always vanishes, why do people bother to define it? But I just can't find out where I am wrong. I put my argument as follows, hoping someone can point out where I am wrong.

Given a manifold V with connection $\nabla^\rm{V}$, if $\bf{X}$ and $\bf{Y}$ are vector fields on the tangent bundle of V, the torsion at a point $\rm{p}\in \rm{V}$ is defined as $T=\nabla^\rm{V}_\bf{X}\bf{Y}-\nabla^\rm{V}_\bf{Y}\bf{X}-[\bf{X},\bf{Y}]$. In this formula, on the one hand, $\nabla^\rm{V}_\bf{X}\bf{Y}$ is the projection of $\bf{X}\bf{Y}$ into the tangent space of V at p, $\rm{V_p}$, and $\nabla^\rm{V}_\bf{Y}\bf{X}$ is the projection of $\bf{Y}\bf{X}$ into $\rm{V_p}$; on the other hand, $[\bf{X},\bf{Y}]=\bf{X}\bf{Y}-\bf{Y}\bf{X}$ must be tangent to V so should be the tangent component of $\bf{X}\bf{Y}$ minus the tangent component of $\bf{Y}\bf{X}$. Thus doesn't $\nabla^\rm{V}_\bf{X}\bf{Y}-\nabla^\rm{V}_\bf{Y}\bf{X}=[\bf{X},\bf{Y}]$ always hold? Then doesn't the torsion $T$ always vanish?

When V is a hypersurface of another manifold M with connection $\nabla^\rm{M}$ (in this case $\nabla^\rm{V}$ is treated as an induced connection), the above argument can also be understood by the Gauss equations: $\textbf{X}\bf{Y}=\nabla^\rm{M}_\textbf{X}\textbf{Y}=\nabla^\rm{V}_\textbf{X}\textbf{Y}+\rm{B}(\textbf{X},\textbf{Y})\textbf{N}...(1)$
$\textbf{Y}\textbf{X}=\nabla^\rm{M}_\textbf{Y}\textbf{X}=\nabla^\rm{V}_\textbf{Y}\textbf{X}+\rm{B}(\textbf{Y},\textbf{X})\textbf{N}...(2)$, in which $\rm{B}$ is the second fundamental form with the property $\rm{B}(\bf{X},\bf{Y})=\rm{B}(\bf{Y},\bf{X})$ and $\textbf{N}$ is the unit normal vector field on V. Then from (1) and (2) we can get $\bf{XY}-\bf{YX}=\nabla^\rm{M}_\bf{X}\bf{Y}-\nabla^\rm{M}_\bf{Y}\bf{X}=\nabla^\rm{V}_\bf{X}\bf{Y}-\nabla^\rm{V}_\bf{Y}\bf{X}$, which says that the torsion always vanishes.

So when on earth can a nonvanishing torsion ever occur?

Or does my argument in the second paragraph only apply to the case when V is embedded in a higher dimensional manifold because only in that case does saying projecting $\bf{X}\bf{Y}$ into $\rm{V_p}$ to get $\nabla^\rm{V}_\bf{X}\bf{Y}$ make sense? If that's the case does that mean a nonvanishing torsion can only occur when V is not embedded in another mainfold, that is, a nonvanihing torsion can only occur to a non-induced connection?

2. Dec 6, 2016

### Orodruin

Staff Emeritus
The point with the embedding is that you are inheriting several properties from the space you embed in. You do not need to do that. The "intuitive" way of defining connections in terms of projections based on embedding spaces does not cover all possible connections.

My favorite example of a space with non-zero torsion is the sphere with the poles removed and the compass directions being parallel fields (this uniquely defines the connection).

3. Dec 7, 2016

### Ben Niehoff

When you define a connection using these projection conditions, you have already picked out the Levi-Civita connection, so of course it will be torsion-free.

But you don't need to define a connection this way. A connection is just a prescription for relating nearby tangent spaces to each other; in principle it can be completely arbitrary, so long as it is smooth. So, e.g., you can choose a connection on $\mathbb{R}^3$ with unit vectors $e_1, e_2, e_3$ given by

$$\nabla_{e_i} e_j = \varepsilon_{ijk} \, e_k$$
This connection has torsion. One way of thinking of torsion is that it measures how frames twist when parallel-propagated along geodesics.

4. Dec 7, 2016

### Ben Niehoff

This is only a unique prescription if you assume the connection is metric-compatible, of course. :D

Which I'm only pointing out to emphasize that the concept of a connection can really be quite arbitrary. It really can be anything; these rules such as metric-compatibility and vanishing-torsion are just things we impose because they pick out a particular connection that is interesting to us. But mathematicians do study objects that are invariant under this completely arbitrary choice of connection (differential topology, for example).

5. Dec 7, 2016

### Orodruin

Staff Emeritus
Right, I assumed normalised compass directions so the assumption of metric compatibility is implicit.