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EDIT: Let me modify the question then: for what submanifolds $N$ does the torsion $T$ preserve tangent vectors to $N$?

If $\nabla$ is a connection on a manifold $M$, then torsion is defined to be the map $$ T(X,Y)=\nabla_XY-\nabla_YX-[X,Y] $$

where $X$ and $Y$ are vector fields on $M$. It can be shown that $T$ is a $2 \choose 1$ tensor on $M$; that is, for all $p\in M$, $$ T:T_pM\times T_pM\longrightarrow T_pM $$

where $T_pM$ is the tangent space to $M$ at $p$.

Suppose $N\subset M$ is a submanifold of $M$. Is it the case that $T$ preserves tangent vectors to $N$? That is, does $$ T:T_pN\times T_pN\longrightarrow T_pN $$

for $p\in N$?

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    $\begingroup$ There's no reason for this to be true, and I'm sure that, for the generic submanifold $N$ of dimension 2 or more (if the dimension of $M$ is at least $3$ and the torsion doesn't satisfy some very special identity) then it won't be true. $\endgroup$
    – Robert Bryant
    Jul 18, 2013 at 0:00
  • $\begingroup$ @Robert: Is it the case that $\nabla_XY-\nabla_YX$ lies in the tangent space of $N$? $\endgroup$
    – Oliver Jones
    Jul 18, 2013 at 1:12
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    $\begingroup$ Pick a point $p$ in $M$ and any subspace $S\subseteq T_pM$. There is a submanifold $N$ of $M$ such that $p\in N$ and $T_pN=S$. If what you want were true, then the torsion tensor would preserve all subspaces of $T_pM$! $\endgroup$
    – Mariano Suárez-Álvarez
    Jul 18, 2013 at 1:14
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    $\begingroup$ @Oliver Jones: Since the Lie bracket respects vector fields which are tangent to $N$, and the torsion does not (in general), so also $\nabla_XY-\nabla_YX$ does not. $\endgroup$
    – Peter Michor
    Jul 18, 2013 at 6:34
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    $\begingroup$ Oliver, you shouldn't accept an answer and then change the question. $\endgroup$
    – Ramiro de la Vega
    Jul 18, 2013 at 22:56

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Edit: Answering Robert, moved to 3-space to give a general example:

A simple example in $M=\mathbb R^3$: Let $N=0\times \mathbb R^2$ and put $$ \nabla_XY = dY(X) + \begin{pmatrix}X^T\,A^1\,Y \\ X^T\,A^2\,Y \\ X^T\,A^3\,Y\end{pmatrix}, \quad A=\begin{pmatrix} a^i_{11} & a^i_{12} & a^i_{13}\\ \dots \\a^i_{31} & a^i_{32} & a^i_{33} \end{pmatrix}, \quad a^2_{kl} = a^3_{kl} = 0 \text{ for } 2\le k,l\le 3. $$ Then $Tor^i_{kl} = A^i_{kl}-A^i_{lk}$ maps $T_{(0,x,y)}(0\times \mathbb R^2)\times T_{(0,x,y)}(0\times \mathbb R^2)$ skew linearly into $T_{(0,x,y)}\mathbb R\times 0$.

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  • $\begingroup$ Perhaps you should point out that this is not an example pertaining to the original question, but the modified one in the comments. After all, the torsion will, by definition, vanish when restricted to any $1$-dimensional subspace. (The modified question doesn't really make much sense anyway, since $\nabla_XY-\nabla_YX$ isn't even a tensorial expression.) $\endgroup$
    – Robert Bryant
    Jul 18, 2013 at 12:15
  • $\begingroup$ @Robert: I meant for the expression $\nabla_XY-\nabla_YX$ to be evaluated at a point in the submaniofld. $\endgroup$
    – Oliver Jones
    Jul 18, 2013 at 21:29

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