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

A simple example in $M=\mathbb R^2$$M=\mathbb R^3$: Let $N=0\times \mathbb R$$N=0\times \mathbb R^2$ and put $$ \nabla_XY = dY(X) + \binom{X^T\,A\,Y}{X^T\,B\,Y}, \quad A=\begin{pmatrix} a_1 & a_2 \\ a_3 & 1 \end{pmatrix}, \quad B=\begin{pmatrix} b_1 & b_2 \\ b_3 & 0 \end{pmatrix}. $$$$ \nabla_XY = dY(X) + \begin{pmatrix}X^T\,A^1\,Y \\ X^T\,A^2\,Y \\ X^T\,A^3\,Y\end{pmatrix}, \quad A=\begin{pmatrix} a^i_{11} & a^i_{12} & a^i_{13}\\ \dots \\a^i_{31} & a^i_{32} & a^i_{33} \end{pmatrix}, \quad a^2_{kl} = a^3_{kl} = 0 \text{ for } 2\le k,l\le 3. $$ 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 $T_{(0,x,y)}\mathbb R\times 0$.

A simple example in $M=\mathbb R^2$: Let $N=0\times \mathbb R$ and put $$ \nabla_XY = dY(X) + \binom{X^T\,A\,Y}{X^T\,B\,Y}, \quad A=\begin{pmatrix} a_1 & a_2 \\ a_3 & 1 \end{pmatrix}, \quad B=\begin{pmatrix} b_1 & b_2 \\ b_3 & 0 \end{pmatrix}. $$

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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Peter Michor
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  • 112

A simple example in $M=\mathbb R^2$: Let $N=0\times \mathbb R$ and put $$ \nabla_XY = dY(X) + \binom{X^T\,A\,Y}{X^T\,B\,Y}, \quad A=\begin{pmatrix} a_1 & a_2 \\ a_3 & 1 \end{pmatrix}, \quad B=\begin{pmatrix} b_1 & b_2 \\ b_3 & 0 \end{pmatrix}. $$