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Studying curves in the Euclidean three dimensional space, one usually defines the curvature and the torsion of a curve. If I am not missunderstanding the thing, I guess that a curve has zero torision if and only if it is contained in a plane.

The curvature is generalised by the second fundamental form of a submanifold.

What it is the generalisation of the torsion for a sub manifold $N$ of a Riemannian manifold $M$? I would like to preserve the property that the torsion vanishes if and only if the submanifold is contaned in a totally geodesic hypersurface.

My guess, if $N$ is a curve, would be something like $\nabla_vII(v,v)-\nabla_{II(v,v)}v-[v,II(v,v)]$, where $\nabla$ is the Levi-Civita connection, $II$ is the seocond fundamental form, and $v$ is a tangent vector to $N$. This guess is insipired by the other use of the word torsion in Riemanninan geometry, and this is likely to be missleading, in view of the question Relating curvature and torsion of a connection to those of a curve

Probably this material is standard, but the references I found are either about torision of a curve in the Euclidean space (and usually they are very computational), or torsion of a connection.

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    $\begingroup$ Note that a general Riemannian manifold has no totally geodesic hypersurfaces, see this MO post. But the question of course makes sense for submanifolds of a Euclidean space. $\endgroup$ Nov 25, 2016 at 12:09

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The torsion of a curve $C$ in $\newcommand{\bR}{\mathbb{R}}$ $\bR^3$ is a a manifestation of two ``freak'' low dimensional accidents: the curve is $1$-dimensional and it lives in a $3$-dimensional Euclidean space.

Being $1$-dimensional allows us, after fixing an orientation, to choose an orthonormal basis of the tangent space $T_pC$ at each point $p\in C$. Just fix the arclength parametrization compatible with the orientation. This allows us to defined the first vector $T(p)$ of the Frenet frame.

If there is a bit of curvature, then we can define the 2nd vector $N(p)$ of the Frenet frame. Since the curve is in $\bR^3$ we can now define the third vector of the Frenet frame $B(p)=T(p)\times N(p)$.

Note that if the curve lived in a higher dimensional space $\bR^n$, $n>3$, the above process would stop after we've defined the normal vector $N(p)$.

For a submanifold $M$ of dimension $m$ living in an Euclidean space $\bR^N$, $N>m$, the 2nd fundamental form $\DeclareMathOperator{\Gr}{\mathbf{Gr}}$ has an alternate description as the shape operator, i.e., the differential of the Gauss map $\Gamma: M\to\Gr_m(\bR^N)$ that associates to a point $p\in M$ the tangent space $T_pM$ viewed as an element of the Grassmannian $\Gr_m(\bR^N)$ of $m$-dimensional subspaces of $\bR^m$. In this sense the shape operator describes the rate of change along $M$ of an orthonormal moving tangent frame.

The Frenet equations of a curve describe a bit more that the shape operator. They describe the motion along $C$ of an orthonormal frame of $\bR^3$ which for some fortunate accident can be chosen more or less canonically. Thus, I do not believe there is a natural simple notion of torsion beyond curves in $\bR^3$.

On the other hand, given $M\subset \bR^N$ we can still ask ask if $M$ is contained in an affine subspace of dimension $n<N$. This corresponds to asking when the image of the Gauss map be contained in a sub-Grassmannian $\Gr_m(\bR^n)$ but, be careful, there are many ways of canonically embedding $\Gr_m(\bR^n)$ in $\Gr_m(\bR^N)$.

In any case, this phenomenon can be expressed in terms of certain constraints on the 2nd fundamental form that you could call torsions. I don't believe they are too useful, though I could be proven wrong.

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  • $\begingroup$ Thanks! Could you please give me a reference about these constaints you mention at the end? $\endgroup$
    – Giulio
    Nov 25, 2016 at 13:05
  • $\begingroup$ Don't know any on the top of my head, but this should be a good exercise for you to try. $\endgroup$ Nov 25, 2016 at 13:21
  • $\begingroup$ Do not even have a key word?? It is possible that they are poorly studied? $\endgroup$
    – Giulio
    Nov 25, 2016 at 13:29
  • $\begingroup$ I honestly don't know. I would recommend to try a simpler case. Ask when $M$ is contained in a hyperplane. This should give you a constraint. $\endgroup$ Nov 25, 2016 at 13:52
  • $\begingroup$ I don't agree that for curves in ${\mathbb R}^n$ one cannot go beyond the normal vector and the curvature. There are analogs of torsion, defined using higher derivatives of the curve. Similarly, there are Frenet formulas in higher dimension, and vanishing of one of these "torsions" identically over the curve means that the curve is contained in an affine subspace. $\endgroup$ Nov 25, 2016 at 18:33
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For curves in the Euclidean space, you can define "generalized curvatures", see Wikipedia, by applying Gram-Schmidt to the vectors $\gamma', \gamma'',\ldots$. The curvature of order $k$ is defined only if the vectors up to $\gamma^{(k)}$ are linearly independent.

Now, if on a curve arc all curvatures up to $(k-1)$-st are defined and don't vanish, but the $k$-th is defined and vanishes identically (which means there is a linear relation between $\gamma', \ldots, \gamma^{(k+1)}$), then the curve is contained in a $k$-dimensional affine subspace. To see this, consider the projection of $\gamma$ to the orthogonal complement of the span of $\gamma', \ldots, \gamma^{(k)}$ and apply the uniqueness theorem for ODE.

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  • $\begingroup$ Izmetiev Thanks for the Wikipedia reference. I now see a (rather complicated) way of answering the same questions for submanifolds of dimension $>1$. $\endgroup$ Nov 25, 2016 at 21:05
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CLARIFICATION: The connection $\nabla$ used below is the Levi-Civita connection for the Riemannian metric on $M$ and not $N$.

First, assume that $N$ is a curve and recall how to construct the Frenet frame of a curve. Let $e_1 \in T_*N$ be a unit tangent vector along the curve. If $\nabla_{e_1}e_1 \ne 0$, then, since $e_1\cdot\nabla_{e_1}e_1 = 0$, there isa a unique unit vector $e_2$ and function $\kappa_1 > 0$ such that $e_2\cdot e_1 = 0$ and $\nabla_{e_1}e_1 = \kappa_{1} e_2$. If, in turn, $\nabla_{e_1}e_2 \ne 0$, then, $e_2\cdot \nabla_{e_1}e_2 = 0$ and $e_1\cdot \nabla_{e_1}e_2 = -e_2\cdot\nabla_{e_1}e_1 = -\kappa_{1}$. Therefore, there exists a unique unit vector $e_3$ and function $\kappa_2 > 0$ such that $e_3\cdot e_1 = e_3\cdot e_2 = 0$ and $$ \nabla_{e_1}e_2 = -\kappa_{1} e_1 + \kappa_{2} e_3. $$ If, at each stage, $\nabla_{e_1}e_k \ne 0$, then this leads to a uniquely defined orthonormal frame of vectors $e_1, \dots, e_n \in T_*M$ along $N$, known as the Frenet frame and corresponding functions $\kappa_1, \dots, \kappa_{n-1}$ such that $$ \nabla_{e_1}e_k = -\kappa_{k-1}e_{k-1} + \kappa_{k}e_{k+1}. $$ If $M$ is flat, then the vanishing of $\kappa_{j}, \dots, \kappa_{n-1}$ implies that $N$ lies in an affine subspace of dimension $j$. The function $\kappa_1$ is usually called the curvature of $N$, and the functions $\kappa_2, \dots, \kappa_{n-1}$ considered as torsion functions.

This can be generalized to higher dimensional submanifolds as described by Liviu. Here is a sketch of one way to do it: If $n_1 = \dim N$, let $e_1, \dots, e_{n_1}$ be an orthonormal frame of vectors tangent to $N$. Suppose at each point in $N$, the vectors $e_1, \dots, e_{n_1}, \nabla_{e_j}e_i$, where $1 \le i, j \le n$, span an $n_2$-dimensional subspace, where $n < n_2$. Extend the orthonormal frame to a basis $e_1, \dots, e_{n_2}$ of this larger subspace. The generalization of $\kappa_1$ for a curve to a higher dimensional submanifold is the second fundamental form $H_{ij} = H^\mu_{ij}e_\mu$, where $H^\mu_{ij} = e_\mu\cdot\nabla_{e_i}e_j$ and $n+1 \le \mu \le n_2$. Now suppose that $e_i, \nabla_{e_j}e_i$, $1 \le i,j \le n_2$ span an $n_3$-dimensional subspace of $T_*M$. Again, extend the orthonormal frame to one that spans this subspace. Then one can define a torsion tensor $T_{\mu,\nu} = T_{\mu\nu}^\eta e_\eta$, where $n+1 \le \mu,\nu \le n_2$ and $n_2+1\le \eta \le n_3$. This process can be continued. If the resulting frame does not span $T_*M$, then one might call $N$ degenerate. If $M$ is flat, this implies that $N$ lies in an affine subspace. Otherwise, one gets a higher dimensional version of a Frenet frame. The frame is not unique, but the the nested sequence of osculating subspaces obtained in this way are.

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  • $\begingroup$ Thanks! this looks great. Do you have a reference? $\endgroup$
    – Giulio
    Nov 26, 2016 at 16:54
  • $\begingroup$ I omitted the normalization step for the second fundamental form at each step, mimicking what is done for a curve. Also, the construction above is done most cleanly using moving frames and differential forms. Unfortunately, I don't know the reference for this. $\endgroup$
    – Deane Yang
    Nov 27, 2016 at 4:49
  • $\begingroup$ If you look carefully, you see the Grassmannians of subspaces in $T_*M$ and their tangent bundles in this formulation. Again, this is all best worked out using moving frames of $1$-forms, rather than frames of tangent vectors. $\endgroup$
    – Deane Yang
    Nov 27, 2016 at 4:52

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