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Suppose an affine connection is given on a smooth manifold $M$ and let $N\subset M$ be an embedded submanifold. Is there a canonical way of defining an induced connection on $N$?

In classical differential geometry of smooth surfaces in Euclidean 3-space, the corresponding construction is that of covariant derivative (cfr. Do Carmo, Differential geometry of curves and surfaces §4-4). The covariant derivative of a vector field along a curve on the surface is defined as the orthogonal projection of the ordinary Euclidean derivative onto the plane tangent to the surfaces.

I wonder how (and if) this can be ported to the language of connections.Wikipedia's entry does something like that by means of the Riemannian structure: I wonder if this extra structure is really necessary.

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    $\begingroup$ A connection more or less tells you how to move tangent vectors consistently from $T_p M$ to $T_q M$. The point about the Riemannian metric is, that it defines orthogonal projections $T_p M \rightarrow T_p N$ if $p\in N$, providing a natural means to induce a connection on $N$. Unless you have some natural projection you will have difficulties to find a canonical way to define a connection on $N$. $\endgroup$
    – user20266
    Jun 7, 2012 at 15:12

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With just an affine structure you will not be able to get an induced connection. (Part of the story is told in Fox's AMS Notices article from March 2012 titled "What is an affine sphere?".)

Instead, you can consider the following for codimension 1 submanifolds: given $\tau:N\to M$ an embedding and let $v$ be a vector field on $M$ along $N$ that is transverse to $N$, then $(\nabla,v)$ on $M$ together induces a connection on $N$. For $(X,Y)$ vector fields on $N$, we can define $$ D^{(v)}_X Y = [\nabla_{\tau_*X}\tau_*Y] $$ where $[W]$ for $W\in T_pM$, $p\in \tau(N)$ is defined by $\tau_*[W] - W = \lambda v$ for some $\lambda\in\mathbb{R}$. For higher codimension case you need more (linearly independent) vector fields. In the Riemannian case, $v$ is canonically chosen to be the unit normal vector to $N$ (or in higher codimension, a family that spans the normal bundle).

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  • $\begingroup$ This is all fine and correct, but due to the fact that your transverse vector field(s) is(are) quite arbitrary it is far from being canonical. $\endgroup$
    – user20266
    Jun 7, 2012 at 19:54
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    $\begingroup$ @Thomas: precisely. The answer is (a) meant to illustrate the fact that the choice of a canonical "induced connection" is quite intimately connected with the choice of a canonical traverse vector field and (b) give reference to the fact that meaningful geometry can still be studied without such a canonical choice. $\endgroup$
    – Willie Wong
    Jun 11, 2012 at 7:23
  • $\begingroup$ So if $M$ is a metric manifold, then $v$ can be canonically chosen to be the normal to the surface, right? $\endgroup$
    – Jackozee Hakkiuz
    Jun 24, 2018 at 10:12
  • $\begingroup$ @JackozeeHakkiuz: I am not sure what you mean by metric manifold. Can you give me a reference? $\endgroup$
    – Willie Wong
    Jun 25, 2018 at 17:30
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    $\begingroup$ In the Riemannian case: yes. In the pseudo-Riemannian case, not necessarily: the "normal vector field" may not be transversal. You have to add the condition that the induced metric on the submanifold $N$ is non-degenerate. @JackozeeHakkiuz $\endgroup$
    – Willie Wong
    Jun 26, 2018 at 1:06

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