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Let $S$ be an n-dimensional manifold and $M$ be an m-dimensional submanifold of $S$.Let $[\xi^i]$ and $[u^a]$ be coordinate system for $S$ and $M$ respectively,and let $X$ and $Y$ be two vector fields on $M$,we may write $\nabla_{X_p}Y$,"the directional derivative of $Y$ along $X_p$",and $\nabla_{X_p}Y$ is a tangent vector of $S$ but is not necessarily a tangent vector of $M$.Why it is not a tangent vector in $M$?

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  • $\begingroup$ Local coordinates are irrelevant here $\endgroup$
    – Didier
    Commented Nov 7, 2022 at 19:08
  • $\begingroup$ @Didier's answer gives a good concrete example, to approach it from another angle: note that the induced (Levi-Civita) connection is not just a restriction to a submanifold, but it needs to be (orthogonally) projected on the tangent bundle of said submanifold. This gives a hint that we don't expect the covariant derivative of two tangent vector fields to stay tangent to the submanifold. $\endgroup$
    – lemon314
    Commented Nov 8, 2022 at 12:17
  • $\begingroup$ @lemon314 Good point, but I think it's more or less the other way around: Since the LeviCivita connection does not stabilizes the tangent bundle of submanifolds, it needs to be projected. $\endgroup$
    – Didier
    Commented Nov 8, 2022 at 12:36
  • $\begingroup$ @Didier, sure, this was more of a comment how the OP may want to remember/change their intuition how the theory fits together. You are of course right that the correct definition of induced connection comes after the observation that the naive restriction does not preserve the tangent bundle. $\endgroup$
    – lemon314
    Commented Nov 8, 2022 at 13:37

1 Answer 1

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Let $S=\Bbb R^2$ endowed with the usual euclidean metric, $M=\Bbb S^1$. Recall that for $p\in \Bbb S^1$, $T_p\Bbb S^1 = \{p\}^{\perp}$.

The Levi-Civita connection is just the directional derivative of the coefficients:

$$ \nabla_X(f \partial _x + g \partial_y) = (Xf)\partial_x + (Xg)\partial_y. $$

Consider $X(x,y)=Y(x,y)=y\partial_x-x\partial_y$. Then:

  1. $X$ and $Y$ are tangent to $\Bbb S^1$, but
  2. $\nabla_XY$ isn't: indeed \begin{align} \nabla_XY &= \nabla_{y\partial_x - x\partial_y}(y\partial_x - x\partial_y)\\ &= y \nabla_{\partial_x}(y\partial_x - x\partial_y) - x \nabla_{\partial_y}(y\partial_x - x\partial_y)\\ &= -x\partial_x - y\partial_y, \end{align} so that for $p\in \Bbb S^1$, $\left(\nabla_XY\right)(p) \perp T_p\Bbb S^1$.

Note: this example does not come out of nowhere. The integral curves of these vector fields are concentric circles. They can be interpreted as the velocity vector $\vec{v}$ of a particle that would move circularly with constant speed (given by $\sqrt{x^2+y^2}$). It turns out that $\nabla_{\vec{v}}{\vec{v}}$ is the acceleration of such a particle, and it is physically clear that it is directed toward the origin, and thus cannot be tangent to the trajectory.

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  • $\begingroup$ As we see we got a vector field $\nabla_{X}Y=-x\partial_{x}-y\partial_{y}=Z(x,y)$ say. I think it is also in $\mathbb{S^1}$. If I choose $(x,y)=(1,0)\in \mathbb{S^1}$ then $Z(1,0)=(-1,0)$ and same for the every vector in $\mathbb{S^1}$ can you explain this? why this is not in $\mathbb{S^1}$ $\endgroup$
    – Andyale
    Commented Dec 1, 2022 at 13:10
  • $\begingroup$ @Andyale It is not tangent to $S^1$. Draw a picture. If $p\in S^1$, the position vector $\vec{p}$ is orthogonal to $T_pS^1$ $\endgroup$
    – Didier
    Commented Dec 1, 2022 at 13:24
  • $\begingroup$ @Andyale Please read the very first line, second sentence. The answer to your late doubt was already written inside. If you don't understand the difference between a point lying in a manifold and a tangent vector to that point, you should read more elementary differential geometry before asking this kind of questions I guess $\endgroup$
    – Didier
    Commented Dec 1, 2022 at 13:31
  • $\begingroup$ O yes sorry sorry, Now I got this the vector $Z(1,0)=(-1,0)$ is orthogonal to $(1,0)$. $\endgroup$
    – Andyale
    Commented Dec 1, 2022 at 16:09
  • $\begingroup$ @Andyale $(-1,0)$ is not orthogonal to $(1,0)$. They are obviously colinear... $(-1,0)$ is orthogonal to $\{(1,0)\}^{\perp}$ $\endgroup$
    – Didier
    Commented Dec 1, 2022 at 16:22

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