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Let $M$ be a $2$-dimensional compact oriented surface in $\mathbb R^3$ with boundary $\partial M$. For any $p\in M \setminus \partial M$ tangent vectors are defined as speed vectors of smooth curves $\Gamma \colon (-1,1) \to M$, $\Gamma(0) = p$ at point $t=0$, i.e. vectors $\dot \Gamma(0)$. They form the linear space $T_p M$ called the tangent space to $M$ at $p$.

Suppose now that $p \in \partial M$. Then if we use the same definition for tangent vectors (i.e. speed vectors at zero of smooth curves $\Gamma \colon (-1,1) \to M$, $\Gamma(0) = p$) we will obtain only vectors, that are tangent to $\partial M$ at point $p$. My question is how to modify the definition of tangent vectors at boundary points to obtain the tangent vectors to $M$ at $p \in \partial M$? Is it possible to define these tangent vectors as speed vectors of curves $\Gamma \colon (-1,1) \to \mathbb R^3$, $\Gamma(-1,0] \subset M$ or $\Gamma[0,1) \subset M$, $\Gamma(0) = p$ at $0$, i.e. vectors $\dot \Gamma(0)$?

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  • $\begingroup$ Almost; the tangent space at a boundary point $p$ is the set of vectors $\dot \Gamma (0)$ of smooth curves $\Gamma : ]-1,1[ \to M$ such that $\Gamma(0) = p$ and $\Gamma([0,1[) \subset M$ (not $\Gamma(]-1,0]) \subset M$, then you get too much) $\endgroup$ – wspin Feb 28 '14 at 13:30
  • $\begingroup$ @wspin But if we consider only $\Gamma([0,1[) \subset M$ (but not $\Gamma(]-1,0]) \subset M$) we will obtain only "inward" pointing vectors, won't we? I think that they won't form a vector space. $\endgroup$ – Appliqué Feb 28 '14 at 14:57
  • $\begingroup$ The tangent space at a boundary point is indeed not a vector space. If the boundary is smooth it is a half space. Compare this en.wikipedia.org/wiki/File:Supporting_hyperplane1.svg $\endgroup$ – wspin Feb 28 '14 at 16:31
  • $\begingroup$ @wspin bizarre, I imagined the tangent space as the tangent plane translated to zero. Could you tell me please in what book have you found your definition? By the way, your image is about supporting hyperplane. $\endgroup$ – Appliqué Feb 28 '14 at 17:38
  • $\begingroup$ This is really a matter of definitions, in my opinion. For instance, if you use the "equivalence class on charts" definition of tangent space, then you get the full vector space. Sometimes this definition is useful, for instance when talking about Stoke's Theorem. $\endgroup$ – Aloizio Macedo Feb 5 '16 at 23:43
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I think the simplest way to define tangent vectors is to not use curves at all. Just say that a vector $v$ is tangent to $M$ at point $p$ if $$\operatorname{dist}(p+tv,M)=o(t),\qquad t\to 0^+\tag{1}$$ This agrees with the usual definition at the non-boundary points. At the boundary points the above definition yields halfspace, as in wspin's comment. If you insist on having a linear space, then either

  • take the linear span; or
  • require (1) to hold for either $t\to 0^+$ or $t\to 0^-$. (The choice may be different for different $t$.)

But you can use curves too, by taking one-sided derivative at the point that is mapped to the boundary. Again, the natural way of doing so (curves begin at $p$) leads to half-plane as the tangent plane. Allowing curve that either begin or terminate at $p$ yields the whole plane.


Aside: from the viewpoint of metric geometry, a tangent space is the [pointed] Gromov-Hausdorff limit of rescaled copies of the surface; as such, it is naturally a halfplane at the boundary points. (And quarter-plane at right-angled corners, etc.)

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    $\begingroup$ Thank you for the answer. From here math.stackexchange.com/a/62808/16273 it follows that if we consider our surface as a manifold with boundary then the tangent space at boundary points is a linear span of your construction (author cites Lee's Introduction to Smooth Manifolds). I think it is better to consider tangent space at boundary as a linear space, no? It leads to natural definition of normal space to boundary at its points (in presence of metric, it is the orthogonal complement of $T_p \gamma$ in $T_p M$, $p \in \gamma$). $\endgroup$ – Appliqué Mar 1 '14 at 7:49
  • $\begingroup$ This answer assumes that the manifold $M$ is an embedded manifold in some linear space (probably $\Bbb R^n$). Since finite-dimensional compact manifolds can in fact be embedded, by Whitney, that's half reasonable, and for OP's question about surfaces in 3-space, it's totally reasonable. For manifolds defined more abstractly, requiring an embedding before one can consider a tangent space may be an awkward way to proceed, and a somewhat more abstract notion of tangent space may be useful. Spivak's Differential Geometry goes through about 5 different equivalent definitions, and is enlightening. $\endgroup$ – John Hughes Jul 23 '18 at 13:00

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