8
$\begingroup$

I’ve avoided doing any manifold (regretting it somewhat) courses, however do have some understanding. Let $p$ be a point on a surface $S:U\to \Bbb{R}^3$, we define:

The tangent space to $S$ at $p$, $T_p(S)=\{k\in\Bbb{R}^3\mid\exists\textrm{ a curve }\gamma:(-ε,ε)\to S\textrm{ with }\gamma(0)=p,\gamma'(0)=k\}$.

The tangent plane to $S$ at $p$ as the plane $p+T_p(S)\subseteq\Bbb{R}^3$.

My current understanding is, in the diagram below the tangent plane is the plane shown, whilst the tangent space would be p minus each element of the plane, hence the corresponding plane passing through the origin. Is this correct or is it incorrect? I’m doing a course called geometry of curves and surfaces and being unsure about this is making understanding later topics difficult.

Edit - can't post images, here's a link instead! http://standards.sedris.org/18026/text/ISOIEC_18026E_SRF/image022.jpg

Thanks!

$\endgroup$
1
  • $\begingroup$ Yes, according to the definition of a vector space, it has to have the $0$ vector. So you have to transfer the tangent plane to the origin in order it to contain the $0$ vector. The tangent plane is not a vector space (in the most cases). $\endgroup$
    – Andreas K.
    Mar 26, 2013 at 23:38

2 Answers 2

5
$\begingroup$

While the definitions you've given are acceptable, I would use different definitions for tangent space and tangent plane that reveal some more mathematical structure.

  • The tangent plane is a geometric object. You can define the tangent plane to a point $p$ on a surface $S$ in $\mathbb{R}^3$ as a plane in $\mathbb{R}^3$ which intersects $p$ and whose normal vector is parallel to the gradient of $f$, if $S$ is represented as the level surface $f(x,y,z) = 0$. But all that formality is simply used to the describe the geometric object that everybody can visualize intuitively.

  • The tangent space is a vector space of linear functionals. Each vector $v \in T_p S$ acts on a function $g \in C^1(S)$ in a way such that $v(g)$ gives you a directional derivative of $g$ in a particular direction (which can be identified with $v$ itself). Thus the tangent space $T_p S$ can be naturally identified with the tangent plane, because the only directions that make sense to take the directional derivative on are the directions that lie on the tangent plane.

$\endgroup$
5
$\begingroup$

Exactly. The tangent space is simply a translation of the tangent plane by taking $p$ to $0$. One nice property that the tangent space always has (but which a tangent plane almost never does) is that it is closed under addition and scalar multiplication--that is, for any $\alpha\in\Bbb R$ and any $x,y$ in the tangent space, we have $x+y$ and $\alpha x$ in the tangent space.

$\endgroup$
2
  • $\begingroup$ Muchos gracias! Thanks to whoever edited, never learnt latex! $\endgroup$
    – user68639
    Mar 26, 2013 at 23:45
  • 4
    $\begingroup$ It should be pointed out that for surfaces, many authors use tangent plane as a synonym for tangent space, and use the term affine tangent plane for the affine plane parallel to the tangent plane and containing the point $p$. This terminology extends easily to higher dimensions: for a smooth submanifold of $\mathbb R^n$, the tangent space at $p$ is a linear subspace of $\mathbb R^n$, and the affine tangent space is a translate of that. $\endgroup$
    – Jack Lee
    Mar 27, 2013 at 17:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .