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I don't understand how to calculate the tangent space of some concrete examples. For example, can someone explain me how to calculate the tangent space in a point $p$ on the sphere? And how to calculate the tangent space in a point on the torus?

g.pomegranate
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  • The basis of the tangent space of a surface for example, just the pushforward of the coordinate basis of the coordinate tangent space under chart map. You can find the examples in elementary differential geometry books such as do Carmo's Curves and Surface. – Kelvin Lois Dec 27 '17 at 20:32

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In general if you have a manifold $M$ (already) embedded in $\mathbb{R}^N$, as $M = \mathbb{S}^n$,then for a given point $p$, you can consider a chart $(U,\varphi:U\to \mathbb{R}^n)$, $p\in U$ and thus $ T_p M = d_{\varphi(p)}\varphi^{-1} (T_{\varphi(p)}(\mathbb{R}^n)) $.

In practice, you choose a local parametrization $\varphi^{-1}:\mathbb{R}^n\to M$ and computing partial derivatives $d_{x}\varphi^{-1}(e_i)\in T_{\varphi^{-1}(x)}M$ will give you $n$ generators of the tangent space.

Since the manifold is given as a subset of $\mathbb{R}^N$, $d_{x}\varphi^{-1}(e_i)$ will be also vectors of $\mathbb{R}^N$, so they are something concrete. If you consider an abstract manifold, you will obtain again some generators but is not so simple to visualize the tangent space.

Another useful remark is the following. If you are dealing with a level set in the form $M = \{ x\in \mathbb{R}^N | \ f(x ) = O \} \ $ where $f:\mathbb{R}^N\to \mathbb{R}^k $ and the point $O$ is regular for $f$ (this is again the case of $\mathbb{S}^n$ or $\mathbb{T}^2$), then it can be proven that $T_pM = \ker d_pf $. So since $\mathbb{S^n} = \{x\in \mathbb{R}^{n+1}| \ ||x||^2 - 1 = 0\}$ and $0$ is regular for $||x||^2 - 1$, it follows that

$$ T_p \mathbb{S}^n= \ker d_p(||\cdot||^2 - 1) = \ker 2\langle p,\cdot\rangle = p^\perp$$

Overflowian
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  • Can you recommend me some books explaining this things, please? I tried to read several books, but I have not found concrete calculations of tangent space in any book. – g.pomegranate Dec 27 '17 at 20:51
  • I suggest Do Carmo's Curves and Surfaces, Pressley's Elementary_Differential_Geometry and maybe you can check out also Milnor's Topology from a differentiable viewpoint which do not make a lot of calculations but is a very concise and effective introduction to submanifolds of $\mathbb{R}^n$. Maybe also in O'Neill's Semi-Riemannian Geometry you can find some calculations. Another one could be Frankel's the Geometry of Physics. – Overflowian Dec 27 '17 at 21:00
  • In the case of the sphere, how do you go about finding $p^{\perp}$? – Euler_Salter Mar 04 '21 at 09:32
  • Given the function $f(x) = (x-x_0)^2+(y-y_0)^+(z-z_0)^2-r^2$ giving rise the the sphere $$ \mathbb{S} := \left{(x,y,z)\in \mathbb{R}^3, :, (x-x_0)^2+(y-y_0)^+(z-z_0)^2-r^2 = 0\right} $$ How do you find its basis for the tangent space? Of course one has $$ \nabla f(x) = \left(2(x-x_0), 2(y-y_0), 2(z-z_0)\right) $$ Using the kernel rule you were talking about we realise that this is the normal vector to the tagent plane. How do you find its basis? – Euler_Salter Mar 04 '21 at 09:35
  • I know that one way is simply to take $$ x = x_0 + r \cos(\phi) \sin(\theta) \ y = y_0 + r \sin(\phi)\sin(\theta)\ z = z_0 + r \sin(\theta) $$ And take its gradient with respect to $(\theta, \phi)$ to get $$ e_{\theta} = (r\cos(\phi)\cos(\theta), r\sin(\phi)\cos(\theta), -r\sin(\theta)) \ e_{\phi} = (-r\sin(\phi)\sin(\theta), r\cos(\phi)\sin(\theta), 0) $$ But how does one do this in general? – Euler_Salter Mar 04 '21 at 09:35
  • @Euler_Salter what you have done in your last comment is an instance of the method I've presentend in the first part of my answer, i.e. in general you take a local parametrization and compute partial derivatives.

    There is not a canonical basis for the tangent space, so is not clear what you are asking for. Anyway, all the possible basis differ from the one you picked by multiplication by an element of $GL$ (beware though that if you consider a local frame i.e. a family of basis over an open subset of the manifold, not all local frames will come from parametrizations)

     – Overflowian Mar 04 '21 at 12:17
  • @Euler_Salter the nice thing about the formulation using the orthogonal to the normal is that once the embedding in R^n is chosen the tangent space is determined (but again you can't hope for finding a special basis), and you can really see it in R^n.

    If your question is instead about how to find a parametrisation, then for level sets they usually come from the implicit function theorem. Another source of parametrizations are group actions (when you deal with homogeneous spaces like spheres). If the equations are polynomial sometimes you can apply some algebraic geometry arguments.

     – Overflowian Mar 04 '21 at 12:27