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My differential geometry professor said that the tangent space of an open set $\Omega$ of $\mathbb{R^n}$ at a point $P$ coincides with $\mathbb{R^n}$ , that is

$T_P(\Omega)=\mathbb{R^n}$

He gave an intuitive example for n=2: if $\Omega$ is any region in $\mathbb{R^2}$, let'say a disc centerend in the origin, and $P$ any point in the disc, the tangent space if the set of all tangent vectors to the curves whose traietories pass by $P$ and are contained in the plane. cleary the $T_P(\Omega)$ consides with the plane $\mathbb{R^2}$ itself.

I tried to do my own example for $\mathbb{R^3}$, according to that is not true that $T_P(\Omega)=\mathbb{R^3}$. If $\Omega$ is half a spheric surface, and $P$ any point on the surface, It is absurd to say that the tangent plane to the sphere is $\mathbb{R^3}$, since is it a PLANE! as it is obvious form the picture, so $T_P(\Omega)=\mathbb{R^2}$

What I am getting wrong? Please refrain from talking about manifolds , varities or technical stuff. These are not part of my course

enter image description here

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    $\begingroup$ A disc in $\Bbb R^2$ is 2D, whereas a spherical hemisphere in $\Bbb R^3$ is not 3D. If you pick an open ball in $\Bbb R^3$, then all of the tangent spaces may be identified with $\Bbb R^3$, just as with an open disc in $\Bbb R^2$. Similarly, if you let $\Omega$ be a semicircle in $\Bbb R^2$, the tangent spaces will just be 1D lines. $\endgroup$
    – anon
    Commented Dec 24, 2020 at 19:15
  • $\begingroup$ @runway44 Ok , but then the tangent spaces may be identified with $\mathbb{R^2}$ as intuition suggests, then how come the definition or property is that $T_{P}\Omega=\mathbb{R^n}$, for $\Omega$ an open set of $\mathbb{R^n}$? $\endgroup$
    – some_math_guy
    Commented Dec 24, 2020 at 19:19
  • $\begingroup$ If $\Omega$ is an open subset of $\Bbb R^n$ then its tangent spaces may be identified with $\Bbb R^n$. Note that for $\Omega$ to be an open subset of $\Bbb R^n$, it must have the same dimension $n$. In general, the tangent spaces of any manifold have the same dimension as the manifold itself. $\endgroup$
    – anon
    Commented Dec 24, 2020 at 19:20
  • $\begingroup$ @runway44 I don't see how taking an open ball in $\mathbb{R}^3$ , the tangent space to a point which I think like the plane tangent to the sphere could possibly be $\mathbb{R}^3$ $\endgroup$
    – some_math_guy
    Commented Dec 24, 2020 at 19:24
  • $\begingroup$ I said ball, not sphere. Do you know the difference? What if someone, say the 2D version of you yourself, told you they don't understand how the tangent space to a point in the unit disk is all of $\Bbb R^2$ and not a tangent line to the circle? $\endgroup$
    – anon
    Commented Dec 24, 2020 at 19:25

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You have embedded a two dimensional object into a three dimensional space however when you move to a higher dimensional space open sets in the lower dimensional space may not be open anymore. A simple example would be the interval $(0,1)$ in $\mathbb{R}$ which is open as it contains all of its interior points, however a line segment is never open in $\mathbb{R}^2$ because it has no interior points.

In your example the open sets would be on the surface of the sphere which are not open in the ambient space. This is for the same reason, every neighborhood around every point will contain points not on the sphere and so it has no interior points, and is therefore not open. As you've correctly deduced this detail is non-trivial otherwise your get a mismatch in dimension.

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  • $\begingroup$ How can I understand intuitively that the tangent space of an open set $\Omega$ of $R^3 $ is $R^3$ and not $R^2$?. Taking an open ball as $\Omega$, I can't help thinking that the tangent space is for a point in the surface still a plane, and for a point inside I can't think of a tangent space. $\endgroup$
    – some_math_guy
    Commented Dec 24, 2020 at 23:54
  • $\begingroup$ My professor has defined the tangent space of a surface at a point as the set of all tangent vectors to the curves contained in the surface at the given point so that for a subset of $R^2$ the tangent space coincides with $R^2$ itself, but I fail to do the same for a subset of $R^3$ $\endgroup$
    – some_math_guy
    Commented Dec 25, 2020 at 0:01
  • $\begingroup$ @mathlover you're trying to visualize a four dimensional embedding of a three dimensional space. This can prove problematic. Consider the tangent space as an approximation of the underlying open set. If the open set is on the surface of the sphere, the approximation should resemble the open set being approximated. Bump it down a dimension. In this case you'd find a tangent line associated with a point defined on some small open neighborhood. The line is one dimensional because the graph is one dimensional, not the plane it's drawn in. $\endgroup$
    – CyclotomicField
    Commented Dec 25, 2020 at 0:09
  • $\begingroup$ If the tangent space is an approximation of the underlying open set, I don't see how all of $R^3$ can be an approximation of an open ball, For one thing $R^3$ is infinite while, the sphere is not, so they don't resemble each other. Furthermore, I am not sure in what sense are you talking about an approximation. $\endgroup$
    – some_math_guy
    Commented Dec 25, 2020 at 0:55
  • $\begingroup$ Somegave an example above " for instance consider the origin. The tangent vectors at the origin in the unit ball can point in any direction at any length, so the tangent space is all of R3" , but I fail to understand it. According to this I picture a vector in the open ball as a needle inserted into a basketball, How can that be a tangent if it going through the ball and not tangent to it? $\endgroup$
    – some_math_guy
    Commented Dec 25, 2020 at 0:59

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