I'm studying differential geometry using De Carmo. To define the notion of a derivative on a manifold we need to associate with each point $x$ on the manifold $M$ a tangent space $TM_x$. In any examples I've encountered the tangent space has ended up being $TM_x=\mathbb{R}^m$ where $m$ is the dimension of the manifold.

I assume this is not always the case but I cannot think of any examples that give a more exotic or unusually tangent space. Can you give an example of a smooth map between smooth manifold with an unusual tangent spaces?

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    $\begingroup$ The tangent space of an $m$-dimensinal manifold is an $m$-dimensional linear space, and hence in general isomorphic to $\mathbb{R}^m$. $\endgroup$ – MisterRiemann May 31 at 11:44
  • $\begingroup$ Each individual tangent space of a smooth manifold $X^n$ is (more or less by the definition of smoothness) an $n$-dimensional space and thus isomorphic to $\mathbb{R}^n$. The more interesting part is the structure of $TX$ as a vector bundle rather than a series of vector spaces; in general, it isn't just $X\times \mathbb{R}^n$. $\endgroup$ – anomaly May 31 at 13:02
  • $\begingroup$ Perhaps you were interested in knowing about the Tangent bundle---how each tangent spaces fit together. A simple example of a non-trivial tangent bundle (i.e., $TM \neq M \times \mathbb R^n$) is a sphere $S^2$. $\endgroup$ – jdoicj Jun 3 at 14:03

The vector space structure of $T_pM$ is a $\dim_\mathbb{R} M$-dimensional vector space over $\mathbb{R}$, so it is isomorphic to $\mathbb{R}^{\dim_\mathbb{R} M}$.

However, since you tag your question with lie-groups, there is another interpretation of your question. The tangent spaces to a Lie group each has a natural Lie algebra structure inherited from the parallelization with left-invariant vector fields. So $T_gG$ is not simply the $\mathbb{R}$-vectorspace $\mathbb{R}^k$, but a Lie algebra $\mathfrak{g}$.


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