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I have read that the tangent vectors on manifolds are differential operators, looking at a particular example, the tangent space to $SO(3)$, ($SO(3)$ is the group of rotations in $\mathbb{R}^3$, which is a Lie group), on the identity matrix, is the set of antisymmetrical matrices, and is called the Lie algebra $so(3)$. If $A\in so(3)$ we can talk about a curve on $SO(3)$ given by: $$ \alpha (s)= e^{sA}, \quad \alpha(0) = e^{0} = \mathbb{I}, \quad \frac{d \alpha}{ds}\Big|_{s=0} = Ae^{sA}\Big|_{s=0} = A. $$ Therefore in this case the matrix $A$ is a tangent vector right?, my conclusion is that differential operators are linear transformations, so they have an associated matrix, the matrix A would be that associated matrix, is that right?, if isn't, how can I see the tangent vector explicitly as a differential operator in this particular case?. If having any other examples on this would be great. Thanks in advance.

Spherk
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Differential operators are linear transformations, yes, but they're linear maps $C^{\infty}(M)\to\Bbb{R}$, the domain is an infinite-dimensional vector space. They don't have a matrix-representation (which is only for maps between finite-dimensional spaces, after choosing bases on the domain and target). The point is that if you have a submanifold $S$ of a vector space $V$ (in your case $S=\text{SO}(3)$ and $V=M_{3\times 3}(\Bbb{R})$), then there is a standard linear injection $T_pS\hookrightarrow V$, so we can identify $T_pS$ with its image under this linear injection.

In the case of $S=\text{SO}(3)$ and $V=M_{3\times 3}(\Bbb{R})$, if you think of $S$ as a manifold, then the tangent space at the identity $T_IS$ is an abstractly-defined vector space, which set-theoretically is very complicated. But, after injecting it into $V$, it is just $\mathfrak{so}(3)$, the space of skew-symmetric matrices $3\times 3$ real matrices, and this is where your $A$ lives.


I have several answers regarding similar issues:

  1. Why is a vector field on the sphere equivalent to $f:S^n\to\Bbb{R}^{n+1}$ such that $f(x)\perp x$? Here I talk about how to relate the abstract definition to the more intuitive notion.
  2. Calculating the derivative of a differentiable map between manifolds. Similar thing where I explain the theory, but also describe how to apply it for the specific computation for a map $f:S^2\to\Bbb{R}$.
  3. How to show that $T_{(1,0)}S^1\cong \text{span}(\{e_2\})$? A similar thing, applied to the special case of the circle $S^1$.
  4. What is meant by part (e) of MTW Exercise 9.13 involving the tangent to a curve on the manifold $\text{SO}(3)$? This is a PhySE answer where I explain similar stuff in the context of the manifold $\text{SO}(3)$. In that link, what I write $\Phi(T_I(SO(3)))$ is precisely equal to $\mathfrak{so}(3)$.

There may even be more answers along these lines, but this should be enough. I suggest reading them in order (though you're of course free to do it in whichever order you like).

peek-a-boo
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  • Thanks so much for your answer. Just for curiosity, does this have something to do with Ado's theorem? – Spherk Aug 08 '22 at 01:43
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    @Sjang never heard of Ado's theorem lol, so I would guess not. – peek-a-boo Aug 08 '22 at 02:05
  • @Sjang Maybe this is what you are looking for: https://mathoverflow.net/q/62234/40570 – J126 Aug 08 '22 at 15:51
  • @Sjang not quite. Ado's theorem tells us that any Lie algebra has at least one faithful representation (and thus can be represented by matrices) but here we are using something simpler because you are starting from a Lie group which you have already assumed to be a matrix group. This makes things easier as it is natural to differentiate a curve in a matrix group and denote the answer by a matrix. Ado's theorem only assumes an abstract Lie algebra. – Callum Aug 09 '22 at 22:08