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This might be a stupid question, but I'm reading Tu's smooth manifold chapter 8 and still couldn't figure out what the tangent vector means and what the above isomorphism tells us intuitively in Euclidean space (so that I can also apply this intuition in the case of manifolds). For me, tangent space at some point $p$ in some object(manifold) is like a tangential line or a plane at that point that intersects trivially. So I want to know, is the "tangent vector" in this context something we can actually calculate for a given manifold? For example, if I'm given a manifold $M$ and the charts, or the equation for $n$ dimensional Euclidean object, we can concretely get n-1 dimensional (please correct me if I'm wrong) tangent "vectors"? And how can I understand these vectors are isomorphic to some differential operators that take some function and spits out real value?

Is it the right way to understand?

able20
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    In abstract setting there is no ambient space that you can place your tangent line or plane or whatever. Thus, you need to be more clever there. In Euclidean space you have vectors in a standard sense and then you have directional derivatives at a point and they satisfy Leibniz rule. You identify tangent vectors with directional derivatives in Euclidean space and in abstract setting you literally define tangent vectors as directional derivatives, in a sense. – Ennar Feb 10 '21 at 09:26
  • I think I kind of get it after reading your comments. Thanks! – able20 Feb 10 '21 at 09:31

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For any finite dimensional vector space we have $T_p(V) \cong V$. To see this, for $v\in V$ and $f\in C^\infty (V)$ define $$D_v(f):=\left.\frac{d}{dt}f(p+tv)\right|_{t=0},$$ which is just directional derivative of $f$ at point $p$ in direction $v$.

You can check that $D_v$ is linear functional on $C^\infty(V)$ since by the chain rule we have $$\left.\frac{d}{dt}f(p+tv)\right|_{t=0} = \nabla f(p)\cdot v = \sum v^i \frac{\partial f}{\partial x^i}(p),$$ where $v^i$ are the coordinates of $v$ in standard basis.

Also, $$D_v(fg) = D_v(f)g(p) + f(p)D_v(g),$$ so $D_v$ is a derivation at point $p$, that is $D_v\in T_p(V)$.

It turns out that $v\mapsto D_v\colon V\to T_p(V)$ is isomorphism of vector spaces. To check injectivity, assume $D_v$ is $0$ and act on coordinate function $x^j\colon V \to \mathbb R$ to get $0 = D_v(x^j) = v^j,$ for all $j = 1,\ldots,n$. So, $v = 0$. Since the dimensions coincide, the map is isomorphism.

Hopefully, this gives enough intuition why the abstract definition of tangent vectors is a good generalization of familiar definitions in $\mathbb R^n$ (or any real finite dimensional vector space).

Ennar
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  • This is what I read in the book but I wanted to know the geometric illustration, but thanks! – able20 Feb 10 '21 at 19:39