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In differential geometry I'm told that directional derivatives can be interpreted as tangent vectors and I'm trying to build some intuition for this using simple cases. If I take a map $f: \Bbb R^2 \to \Bbb R$, $f(x,y)=2-x^2-y^2$ and a vector $v=\langle \frac{1}{\sqrt2}, -\frac{1}{\sqrt2}\rangle$, then the directional derivative is the gradient dotted with $v$ $$D_vf= \nabla f \cdot v = -\sqrt2x + \sqrt2y.$$ I fail to see how is this a vector? It's a real number in $\Bbb R$? So in a sense a vector in $\Bbb R$, but the only thing it's tangent to is the real line?

I'm trying to build up to the tangent space which uses derivations which seem to be in a sense some kind of directional derivative?

Arctic Char
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Jasper
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  • What is a vector to you? – Arctic Char Feb 02 '22 at 19:31
  • A point in space. I think it depends on the context whether to intrepret say $(1,1)$ as a vector or a point? – Jasper Feb 02 '22 at 19:34
  • However in this case I was initially thinking that these directional derivatives would be tangent vectors to the surface $f(x,y)$ here in which the tangent space would be constructed, but it seems that they're tangent vectors in the codomain? That doesn't make sense. I'm not interested in points/vectors in $\Bbb R$. I'm interested in the tangent space on the manifold I'm in. – Jasper Feb 02 '22 at 19:35
  • If you imagine the the universe as a 3 dimensional manifold, what is the tangent vector at a point to you? – Arctic Char Feb 02 '22 at 20:00
  • If the point lies in some kind of vector valued curve, then I guess I would say the derivative of the curve? – Jasper Feb 02 '22 at 20:12
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    Yes, that's one way. But you will need to introduce a local coordinate system in order to define that. Another definition is to define the tangent space as the space of derivation (that is, it acts on functions and satisfies the product rule). So $D_v$ is a derivation and one can think of it as a vector. – Arctic Char Feb 02 '22 at 20:25
  • What is the motivation for this "acting on functions" it seems unnecessary to act on something since we already have what we were trying to get? That is the set of all tangent vectors at a point? – Jasper Feb 02 '22 at 21:09
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    I will refer you what I read from "Introduction to Smooth Manifold" of John Lee about the definition of tangent vectors as derivations, not verbatim. The choice of which definition to adopt is a matter of taste. The one with derivations may seem abstract at first but it has several advantages, it makes the vector space structure of the tangent space obvious; and it leads to straightforward coordinate-indipendent definitions of differentials, velocities, and many other objects you will be studying. – Matteo Aldovardi Feb 02 '22 at 21:58
  • Let me point two misunderstandings here: 1) you should start with a vector $v$ tangent to $\mathbb{R}^2$ at a specific point $p$ (just choose whichever you like), then only consider the directional derivative $D_v f(p)$ at that point, 2) the number $D_v f(p)$ doesn't correspond to a vector, but the map $f \mapsto D_v f(p)$ does. – Michał Miśkiewicz Feb 02 '22 at 22:43
  • I think it's worth mentioning that the statements "directional derivatives can be interpreted as vectors" and "vectors can be interpreted as directional derivatives" are logically equivalent, but have very different connotations. – Jacob Manaker Feb 03 '22 at 11:19

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