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I apologize in advance because I don't know how to enter code to format equations, and I apologize for how elementary this question is. I am trying to teach myself some differential geometry, and it is helpful to apply it to a simple case, but that is where I am running into a wall.

Consider $M=\mathbb{R}^2$ as our manifold of interest. I believe that the tangent space is also $\mathbb{R}^2$. From linear algebra, we know that a basis set for $\mathbb{R}^2$ is $$\left\{\left[\matrix{1\\0}\right], \left[\matrix{0\\ 1}\right] \right\}\;.$$

Now, from differential geometry, we are told that basis vectors are $\frac{d}{dx}$ and $\frac{d}{dy}$ where the derivatives are partial deravatives.

So my question is how does one obtain a two-component basis vector of linear algebra from a simple partial derivative?


EDIT: Thanks to everyone for the replies. They have been very helpful, but thinking as a physicist, I would like to see how the methods of differential geometry could be used to derivive the standard basis of linear algebra. It seems that there must be more to it than saying that there is an isomorphism between the space of derivatives at a point and R^n which sets up a natural correspondence between the basis vectors.

I may be completely off-the-wall wrong, but somehow I think that the answer involves partial derivatives of a local orthognal coordinate system at point p.

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    It's a little unclear to me what you're asking. Do you want to know why that is a basis? How do you think of the tangent space?2012-01-21
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    It sounds like perhaps this is more to do with interpreting what differential geometers mean when they say $d/dx$ is a "basis vector". The interpolation between the two points of view is $d/dx$ is the directional derivative in the $(1,0)$ direction. i.e. you are essentially conflating the notion of "vector" with "directional derivative in the direction of a vector". They're equivalent ideas, since the latter is determined by its vector input.2012-01-21
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    @Sam: Do you understand the difference between an abstract vector space of dimension 2 over $\mathbb{R}$ and the concrete vector space $\mathbb{R}^2$?2012-01-21
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    One problem with your question is the statement «the tangent space [of $M=\mathbb R^2$] is also $\mathbb R^2$.» At most, the tangent space to $M$ *at a specific point* is isomorphic to $\mathbb R^2$ (in this peculiar situation, there is a somewhat canonical isomorphism...)2012-01-22
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    Yes, I agree that I should have been more careful and state that the tangent space to M at a specific point is isomorphic to R^2 (I am coming at this as a physicist and not a mathematician).2012-01-22

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