# How to obtain the standard basis for $\mathbb{R}^2$ using differential geometry?

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.

-
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? – Dylan Moreland Jan 21 '12 at 20:08
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. – Ryan Budney Jan 21 '12 at 20:12
@Sam: Do you understand the difference between an abstract vector space of dimension 2 over $\mathbb{R}$ and the concrete vector space $\mathbb{R}^2$? – Zhen Lin Jan 21 '12 at 20:13
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...) – Mariano Suárez-Alvarez Jan 22 '12 at 5:10
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). – Sam Jan 22 '12 at 15:49

This isomorphism is usually established in elemantary textbooks (see Schutz. Geometrical Methods or Wheeler.Gravitation) via the directional derivative. I could go through the argument, but you can find it here:

http://en.wikipedia.org/wiki/Vector_(geometric)

under the section "The Vector as a Directional Derivative"

and ending with "Therefore any directional derivative can be identified with a corresponding vector, and any vector can be identified with a corresponding directional derivative."

-

Yes, the fact that the tangent space at a point isn't actually R^2 cannot be emphasized enough. Another way to think of the tangent space at a point is through equivalence classes of differentiable curves through the given point, with the relation being that two curves are equivalent if the curve composed with your coordinate chart (in this case your chart is trivial) have the same velocity at this point. So in order to obtain a basis for the tangent space all you need to do is take two curves through your given point, and then differentiate them and evaluate. These curves then act on real-valued differentiable functions, say f:M->R^2 so then tangent vector is a function $\alpha'(0):D\rightarrow\mathbb{R}$ $$\alpha'(0)f=\frac{d}{dt}(f\circ\alpha) |_{t=0}$$ Do Carmo's book has a good explanation of this.

-

Here is, for what it's worth, how I see this.

In differential geometry, it is sometimes convenient to denote the canonical basis of $\mathbb R^n$ by $$\left(\frac{\partial}{\partial x_i}\right)_{i=1,\dots,n}\quad,$$ and the dual basis (which is a basis of $(\mathbb R^n)^*$) by $$(dx_i)_{i=1,\dots,n}.$$ So, in my modest opinion, it's just a matter of notation, nothing more.

EDIT. Here is how I understand the question asked by the OP:

Why is the $i$-th vector of the canonical basis of $\mathbb R^n$ equal to $$\frac{\partial}{\partial x_i}\quad?$$ The answer is: By definition.

Of course, this answer prompts a second question, which is:

What's the reason for such a bizarre definition?

Dylan answered this (deeper) question perfectly, but I think it is important that the OP understands the difference between these two questions. This is why I tried to insist (perhaps awkwardly) on the aspect that was not explicitly covered by Dylan's answer (which I upvoted --- I also upvoted the question).

-
Thank you. Finally someone has stated in one paragraph the basis and dual basis. – Sam Jan 22 '12 at 18:24
@Sam: your question does not ask about a dual basis. Perhaps you could adjust your question to be more specific? – Ryan Budney Jan 22 '12 at 19:02
@Ryan Budney: I understand that my question does not ask about a dual basis and that was not the thrust of my question. I was just happy to see in Pierre's answer the bases listed together because in most books they are in different chapters. – Sam Jan 23 '12 at 15:22

$\newcommand{\bR}{\mathbf{R}}$We can view the tangent space of $\bR^2$ at a point $P$ as the space of all derivations at $P$; that is, $T_P(\bR^2)$ is the set of linear maps $X\colon C^\infty(\bR^2) \to \bR$ satisfying a Leibniz rule $$X(fg) = (Xf)g(P) + f(P)Xg.$$ This might seem strange, but there are some familiar elements in here: the directional derivatives $D_v|_P$ at $P$ for various $v \in \bR^2$, which include $(\partial/\partial x)|_P$ and $(\partial/\partial y)|_P$. Moreover, one can show that the canonical map $$\bR^2 \to T_P(\bR^2) \quad \text{given by} \quad v \mapsto D_v|_P$$ is an isomorphism. This gives a natural correspondence between $(1, 0)$ and $(\partial/\partial x)|_P$. Injectivity is easy, and surjectivity follows from Taylor's theorem in several variables; the details are in, for example, Lee's book.

-