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The concrete setting: Let $M\subset \mathbb{R}^3$ be a regular surface. A vector field $X$ in $M$ is a differentiable function $X: M\to \mathbb{R}^3$ such that $X(p)\in T_pM$.

Here we are taking the following "immersed" definition of $T_pM$: it is the image of $d\varphi_0$, where $\varphi:U\subset \mathbb{R}^2\to M$ is a diffeomorphism onto its image, such that $\varphi(0)=p$.

Then the standard connection $\nabla$ in $M$ is defined as $\nabla_X Y(p)=P_{T_pM}(dY_p(X(p)))$

where $P_{T_pM}:\mathbb{R}^3\to T_pM$ is the orthogonal projection.

The problem: I'm trying to verify that this connection is symmetric, but I'm having a clash of definitions.

On one hand, $$(\nabla_X Y-\nabla_y X)(p)=P_{T_pM}(dY_p(X(p))-dX_p(Y(p)))\hspace{2cm} \mbox{(1)}$$

Now I'd like to calculate the Lie bracket, and here is where I run into trouble.

The abstract setting: The definition of the Lie bracket I have is for two vector fields on an abstract manifold, where the tangent space at $p$ is the vector space of derivations at $p$. In this setting, a vector field is a section of the tangent bundle. If $X,Y$ are vector fields in $M$ and $f\in C^\infty(M)$, we have: $$[X,Y](p)(f):= X(p)(Yf) - Y(p)(Xf)$$

where $(Xf)(p):= (X(p))(f)=df_p(X(p))(\operatorname{id}_{\mathbb{R}})$.

Thus, $$[X,Y](p)(f)=d(Yf)_p(X(p))(\operatorname{id}_\mathbb{R}) - d(Xf)_p(Y(p))(\operatorname{id}_\mathbb{R}) \hspace{2cm} \mbox{(2)}$$

Back to the problem: I should prove that (1) and (2) are... "equal", but the concrete setting vs. the abstract setting clash: (1) is a vector in $T_pM$ while (2) is a real number.

I think I should reinterpret one in terms of the other, but I'm failing to do so. If I could interpret (1) in the abstract setting, maybe when I evaluate it at a function $f$ I would get (2). But I don't know how to properly justify it.

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Just act the vector field you get in (1) on arbitrary functions $f$. Two vectors $v,w\in T_pM$ are equal iff for every neighborhood $U\ni p$ and every function $f\in C^1(U)$ that $v(f) = w(f)$. –  Willie Wong Feb 5 '13 at 16:48
Ah, are you worried that the $T_pM$ used in the first definition is not the same space as $T_pM$ in the intrinsic definition? –  Willie Wong Feb 5 '13 at 16:53
@WillieWong: yes, that's it. I see (1) as an element of $T_pM\subset \mathbb{R}^3$, and I don't know how to translate it into a derivation. –  lentic catachresis Feb 5 '13 at 17:07

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