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Let $\Gamma \subset \mathbb{R}^2$ be a curve. Define for a smooth function $f$, $$\nabla_\Gamma f = \nabla f - (\nabla f \cdot N)N$$ where $N$ is the unit normal.

Let $X:S \to \Gamma$ be a smooth regular parameterisation with $|\partial_s X(s) | > 0$.

Let $\tilde{f}(s) = f(X(s))$. How do I show that $$\nabla_\Gamma f = \frac{1}{|\partial_s X|} \partial_s \tilde{f}\frac{\partial_s X}{|\partial_s X(s)|}$$ ?

I don't know where to start. The notation is confusing..

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This is called "covariant derivative" along the curve. If all else fails, try looking on Do Carmo's Differential geometry of curves and surfaces. –  Giuseppe Negro Jun 29 '12 at 12:04

1 Answer 1

up vote 2 down vote accepted

We don't have to look far: It's the chain rule in disguise.

Let $$T:={\partial_s X\over|\partial_s X|}$$ be the unit tangent vector. Then the definition of $\nabla_\Gamma f$ amounts to $$\nabla_\Gamma f=(\nabla f\cdot T)T\ .$$ This is just two dimensional vector algebra: For any two orthogonal unit vectors $T$, $N$ and an arbitrary vector $V$ one has $V=(V\cdot T)T+(V\cdot N)N$. It follows that $$\nabla_\Gamma f=(\nabla f\cdot\partial_s X){\partial_s X\over|\partial_s X|^2}\ .$$ Now by the chain rule $\partial_s\tilde f=\nabla f\cdot\partial_s X$, and plugging this into the last formula you get the claim.

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Thanks a lot... –  blahb Jun 29 '12 at 14:28

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