# Covariant derivative and orthogonal projection

Let $V$ be a smooth vector field along a curve $c$. The covariant derivative of $V$ along $c$ at the point $c(t)$ us given by

$\frac{D_cV}{dt}(c(t))=\lim_{s\to t}\frac{P_{c(t),c(s)}V_{c(s)}-V_{c(t)}}{s-t},$ where we define $P_{c(t),c(s)}$ to be the parallel transport of the tangent vector $V_{c(s)}$ to a tangent vector at the point $v(t)$.

Geometrically, the covariant derivative $\frac{D_cV}{dt}(c(t))$ represents the projection of $dV/dt$ onto the tangent plane of the surface, but I can't prove it clearly, how to give a intuitive way to see this?

Here we consider $c$ as a curve from a surface in $\mathbb{R}^3$

• "projection . . . onto the surface" . . . the surface? Which surface? Inquiring minds want to know! Cheers! – Robert Lewis Jun 11 '18 at 1:37
• You should specify your setting. You're doing a submanifold of $\Bbb R^n$? A surface in $\Bbb R^3$? A submanifold of a Riemannian manifold with the induced Riemannian connection? ... You have to understand the connection geometrically. How is parallel transport defined? – Ted Shifrin Jun 11 '18 at 17:09
• @TedShifrin let's say it's a surface in $\mathbb{R}^3$. – 6666 Jun 11 '18 at 19:51
• You didn't answer my other question. I define parallel transport in terms of the covariant derivative. Of course, one can do it in the other way, but you have to tell me how parallel transport is defined for you. – Ted Shifrin Jun 11 '18 at 20:04
• @TedShifrin the tangent vector maintains constant length and constant angle with the geodesic. Is that what you want? – 6666 Jun 11 '18 at 20:35