I'm having trouble to understand the concept of Covariant Derivative of a vector field.

The definition from doCarmo's book states that the Covariant Derivative $(\frac{Dw}{dt})(t), t \in I$ is defined as the orthogonal projection of $\frac{dw}{dt}$ in the tangent plane.

Does that mean that if $w_0 \in T_pS$ is a vector in the tangent plane at point $p$, then its covariant derivative $Dw/dt$ is always zero? Since $dw_0/dt$ will be parallel to the normal $N$ at point $p$.

Is that correct?

If so, then for a vector field to be parallel, then every vector must be in the tangent plane.

Is that also correct?

Could you explain without using tensors and Riemannian Manifolds? Thank you


2 Answers 2


The vector fields you are talking about will all lie in the tangent plane. However the (ordinary) derivative of a vector field (in the tangent plane) does not necessary lie in the tangent plane. Remember that the tangent plane may vary from point to point. This (ordinary) derivative does not belong to the intrinsic geometry of a surface, however its projection back onto the tangent plane will again be an intrinsic concept. Does this answer you concerns ?

  • $\begingroup$ You mean that $Dw/dt$ lie in the tangent plane, but $dw/dt$ does not necessarily lies in the tangent plane, correct? Can I say that if a vector $w_0$ in this vector field $w$ lies in the tangent plane, that is $w_0 \in T_pS$, then its covariant derivative (at this point $p$) is zero? $\endgroup$
    – cryptow
    Jun 30, 2014 at 1:22
  • $\begingroup$ To the first part, yes. For the second I dont understand, are you taking the derivative of a single vector ? Dont you just differentiate fields ? $\endgroup$ Jun 30, 2014 at 1:28
  • $\begingroup$ What I mean is, for each point $p \in S$, i have a vector determined by this vector field $w$. At this point p, $Dw/dt$ is the projection of $dw/dt$ in the tangent plane. My question is: if the vector at $p$, determined by my vector field $w$ lies (the vector) in the tangent plane, does that mean the covariant derivative at this point will be zero? Can I even ask that? Or is it totally out of sense? $\endgroup$
    – cryptow
    Jun 30, 2014 at 1:42
  • 1
    $\begingroup$ ALL of the vectors of the field lie in the tangent plane. And no the derivative may not be zero, it depends on how the neighbouring vectors (also in the tangent plane) are situated. $\endgroup$ Jun 30, 2014 at 1:56
  • $\begingroup$ I think I understand now: $dw/dt$ is the "rate" of change of the vector field $w$ along the tangent vector $\alpha'(0)$ at $p$. And $Dw/dt$ is the projection of this rate to the tangent plane. So, $Dw/dt = 0$ means the vector field doesn't change (locally) along side the direction defined by the tangent vector $y$(for a curve $\alpha$ and $\alpha'(0) = y$). It is also proved that the covariant derivative does not depend on this curve, only on the direction $y$. $\endgroup$
    – cryptow
    Jun 30, 2014 at 3:31

Consider that the surface is the plane $OXY.$ Consider the curve $(t,0,0)$ and the vector field $V(t)=t\partial_x.$ You have that its covariant derivative $\frac{dV}{dt}=\partial_x$is not zero. Note that, even being $N$ constant, the length of $V$ changes. This is the reason, in this case, to have non-zero covariant derivative.

Now, when we say that a vector field is parallel we assume it is tangent to the surface. In any case, if you consider that the orthogonal projection is zero without being tangent, think of the above case of the plane and $V=\partial_x+\partial_z.$


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