In differential geometry the covariant basis vectors are defined as $$\hat{e}_\alpha = \frac{\partial \vec{R}}{\partial x^{\alpha}}, $$ and tangent vectors to a curve $x^\alpha = x^\alpha(\lambda) $ are given by

$$ t^\alpha = \frac{\partial x^\alpha}{\partial \lambda}. $$

Well I have just been reading about covariant derivatives and how partial derivatives no longer work because they are comparing vectors defined at different locations which transform differently, well how can the tangent vectors and basis vectors be defined as partial derivatives then?

  • $\begingroup$ What is the vector R? $\endgroup$ Aug 9, 2021 at 23:55
  • $\begingroup$ @VeronicaNoordzee the position vector $\endgroup$
    – Matt0410
    Aug 29, 2021 at 9:21

3 Answers 3


Your confusion is due to the fact that your first definition is in general not correct. It is the correct definition for a manifold embedded in $\Bbb R^N$. For an abstract manifold, the tangent space at a point is defined differently, and it is somewhat complicated.

There is no problem in taking the derivative of a smooth function on a manifold. Given a manifold $M$, a curve $c:(-\epsilon,\epsilon)\to M$, $c(0)=p$, and a function $f:M\to\Bbb R$, the limit $$\lim_{t\to 0}\frac{f(c(t))-f(p)}{t}=:v(f)$$ makes perfect sense. This defines an operator $v$ on the space of smooth functions, which is the tangent vector to $v$ at $p$. Now, giving $c$ the coordinates $(x^i(t))$, we have $$v(f)=\frac{dx^i}{dt}\frac{\partial f}{\partial x^i},$$ using the chain rule. Here $f$ denotes the coordinate representation of the abstract $f$. Once again, $\frac{dx^i}{dt}$ make sense as each $x^i$ is just a single variable function.

The issue with covariant derivatives is vector fields, and that $X(p)-X(q)$ does not make sense for $p,q$ different points in $M$ ($X$ is a vector field). As derivatives require differences, the usual difference quotient fails to be defined. The reason is that $X(p)$ lives in $T_pM$ and $X(q)$ lives in $T_qM$. These vector spaces are not canonically isomorphic, so there is no way to compare them. A connection is basically just a rule for comparing tangent spaces.


In differential geometry the basis vectors are derivatives. For coordinate basis one has

\begin{equation}e_i=\partial_i. \end{equation}

The tangent vector to a curve is just a particular case of this concept: \begin{equation}t=\frac{\mathrm{d}}{\mathrm{d}\lambda}=\frac{\mathrm{d}x^i}{\mathrm{d}\lambda}\partial_i, \end{equation}

where $\lambda$ is a parametrization along the curve, so that

\begin{equation}t^i=\frac{\mathrm{d}x^i}{\mathrm{d}\lambda}. \end{equation}

See, for example, the brilliant textbook by Nakahara "Geometry, Topology and Physics".

  • $\begingroup$ You wrote "basis vectors are derivatives". Derivatives of what? $\endgroup$ Aug 9, 2021 at 23:51
  • $\begingroup$ Of any function on the manifold. $\endgroup$ Aug 12, 2021 at 14:27

Given a vector V = x . i + y . j the derivative dV/dx = i (in Cartesian coordinates). So a derivative of a vector w.r.t. a component being a vector should not surprise you, I think. For tangent-vectors, just think of the velocity-vector along a curve C over some surface S.


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