0
$\begingroup$

In wikipedia page about covariant derivative, it is said:

The covariant derivative is a generalization of the directional derivative from vector calculus. As with the directional derivative, the covariant derivative is a rule, $ \nabla _{\mathbf {u} }{\mathbf {v} } $, which takes as its inputs: [...]. The output is the vector $\nabla_{\mathbf u}{\mathbf v}(P)$, also at the point 'P'.

this phrase surprises me, because it is said covariant derivative extends the directional derivative, but directional derivative result can be and scalar (when applied to a function $R^n \rightarrow R$), a vector (when applied to a function $R^n \rightarrow R^m$), ...

However, notation $\nabla_{\mathbf u}{\mathbf v}(P)$ remembers the one of gradient, that is always a vector, but only applicable to a function $R^n \rightarrow R$.

Could someone please clarify ? Thanks.

$\endgroup$
0
$\begingroup$

When you take the directionnal derivative of a function, you still get a function $$\nabla_Uf:x\mapsto (\nabla_Uf)(x)= df_x(U)\in\mathbb{R}.$$ What allows the covariant derivative is to "differentiate an object along a direction", and still get an object of the same nature. Here if $V$ is a vector field on your manifold $M$, $\nabla_UV$ will still be a vector field $$\nabla_UV:x\mapsto(\nabla_UV)|_x\in T_xM.$$ About the notation $\nabla$: if you are endowed with a metric $(\cdot,\cdot)$ you can associate to your function $f$ a vector field $\mathrm{grad}\,f$ checking the relation $(\mathrm{grad}\,f,X)=df(X)=\nabla_X f$ for all vector field $X$, and if so you usually identify $\nabla f$ and $\mathrm{grad}\,f$ even if these are two different mathematical things, namely a $1$-form and a vector field.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.