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.
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.