# result of covariant derivative

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 $$R^n \rightarrow R$$.
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.