# Gradient and Directional Derivative in Riemannian Manifold

Probably this question is too dumb to be asked, but I am an engineer trying to learn differential geometry, please go easy on me.

I am trying to understand that, in Riemannian space, gradient satisfies the property $$g(\nabla f, v) = df(v)$$. Here, $$f$$ is a scalar-valued function, $$v$$ is a vector in the tangent space at point $$P$$ on the manifold $$\mathcal{M}$$ and the directional derivative along $$v$$ is $$df(v)$$.

In Orthonormal Euclidean Space $$\{x,y,z\}$$, the gradient is $$\nabla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y},\frac{\partial f}{\partial z} \right)$$. So if one wants to know how the function $$f$$ changes along a particular direction $$\mathbf{v} = v_xx + v_yy + v_zz$$, we take the inner product of $$\langle \mathbf{v} , \nabla f \rangle$$ divide by the norm of $$\mathbf{v}$$.
• Only in most beginning calculus courses are students required to normalize the vector $v$ to compute a directional derivative. The tern "directional derivative" suggests it should depend only on the direction of $v$. What we should ask for is the rate of change of $f$ as one moves with velocity vector $v$. This varies linearly with $v$ and is the proper notion. – Ted Shifrin May 8 at 17:48
The directional derivative along a direction $$v$$ in Euclidean space, as usually defined in calculus is, provided everything is sufficiently smooth, $$$$v \cdot \mathrm{grad}(f).$$$$ It is not required for $$v$$ to be a unit vector, I disagree with your last sentence. Of course you may prefer to consider unit vectors, and you can do that in Euclidean space as well as on a Riemannian manifold. No big deal, at the end you are just rescaling by a non-zero constant.