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 $df(v)$.

Essentially, this property links the gradient to the directional derivative.

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 $\mathbf{v}$.

Question: Why are we not normalizing in the Riemannian space similarly?

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    $\begingroup$ 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. $\endgroup$ – 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, \begin{equation} v \cdot \mathrm{grad}(f). \end{equation} 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.


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