# Exponential of the directional derivative on a surface

The exponential of the derivative operator

$$\exp\Big(a\frac{d}{dx}\Big) = \sum_k \frac{a^k\big(\frac{d}{dx}\big)^k}{k!}$$

$$\exp\Big(a\frac{d}{dx}\Big) f(x) = \sum_k f^{(k)}(x)\frac{a^k}{k!}$$

gives the Taylor series of a function, which is $f(x+a)$ if the function is analytic. More generally,

$$\exp\Big(\vec a\cdot\vec\nabla\Big) f(\vec x) = f(\vec x + \vec a).$$

I'm wondering if this generalizes to surfaces. The gradient on a surface is

$$\nabla = x^u\frac{\partial}{\partial u} + x^v\frac{\partial}{\partial v},$$

where $(x^u,x^v)$ is the reciprocal basis to $(x_u,x_v)=(\frac{\partial x}{\partial u},\frac{\partial x}{\partial v})$. If $a$ is a tangent vector, then the directional derivative ($a\cdot\nabla$) should be the same whether $\nabla$ is the gradient for the surface or the external (Euclidean) space.

If $f$ is defined only on the surface, then $f(x+a)$ is undefined in general. So what is the meaning of $\exp(a\cdot\nabla)f(x)$?

Is this related to the "exponential map" from tangent vectors to geodesics?

EDIT1

I worked a few examples, and it seems that the most important question is how $a$ varies. (If it's constant, then it does not remain tangent.) This doesn't affect the first derivative, but it affects all higher derivatives of $f$. I guess that if $a$'s derivative is orthogonal to the surface, then the exponential gives the value of $f$ along the geodesic in $a$'s direction. Is this correct?

Take $a$ to be a tangent vector for some curve (on a surface or in empty space):

$$a = \lVert a\rVert x_s$$

where $s$ is the arclength parameter. The directional derivative only depends on the part of the gradient parallel to the curve; the other components are projected away.

$$a\cdot\nabla = \big(\lVert a\rVert x_s\big)\cdot\Big(x^s\frac{\partial}{\partial s}\Big) = \lVert a\rVert\frac{\partial}{\partial s}$$

$$(a\cdot\nabla)^2 = \lVert a\rVert\Big(\frac{\partial\lVert a\rVert}{\partial s}\frac{\partial}{\partial s}+\lVert a\rVert\frac{\partial^2}{\partial s^2}\Big)$$

If $a$ has constant magnitude, then the second derivative simplifies to

$$(a\cdot\nabla)^2 = \lVert a\rVert^2\frac{\partial^2}{\partial s^2},$$

and all higher derivatives simplify, so the exponential is

$$\exp(a\cdot\nabla)f(s) = f(s+\lVert a\rVert).$$

This gives the value of $f$ at a point along the curve, from the derivatives of $f$ at the base point. The curve itself should be determined by $a$, which is essentially a vector field.

Constant magnitude implies that the derivative $(a\cdot\nabla)a$ is orthogonal to $a$. If it's also orthogonal (everywhere, not just at the base point) to a surface, then the curve is indeed a geodesic on the surface.

If the function $f$ is the position vector itself, then this derivative exponential is very similar to the "exponential map", though that takes only a single vector (not a curve or field), and gives a point on a geodesic. This is more general, not restricted to geodesics or surfaces.