The derivative is the generator of the translation group

Take the group of all Translation in

R^{1}

Let's see how the generator of this group is the derivative operator:

\frac{\partial}{\partial x}

The way to think about this is:

So let's take the exponential map:

e^{x_{0} \frac{\partial}{\partial x}} f (x) = (1 + x_{0} \frac{\partial}{\partial x} + x_{0}^{2} \frac{\partial ^{2}}{\partial x ^{2}} + \dots) f (x)

and we notice that this is exactly the Taylor series of

f (x)

around the identity element of the translation group, which is 0! Therefore, if

f (x)

behaves nicely enough, within some radius of convergence around the origin we have for finite

x_{0}

e^{x_{0} \frac{\partial}{\partial x}} f (x) = f (x + x_{0})

This example shows clearly how the exponential map applied to a (differential) operator can generate finite (non-infinitesimal) Translation!

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