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I was reading the Wikipedia article about Noether's theorem and this thing popped out:

Then the resultant perturbation can be written as a linear sum of the individual types of perturbations

$$\delta t = \sum_r \epsilon_r T_r \!$$ $$\delta \mathbf{q} = \sum_r \epsilon_r \mathbf{Q}_r ~$$

where $\epsilon_r$ are infinitesimal parameter coefficients corresponding to each:

  • generator $T_r$ of time evolution, and
  • generator $Q_r$ of the generalized coordinates.

But, when I clic on "generator" it leads me to the article about Lie Groups which is by itself a topic in which I could spend weeks, to say the least.

I had once in my past discrete groups in a lecture and I am more or less conversant with elementary lagrangian mechanics, matrix calculus and elementary non-relativistic quantum mechanics (however not with that SU(?) stuff about rotation invariances yet). Is it possible that someone here provides a very short explanation of what those generators are, and the simplest mathematical example you can think of?

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up vote 0 down vote accepted

Let us for simplicity address OP's question in the context of point mechanics where $q^i$ are generalized position coordinates on some manifold $M$ [instead of considering field theory with fields $\phi^{\alpha}(x)$]. Let us also for simplicity consider the special case where there are no horizontal/time transformations

$$\delta t ~=~ \sum_a \epsilon_{(a)} X_{(a)} ~=~ 0,$$

but only vertical transformations

$$\delta q^i ~=~ \sum_a \epsilon_{(a)} Y^i_{(a)}, $$

where $\epsilon_{(a)}$ are (infinitesimal) parameters.$^1$

In a nutshell, the generators $Y_{(a)}$ are vector fields

$$ Y_{(a)}~=~ \sum_i Y_{(a)}^i\frac{\partial}{\partial q^i}. $$

The Lie bracket of two vector fields $[Y_{(a)},Y_{(b)}]$ is again a vector field. In certain cases, the generators/vector fields form a Lie algebra.

Example: The generator for a translation in the $a$'th direction of position space $\mathbb{R}^3$ is

$$Y_{(a)}~=~\frac{\partial}{\partial q^a}, $$

with constant vector field components

$$ Y^i_{(a)}~=~\delta^i_a. $$

In this case, the generators $Y_{(a)}$ commute, so they form an Abelian Lie algebra.


$^1$ See also this Phys.SE answer.

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Thanks very much, it is a nice answer. – Mephisto Feb 20 '13 at 12:07

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