Ciro Santilli @cirosantilli 37

 Incoming links: Generator of a Lie algebra

Infinitesimal generator Updated 2025-07-16

Elements of a Lie algebra can (should!) be seen a continuous analogue to the generating set of a group in finite groups.

For continuous groups however, we can't have a finite generating set in the strict sense, as a finite set won't ever cover every possible point.

But the generator of a Lie algebra can be finite.

And just like in finite groups, where you can specify the full group by specifying only the relationships between generating elements, in the Lie algebra you can almost specify the full group by specifying the relationships between the elements of a generator of the Lie algebra.

This "specification of a relation" is done by defining the Lie bracket.

The reason why the algebra works out well for continuous stuff is that by definition an algebra over a field is a vector space with some extra structure, and we know very well how to make infinitesimal elements in a vector space: just multiply its vectors by a constant

c

that cana be arbitrarily small.

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Local symmetries of the Lagrangian imply conserved currents Updated 2025-07-16

TODO. I think this is the key point. Notably,

U (1)

symmetry implies charge conservation.

More precisely, each generator of the corresponding Lie algebra leads to one separate conserved current, such that a single symmetry can lead to multiple conserved currents.

This is basically the local symmetry version of Noether's theorem.

Then to maintain charge conservation, we have to maintain local symmetry, which in turn means we have to add a gauge field as shown at Video "Deriving the qED Lagrangian by Dietterich Labs (2018)".

Forces can then be seen as kind of a side effect of this.

Bibliography:

photonics101.com/relativistic-electrodynamics/gauge-invariance-action-charge-conservation#show-solution has a good explanation of the Gauge transformation. TODO how does that relate to $U (1)$ symmetry?
physics.stackexchange.com/questions/57901/noether-theorem-gauge-symmetry-and-conservation-of-charge

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Quantum field theory lecture by Tobias Osborne (2017) / Lecture 3 Updated 2025-07-16

www.youtube.com/watch?v=cj-QpsZsDDY&list=PLDfPUNusx1EpRs-wku83aqYSKfR5fFmfS&index=3

symmetry in classical field theory
from Lagrangian density we can algorithmically get equations of motion, but the Lagrangian density is a more compact way of representing the equations of motion
definition of symmetry in context: keeps Lagrangian unchanged up to a total derivative
Noether's theorem
youtu.be/cj-QpsZsDDY?list=PLDfPUNusx1EpRs-wku83aqYSKfR5fFmfS&t=3062 Lagrangian and conservation example under translations
youtu.be/cj-QpsZsDDY?list=PLDfPUNusx1EpRs-wku83aqYSKfR5fFmfS&t=3394 same but for Poincaré transformations But now things are harder, because it is harder to describe general infinitesimal Poincare transforms than it was to describe the translations. Using constraints/definition of Lorentz transforms, also constricts the allowed infinitesimal symmetries to 6 independent parameters
youtu.be/cj-QpsZsDDY?list=PLDfPUNusx1EpRs-wku83aqYSKfR5fFmfS&t=4525 brings out Poisson brackets, and concludes that each conserved current maps to a generator of the Lie algebra
This allows you to build the symmetry back from the conserved charges, just as you can determine conserved charges starting from the symmetry.

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The derivative is the generator of the translation group Updated 2025-07-16

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:

the translation group operates on the argument of a function $f (x)$
the generator is an operator that operates on $f$ itself

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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