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I see this term used a lot, especially in the context of Lie algebras. I have seen it used (though not defined) in both math and physics books, and I could not find a good definition anywhere on the web. One source mentioned that refers to any element of the Lie algebra to a Lie group.

Is there a rigorous definition of an infinitesimal symmetry (and if so, what is it?) or is it used informally akin to $dx$ being an infinitesimal displacement?

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CBBAM
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    Rigorous? Is your high-school expression $f(x+\epsilon)=f(x) + \epsilon f'(x) + O(\epsilon^2)$ rigorous enough for you? – Cosmas Zachos May 08 '23 at 21:39
  • @Cosmas: That expression should be rigorous enough for anybody (together with explanations, definitions etc., with context!), right? I don't get your point. – kricheli May 08 '23 at 23:05
  • @kricheli Where do you reckon is the trouble, then? – Cosmas Zachos May 09 '23 at 00:23
  • @CosmasZachos I agree that your expression gives a small change in $f$, but how does that relate to symmetries? Putting aside rigor for now, does it mean that we can think of infinitesimal symmetries as an arbitrary small change in the symmetry parameter? For example, if our symmetry is rotation about an angle $\theta$ then we are moving an angle $d\theta$? – CBBAM May 09 '23 at 01:18
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    Of course, as most of you sources stress! You seem uncomfortable with differentials, so I bypassed them with ε…. – Cosmas Zachos May 09 '23 at 01:49
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    @CosmasZachos Thank you for your help! – CBBAM May 09 '23 at 01:56

1 Answers1

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  1. In a basic setting a flow $$\Delta x^i~:=~ x^{\prime i}- x^i~=~ \epsilon X^i +o(\epsilon)\tag{1} $$ is generated by a vector field $$X~=~X^i\partial_i.\tag{2}$$ In eq. (1) we have used the little-$o$ notation.

  2. In the physics literature, this situation is often transcribed as an infinitesimal transformation (IT) or an infinitesimal vector field (IVF), $$\delta x^i~:=~ x^{\prime i}- x^i~=~ \epsilon X^i,\tag{3} $$ where $\epsilon$ is an infinitesimal parameter. Now it turns out that the notion of infinitesimals in non-standard analysis is not easy to make rigorous, cf. e.g. this & this Phys.SE posts, so when pressed what eq. (3) means, a physicist will typically retreat to eq. (1).

  3. That the infinitesimal transformation (3) is an infinitesimal symmetry (IS) for a quantity $Q(x)$ merely means that $$\Delta Q~:=~Q(x+\Delta x)-Q(x)~=~o(\epsilon), \tag{4}$$ where $\Delta x$ is given in eq. (1). Equivalently, the Lie derivative of the vector field (2) vanishes $${\cal L}_X Q~=~0.\tag{5}$$

  4. A Lie algebra $\mathfrak{g}$ for a Lie group $G$ can be viewed as the set of left-invariant vector fields on $G$ equipped with the Lie bracket of vector fields. Therefore a Lie algebra element is a vector field $X$, which in turn is associated with an infinitesimal transformation (3).

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