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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?

Qmechanic
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CBBAM
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2 Answers2

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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).

Qmechanic
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I'm not a guy from physics. Let me explain in a simpler case.

Consider $V$ a vector space which endows with a (continuous) $G$-symmetry, that is there is a $G$-action on $V$: $$G\times V\longrightarrow V$$

We know the identity element $1\in G$ acts as the identity transformation on $V$. Now, what about the actions of elements which are near to $1$?

Those elements are near to $1$ can be described by the tangent space of $G$ at $1$, i.e., the Lie algebra $\mathfrak{g}$. Such an element can be write as $\exp(\epsilon X)$, where $X\in\mathfrak{g}$ and $\epsilon$ a small real number.

To measure the difference between the identity action and the actions of the elements which are near to 1, we can use $$\lim_{\epsilon\to 0}\frac{\exp (\epsilon X)\cdot v-v}{\epsilon}$$ That quantity is supposed to be a vector in $V$, denoted by $\underline{X}(v)$.It means if we perturb the identity element along direction $X$ by a small $\epsilon$, then the perturbed action takes $v$ to $v+\epsilon \underline{X}(v)$.

The above action: $$\mathfrak{g}\times V\longrightarrow V, (X,v)\mapsto \underline{X}(v)$$ is referred as the infinitesimal action, or the infinitesimal symmetry.

If we replace $V$ by a bended manifold $M$, the corresponding infinitesimal action becomes to $$\mathfrak{g}\times M\longrightarrow TM$$ as $\underline{X}(x)=\left.\frac{d}{d\epsilon}\right|_{\epsilon=0}\exp(\epsilon X)\cdot x$ is now a vector field, called the fundamental vector field.

Alexander
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