What is the meaning of this operator? The problem I'm solving involves a proof using the operator $ad A$, which is defined as follows:
$ad$ $A $ $\cdot =[A,\cdot]$
What does this notation mean?
 A: In Lie theory, variously $\mathrm{ad}_A$, $\mathrm{ad}(A)$ or $\mathrm{ad}\,A\,$ all refer to the adjoint action of an endomorphism $A$; we will choose $\mathrm{ad}_A$. Note this is distinct from the adjoint representation of the Lie group, which is denoted by $\mathrm{Ad}_g$ for $g\in G$, with 'Ad' capitalized, although $\mathrm{ad}$ and $\mathrm{Ad}$ are highly analogous (indeed, the former is a linearization of the latter, and the latter is a smooth group conjugation action).
Given a vector space $V$, the set of endomorphisms $\mathrm{End}(V)$ is a vector space itself under pointwise addition and scalar multiplication. A Lie algebra is defined as a vector space $L$ with a bilinear operation $[\cdot,\cdot]$ (called the Lie bracket) satisfying skew-symmetry and the Jacobi identity. $\mathrm{End}(V)$ is a Lie algebra if we define $[X,Y]=XY-YX$, where writing endomorphisms next to each other means functional composition. This particular Lie bracket is called the commutator bracket.
Given $A\in\mathrm{End}(V)$, the adjoint $\mathrm{ad}_A:X\mapsto [A,X]$ is an endomorphism of $\mathrm{End}(V)$. Thus, the adjoint Lie algebra representation $\mathrm{ad}:\mathrm{End}(V)\to \mathrm{End}(\mathrm{End}(V)):A\to\mathrm{ad}_A$ is a linear operator. Moreover, $\mathrm{ad}_A$ is a derivation on $\mathrm{End}(V)$ for each $A$ and $\mathrm{ad}$ is a Lie algebra homomorphism; both of these facts are equivalent to the Jacobi identity axiom.
Since each $\mathrm{ad}_A$ is an endomorphism of the space $\mathrm{End}(V)$, it makese sense to consider powers in the operator $\mathrm{ad}_A$, and more generally power series in it. If the ground field is $\Bbb R$ or $\Bbb C$, then
$$\exp(t\,\mathrm{ad}_A)=\sum_{n=0}^\infty\frac{(t\,\mathrm{ad}_A)^n}{n!}:~~ X\to\mathrm{Id}_V+\sum_{n=1}^\infty\frac{t^n}{n!}[\underbrace{A,[A,\cdots[A}_n,X]\cdots]].$$
This wraps up the comment section pretty much.
