# 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?

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It is the adjoint action. Namely, if $A\in\mathrm{End}(V)$, then $\mathrm{ad}_A:X\mapsto [A,X]:=A\circ X-X\circ A$ is, for all $A$, a linear map on $\mathrm{End}(V)$. When working with the endomorphism ring as something called a Lie algebra, the bracket notation by convention refers to the commutator bracket I just wrote. We can also define this Lie bracket in abstract settings where the bracket satisfies the axioms of bilinearity, skew-symmetry, and the Jacobi identity, but you don't need to know these things for just linear algebra. – anon Nov 15 '12 at 2:11
I'm not quite understanding. I'm given a matrix A, and the question involves e^(ad A). What does this mean? – abc Nov 15 '12 at 2:20
$\mathrm{ad}_A$ is an operator on the vector space $\mathrm{End}(V)$. That is, it is an endomorphism of the space of endomorphisms. Thus it makes sense to talk about powers of $\mathrm{ad}_A$ e.g. $$\mathrm{ad}_A^2:X\to [A,[A,X]]=A^2X-2AXA+XA^2.$$ Then, when the ground field has char zero, we define the maps $\mathrm{End}(V)\to\mathrm{End}(V)$ $$\exp(t~\mathrm{ad}_A):X\mapsto \mathrm{Id}+\sum_{n\ge1}\frac{t^n}{n!}[\underbrace{A,[A,\cdots,[A}_n,X]\cdots]]‌​ when t is a scalar. Are you sure the context of the question is just linear algebra, and not more informatively Lie theory? – anon Nov 15 '12 at 2:25 Since mathjax is buggy:$$\exp(t~\mathrm{ad}_A)=\sum_{n=0}^\infty\frac{(t\,\mathrm{ad}_A)^n}{n!}:~~ X\mapsto \mathrm{Id}+\sum_{n=1}^\infty\frac{t^n}{n!}[\underbrace{A,[A,\cdots,[A}_n,X] \cdots]].$$– anon Nov 15 '12 at 2:30 Thanks, that makes a lot more sense now. – abc Nov 15 '12 at 6:17 ## 1 Answer In Lie theory, variously \mathrm{ad}_A, \mathrm{ad}(A) or \mathrm{ad}\,A\, all refer to the adjoint action of an endomorphism A; we will choose \mathrm{ad}_A. Note this is distinct from the adjoint representation of the Lie group, which is denoted by \mathrm{Ad}_g for g\in G, with 'Ad' capitalized, although \mathrm{ad} and \mathrm{Ad} are highly analogous (indeed, the former is a linearization of the latter, and the latter is a smooth group conjugation action). Given a vector space V, the set of endomorphisms \mathrm{End}(V) is a vector space itself under pointwise addition and scalar multiplication. A Lie algebra is defined as a vector space L with a bilinear operation [\cdot,\cdot] (called the Lie bracket) satisfying skew-symmetry and the Jacobi identity. \mathrm{End}(V) is a Lie algebra if we define [X,Y]=XY-YX, where writing endomorphisms next to each other means functional composition. This particular Lie bracket is called the commutator bracket. Given A\in\mathrm{End}(V), the adjoint \mathrm{ad}_A:X\mapsto [A,X] is an endomorphism of \mathrm{End}(V). Thus, the adjoint Lie algebra representation \mathrm{ad}:\mathrm{End}(V)\to \mathrm{End}(\mathrm{End}(V)):A\to\mathrm{ad}_A is a linear operator. Moreover, \mathrm{ad}_A is a derivation on \mathrm{End}(V) for each A and \mathrm{ad} is a Lie algebra homomorphism; both of these facts are equivalent to the Jacobi identity axiom. Since each \mathrm{ad}_A is an endomorphism of the space \mathrm{End}(V), it makese sense to consider powers in the operator \mathrm{ad}_A, and more generally power series in it. If the ground field is \Bbb R or \Bbb C, then$$\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.

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