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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 $\mathrm{End}(V)\to\mathrm{End}(V)$ $$\exp(t~\mathrm{ad}_A):X\mapsto \mathrm{Id}+\sum_{n\ge1}\frac{t^n}{n!}[\underbrace{A,[A,\cdots,[A}_n,X]\cdots]]$‌​$ when $t$ is a scalar. Are you sure the context of the question is just linear algebra, and not more informatively Lie theory? –  anon Nov 15 '12 at 2:25
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Since mathjax is buggy: $$\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 1

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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