# The $\text{ad}$ Homomorphism Consider an arbitrary [[Lie algebra]] $\mathfrak{g}$, then the [[adjoint action]] is: $ \text{ad}_X(Y) \equiv [X, Y], \quad X, Y \in \mathfrak{g} $ Note that we have turned $X$ into a [[linear operator]] on the [[vector space]] it lives in, i.e., $\text{ad}_X(Y) \in \mathcal{L}(\mathfrak{g})$ technically speaking. However, to be more precise, $\text{ad}_X(Y) \in \mathfrak{gl}(\mathfrak{g})$ since we would like to equip $\mathcal{L}(\mathfrak{g})$ with a Lie bracket and make it into a [[Lie algebra]]. See [[General Linear Group#Lie Algebra|here]] for more details on $\mathfrak{gl}(V)$ Note that this construction introduces a [[linear map]] between Lie algebras. $ \begin{align} \text{ad}: \mathfrak{g} & \rightarrow \mathfrak{gl}(\mathfrak{g})\\ X &\mapsto \text{ad}_X \end{align} $ For this map to be a [[Lie Algebra Homomorphism]], we must have $ \text{ad}_{[X,Y]} = [\text{ad}_X, \text{ad}_Y] $ Note that the [[commutator]] on the left is taken on $\mathfrak{g}$ and the one on the right is on $\mathfrak{gl}(\mathfrak{g})$. This identity is easy to verify: $ \text{ad}_{[X,Y]}(Z) = [[X,Y], Z] $ $ \begin{align} [\text{ad}_X, \text{ad}_Y]Z &= \text{ad}_X( \text{ad}_Y(Z)) - \text{ad}_Y( \text{ad}_X(Z)) \\ &= [X, [Y,Z]] - [Y, [X, Z]]\\ &= [[X, Y], Z], \end{align} $ where the last equality follows from the [[Jacobi Identity]]. Thus, $\text{ad}_{\cdot}$ is a [[Lie Algebra Homomorphism]]. What this tells us is that the whole purpose of the [[Jacobi Identity]] is to guarantee that the adjoint action $\text{ad}_\cdot$ is homomorphism for any abstract Lie algebra $\mathfrak{g}$. # The $\text{Ad}$ Homomorphism Consider $A \in G$ and $X \in \mathfrak{g}$, where $G$ is a [[Matrix Lie Group]] and $\mathfrak{g}$ is its [[Lie algebra]]. We know that $A X A^{-1} \in \mathfrak{g}$. Thus, let's define a linear operator on $\mathfrak{g}$: $ \text{Ad}_A(X) = A X A^{-1}, \quad X \in \mathfrak{g} $ This is essentially a [[linear operator]] which takes a matrix $X$ and applies the [[similarity transformation]] by the matrix $A$. This leads us to define a [[group homomorphism]]: $ \begin{align} \text{Ad}: G & \rightarrow GL(\mathfrak{g})\\ A &\mapsto \text{Ad}_A \end{align} $ Here, $GL(\mathfrak{g})$ is the [[general linear group]] of the [[vector space]] $\mathfrak{g}$. Note that $\text{Ad}_A \text{Ad}_B = \text{Ad}_{AB}$ so this is indeed a homomomorphism. Since we have a [[group homomorphism]] between 2 [[matrix Lie group|matrix Lie groups]] $G$ and $GL(G)$, then we have an [[Induced Lie Algebra Homomorphism]] $ \phi: \mathfrak{g} \rightarrow \mathfrak{gl}(\mathfrak{g}) $ For a given $X \in \mathfrak{g}$, we have $\phi(X) \in \mathfrak{gl}(\mathfrak{g})$ given by: $ \phi(X) = \left.\frac{d}{dt} \text{Ad}_{e^{tX}}\right|_{t=0} $ To figure out what $\phi$ looks like, we have to evaluate it on some $Y \in \mathfrak{g}$ (recall that $\mathfrak{gl(g)}$ is the [[vector space]] of all [[linear operator|linear operators]] equipped with a Lie bracket, so to get an idea of what a linear operator does, we have to evaluate it on an element of its [[Domain, Co-domain, Image|domain]]). $ \begin{align} \phi(X)(Y) &= \left.\frac{d}{dt} \text{Ad}_{e^{tX}}Y\right|_{t=0}\\ &= \left.\frac{d}{dt} e^{tX}Ye^{-tX}\right|_{t=0}\\ &= XY - YX \\ &= [X, Y]\\ &= \text{ad}_{X}(Y) \end{align} $ thus: $ \phi(X) = \text{ad}_X $ which is just the $\text{ad}$ homomorphism discussed [[#The text ad Homomorphism|above]]. Thinking of [[Induced Lie Algebra Homomorphism|induced Lie algebra homomorphisms]] as an infinitesimal version of [[matrix Lie group]] [[Group Homomorphism|homomorphisms]], we conclude that the [[commutator]] is the infinitesimal version of the [[similarity transformation]]. # Identities We have two identities here. The first follows from the definition of induced Lie algebra homomorphisms: $ \text{Ad}_{e^{tX}} = e^{t\, \text{ad}_X} $ The second follows from the first and a bit of math: $ e^{tX} Y e^{-tX} = Y + t[X,Y] + \frac{t^2}{2}[X,[X,Y]] + \frac{t^3}{3!}[X,[X,[X,Y]]]+\ldots $ Moreover, we can show that, given $A \in G$ and $X, Y \in \mathfrak{g}$ $ \begin{align} \text{ad}_{\text{Ad}_{A}(X)}(Y) &= [A X A ^{-1}, Y] \\ &= AXA^{-1}Y - Y AXA^{-1}\\ &= (A A^{-1})AXA^{-1}Y - Y AXA^{-1}(A A^{-1})\\ &= A(XA^{-1}YA)A^{-1} - A(A^{-1} Y A X)A^{-1}\\ &= A[X, A^{-1}YA]A^{-1}\\ &= A[X, \text{Ad}_{A^{-1}}(Y)]A^{-1}\\ &= A[\text{ad}_X(\text{Ad}_{A^{-1}}(Y)]A^{-1}\\ &= \text{Ad}_{A}(\text{ad}_X(\text{Ad}_{A^{-1}}(Y)) \end{align} $ So we can conclude: $ \text{ad}_{\text{Ad}_A(X)} = \text{Ad}_{A} \circ \text{ad}_X \circ \text{Ad}_{A^{-1}} $