In tying together [Lie algebras](Lie%20algebras.md) and [Lie groups](Lie%20groups.md) we find that $\forall A\in \mbox{G}$, $AXA^{-1}\in\mathfrak{g}$ where $X\in\mathfrak{g}$. This follows directly from [property 1](Lie%20group%20homomorphisms.md#^80fb3a) of the list of properties which tie Lie Group and Lie Algebra homomorphisms. Using this fact, we define the [linear map](Linear%20map.md), $\mbox{Ad}_A:\mathfrak{g}\rightarrow\mathfrak{g}$ as: $\mbox{Ad}_A(X) = AXA^{-1}$ ^f1f901 This is called the _adjoint map_ or _adjoint [representation](Lie%20algebra%20representation.md)_. The linear map, $A\rightarrow\mbox{Ad}_A$ is a [Lie group homomorphism](Lie%20group%20homomorphisms.md), $\mbox{Ad}:\mbox{G}\rightarrow\mbox{GL}(\mathfrak{g})$, where $\mbox{GL}(\mathfrak{g})$ is the [matrix Lie group](Matrix%20Lie%20group.md) of [invertible linear transformations](Operator%20inverse.md) of $\mathfrak{g}$. ^df36e4 # Adjoint map of [[Lie algebras]] [Lie group homomorphisms are continuous](Lie%20group%20homomorphisms.md) therefore there's also a linear map $X\rightarrow \mbox{ad}_X$, which is from $\mathfrak{g}$ to $\mathfrak{gl}(\mathfrak{g})$ (i.e the map from $\mathfrak{g}$ to the space of all linear maps from $\mathfrak{g}$ to itself) that is written as $e^{\mbox{ad}_X}=\mbox{Ad}_{e^X}.$ # Types of Adjoint Maps ## Commutators The [commutator](Commutators.md) is an adjoint map from a Lie Algebra to itself, since $\forall X,Y \in \mathfrak{g}$, $\mbox{ad}_X(Y) = [X,Y]$ ([proof](Adjoint%20map.md#Proof%20that%20the%20commutator%20is%20an%20adjoint%20map%20of%20a%20Lie%20Algebra)). --- # Proofs and Examples ## Proof that the commutator is an adjoint map of a Lie Algebra     #MathematicalFoundations/Algebra/AbstractAlgebra/GroupTheory/Lie/LieGroups/Algebras/LieAlgebras