25 Apr 2014 No Comments

# Chapter 7: The Lie Bracket and the Little Adjoint Representation

We have seen that the Lie algebra of the Lie group $\G$ is a vector space $\g$ over the reals $\R$. However, it is not simply any old vector space. It has further structure wrought by a bilinear binary operation called the Lie bracket. In this post I shall define the Lie bracket and derive all of its most important properties. I shall emphasise the Lie bracket as a Lie algebraic version of a commutation detector.

When we studied the exponential map we saw how each Lie algebra member uniquely defined a very special path through the connected Lie group, the one parameter group $\mathfrak{E}_X=\{\exp(\tau\,X):\,\tau\in\R\}$. We saw, through Theorem 5.11, also how this path, under the Adjoint representation, became a one parameter matrix group:

\begin{equation}

\label{AdjointExponentiation}

\Ad(\mathfrak{E}_X) = \{\Ad\left(e^{\tau\,X}\right)=\exp(\tau\,\mathscr{X}):\,\tau\in\R\}

\end{equation}

because **( i) **$\Ad$ is a homomorphism which replicates the flow equation fulfilled by $e^{\tau\,X}$,

**(**$\Ad\left(e^{\tau\,X}\right)$ is a continuous matrix function of $\tau$ (with respect to the wonted topology on the linear space of $N\times N$ matrices) and

*ii*)**(**the only continuous matrix function of $\tau$ fulfilling the flow equation is a matrix exponential function of $\tau$. Therefore, $\Ad\left(e^{\tau\,X}\right)=\exp(\tau\,\mathscr{X})$, for some constant $N\times N$ matrix $\mathscr{X}$ which depends only on the original tangent $X$.

*iii*)We can bring the meaning of $\Ad$ into sharper focus by studying its kernel some more. Indeed the kernel is the group’s centre:

**Theorem 7.1**** (Central Kernel of Adjoint Representation):**

For a connected Lie group $\G$ the the kernel of the Adjoint reprensentation $\Ad:\G\to GL(\g)$ is precisely the group’s centre, *i.e.* $\ker(\Ad) = \mathscr{Z}(\G)$.

**Proof: **Show Proof

Clearly if $\gamma\in\mathscr{Z}(\G)$ then $\gamma\,\sigma(\tau)\,\gamma^{-1} = \gamma\,\gamma^{-1}\,\sigma(\tau)=\sigma(\tau)$ for any $C^1$ path $\sigma:[-\tau_{max},\,\tau_{max}]\to\Nid$ passing through the identity at $\tau=0$ so that the tangent to the path is unchanged under conjugation by $\gamma$. Therefore $\gamma\in\ker(\Ad)$ and we have $\mathscr{Z}(\G)\subseteq\ker(\Ad)$.

For the converse, choose a basis $\{\hat{X}_j\}_{j=1}^N$ for the Lie algebra $\g$, thus laying down exponential canonical co-ordinates of the second kind that uniquely label a small enough nucleus $\K\subseteq\Nid$, so that every $\zeta\in\G$ can be written as a (not needfully unique) product of the form:

\begin{equation}

\label{CentralKernelOfAdjointRepresentationTheorem_1}

\zeta = \prod\limits_{k=1}^M\,e^{\tau_k\,\hat{X}_j(k)}

\end{equation}

so that:

\begin{equation}

\label{CentralKernelOfAdjointRepresentationTheorem_2}

\begin{array}{lcl}

\gamma\,\zeta\,\gamma^{-1} &=& \gamma\,\prod\limits_{k=1}^M\,e^{\tau_k\,\hat{X}_j(k)}\,\gamma^{-1}\\&=& \gamma\,e^{\tau_1\,\hat{X}_j(1)}\,\gamma^{-1}\,\gamma\,\,e^{\tau_2\,\hat{X}_j(2)}\,\gamma^{-1}\,\cdots\\

&=&\prod\limits_{k=1}^M\,\left(\gamma\,e^{\tau_k\,\hat{X}_j(k)}\,\gamma^{-1}\right)\\

&=&\prod\limits_{k=1}^M\,\exp\left(\tau_k\,\Ad(\gamma)\,\hat{X}_j(k)\right)

\end{array}

\end{equation}

the last line following from Lemma 5.12. Therefore, if $\gamma\in\ker(\Ad)$ then, by definition $\Ad(\gamma)\,\hat{X}_j=\hat{X}_j,\,\forall\,j=1,\,2,\,\cdots,\,N$ and $\eqref{CentralKernelOfAdjointRepresentationTheorem_2}$ becomes:

\begin{equation}\label{CentralKernelOfAdjointRepresentationTheorem_3}\gamma\,\zeta\,\gamma^{-1} =\prod\limits_{k=1}^M\,\exp\left(\tau_k\,\Ad(\gamma)\,\hat{X}_j(k)\right)=\prod\limits_{k=1}^M\,\exp\left(\tau_k\,\hat{X}_j(k)\right) = \zeta\end{equation}

and so $\gamma^{-1}\,\zeta\,\gamma = \zeta\,\Leftrightarrow\,\zeta\,\gamma=\gamma\,\zeta,\,\forall\,\zeta\in\G$ and so $\gamma \in\G$ and we have $\ker(\Ad)\subseteq\mathscr{Z}(\G)$. This proves the theorem, but it is well to witness that we can use geodesic co-ordinates with the appropriate nucleus $\K\subseteq\Nid$ just as well, in which case:

\begin{equation}

\label{CentralKernelOfAdjointRepresentationTheorem_4}

\begin{array}{lcl}

\gamma\,\zeta\,\gamma^{-1} &=& \gamma\,\prod\limits_{k=1}^M\,e^{X_k}\,\gamma^{-1}\\

&=& \gamma\,e^{X_1}\,\gamma^{-1}\,\gamma\,\,e^{X_2}\,\gamma^{-1}\,\cdots\\

&=&\prod\limits_{k=1}^M\,\left(\gamma\,e^{X_k}\,\gamma^{-1}\right)\\

&=&\prod\limits_{k=1}^M\,\exp\left(\Ad(\gamma)\,X_k\right)

\end{array}

\end{equation}

where $X_k\in\g$. Then we use the linearity of $\Ad(\gamma)$ to show that:

\begin{equation}

\label{CentralKernelOfAdjointRepresentationTheorem_5}

\Ad(\gamma)\,X_k = \Ad(\gamma)\,\left(\sum\limits_{j=0}^N x_{k,j}\,\hat{X}_j\right) = \sum\limits_{j=0}^N\left( x_{k,j}\,\Ad(\gamma)\,\hat{X}_j\right)=\sum\limits_{j=0}^N\left( x_{k,j}\,\hat{X}_j\right)=X_k

\end{equation}

if $\gamma\in\ker(\Ad)$ for each $X_k$, again showing that $\zeta\,\gamma=\gamma\,\zeta,\,\forall\,\zeta\in\G$ and therefore $\gamma\in\mathscr{Z}(\G)$.$\qquad\square$

Therefore we see that the image $\Ad(\gamma)$ under the Adjoint representation of any $\gamma\in\G$ is nontrivial if and only if $\gamma$ does not commute with at least one Lie group element. $\Ad$ is all about “Abelianhood detection”; the Adjoint representation is trivial if and only if the group is Abelian. Likewise, since $\gamma\,e^{\tau\,X}\,\gamma^{-1} = \exp(\tau\,\Ad(\gamma)\,X)$, $\gamma$ commutes with an element of the form $e^{\tau\,X}$ if and only if $\Ad(\gamma)\,X = X$ when $\left\|\tau\,X\right\|$ is small enough to make geodesic co-ordinates unique.

So the $N\times N$ matrix $\Ad(\gamma)$ encodes information about which elements of $\G$ commute with $\gamma$. Since each member $\gamma$ of a connected Lie group $\G$ can be written $\gamma = \prod\limits_{k=1}^M e^{X_k}$ where $X_k\in\g$ (*i.e.* we have written $\gamma$ as a product of a nucleus $\K\subseteq\Nid$ small enough that geodesic co-ordinates are unique) and since $\Ad$ is a homomorphism, any image of a connected Lie group member under the Adjoint representation can be written as the product of square nonsingular matrices $\Ad(\gamma)=\left.\prod\limits_{k=0}^M e^{\tau\,\mathscr{X}_k}\right|_{\tau=1}$, where $\mathscr{X}_k = \left.\d_\tau \Ad(e^{\tau\,X_k})\right|_{\tau=0}=\log\,\Ad(e^{X_k})$, $\log$ is the matrix logarithm and the nucleus $\K\subseteq\Nid$ is assumed small enough that the Taylor series for the matrix logarithm:

\begin{equation}

\label{LogartihmTaylorSeries}

\log A = A-\id_N – \frac{1}{2}(A-\id_N)^2 + \frac{1}{3}(A-\id_N)^3-\frac{1}{4}(A-\id_N)^4+\,\cdots

\end{equation}

is uniformly convergent, which will be true as long as $\left\|\Ad(\gamma)-\id_N\right\|<1$. This all shows that the $N\times N$ matrix function of a Lie algebra member $X$ defined by:

\begin{equation}

\label{LittleAdjointDefinitionEquation}

\ad(X)\stackrel{def}{=} \left.\d_\tau \Ad(e^{\tau\,X})\right|_{\tau=0}

\end{equation}

also encodes information about which elements of $\G$ an element of the kind $e^{\tau\,X},\,\tau\in\R$ commutes with. In particular, if $e^{s\,X},\,e^{\tau\,Y}\in\K$ commute for any $\tau$ in an arbitrarily small interval $\mathcal{I}$ (or even for every $\tau$ in any (maybe countably) infinite set $\mathcal{I}$ with a limit point) then:

\begin{equation}

\label{LittleAdjointMotivationEquation}

\begin{array}{cl}

&e^{s\,X}\,e^{\tau\,Y}=e^{\tau\,Y}\,e^{s\,X},\;\forall\,\tau\in\mathcal{I}\\

\Leftrightarrow&\exp\left(s\,\Ad\left(e^{\tau\,Y}\right)\,X\right) = e^{s\,X}\\

\Leftrightarrow&\Ad\left(e^{\tau\,Y}\right)\,X = X\\

\Leftrightarrow&\tau\,\ad(Y)\,X+\frac{\tau^2}{2!}\,\ad(Y)^2\,X + \cdots = 0\\

\Leftrightarrow&\ad(Y)\,X = 0

\end{array}

\end{equation}

This suggests the following binary operation as an “Abelianhood detector” within the Lie algebra of a connected Lie group.

The Lie Bracket or Lie product $\g\times\g\to\g$ is the Lie algebra binary operation defined by:

\begin{equation}

\label{LieBracketDefinition_1}

[X,\,Y] \stackrel{def}{=} \left.\d_\tau \left(\Ad\left(e^{\tau\,X}\right)\, Y\right)\right|_{\tau=0} = \ad(X)\,Y

\end{equation}

where the “little a adjoint representation of the Lie algebra $\g$” is defined by:

\begin{equation}

\label{LieBracketDefinition_2}

\ad:\g\to \mathcal{M}_N;\,\ad(X) = \left.\d_\tau \Ad\left(e^{\tau\,X}\right)\right|_{\tau=0}

\end{equation}

so that we have the *“braiding relationship”*:

\begin{equation}

\label{LieBracketDefinition_3}

\Ad\left(e^{\tau\,X}\right) = e^{\tau\,\ad(X)}

\end{equation}

Here $\mathcal{M}_N$ is the linear space of $N\times N$ square matrices. We shall soon prove that indeed the set of matrices $\Ad(\G)=\{\Ad(\gamma):\,\gamma\in\G\}$ together with the everyday matrix product is itself a connected matrix Lie group fulfilling our five basic axioms, and $\ad$ is a Lie algebra homomorphism between the Lie algebra $\g = \mathrm{Lie}(\G)$ of the Lie group $\G$ and the Lie algebra $\mathrm{Lie}(\Ad(\G))$ of the connected Lie group $\Ad(\G)$ of big-A adjoint matrices. Take heed that there is no problem defining $\ad(X)=\left.\d_\tau \Ad\left(e^{\tau\,X}\right)\right|_{\tau=0}$ because, as we have seen, $\Ad\left(e^{\tau\,X}\right)$ is a matrix exponential function defined by the universally convergent (for finite $N$) series:

\begin{equation}

\label{ExponentialTaylorSeries}

\exp(\tau\,\mathscr{X}) = \id_N + \tau\,\mathscr{X} + \frac{\tau^2}{2!}\,\mathscr{X}^2+ \frac{\tau^3}{3!}\,\mathscr{X}^3+\cdots

\end{equation}

and thus *has derivatives with respect to $\tau$ of all orders (is $C^\infty$), indeed is analytic (is $C^\omega$) in $\tau$*.

The Lie product $\ad_X Y$ for $X,\,Y\in\g$ also belongs to the Lie algebra $\g$. This is because $\Ad\left(e^{s\,X}\right)\,Y = e^{s\,\ad(X)}\,Y$ clearly belongs to the Lie algebra (it is the tangent to the path $\sigma(s) = e^{s\,X}\, e^{\tau\,Y}\,e^{-s\,X}$ at $\tau=0$) for all real $s$. Sums, differences and scalar multiples of Lie algebra members are also members (since $\g$ is a vector space) and so therefore:

\begin{equation}

\label{LieBracketLimitPrototype}

\left(e^{s\,\ad(X)}\, Y – Y\right)\,s^{-1}\in\g,\,\forall\,s\in\R\setminus\{0\}

\end{equation}

as well. But vector spaces over $\R$ are closed and complete: therefore, since $\lim\limits_{s\to0}\left(e^{s\,\ad(X)}\, Y – Y\right)\,s^{-1} = \ad(X)\,Y$ exists and $\left(e^{s\,\ad(X)}\, Y – Y\right)\,s^{-1}\in\g,\,\forall\,s\in\R\setminus\{0\}$, the limit must also be in $\g$ (choose $\{s_n\}_{n=1}^\infty$ with $s_n\to0$ to be any Cauchy sequence converging to nought and $\left(e^{s_n\,\ad(X)}\, Y – Y\right)\,s_n^{-1}$ is also a Cauchy sequence in $\g$).

*The Lie algebra, as well as being a vector space, is also closed under the Lie bracket.*

The collection $\left\{ \ad_{X_j } \right\}_{j=1}^N $ is known as the Lie algebra’s “structure co-efficients”. Often the structure co-efficients are written $[\ad_{X_j } ]_{k,\ell} = c_{j,k} ^\ell$ when the Lie group and algebra are defined conventionally in the language of manifolds; the indices then show how the structure co-efficients must transform under mappings between manifolds if these latter have a metric,* i.e.* by pullback (covariantly, like covectors) for the bottom indices $j$,$ k$ and by pushforward (contravariantly, like vectors) for $\ell$. (If you haven’t studied differential geometry, don ‘t worry about the last paragraph.

Now we give a quick proof that the image of the big A adjoint representation is indeed a Lie group. It is then of course a matrix Lie group. The final piece to the jigsaw will be the Campbell Baker Hausdorff theorem, which I am yet to prove. Although we won’t use the Campbell Baker Hausdorff in the following (which is why I can justify leaving its completion till later when I prove the Campbell Baker Hausdorff theorem (Theorem 8.1), it does make the rest of the discussion in this post smoother. Note that part of the proof of the Campbell Baker Hausdorff theorem will be that:

**Theorem 7.3**** (Matrix Campbell Baker Hausdorff Theorem):**

For square matrices $X$ and $Y$ of small enough norm, $\log\left(e^X\,e^Y\right)$ is a convergent series of terms involving only linear combinations of $X$, $Y$ and Lie brackets thereof, *i.e.* if $X$ and $Y$ belong to a matrix Lie algebra, then $\log\left(e^X\,e^Y\right)$ belongs to the same matrix Lie algebra.

**Proof:** See full proof (Theorem 8.1) of the Campbell Baker Hausdorff Theorem and the discussion thereafter. $\quad\square$

**Theorem 7.4**

**(Full Campbell Baker Hausdorff Theorem):**

There is a nucleus $\K\subseteq\Nid\subseteq\G$ within a connected Lie group $\G$ such that $\K$ can be kitted with geodesic co-ordinates, any two members of $\K$ can be written $e^X,\,e^Y$ for $X,\,Y\in\g$ and such that their product is of the form $e^Z = e^X\,e^Y$, the logarithm $\log(e^Z)$ is uniquely defined (so that $\log(e^Z) = Z$ and $Z$ is a series of terms comprising $X$, $Y$ and entities gotten from $X$ and $Y$ by a finite number of linear operations and Lie brackets. Otherwise put, $Z$ belongs to the smallest Lie algebra generated by $X$ and $Y$.

**Proof:** See full proof (Theorem 8.1) of the Campbell Baker Hausdorff Theorem and the discussion thereafter. $\quad\square$

This matrix version of the theorem depends only on $X$ and $Y$ being square matrices of small enough norm; it does not depend on their belonging to the Lie algebra of a Lie group. Therefore this part of the theorem can be used in the following, even though the theorem is proving that $\Ad(\G)$ is a Lie group, *i.e.* we cannot assume that the matrices belong to the Lie algebra of a Lie group.

**Theorem 7.5**** (Image of $\Ad$ is a Lie Group):**

The image $\Ad(\G)$ of a connected Lie group $\G$ under the big-A adjoint representation together with the matrix product is a connected matrix Lie group.

**Proof:** Show Proof

The image is clearly a group, being closed under matrix products and inverses. Now we choose a $\g$-basis $\{\hat{X}_j\}_{j=1}^N$ and an open nucleus $\K\subseteq\Nid\subset\G$ in the original Lie group $\G$ small enough that (Theorem 6.2) geodesic co-ordinates uniquely label $\K$. So every element of $\Ad(\K)$ is a *matrix* exponential of the form $\exp\left(\sum\limits_{j=1}^N\tau_j\,\ad(\hat{X}_j)\right)$. If the kernel of $\Ad$ is nontrivial, then some of the $\ad(\hat{X})_j$ are linearly dependent, so we instead choose a set of square matrices $\{\hat{Y}_j=\sum\limits_{k=1}^N y_{j\,k} \,\ad(\hat{X})_j\}_{j=1}^M$ where $M<N$ that are linearly independent and span the same $M$-dimensional vector space as the $\ad(\hat{X})_j$. So now every element of $\Ad(\K)$ is a *matrix* exponential of the form $\exp\left(\sum\limits_{j=1}^M\tau_j\,\hat{Y}_j\right)$.

Clearly the Homgeneity Axiom 5 holds if we set our namespace set to be $\Ad(\K)$; every element of $\G$ is a finite product $\prod_{j=1}^R \gamma_R$ of $\gamma_R\in\K$, therefore by the homomorphism law, the corresponding element of $\Ad(\G)$ is $\prod_{j=1}^R \Ad(\gamma_R)$ and every element of $\Ad(\G)$ is of this kind.

The product in $\Ad(\K)$ of the form:

\begin{equation}

\label{AdjointRepresentationLieGroupHomomorphismTheorem_1}

\prod\limits_{j=1}^M \exp\left(\sum\limits_{j=1}^M\alpha_j\,\hat{Y}_j\right)\,\exp\left(-\sum\limits_{j=1}^M\beta_j\,\hat{Y}_j\right)

\end{equation}

can also be written in the form $\prod\limits_{j=1}^M \exp\left(\sum\limits_{j=1}^M\omega_j\,\hat{Y}_j\right)$, where the $\omega_j$ are analytic functions of the $\alpha_j,\,\beta_j$ for small enough $\alpha_j,\,\beta_j$ by Theorem 7.3. Therefore, we simply choose $\K$ to be small enough for the matrix exponential Campbell Baker Hausdorff theorem to hold. This then proves the Group Product Continuity Axiom 3 and the Nontrivial Continuity Axiom 4 for the group $\Ad(\G)$ with the namespace set chosen to be $\Ad(\K)$ and $\lambda$ chosen to be the matrix logarithm. The Labeller Axiom 1 and the Connectedness Axiom 2 are trivially fulfilled in this case as well. $\qquad\square$

The above theorem is not used to derive the Campbell Baker Hausdorff theorem, so no begging of the question arises.

**Definition 7.6** **(Matrix Lie Group):**

A matrix Lie group’s members are all square matrices and they are subgroups of the group $GL(N,\,\R)$ of invertible square matrices of a given size $N\times N$. Here $N$ is *not* needfully the dimension of the Lie algebra. We have an example of a matrix Lie group, to wit, the image $\Ad(\G)$, under the big A adjoint representation, of any connected Lie group $\G$. We shall soon see many more; indeed almost all Lie groups are in fact matrix groups, or isomorphic as Lie groups thereto. In a matrix Lie group, the exponential function is, as we have seen, the matrix exponential defined by the universally convergent exponential Taylor series. We can calculate the Lie bracket straight from the definition since we can write down the action of the group $\G$ on its own Lie algebra with literal matrix products: here is one case where “products” between Lie group and algebra members are meaningful, although a general product $\Gamma\, X$ where $\Gamma\in\G$ and $X\in\g$ lies an $\mathcal{M}_N$ but may not lie in either the Lie group $\G$ or its Lie algebra $\g$. We can measure tangents in the space $\mathcal{M}_N$, so that the Lie algebra must be a linear subspace of the set of $N\times N$ matrices. So now let us bring this luxury of being able to measure tangents in the superset $\mathcal{M}_N$ to bear on the definition of the Lie bracket. We have the $C^1$ (indeed $C^\omega$) path $\sigma_s(\tau)=e^{s\,X}\,e^{\tau\,Y}\,e^{-s\,X}$ passing through the identity at $\tau=0$ defined by the product of matrix exponentials, so that its tangent at $\tau=0$, measured in $\mathcal{M}_N$ is:

\begin{equation}

\label{MatrixLieGroupLieBracketExample_1}

\Ad\left(e^{s\,X}\right) \,Y\stackrel{def}{=}\left.\d_\tau e^{s\,X}\,e^{\tau\,Y}\,e^{-s\,X}\right|_{\tau=0} = e^{s\,X}\,Y\,e^{-s\,X}

\end{equation}

another $C^\omega$ function of $s$. Thus we can readily calculate the Lie bracket:

\begin{equation}

\label{MatrixLieGroupLieBracketExample_2}

\ad(X)\,Y\stackrel{def}{=}\left.\d_s e^{s\,X}\,Y\,e^{-s\,X}\right|_{s=0} = X\,Y – Y\,X

\end{equation}

and so, in a matrix Lie group, the Lie bracket is simply the wonted commutator bracket.

Another, equivalent, definition of the Lie bracket is:

\begin{equation}

\label{SecondLieBracketDefintion_1}

[\,,\,]:\g\times\g;\;[X,\,Y] = \lim\limits_{m\to\infty}\left(m^2\,\log\left(e^{\frac{1}{m}\,X}\,e^{\frac{1}{m}\,Y}\,e^{-\frac{1}{m}\,X}\,e^{-\frac{1}{m}\,Y}\right)\right)

\end{equation}

**Proof of Soundness of Definition:** Show Proof

We must prove that the limit exists and is the same as our other definition of the Lie bracket. Choosing $m$ big enough that the Dynkin formula for the Campbell Baker Hausdorff theorem applies to the product of all members of the product below we have:

\begin{equation}

\label{SecondLieBracketDefintion_2}

\begin{array}{rl}

&m^2\,\log\left(e^{\frac{1}{m}\,X}\,e^{\frac{1}{m}\,Y}\,e^{-\frac{1}{m}\,X}\,e^{-\frac{1}{m}\,Y}\right) \\

=& m^2\log\left(\exp\left(\frac{1}{m}\,e^{\frac{1}{m}\ad(X)}\,Y\right)\,e^{-\frac{1}{m}\,Y}\right)\\

=&m^2\log\left(\exp\left(\frac{1}{m}\,Y+\frac{1}{m^2}\,\ad(X)\,Y + \O\left(\frac{1}{m^3}\right)\,Y\right)\,e^{-\frac{1}{m}\,Y}\right)\\

=&m^2\left(\frac{1}{m}\,Y+\frac{1}{m^2}\,\ad(X)\,Y + \O\left(\frac{1}{m^3}\right)\,Y – \frac{1}{m}\,Y + \O\left(\frac{1}{m^2}\right) \O\left(\frac{1}{m}\right)\right) \to \ad(X)\,Y\,\text{as}\,m\to\infty

\end{array}

\end{equation}

$\square$

Again, the above theorem is not used to derive the Campbell Baker Hausdorff theorem, so there is no begging of the question here.

Let us now look at some of the fundamental properties of the Lie bracket.

**Theorem 7.8**

**(Lie Algebra Fundamental Properties):**

The Lie Bracket $\g\times\g\to\g$ has the following fundamental properties:

**1. Billinearity**, to wit

\begin{equation}

\label{LieBracketFundamentalPropertiesTheorem_1}

\begin{array}{ll}

&[x\,X+y\,Y,\,Z] = x\,[X,\,Z] +y\,[Y,\,Z]\,\forall\,X,\,Y,\,Z\in\g,\,x,\,y\in\R\\

\Leftrightarrow&\ad(x\,X+y\,Y) = x\,\ad(X) + y\,\ad(Y)\,\forall\,X,\,Y\in\g,\,x,\,y\in\R\\

&\\

\text{and}&\\&

\\&[Z,\,x\,X+y\,Y] = x\,[Z,\,X] +y\,[Z,\,Y]\\

\Leftrightarrow&\ad(Z) (x\,X+y\,Y) = x\,\ad(Z) \,X+ y\,\ad(Z)\,Y\,\forall\,X,\,Y,\,Z\in\g,\,x,\,y\in\R

\end{array}

\end{equation}

**2. Skew-Symmetry:**

\begin{equation}

\label{LieBracketFundamentalPropertiesTheorem_2}

\begin{array}{ll}

&[X,\,Y] = -[Y,\,X]\\

\Leftrightarrow&\ad(X)\,Y = -\ad(Y)\,X\,\forall\,X,\,Y\in\g

\end{array}

\end{equation}

**3. Fulfills the Jacobi Identity:**

\begin{equation}

\label{LieBracketFundamentalPropertiesTheorem_3}

\begin{array}{ll}

&[X,\,[Y,\,Z]] + [Z,\,[X,\,Y]] + [Y,\,[Z,\,X]] = 0\,\forall\,X,\,Y,\,Z\in\g\\

\Leftrightarrow&\ad([X,Y]) = \ad(X)\,\ad(Y)-\ad(Y)\,\ad(X) = [\ad(X),\,\ad(Y)]\,\forall\,X,\,Y\in\g

\end{array}

\end{equation}

**Proof:** Show Proof

*Billinearity: *The second relationship $\ad(Z) (x\,X+y\,Y) = x\,\ad(Z) \,X+ y\,\ad(Z)\,Y$ is clear, since $\Ad(e^{\tau\,Z})$ is a homogeneous linear map $\g\to\g$, therefore its differential $\ad(Z)$ at $\tau=0$ is too. So we need to show $\ad(x\,X+y\,Y) = x\,\ad(X) + y\,\ad(Y)$. To this end, set $u=v=\tau$ in the following so that:

\begin{equation}

\label{LieBracketFundamentalPropertiesTheorem_4}

\begin{array}{lcl}

\ad(x\,X+y\,Y)&=&\left.\d_\tau \Ad\left(e^{\tau\,(x\,X+y\,Y)}\right)\right|_{\tau=0}\\

&=&\left.\partial_u \Ad\left(e^{u\,x\,X+v\,y\,Y}\right)\right|_{u=v=0}\,\frac{\d\,u}{\d\,\tau} + \left.\partial_v \Ad\left(e^{u\,x\,X+v\,y\,Y}\right)\right|_{u=v=0}\,\frac{\d\,v}{\d\,\tau}\\

&=&\left.\d_\tau\Ad\left(e^{\tau\,x\,X}\right)\right|_{\tau=0}+\left.\d_\tau\Ad\left(e^{\tau\,y\,Y}\right)\right|_{\tau=0}\\

&=&x\,\ad(X) + y\,\ad(Y)\,\forall\,X,\,Y\in\g,\,x,\,y\in\R

\end{array}

\end{equation}

*Skew-Symmetry: *$e^{-s\,X}\,e^{\tau\,X}\,e^{s\,X} = e^{(-s+\tau+s)\,X} = e^{\tau\,X}$ by the flow equation and so:

\begin{equation}

\label{LieBracketFundamentalPropertiesTheorem_5}

\Ad\left(e^{s\,X}\right)\,X=\left.\d_\tau \left(e^{s\,X}\,e^{\tau\,X}\,e^{-s\,X}\right)\right|_{\tau=0} =X

\end{equation}

whence $\left.\d_s \Ad\left(e^{s\,X}\right)\,X\right|_{s=0} = \ad(X)\,X = 0,\,\forall X\in\g$. So now we have, from linearity $\ad(Z) (x\,X+y\,Y) = x\,\ad(Z) \,X+ y\,\ad(Z)\,Y$:

\begin{equation}

\label{LieBracketFundamentalPropertiesTheorem_6}

\begin{array}{lcl}

0=\ad(X+Y)\, (X+Y) &=& \ad(X)\,X + \ad(Y)\,Y + \ad(X)\,Y + \ad(Y)\,X \\

&=& \ad(X)\,Y + \ad(Y)\,X\,\forall\,X,\,Y\in\g

\end{array}

\end{equation}

and so we have $\ad(X)\,Y = -\ad(Y)\,X\,\forall\,X,\,Y\in\g$, which is the skew-symmetry relationship.

*Jacobi Identity: *Behold the entity $\mathcal{Q}=e^{+s_1\,X}\,e^{+s_2\,Y}\,e^{-s_1\,X}\,e^{\tau\,Z}\,e^{+s_1\,X}\,e^{-s_2\,Y}\,e^{-s_1\,X}\in\G$ for any $X,\,Y,\,Z\in\g$ and $s_1,\,s_2,\,\tau\in\R$. Firstly, group and multiply this product out as follows:

\begin{equation}\label{LieBracketFundamentalPropertiesTheorem_7}

\begin{array}{lcl}

\mathcal{Q}&=& \left(e^{+s_1\,X}\,e^{+s_2\,Y}\,e^{-s_1\,X}\right)\;\;e^{\tau\,Z}\;\;\left(e^{+s_1\,X}\,e^{-s_2\,Y}\,e^{-s_1\,X}\right)\\

&=&\exp\left(+s_2\,\Ad\left(e^{s_1\,X}\right)\,Y\right)\;e^{\tau\,Z}\;\exp\left(-s_2\,\Ad\left(e^{s_1\,X}\right)\,Y\right)\\

&=&\exp\left(+s_2\,e^{s_1\,\ad(X)}\,Y\right)\;e^{\tau\,Z}\;\exp\left(-s_2\,e^{s_1\,\ad(X)}\,Y\right)\\

&=&\exp\left(\tau\,\exp\left(s_2\,\ad\left(e^{s_1\,\ad(X)}\,Y\right)\right)\,Z\right)

\end{array}

\end{equation}

then group it as:

\begin{equation}

\label{LieBracketFundamentalPropertiesTheorem_8}

\begin{array}{lcl}

\mathcal{Q}&=& e^{+s_1\,X}\,e^{+s_2\,Y}\;\;\left(e^{-s_1\,X}\,e^{\tau\,Z}\,e^{+s_1\,X}\right)\;\;e^{-s_2\,Y}\,e^{-s_1\,X}\\

&=&e^{+s_1\,X}\,e^{+s_2\,Y}\;\exp\left(\tau\,\Ad\left(e^{-s_1\,X}\right)\,Z\right)\;e^{-s_2\,Y}\,e^{-s_1\,X}\\

&=&\exp\left(\tau\,\Ad\left(e^{s_1\,X}\right)\,\Ad\left(e^{s_2\,Y}\right)\,\Ad\left(e^{-s_1\,X}\right)\,Z\right)\\

&=&\exp\left(\tau\,e^{s_1\,\ad(X)}\,e^{s_2\,\ad(Y)}\,e^{-s_1\,\ad(X)}\,Z\right)

\end{array}

\end{equation}

As $\tau$ varies, $\mathcal{Q}$ follows a $C^1$ path through the Lie group $\G$. The tangent to the path at $\tau=0$ can be readily calculated for both the above groupings, so we see that:

\begin{equation}

\label{LieBracketFundamentalPropertiesTheorem_9}

\begin{array}{lcl}

\exp\left(s_2\,\ad\left(e^{s_1\,\ad(X)}\,Y\right)\right) &=& e^{s_1\,\ad(X)}\,e^{s_2\,\ad(Y)}\,e^{-s_1\,\ad(X)} \\

&=& \exp\left(s_2\, e^{s_1\,\ad(X)}\,\ad(Y)\,e^{-s_1\,\ad(X)}\right)\,\forall\,X,\,Y\in\g,\,s_1,\,s_2\in\R

\end{array}

\end{equation}

This is an equation describing a matrix following a $C^1$ (indeed $C^\omega$) path in $GL(\g)$ as $s_2$ varies. The path is an exponential function of $s_2$ and is thus wholly determined by its tangent at $s_2=0$; on calculating this tangent for both forms this exponential function we find:

\begin{equation}

\label{LieBracketFundamentalPropertiesTheorem_10}

\ad\left(e^{s_1\,\ad(X)}\,Y\right) = e^{s_1\,\ad(X)}\,\ad(Y)\,e^{-s_1\,\ad(X)}\,\forall\,X,\,Y\in\g,\,s_1\in\R

\end{equation}

this equation describes a vector following a $C^\omega$ path in $\g$ and holds for all $s_1\in\R$. In particular, the first derivative at $s_1=0$ of both sides of this equation must be the same; on calculating this derivative:

\begin{equation}

\label{LieBracketFundamentalPropertiesTheorem_11}

\ad\left(\ad(X)\,Y\right) = \ad\left([X,\,Y]\right)=\ad(X)\, \ad(Y) – \ad(Y) \,\ad(X) = [\ad(X),\,\ad(Y)]
\end{equation}

since, in the Lie algebra of a matrix group, the Lie bracket is simply the “commutator” $[U,\,V] = U\, V – V\, U$. On letting both sides of $\eqref{LieBracketFundamentalPropertiesTheorem_11}$ operate on an arbitrary vector $Z\in\g$ we get:

\begin{equation}

\label{LieBracketFundamentalPropertiesTheorem_12}

\begin{array}{ll}

&[[X,\,Y],\,Z] = [X,\,[Y,\,Z]] – [Y,\,[X,\,Z]]\\

\Leftrightarrow& [X,\,[Y,\,Z]] + [Z,\,[X,\,Y]] + [Y,\,[Z,\,X]]= 0

\end{array}

\end{equation}

where we have used the skew-symmetry relationships to rearrange the double Lie brackes and conversely we can derive $\eqref{LieBracketFundamentalPropertiesTheorem_11}$ from $\eqref{LieBracketFundamentalPropertiesTheorem_12}$ on noting that the latter holds for arbitrary $Z\in\g$. Thus $\eqref{LieBracketFundamentalPropertiesTheorem_11}$ and $\eqref{LieBracketFundamentalPropertiesTheorem_12}$ are logically equivalent and are the sought Jacobi identity. $\qquad\square$

A Lie algebra of a Lie group must be closed under the Lie bracket operation (*i.e.* brackets can’t give anything outside the algebra). An *abstract Lie algebra* is defined as follows:

**Definition 7.9****: (Abstract Lie Algebra)**

An *abstract Lie algebra over a field $\mathbb{K}$ *is a vector space $\g(\mathbb{K})$ over the field $\mathbb{K}$ kitted with a binary mapping $\left[\cdot,\,\cdot\right]:\g(\mathbb{K})\to\g(\mathbb{K})$ such that $\left[\cdot,\,\cdot\right]$ is billinear (linear in both arguments), skew-symmetric and fullfils the Jacobi identity.

**Definition 7.10**

**: (Lie Algebra Homomorphism)**

Let $\g,\,\h$ be abstract Lie algebras over the same field $\mathbb{K}$. Then a mapping $\rho:\g\to\h$ is called a *Lie Algebra Homomorphism *if it is a vector space homomorphism (*i.e.* a linear map) that also respects Lie brackets, *i.e.*:

\begin{equation}\label{LieAlgebraHomomorphismDefinition_1}\begin{array}{lcll}\rho(x\,X+y\,Y) &=& x\,\rho(X) + y\,\rho(Y)&\\\rho(\left[X,\,Y\right]) &=& \left[\rho(X),\,\rho(Y)\right]&\forall\,X,\,Y\in\g;\,\forall\,x,\,y\in\mathbb{K}\end{array}\end{equation}

If $\ker(\rho)=\{\Or\}$, then, naturally, $\rho$ is called a *Lie Algebra Isomorphism*.

Clearly, in the light of what has just been discussed, the Lie algebra of a Lie group is a general, abstract Lie algebra. It can also be shown that all abstract Lie algebras over the field of real numbers are the Lie algebras of some connected Lie group of nonsingular matrices,* i.e.* $\G\subseteq GL(N,\,\R)$. This is Lie’s third theorem, which will be proven later. It is not a hard theorem to prove, but the simplest proof rests on the rather tricky (but whose meaning is very clear and intuitive) Ado theorem, which Terence Tao sketches out on his blog.

Every abstract finite dimensional Lie algebra $\g$ over the reals $\R$ has a faithful finite dimensional linear representation. That is, there is a onto (surjective) Lie algebra isomorphism (in the sense of Definition 7.10) $\rho:\g\to\h$ where $\h$ is a Lie algegra of square matrices.

Generalisation ([Ado, 1949]): the above holds if the field $\R$ is replaced by *any* field $\mathbb{K}$ with characteristic 0.

Generalisation 2 ([Iwasawa, 1948], [Harish-Chandra, 1949]): the above holds without restriction on the characteristic of the field$\mathbb{K}$.

**Proof:** See [Tao, 2011] or [de Graaf, 2014]. $\quad\sqaure$

Ado’s theorem also holds for every field $\mathbb{K}$ of characteristic nought, although this latter generalisation is not needed for our present exposition. This was the full work of Ado; later, even the field characteristic restriction was dropped in a further generalisation by [Iwasawa, 1948] and [Harish-Chandra, 1949]. Ado’s theorem reminds me of a piece of computer software that is essentially simple, but which has many fiddly bits bolted onto it to cope with bothersome special cases which seem as though they should be trivial yet prove surprisingly hard to nail down thoroughly, so that the end product comes out rather bedraggled and messy. Terence Tao’s blog article gives a good feeling for this: Ado’s theorem “*should*” be simple, as, for a centreless algebra, the adjoint representation is indeed an isomorphism, so, in particular, for simple and semisimple Lie algebras Ado’s theorem is trivial: you simply work out the images of any basis of a semisimple algebra’s direct summands under the adjoint representation (to be talked about below) and these span a matrix version isomorphic to each direct summand. Then you form the direct sum and there’s your matrix Lie algebra isomorphic to the original. I myself really only “understood” Ado’s theorem by writing the procedure out and testing it as an actual piece of software given the prescription of the constructive proof in [de Graaf, 2014], something which I do not recommend doing unless you have a great deal of time on your hands. Terence Tao’s exposition is for me still the clearest, although still very hard work.

The Jacobi identity warrants a few words. Most texts present it as simply a property of a Lie algebra, a pretty-looking cyclically symmetric identity (*i.e.* in the sense that one can cyclically permute the roles of $X,\,Y,\,Z$ in $\eqref{LieBracketFundamentalPropertiesTheorem_11}$ in Theorem 7.8 and still have the same identity) that one might abstractly define simply for the sake of symmetry to beget an abstract object – a “Lie algebra” for study for its own sake. Otherwise it is presented as a handy curlicue for doing calculations with. However, Lie algebras historically came from the study of Lie groups, and the Jacobi identity has a deep and beautiful direct meaning:

*The big A adjoint representation induces a homomorphism of the corresponding Lie algebras that respects Lie brackets, not simply the vector space properties of the Lie algebra*.

One would do well to ponder thoroughly on $\eqref{LieBracketFundamentalPropertiesTheorem_11}$ in Theorem 7.8, which is logically equivalent to the Jacobi identity and makes this respect of Lie brackets clear. So, if $\sigma_X$ and $\sigma_Y$ are $C^1$ paths through the identity in $\G$, then there must also be a path $\sigma_{[X,\,Y]}$ which passes through the identity has the tangent $[X,\,Y]\in\g$ there. Later I shall give an explicit construction of this path. In the adjoint representation, the paths $\Ad(\sigma_x),\,\Ad(\sigma_Y)$ have tangents $\ad(X)$ and $\ad(Y)$ to the identity. The Jacobi identity means that $\Ad(\sigma_{[X,\,Y]})$ has the tangent $\ad([X,\,Y]) = [\ad(X),\,\ad(Y)]$ there.

Indeed the Jacobi identity runs deeper than simply the Adjoint representation; the Jacobi identity as talked about above is only a special case of a much more general beautiful and intuitive behaviour:

**Theorem 7.12**** (Induced Homomorphism of Lie Algebras):**

Every continuous homomorphism $\rho:\G\to\H$ from connected Lie group $\G$ to connected Lie group $\H$ induces a corresponding homomorphism $d\rho:\g\to\h$ between the corresponding Lie algebras $\g$ and $\h$ that respects Lie brackets.

**Note on Terminology:** A “homomorphism between Lie algebras” is wontedly understood to respect Lie brackets so that $d\rho([X,\,Y])=[d\rho({X}),\,d\rho({Y})],\,\forall\,X,\,Y\in\g$ as well as being a “bare” homomorphism (*i.e.* homogeneous linear map between vector spaces) as implied by the word.

**Proof:** Show Proof

Let $\mathfrak{E}_X = \left\{e^{\tau\,X}:\,\tau\in\R\right\}\subset\G$ be a $C^1$ one parameter subgroup of $\G$, and let $\K\subset\Nid\subset\G$ be an open nucleus in $\G$. The homomorphism is continuous, so that $\rho(\K)$ contains an open neighbourhood of $\id$ in $\H$ in the corresponding namespace set of $\H$, *i.e.* $\rho(\K)\subset\H$ contains an opennucleus $\mathcal{L}\subset\H$ in $\H$. Moreover:

\begin{equation}

\label{InducedLieAlgebraHomomorphismTheorem_1}

\rho(\mathfrak{E}_X) = \left\{\zeta_X(\tau)\stackrel{def}{=}\rho\left(e^{\tau\,X}\right):\,\tau\in\R\right\}\subset\H

\end{equation}

must pass through the nucleus $\mathcal{L}\subset\H$ and must be continuous there, *i.e. *$\tau\mapsto \zeta_X(\tau)=\rho\left(e^{\tau\,X}\right)$ must map an open neighbourhood $(-\tau_{max},\,\tau_{max})$ of $0$ in $\R$ to a $C^0$ path in $\mathcal{L}\subset\H$, by continuity of the homomorphism. Furthermore, by the flow equation and by definition of a homomorphism:

\begin{equation}

\label{InducedLieAlgebraHomomorphismTheorem_2}

\zeta_X(\tau+s) = \rho\left(e^{(\tau+s)\,X}\right) = \rho\left(e^{\tau\,X}\,e^{s\,X}\right) = \rho\left(e^{\tau\,X}\right)\,\rho\left(e^{s\,X}\right) = \zeta_X(\tau)\,\zeta_X(s)

\end{equation}

so that $\rho(\mathfrak{E}_X)$ is a $C^0$ one parameter (possibly trivial) subgroup of $\H$. However, (Theorem 6.10) we have seen that the only $C^0$ one parameter subgroups of a connected Lie group are indeed $C^1$ and must be of the form :

\begin{equation}

\label{InducedLieAlgebraHomomorphismTheorem_3}

\rho(\mathfrak{E}_X) = \left\{e^{d\rho(X)\,\tau}:\,\tau\in\R\right\}

\end{equation}

for some tangent $d\rho(X) = \log\left(\rho\left(e^X\right)\right)\in\h$ which depends only on $X$ (since the original one parameter group depends wholly on $X$). We must now check that the mapping $d\rho(X)$ is defined for all tangents to $C^1$ paths in $\G$, is linear and respects Lie brackets.

So now, kit the open nucleus $\K\subset\G$ with second kind canonical co-ordinates so that $\gamma = \prod\limits_{j=1}^N\,e^{\tau_j(\gamma)\,\hat{X}_j},\,\forall\,\gamma\in\K$ and the $\tau_j(\gamma)$ are unique. Then

\begin{equation}

\label{InducedLieAlgebraHomomorphismTheorem_4}

\rho(\gamma) = \prod\limits_{j=1}^N\,\rho\left(e^{\tau_j(\gamma)\,\hat{X}_j}\right) = \prod\limits_{j=1}^N\,\rho\left(e^{\tau_j(\gamma)\,d\rho(\hat{X}_j)}\right)

\end{equation}

labels all $\rho(\gamma)\in\mathcal{L}$ (possibly redundantly, if the homomorphism’s kernel is nontrivial); then any $C^1$ path in $\K$ is described by the a path followed by $\gamma$ when the second kind canonical co-ordinates vary as $C^1$ functions of “time” $\tau$* i.e.* $\tau_j = \alpha_j(\tau)$ path in $\K$. The path in $\mathcal{L}$ described by $\eqref{InducedLieAlgebraHomomorphismTheorem_4}$ is clearly $C^1$ because all its terms are so that their product is also by the Group Product Continuity Axiom 3.Therefore we see that the homomorphism must in fact automatically be $C^1$ if $C^0$. So if we choose $\tau_j(\tau) = \alpha_j(\tau)$ with $\alpha_j(0) = 0$, then, by differentiating both sides of $\eqref{InducedLieAlgebraHomomorphismTheorem_4}$ when $\tau_j=0$, we find that an arbitrary $C^1$ path with tangent $\sum\limits_{j=1}^N\dot{\alpha}_j\,\hat{X}_j$ to the identity is mapped to a path with tangent $\sum\limits_{j=1}^N\dot{\alpha}_j\,d\rho(\hat{X}_j)$ to the identity. This proves definition of $d\rho$ for all $C^1$ paths and all tangents $X\in\g$ and also proves linearity.

Now let us look at the path $\tau\mapsto e^{s\,X}\,e^{\tau\,Y}\,e^{-s\,X}\in\K$. Reasoning as above, this maps to the path $\tau\mapsto e^{s\,d\rho(X)}\,e^{\tau\,d\rho(Y)}\,e^{-s\,d\rho(X)}\in\K$, so that the image of the tangent $e^{s\,\ad(X)}\,Y\in\g$ is the tangent $e^{s\,\ad(d\rho(X))}\,d\rho(Y)\in\h$. Differentiating these tangents at $s=0$ shows that $d\rho$ maps the tangent $\ad(X)\,Y\in\g$ to tangent $\ad(\ad\rho(X))\,\ad\rho(Y)\in\g$. $\quad\square$

The Lie algebra homomorphism $d\rho:\g\to\h$ induced by the continuous homomorphism $\rho:\G\to\H$ between the two connected Lie groups $\G$ and $\H$ with Lie algebras $\g$ and $\h$, respectively, is called the *Lie map* of the homomorphism $\rho$.

An important special case is any inner automorphism of a Liue group $\G$ onto itself:

**Corollary 7.14**** (Lie Map of Lie Group Inner Automorphism):**

Let $\G$ be a connected Lie group, $\gamma\in\G$ and $\Ad(\gamma):\g\to\g$. Then $\Ad(\gamma)$ is a Lie bracket respecting automorphism (invertible linear map) of the Lie algebra $\g$.

**Proof:** Consider the inner automorphism $\operatorname{In}(\gamma):\G\to\G;\;\operatorname{In}(\gamma)(\zeta) = \gamma\,\zeta\,\gamma^{-1}$ on the Lie group. It is a continuous automorphism and its Lie map $\g\to\g$ is $\Ad(\gamma)$; $\Ad(\gamma)$ is then, by Theorem 7.12, a Lie bracket respecting automorphism $\g\to\g$.$\quad\square$

So now the mapping $X\in\g\mapsto \ad(X)$ has indeed been shown to be a Lie-bracket-respecting homomorphism of Lie algebras, because $\Ad$ is the Lie group homomorphism that induces it. The terminology “little a adjoint representation of the Lie algebra $\g$” for this mapping is therefore altogether justified: it is indeed a Lie-bracket-respecting representation of the Lie algebra. The question now arises: what is this Lie algebra homomorphism’s kernel? It is precisely $\{X\in\g:\ad(X) = 0\}$; the set $\mathscr{Z}(\G)$ of all Lie algebra elements whose brackets with all other algebra members vanish. It is a subalgebra of $\g$: it is a vector space closed under the Lie bracket (which is of course nought between all pairs of members of the subalgebra). Not surprisingly we call the subalgebra the Lie algebra’s “centre”:

**Definition 7.15**** (Centre of a Lie Algebra):**

The centre $\mathscr{Z}(\G)$ of a Lie algebra $\g$ is the set $\{X\in\g:\ad(X) = 0\}$ of all elements of $\g$ that commute with all elements of $\g$.

Representing any connected Lie group member $\gamma$ as a finite product $\gamma = \prod\limits_{j=1}^M e^{X_j},\,X_j\in\mathcal{g}$ we therefore have $e^Y\,\gamma\,e^{-Y} = \prod\limits_{j=1}^M \exp\left(e^{\ad(Y)}\,X_j\right)=\prod\limits_{j=1}^M e^{X_j} = \gamma$ for every $Y\in\mathscr{Z}(\G)$, so we have $\exp(\mathscr{Z}(\G))\subseteq\mathscr{Z}(\G)$, *i.e.* the centre of the Lie algebra exponentiates *into* the centre of the Lie group. In general, however, there are central elements of the group that are not in $\exp(\mathscr{Z}(\G))$. But the whole set of central elements is at most the direct product of $\exp(\mathscr{Z}(\G))$ and a nontrivial *discrete* subgroup of $\G$, if $\exp(\mathscr{Z}(\G))$ is not the whole centre.

**Definition 7.16**** (Continuous Centre of a Connected Lie Group):**

The set $\mathscr{Z}_C(\G) = \exp(\mathscr{Z}(\G)) = \{e^X:\,X\in\mathscr{Z}(\G)\}$ is called the *continuous centre* of the connected Lie group $\G$ where $\mathscr{Z}(\G)$ is the centre of the Lie algebra $\g$ of the group $\G$.

**Lemma 7.17**

**(Commutative Index Law):**

In a connected Lie group, two elements of the form $e^X$ and $e^Y$ commute if and only if $e^X\,e^Y=e^{X+Y}$.

**Proof:** Show Proof

If $e^X$ and $e^Y$ commute, then both the entities **( i) **$\sigma_1(\tau) = e^{\tau X}\,e^{\tau\,Y}$ and

**(**$\sigma_2(\tau)=e^{(X+Y)\,\tau}$ are readily shown to fulfill the flow equation $\sigma(\tau)\,\sigma(s) = \sigma(\tau+s)$ as well as both fulfilling the same Cauchy initial value problem $\left.\d_\tau\,\sigma^{-1}(s)\,\sigma(s+\tau)\right|_{\tau=0} = X+Y;\, \sigma(0)=\id$. By Theorem 5.5, they must be the same function of $\tau$ and, setting $\tau=1$, we have $e^X\,e^Y = e^{X+Y}$. Conversely, if $e^X\,e^Y = e^{X+Y}$ then, swapping roles of $X$ and $Y$ shows $e^X\,e^Y = e^Y\,e^X$.

*ii*)**Theorem 7.18**** (Continuous Centre is a Group):**

The continuous centre $\mathscr{Z}_C(\G)$ of a connected Lie group $\G$ is a group, a normal subgroup of both $\G$ and of $\mathscr{Z}(\G)$.

**Proof:** Show Proof

By Lemma 7.17, $e^X\,e^Y = e^{X+Y}$ and so $\mathscr{Z}_C(\G)$ is closed under group multiplication. $e^X\in \mathscr{Z}_C(\G)\,\Leftrightarrow\,(e^{-X})^{-1} = e^{-X}\in \mathscr{Z}_C(\G)$ so that $\mathscr{Z}_C(\G)$ is closed under all the group operations. $\mathscr{Z}(\G)$, by definition of the centre, is Abelian, hence $\mathscr{Z}_C(\G)$ is a normal subgroup of $\mathscr{Z}(\G)$. $\quad\square$

**Theorem 7.19**** (Discrete Centre):**

The quotient group $\mathscr{Z}(\G) / \mathscr{Z}_C(\G)$ is discrete, *i.e.* there is an open neighbourhood $\K$ of $\id$ such that $\gamma_c \K$ holds no other elements of $\mathscr{Z}(\G) / \mathscr{Z}_C(\G)$ if $\gamma_c \in \mathscr{Z}(\G) / \mathscr{Z}_C(\G)$.

**Proof:** Show Proof

Consider any nucleus $\K\subseteq\Nid\subseteq\G$ small enough that geodesic co-ordinates are unique and thus the logarithm uniquely defined. Then if $\gamma_c\in\K\cap\mathscr{Z}(\G)$ we have $\gamma_c=\exp(X)$ where $X=\log(\gamma_c)$ and $\gamma_c\,\gamma\gamma_c^{-1}=\gamma$ for all $\gamma\in\mathcal{K}$, but the latter relationship can hold only if $[X\,\hat{X}_j] = 0,\,j=1\cdots N$. Thus $\gamma_c$ is in the continuous centre $\mathscr{Z}_C(\G)$. Thus we see that there is a nucleus $\K$ which contains no elements of $\mathscr{Z}(\G)\setminus \mathscr{Z}_C(\G)$. Therefore, suppose we have two elements $\gamma_{c1},\,\gamma_{c2}$ of the quotient group $\mathscr{Z}(\G)/\mathscr{Z}_C(\G)$: it cannot be that $\gamma_{c1},\,\gamma_{c2}^{-1}\in\mathcal{K}$, for then $\gamma_{c1},\,\gamma_{c2}^{-1}$, being central, would have to belong to the continuous centre, which becomes the identity under the homomorphism $\mathscr{Z}(\G)\to \mathscr{Z}(\G) / \mathscr{Z}_C(\G)$, so $\gamma_{c1}, \gamma_{c2}$ would have to be the same element of $\mathscr{Z}(\G) / \mathscr{Z}_C(\G)$, so that the lemma follows. $\quad\square$

Indeed the only kinds of discrete normal subgroups of a connected Lie group are subgroups of the centre, a kind of converse to the lemma above that can be simply proven by an ingenious demonstration owing to Otto Schreier:

**Theorem 7.20**** (Schreier, 1925; Discrete Normal Subgroups are Central):**

Any discrete normal subgroup $\mathcal{D}\subset\G$ of a connected Lie group is a subgroup of the centre $\mathscr{Z}(\G)$

**Proof:** Show Proof

$\forall\,\gamma\in\G\,\gamma=\prod\limits_{j=0}^M e^{X_j},\,X_j\in\g$ so that $\sigma:[0,\,1]\to\G;\,\sigma(\tau) = \prod\limits_{j=0}^M e^{\tau\,X_j}$ is a $C^1$ path linking $\sigma(0)=\id$ and $\sigma(1) = \gamma$. Now suppose $\gamma_d\in \mathcal{D}$ so that there is some nucleus $\K\subseteq\G$ such that $(\gamma_d\, \K)\cap\mathcal{D} = \{\gamma_d\}$ and consider $\zeta:[0,\,1];\,\zeta(\tau) = \sigma(\tau)\,\gamma_d\,\sigma(\tau)^{-1}$ a continuous function of $\tau$ over the whole interval $[0,\,1]$. Since $\mathcal{D}$ is normal, $\gamma\,\gamma_d\,\gamma^{-1}\in\mathcal{D},\,\forall\,\gamma\in\G$, so $\zeta(\tau) = \sigma(\tau)\,\gamma_d\,\sigma(\tau)^{-1}$ must take on discrete values, *i.e.* can only jump from $\gamma_d = \zeta(0)$ to another value outside $\zeta(0)\,\K$, which behaviour gainsays the continuity of $\zeta$ unless $\zeta(\tau) = \zeta(0)$ and $\zeta$ is constant, *i.*e.$\zeta(1) = \zeta(0) = \gamma_d\,\Rightarrow\, \gamma\,\gamma_d = \gamma_d\,\gamma$. Since this reasoning works for any $\gamma\in\G$, $\gamma_d$ must commute with every $\gamma\in\G$, *i.e.* $\gamma_d\in\mathscr{Z}(\G)$ and $\mathcal{D} \subseteq \mathscr{Z}(\G)$. $\quad\square$

If we work through the proof of Theorem 7.5 and make use of the general Campbell Baker Hausdorff theorem that I shall derive in my next post, we have:

**Theorem 7.21**** (Lie Group Image Under Homomorphism):**

The image $\rho(\G)$ of any connected Lie group $\G$ under a general (abstract group) homomorpism $\rho$ can be made in to another connected Lie group and the resulting Lie group homomorphism is continuous.

**Proof:** Show Proof

Wholly analogous to the proof of Theorem 7.5 when we replace $\Ad$ by a general, abstract group homomorphism $\rho$, although now we must make use of the general Campbell Baker Hausdorff (CBH) theorem yet to be derived (the present theorem is not used to derive the CBH theorem, so no begging of the question arises here). The homomorphism clearly is continuous if we take an open nucleus $\K\subset\G$ small enough that the CBH theorem applies to all products of elements of $\K$ and use its image $\rho(\K)$ as the namespace set $\Nid$ in the image group $\rho(\G)$.

We now have a second proof of the Jacobi identity: the above theorem shows that the image of the abstract homomorphism $\Ad$ is a Lie group, thus $\Ad$ can be made into a continuous Lie group homomorphism, which always induces a corresponding homomorphism of Lie algebras. The induced homomorphism $\ad$ respects Lie brackets, therefore $\ad([X,\,Y]) = [\ad(X),\,\ad(Y)]$, which, we have seen, is logically equivalent to the Jacobi identity.

Example 7.22 **(Lie Map of the Big A Adjoint)**

Let $\G$ be a connected Lie group and $\g$ its Lie algebra. The map $\Ad:\G\to\Ad(G)$ is a continuous homomorphism of Lie groups. Its Lie map is $\ad:\g\to\ad(\g)$, which has been shown to respect Lie brackets. This is proven either by the method of proving the form $\ad(\left[X,\,Y\right]) = \left[\ad(X),\,\ad(Y)\right]$ of the Jacobi identity shown in Theorem 7.8 or it can be proven by Theorem 7.12, which thus becomes a second way to prove the Jacobi identity.

Our continuous Lie group homomorphisms induce Lie bracket respecting homomorphisms of the Lie algebras. If any Lie algebra homomorphism (whether the Lie map of a Lie group homomorphism or not) has trivial kernel and is thus an isomorphism, we can find a handy formula for how the matrix of the mapping $\ad(X):\g\to\g$ transforms under this Lie algebra isomorphism.

**Lemma 7.23** **(Transformation Law for Matrix of $\ad$):**

Let $\g,\,\h$ be Lie algebras and $\rho:\g\to\h$ a Lie algebra isomorphism, *i.e.* is bijective, linear and respects Lie brackets. Then:

\begin{equation}\label{TransformationOfLittleAdjointLaw_1}\ad(\rho\,X) = \rho\,\ad(X)\,\rho^{-1};\;\forall\,X\in\g\end{equation}

**Proof: **By assumption $\rho([X,\,Y]) = [\rho(X),\,\rho(Y)];\;\forall\,X,\,Y\in\g$. Now let $Z=\rho(Y)\in\h\,\Leftrightarrow\,Y=\rho^{-1}(Z)\in\g$ and we can choose a $Y\in\g$ such that $Z=\rho(Y)\in\h$ for any $Z\in\h$. The Lie bracket respect equation then becomes $\rho\,\ad(X)\,\rho^{-1}\,Z = \ad(\rho(X))\,Z$. Since this holds for all $Z\in\h$, the claimed result follows.$\quad\square$

Example 7.24 **(Transformation of $\ad(X)$ under $\Ad(\gamma)$)**

Let $\G$ be a Lie group and $\g$ its Lie algebra. Then $\Ad(\gamma):\g\to\g$ is a Lie bracket respecting automorphism of $\g$ for any $\gamma\in\G$, by Corollary 7.14. Thus we apply Lemma 7.23 to $\Ad(\gamma)$ and we find:

\begin{equation}\label{TransformationOfLittleAdUnderAdExample_1}\ad(\Ad(\gamma)\,X) = \Ad(\gamma)\,\ad(X)\,\Ad(\gamma)^{-1};\;\forall\,X\in\g,\,\gamma\in\G\end{equation}

Even though Lie algebras over $\R$ and $\mathbb{C}$ can be defined abstractly and divorced from the theory of Lie groups, we can, by Ado’s theorem, construct a faithful matrix representation of any Lie algebra, that is, one can construct an algebra of matrices which is a Lie algebra with the same structure co-efficients as any given Lie algebra. This algebra, through the matrix exponential, can be exponentiated into a matrix Lie group, whose Lie algebra which we shall show to be the same as the beginning, Ado-theorem-constructed one. So every Lie algebra over $\R$ and $\mathbb{C}$ is the Lie algebra of *some* matrix Lie group. I shall write this proof sketch out in more detail in a later chapter.

Having seen that every continuous Lie group homomorphism begets a corresponding homomorphism between the corresponding Lie algebras, it is natural to wonder whether the converse holds, *i.e.* does every homomorphism between the Lie algebras of two Lie group “lift” to a homomorphism of Lie groups. The answer in general is no: there is a final ingredient to a Lie group aside from its Lie algebra that one needs to fully define the former. This last ingredient is the group’s global topology: two different Lie groups can have the same Lie algebra, but only one *simply* connected Lie group can have a given Lie algebra. So the “lift” cannot in general be made unless the Lie algebra homomorphism concerned is between *simply* connected Lie groups. I shall study the topic of lifting in a later chapter.

Lastly, we think about the possibility of iterating the big-A adjoint representation. So $\G\to\Ad(\G)$ annihilates the group centre, $\mathscr{Z}(\G)$. With the centre annihilated (*i.e.* sent to the identity by the homomorphism), we have the possibility of a new centre in the image $\Ad(\G)$. That is, suppose:

\begin{equation}

\label{GroupOfCommutatorsDefinition}

\mathcal{C}(\zeta) = \{\gamma\,\zeta\,\gamma^{-1}\,\zeta^{-1}:\,\gamma\in\G\}\subset \mathscr{Z}(\G)

\end{equation}

but that $\mathcal{C}(\zeta)$ includes central values other than the identity. Then, under the Adjoint representation, $\Ad(\mathcal{C}(\zeta)) = \{\id\}$ and therefore $\gamma^\prime\,\Ad(\zeta) = \Ad(\zeta)\,\gamma^\prime\,\forall\,\gamma^\prime\in\Ad(\G)$, so that $\Ad(\zeta)$ belongs to the centre $\mathscr{Z}(\Ad(\G))$ even though $\zeta$ is not central in $\G$.

However, when the group has a discrete center:

**Lemma 7.25**

**($\Ad$ and $\ad$ idempotent for Discrete Centered Connected Lie Groups):**

Suppose connected Lie group $\G$ has a discrete center. Then $\ad(\ad(\g)) = \ad(\g)$ and $\Ad(\Ad(\G))=\Ad(\G)$.

**Proof: **$\ker(\Ad)$ is the discrete center $\mathscr{Z}(\G)$ and $\ker(\ad)=\{\Or\}$, {\it i.e.} the map $\ad$ is an isomorphism. The matrix $\ad(X)$ for $X\in\G$ encodes the Lie bracket relationships between $X\in\G$ and other members of $\G$, so, given the Jacobi identity $\ad([X,\,Y])=[\ad(X),\,\ad(Y)]$, these are precisely the same Lie bracket relationships as those between $\ad(X)\in\ad(\g)$ and other members of $\ad(\g)$ and therefore we see that $\ad(\ad(X))=\ad(X),\,\forall\,X\in\g$. So $\ad(\g)$ is precisely the same matrix Lie algebra as $\ad(\ad(\g))$. Therefore it exponentiates to precisely the same connected Lie group component as $\ad(\g)$, namely, the identity connected component of $\Ad(\G)$.$\quad\square$

Later on we shall defined Lie groups with more than one connected component. In this case, one still has the possibility that there is a $\gamma\in\G$ noncentral in $\Ad(\G)$ such that $\Ad(\gamma)$ is central in $\Ad(\G)$ and such that $\gamma\,\G$ is not an identity connected component. This happens if the commutators of noncentral $\gamma$ and all other $\mathfrak{G}$ members were central.

**References:**

- Terence Tao, “Ado’s Theorem”; a proof of Ado’s theorem sketched out on Terrence Tao’s Personal Blogblog, 10
^{th}May, 2011 - Willem A. De Graaf, “Constructing Faithful Matrix Representations of Lie Algebras”, Downloaded from ResearchGate 16th June 2014
- I. D. Ado, “
*The representation of Lie algebras by matrices*“, Amer. Math. Soc. Trans.**1949**#2,*p*21, 1949 - Iwasawa, Kenkichi, “On the representation of Lie algebras”, Japanese J. Math.
**19**,*pp*405–426, 1948 - Harish-Chandra, “Faithful representations of Lie algebras”, Annals of Mathematics. Second Series
**50**,*pp*68–76, 1949 - Otto Schreier. “Abstrakte kontinuierliche Gruppen”.
*Abhandlungen aus dem Mathematischen Seminar der Universität Hamburg*,**4**:15–32, 1925.

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