automorphism of a Lie algebra

Let $\mathfrak{g} = (\mathfrak{g},[,])$ be a Lie algebra over a field $k$. A $k$-linear map $\psi:\mathfrak{g}\to\mathfrak{g}$ is a **Lie algebra automorphism** of $\mathfrak{g}$ if it preserves the Lie bracket, i.e. for all $X,Y\in\mathfrak{g}$ $\psi([X,Y]) = \psi(X),\psi(Y)]$.

Let now $\mathfrak{g}$ be finite dimensional. Let $e_1,\ldots,e_n$ be a basis of $\mathfrak{g}$ and

$[e_i,e_j] = C^k_{i j} e_k$

determine the structure constants $C^k_{i j}$. Then $\psi(e_i) = e_j M^j_i$; for some invertible matrix $M$. Denote $e'_i = \psi(e_i)$ the new basis. Then

$[e_i', e_j'] = [e_l,e_m] M^l_i M^m_j = C^p_{l m} M^l_i M^m_j e_p$

$C^k_{i j} e'_k = C^k_{i j} e_p M^p_k$

Therefore the condition that $\psi$ is an isomorphism is that its matrix $M$ is invertible and satisfies

$C^k_{i j} e_p M^p_k = C^p_{l m} M^l_i M^m_j e_p$

hence, for all $i,j,p$,

$C^k_{i j} M^p_k = C^p_{l m} M^l_i M^m_j$

For a finite dimensional Lie algebra over reals or complexes, one says that an automorphism $\psi$ is an inner automorphism if it is of the form $Ad(g)$ for some $g\in G$. Recall that $Ad(g)(X) = \frac{d}{d t} g exp(t X) g^{-1} |_{t = 0}$. If $g$ is not in the component of the unit element then there is $g'$ in that component such that $Ad(g') = Ad(g)$, so we can take $g$ in the unit component. This $g$ is not necessary in the image of the exponential map, but it is always a product of elements in the image; therefore the inner automorphisms are generated by elements which are in the image of the exponential map. In fact one can use the Hadamard formula

$e^X Y e^{-X} = e^{ad(X)} Y$

and consider generators of the form $e^{ad(X)}$ instead as being infintesimal automorphisms (by the definition); again if we need a product of several ones then one could use the Hausdorff formula to reduce to one, but that one has its own convergence limits. The expressions $e^{ad(X)}$ sometimes make sense in infinite dimensional situations, with other definitions of the exponential.

For example the following formal computation may make sense. Let $D$ be a real linear derivation and $\mathfrak{g}$ also over reals. Define the exponential of the operator on underlying (topological) vector space of $\mathfrak{g}$ by $exp(D) = \underset{N\to\infty}{lim} (1 +\frac{D}{N})^N$. Then

$exp(D)$ is an automorphism of $\mathfrak{g}$.

Sketch of the proof.

Let $t$ be a real parameter in some neighborhood of zero.

$\left[ \underset{N\to\infty}{lim} (1 +\frac{D}{N})^N X, \underset{K\to\infty}{lim}(1 +\frac{D}{K})^K Y\right]$

$= \underset{N\to\infty}{lim} \left[ (1 +\frac{D}{N})^N X, (1 +\frac{D}{N})^N Y\right]$

$= \underset{N\to\infty}{lim} \left[ (1 +\frac{D}{N})^N X, (1 +\frac{D}{N})^N Y\right]$

Now define $\tilde{X} = (1+\frac{D}{N})^{N-1} X$ and similarly $\tilde{Y}$. Then

$= \underset{N\to\infty}{lim} \left[ \tilde{X} + D(\tilde{X})/N, \tilde{Y}+D(\tilde{Y})/N\right]$

$= \underset{N\to\infty}{lim} ([\tilde{X},\tilde{Y}] + [D(\tilde{X}),\tilde{Y}]/N, \tilde{Y}+[\tilde{X},D(\tilde{Y})]/N + O(N^{-2}))$

Continue with $\tilde{\tilde{X}}$ etc. and up to $N$ terms of size $O(N^{-2})$, that is a sum of size $O(N^{-1})$ we get the same as from

$\underset{N\to\infty}{lim}(1+D/N)^N[X,Y]$

as the latter can be the same way transformed to

$\underset{N\to\infty}{lim}
(\widetilde{[X,Y]} + \frac{\widetilde{D([X,Y])}}{N})$

and so on and $D$ is derivation hence

$D([\tilde{X},\tilde{Y}])/N
= [D(\tilde{X}),\tilde{Y}]/N +[\tilde{X},D(\tilde{Y})]/N.$

End of proof.

Now we can take $D = ad(t X) = t ad(X)$ for $X\in\mathfrak{g}$ and $t$ n some neighborhood of zero in real numbers.

Consider the algebra of functions $Fun(Aut(\mathfrak{g}))$. It is a Hopf algebra and $Aut(g)$ is an affine algebraic group cut out in $GL(n)$ by the relations

$C^k_{i j} M^p_k = C^p_{l m} M^l_i M^m_j$

If we choose another basis we get another embedding of $Aut(\mathfrak{g})$ into $GL(n)$ as it is easy to check. It is instructive to check that the above relations determine a group (the relations for $M$ are satisfied for the products and inverses) or in Hopf algebra language that if $M=\mathcal{O}$ denotes the generic invertible matrix of taking entries functions, then the above relations determine a Hopf ideal in $Fun(GL(n))$.

Take now an element $X\in\mathfrak{g}$, and $f\in Fun(GL(n))$ or $f\in Fun(Aut(\mathfrak{g}))$. Then

$\frac{d}{d t} f(exp(t ad(X))|_{t = 0}$

determines a number.

Theorem. The above formula extends to a unique (degenerate) Hopf pairing between the enveloping algebra $U(\mathfrak{g})$ and the Hopf algebra of regular functions on the automorphism group $Fun(Aut(\mathfrak{g}))$.

This Hopf pairing may be used to define the structure on the enveloping algebra of a braided commutative monoid in the category of Yetter-Drinfeld modules over the Hopf algebra $Fun(Aut(\mathfrak{g}))$.

Created on August 16, 2014 at 06:33:00. See the history of this page for a list of all contributions to it.