Zoran Skoda
automorphism of a Lie algebra

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

Let now 𝔤\mathfrak{g} be finite dimensional. Let e 1,,e ne_1,\ldots,e_n be a basis of 𝔤\mathfrak{g} and

[e i,e j]=C ij ke k [e_i,e_j] = C^k_{i j} e_k

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

[e i,e j]=[e l,e m]M i lM j m=C lm pM i lM j me p [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 ij ke k=C ij ke pM k p C^k_{i j} e'_k = C^k_{i j} e_p M^p_k

Therefore the condition that ψ\psi is an isomorphism is that its matrix MM is invertible and satisfies

C ij ke pM k p=C lm pM i lM j me p 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,pi,j,p,

C ij kM k p=C lm pM i lM j m 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 ψ\psi is an inner automorphism if it is of the form Ad(g)Ad(g) for some gGg\in G. Recall that Ad(g)(X)=ddtgexp(tX)g 1| t=0Ad(g)(X) = \frac{d}{d t} g exp(t X) g^{-1} |_{t = 0}. If gg is not in the component of the unit element then there is gg' in that component such that Ad(g)=Ad(g)Ad(g') = Ad(g), so we can take gg in the unit component. This gg 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 XYe X=e ad(X)Y e^X Y e^{-X} = e^{ad(X)} Y

and consider generators of the form e ad(X)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)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 DD be a real linear derivation and 𝔤\mathfrak{g} also over reals. Define the exponential of the operator on underlying (topological) vector space of 𝔤\mathfrak{g} by exp(D)=limN(1+DN) Nexp(D) = \underset{N\to\infty}{lim} (1 +\frac{D}{N})^N. Then
exp(D)exp(D) is an automorphism of 𝔤\mathfrak{g}.

Sketch of the proof.

Let tt be a real parameter in some neighborhood of zero.

[limN(1+DN) NX,limK(1+DK) KY] \left[ \underset{N\to\infty}{lim} (1 +\frac{D}{N})^N X, \underset{K\to\infty}{lim}(1 +\frac{D}{K})^K Y\right]
=limN[(1+DN) NX,(1+DN) NY] = \underset{N\to\infty}{lim} \left[ (1 +\frac{D}{N})^N X, (1 +\frac{D}{N})^N Y\right]
=limN[(1+DN) NX,(1+DN) NY] = \underset{N\to\infty}{lim} \left[ (1 +\frac{D}{N})^N X, (1 +\frac{D}{N})^N Y\right]

Now define X˜=(1+DN) N1X\tilde{X} = (1+\frac{D}{N})^{N-1} X and similarly Y˜\tilde{Y}. Then

=limN[X˜+D(X˜)/N,Y˜+D(Y˜)/N] = \underset{N\to\infty}{lim} \left[ \tilde{X} + D(\tilde{X})/N, \tilde{Y}+D(\tilde{Y})/N\right]
=limN([X˜,Y˜]+[D(X˜),Y˜]/N,Y˜+[X˜,D(Y˜)]/N+O(N 2)) = \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 X˜˜\tilde{\tilde{X}} etc. and up to NN terms of size O(N 2)O(N^{-2}), that is a sum of size O(N 1)O(N^{-1}) we get the same as from

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

as the latter can be the same way transformed to

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

and so on and DD is derivation hence

D([X˜,Y˜])/N=[D(X˜),Y˜]/N+[X˜,D(Y˜)]/N. 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(tX)=tad(X)D = ad(t X) = t ad(X) for X𝔤X\in\mathfrak{g} and tt n some neighborhood of zero in real numbers.

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

C ij kM k p=C lm pM i lM j m 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(𝔤)Aut(\mathfrak{g}) into GL(n)GL(n) as it is easy to check. It is instructive to check that the above relations determine a group (the relations for MM are satisfied for the products and inverses) or in Hopf algebra language that if M=𝒪M=\mathcal{O} denotes the generic invertible matrix of taking entries functions, then the above relations determine a Hopf ideal in Fun(GL(n))Fun(GL(n)).

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

ddtf(exp(tad(X))| t=0 \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(𝔤)U(\mathfrak{g}) and the Hopf algebra of regular functions on the automorphism group Fun(Aut(𝔤))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(𝔤))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.