Structure of the group of automorphisms of the spectral \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$2$$\end{document}-ball

We show that the group generated by triangular and diagonal conjugations is dense in \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathrm Aut (\Omega _2)$$\end{document} (in the compact-uniform topology). Moreover, it is shown that any automorphism of \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\Omega _2$$\end{document} is a local holomorphic conjugation (it extends the results from (Rostand, Studia Math 155:207–230, 2003, Thomas, Collect Math 59(3):321–324, 2008)).


Introduction and statement of the result
Let n denote the spectral ball in C n 2 , that is a domain composed of n × n complex matrices whose spectral radius is <1.
The natural question that arises is to classify its group of automorphisms. It may be easily checked that among them there are the following three forms: (i) Transposition: τ : x → x t , (ii) Möbius maps: m α,γ : x → γ (x − α)(1 − αx) −1 , where α lies in the unit disc and |γ | = 1, (iii) Conjugations: n is a conjugate invariant holomorphic map, i.e. u(q −1 xq) = u(x) for each x ∈ and q ∈ M −1 n (throughout the paper M −1 n denotes the group of invertible n × n complex matrices). Ransford and White have asked in [9] whether the compositions of the three above forms generate the whole Aut( )-the group of automorphisms of the spectral ball. In [6] we have shown that the answer to this question is negative. Nevertheless the question about the description of Aut( ) remains open. In this note we deal with this problem. Let us introduce some notation. M n denotes the algebra of n × n matrices with complex coefficients. Moreover, T n is the set of non-cyclic matrices lying in n . Recall that a matrix Ł. Kosiński  M is cyclic if there exists a cyclic vector v ∈ C n for M, i.e. v such that (v, Mv, . . . , M n−1 v) spans C n . It is well known that M is cyclic if and only if for any λ ∈ C dim Ker(M − λ) ≤ 1.
In particular, any matrix having n distinct eigenvalues is cyclic.
In the case n = 2 we shall simply write = 2 , M = M 2 and T = T 2 .
Definition 1 Let U be an open subset of the spectral ball n . We shall say that a mapping ϕ : U → M n is a holomorphic conjugation if there is a holomorphic mapping p : U → Moreover, we shall say that ϕ is a local holomorphic conjugation if any x ∈ U has a neighborhood V such that ϕ restricted to it is a holomorphic conjugation.
The paper is organized as follows. Recall that we presented in [6] two counterexamples to the question on the description of the group of automorphisms of the spectral ball. We focus on them in the sense that we classify all automorphisms of the spectral ball of the form is either diagonal or triangular, x ∈ (we shall call them diagonal and triangular conjugations respectively).
It is known that any automorphism of the spectral ball fixing the origin preserves the spectrum (see [9]). Therefore, trying to describe the group Aut( ) it is natural to investigate the behavior of automorphisms restricted to the fibers of , i.e. sets of the form F (λ 1 ,λ 2 ) := {x ∈ : σ (x) = {λ 1 , λ 2 }}, λ 1 , λ 2 ∈ D, where σ (x) denotes the spectrum of x ∈ M. If λ 1 = λ 2 , the fiber F (λ 1 ,λ 2 ) forms a submanifold known as the Danielewski surfaces. Recall that the Danielewski surface associated with a polynomial p ∈ C[z] is given by and its complex structure is naturally induced from C 3 . Algebraic properties of Danielewski surfaces have been intensively studied in the literature. Anyway little is known about their holomorphic automorphisms. It was lastly shown (see [5,7]) that the group generated by shears and overshears (for the definition see e.g. [8]) is dense in the group of holomorphic automorphisms. It was a little surprise to us that shears are just the restriction to the fiber of diagonal and triangular conjugations.
Following the idea from [8] (we recall all details for the convenience of the reader) we shall show that the spectral ball satisfies the property obtained by Andersén and Lempert for holomorphic automorphisms of C n (see [1,2,4]). To be more precise we shall show that the group generated by triangular and diagonal conjugations is dense in Aut( ) in the local-uniform topology (see Theorem 4).
Finally, in Lemma 6 we shall show that the uniform limit of conjugations is a local conjugation in a neighborhood of T . This, together with the density of the the group generated by triangular and diagonal conjugations and results obtained by Thomas and Rostand (see [10,11]) imply that any automorphism of the spectral ball is a local conjugation (see Theorem 7).
The approach presented in the proof seems to be new and we do believe that it is an important step toward solving the problem of finding the description of the group of automorphism of the spectral unit ball.

Diagonal and triangular conjugations
Throughout the paper G 2 denotes the symmetrized bidisc, i.e. a domain in C 2 given by the formula G 2 := {(tr x, det x) : x ∈ } (see [3] for its basic properties).
Let us focus on conjugations of the following form: and ( Our aim is to describe holomorphic functions a and b such that the above mappings are automorphisms of the spectral unit ball. The case (1) is easy. First note that the simply-connectedness of imply that there is a ∈ O( ) such that a = exp(ã/2).

Some easy computations givẽ
Therefore, for fixed x 11 and x 22 the mapping (x 12 , x 21 ) → (exp(ã(x))x 12 , exp(−ã(x))x 21 ) is injective on its domain. In particular, putting t = x 21 x 12 we see that the mapping z → exp(ã(x 11 , z, t/z, x 22 ))z is an automorphism of C * . Thus z →ã(x 11 , z, t/z, x 22 ) is constant, whenceã depends only on x 11 , x 22 and x 12 x 21 . Now we focus our attention on (2). We want to find b such that is an automorphism of . Put (s, p) := (tr x, det x) ∈ G 2 . Looking at the automorphism restricted to the fibers (it is obvious that conjugations preserve fibers) intersected with {x 12 = 0} we get that the mapping Applying this reasoning to b we find that there are c Putting Using this we see that c may be extended holomorphically through x 12 = 0. Moreover, one may easily check that c( Summarizing we get the following x ∈ , and α and β are as in Proposition 2.

Vector fields generated by triangular and diagonal conjugations and relations between them
To describe vector fields generated by (1) and (2) it is convenient to introduce the following notation: Note that the mappings D a , T β and S α generate the following vector fields , , .
Let us write the above vector fields in "spectral" coordinates Observe that for any vector field V on orthogonal to det x and tr x its divergence div V (in Euclidean coordinates) is equal to We get: A straightforward calculation leads to: Proof Let ϕ be an automorphism of . Composing it, if necessary, with a Möbius map one may assume that ϕ(0) = 0 (see [3,9]). Then ϕ (0) is also an automorphism of the spectral ball. By [9] there is an invertible matrix a such that either Losing no generality assume that the first possibility holds. Since any invertible matrix may be represented as a finite product of triangular and diagonal matrices ϕ (0) satisfies trivially the assertion.
Note that t → t := t −1 ϕ(t·) is a well defined mapping from R into Aut( ), 0 = ϕ (0). Since any automorphism of fixing the origin preserves the spectrum we see that σ ( t (x)) = σ (x), t ∈ R, x ∈ (σ (x) denotes the spectrum of a matrix x).
Consider the following time-dependent vector field: It is clear that t 0 • −1 0 is obtained by integrating X t from 0 to t 0 . Since t preserves the spectrum it follows that X t 0 (det x) = 0 and X t 0 (tr x) = 0 (in other words X t 0 is orthogonal to tr x and det x).
We proceed as in [2] and [8]. X t may be approximated on compact sets by integrating the time dependent vector field X k/N from time k/N to (k + 1)/N . Note that every X k/N may approximated by polynomial vector fields X such that X (det x) = 0 and X (tr x) = 0. Actually, write X k/N as Since X k/N is orthogonal to tr x we find that w 1 + w 4 = 0. Similarly, the orthogonality of det x means that w 1 (x 22 − x 11 ) = w 2 x 21 + w 3 x 12 . Now observe that the spectral ball is a pseudoconvex balanced domain, so any holomorphic function on it may be expanded into a series of homogenous polynomials (in particular the spectral ball is a Runge domain). Expanding coefficients of X k/N in a series of homogenous polynomials and looking at degrees of the polynomials appearing there one can easily get the desired claim. We shall show that X is a Lie combination of polynomial vector fields generated by diagonal and triangular conjugations and the transposition. Recall that a Lie combination of the subset S is an element which can be written as a finite sum of terms of the form [. .[[a 1 , a 2 ], a 3 ], . . . , a μ ], where a 1 , . . . , a μ ∈ S. Then the standard argument (sometimes called Euler's method) will imply that we can approximate 1 • −1 0 by compositions of the transposition and diagonal and triangular conjugations. Whence we will be able to approximate 1 , as well.
Therefore it remains to show that X is a sum of vector fields appearing in Sect. 3. Let us denote Recall that assumptions on X t 0 imply that (use the orthogonality of tr x and det x). Note that v 1 may be written as (note that they are of zero divergence) we may assume that v 1 = 0.
Thus, up to adding Lie combinations of the vector fields generated by triangular and diagonal conjugations we may assume that and X (det x) = 0 and div X = ψ(x 12 x 21 , x 11 , x 22 ). The second condition means that x 21 v 2 + x 12 v 3 = 0 so v 2 = x 12 w, v 3 = −x 21 w for some polynomial w. In particular, div X = x 12  w(x, s, p). In particular X is of the form H D a with a depending only on (x 12 x 21 , x 11 , x 22 ). This finishes the proof.

Limit of conjugations
Remark 5 Note that the holomorphic mapping p occurring in Definition 1 is defined up to a multiplication with x → a(x) The proof of this fact is quite elementary and we leave it to the reader. We would like to point out that the comutant of x coincides with the polynomials in x for a dense subset of matrices. Moreover, those polynomials can be taken of degree no more than 1 (recall that n = 2).
. Then there is a neighborhood U of T and diagonal mappings a n , b n ∈ O(U, M) such that p n (x)(a n (x) + xb n (x)) is locally uniformly convergent on U to u ∈ O(U, M −1 ) and det(a n (x) + xb n (x)) = 1.
The proof presented below is elementary and relies upon purely analytic methods. We do not know whether the lemma would follow from more general algebraic properties. Note that the main difficulty lies in the fact that standard methods do not work for non-cyclic matrices.
Proof To simplify the notation we will omit subscript n. Composing ϕ with Möbius maps from both sides (more precisely taking m −a • ϕ • m a , where m a = m a,1 ) we easily see that q(a, x) := p(m a (x))x p(m a (x)) −1 is convergent locally uniformly with respect to (a, x) in a neighborhood of D × T . Therefore ∂q ∂ x (a, x)(h) converges locally uniformly in an open neighborhood of D × T for any h. Putting x = 0 we find that whence p(a) is convergent locally uniformly with respect to a ∈ D. Therefore, replacing ϕ with we may assume that p(x) = 1 for all non-cyclic matrices x ∈ .
Put := x 11 0 x 21 x 11 : x 11 ∈ D, x 21 ∈ C . First we show that there are diagonal map- lying in a neighborhood of T . Multiplying p(x)x p(x) −1 out we get Using the fact that p(x)x p(x) −1 converges locally uniformly on W ∩ , we get that p 21 and p 22 converge uniformly on compact subsets of W ∩ . Since p 21 ≡ 0 on T we get that there is q ∈ O(W ∩ ) such that p 21 (x) = x 21 q(x). Moreover x 21 ∈ C such that p(x)(a (x) + xb (x)) is convergent locally uniformly and det(a (x) + xb (x)) = 1 for x ∈ in a small neighborhood of T . Let us consider the following projection j 1 : and note that V := j −1 1 (U ) is a neighborhood of T in . It follows from the previous step that is convergent locally uniformly on V and det u = 1 there. Therefore, replacing ϕ with Since p(x)x p(x) −1 = x on U we deduce from the formula above that p 12 = 0 and p 2 22 = 1 and thus p 11 = p 22 = 1 on U .
In particular, there is a holomorphic function q 12 on V such that p 12 , so it is equal to 1 when x ∈ T . Therefore there is a simply-connected neighborhood U of T in (uniform with respect to subscripts n), U ⊂ V such that det(ã(x) + xb(x)) does not vanish there. Let us take the branch of the square root s(x) := det(ã(x) + xb(x)) 1/2 preserving 1. Observe that a :=ã/s, b :=b/s satisfy the claim. Now we prove the existence of a and b satisfying the assertion of the lemma. Put Note that V := ∩ j −1 2 (U ) is a neighborhood of T in so we may repeat the previous reasoning: since v := p • j 2 · (a • j 2 + j 2 · b • j 2 ) is convergent on V and det v = 1 there, replacing ϕ with v −1 ϕv we may assume that ϕ(x) = x for x ∈ U . Comparing the coefficients of p(x)x p(x) −1 for x ∈ U :

Local form of the automorphisms of the spectral unit ball
It is well known (see e.g. [3]) that for any ϕ ∈ Aut( ) there is a Möbius map m such that σ (ϕ(x)) = σ (m(x)). Therefore, composing ϕ with a Möbius map, a transposition and a linear automorphism of we may always assume that ϕ is normalized, i.e. σ (ϕ(x)) = σ (x), x ∈ and ϕ (0) = id. As a consequence of our considerations we get the following Proof It follows from Theorem 4 that there is a sequence ( p n ) ⊂ O( , M −1 ) such that p n (x)x p n (x) −1 converges locally uniformly to ϕ(x). It follows from Lemma 6 that there is a neighborhood U of T and u ∈ O(U, On the other hand it is well known (see [11]) that ϕ is a local conjugation on \ T (we would like to note that this fact may also deduced from Theorem 4).
On the other hand, suppose that { αβ , a αβ , b α,β } are data for the above problem. Then, solving the second Cousin problem for matrices we get u α ∈ O( α , M −1 ) such that a αβ (x)+ xb αβ (x) = u α (x)u β (x) on α ∩ β . Putting ϕ(x) := u α (x)xu α (x) −1 we get a local holomorphic conjugation on U. If it were a holomorphic conjugation, we would get u ∈ O(U, M −1 ) such that ϕ(x) = u(x)xu(x) −1 . Making use of Remark 5 again we get a α , b β such that u α (x) = (a α (x) + xb α (x))u(x), x ∈ α . Then it is a direct to observe that a α , b α solve the above problem.
Finally, we present a list of open questions: Open problems • Does the similar results holds for n ≥ 3? We conjecture that the group of autmorphism generated by conjugations x → a(x)xa(x) −1 , where a(x) is an elementary matrix, is dense in Aut( n ).
• More generaly, one may conjecture that the higher-dimensional counterparts of the Danielewski surfaces, i.e. sets of the form {x ∈ M n×n (C) : σ (x) = {λ 1 , . . . , λ n }}, where λ i = λ j , i = j, satisfy the density property. • Is every automorphism of a holomorphic conjugation? • In the case n = 2 the affirmative answer to the question stated above would follow from the following one: Is a local holomorphic conjugation a holomorphic conjugation? • Do Danielewski surfaces have the Alexander property i.e. is every proper holomorphic selfmapping of D P an automorphism? Note that a similar result holds for (see [12]).
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