Definition

Let E be a finite-dimensional real vector space with a symmetric bilinear product operation

${\displaystyle \displaystyle {E\times E\rightarrow E,\,\,\,a,b\mapsto ab=ba,}}$

with an identity element 1 such that a1 = a for a in A and a real inner product (a,b) for which the multiplication operators L(a) defined by L(a)b = ab on E are self-adjoint and satisfy the Jordan relation

${\displaystyle \displaystyle {L(a)L(a^{2})=L(a^{2})L(a).}}$

As will turn out below, the condition on adjoints can be replaced by the equivalent condition that the trace form Tr L(ab) defines an inner product. The trace form has the advantage of being manifestly invariant under automorphisms of the Jordan algebra, which is thus a closed subgroup of O(E) and thus a compact Lie group. In practical examples, however, it is often easier to produce an inner product for which the L(a) are self-adjoint than verify directly positive-definiteness of the trace form. (The equivalent original condition of Jordan, von Neumann and Wigner was that if a sum of squares of elements vanishes then each of those elements has to vanish.^[3])

Power associativity

From the Jordan condition it follows that the Jordan algebra is power associative, i.e. the Jordan subalgebra generated by any single element a in E is actually an associative commutative algebra. Thus, defining aⁿ inductively by aⁿ = a (aⁿ⁻¹), the following associativity relation holds:

${\displaystyle \displaystyle {a^{m}a^{n}=a^{m+n},}}$

so the subalgebra can be identified with R[a], polynomials in a. In fact polarizing of the Jordan relation—replacing a by a + tb and taking the coefficient of t—yields

${\displaystyle \displaystyle {2L(ab)L(a)+L(a^{2})L(b)=2L(a)L(b)L(a)+L(a^{2}b).}}$

This identity implies that L(a^m) is a polynomial in L(a) and L(a²) for all m. In fact, assuming the result for lower exponents than m,

${\displaystyle \displaystyle {a^{2}a^{m-1}=a^{m-1}(a^{2})=L(a^{m-1})L(a)a=L(a)L(a^{m-1})a=L(a)a^{m}=a^{m+1}.}}$

Setting b = a^{m – 1} in the polarized Jordan identity gives:

${\displaystyle \displaystyle {L(a^{m+1})=2L(a^{m})L(a)+L(a^{2})L(a^{m-1})-2L(a)^{2}L(a^{m-1}),}}$

a recurrence relation showing inductively that L(a^{m + 1}) is a polynomial in L(a) and L(a²).

Consequently, if power-associativity holds when the first exponent is ≤ m, then it also holds for m+1 since

${\displaystyle \displaystyle {L(a^{m+1})a^{n}=2L(a)L(a^{m})a^{n}+L(a^{2})L(a^{m-1})a^{n}-2L(a)^{2}L(a^{m-1})a^{n}=a^{m+n+1}.}}$

Idempotents and rank

An element e in E is called an idempotent if e² = e. Two idempotents are said to be orthogonal if ef = 0. This is equivalent to orthogonality with respect to the inner product, since (ef,ef) = (e,f). In this case g = e + f is also an idempotent. An idempotent g is called primitive or minimal if it cannot be written as a sum of non-zero orthogonal idempotents. If e₁, ..., e_m are pairwise orthogonal idempotents then their sum is also an idempotent and the algebra they generate consists of all linear combinations of the e_i. It is an associative algebra. If e is an idempotent, then 1 − e is an orthogonal idempotent. An orthogonal set of idempotents with sum 1 is said to be a complete set or a partition of 1. If each idempotent in the set is minimal it is called a Jordan frame. Since the number of elements in any orthogonal set of idempotents is bounded by dim E, Jordan frames exist. The maximal number of elements in a Jordan frame is called the rank r of E.

Spectral decomposition

The spectral theorem states that any element a can be uniquely written as

${\displaystyle \displaystyle {a=\sum \lambda _{i}e_{i},}}$

where the idempotents e_i's are a partition of 1 and the λ_i, the eigenvalues of a, are real and distinct. In fact let E₀ = R[a] and let T be the restriction of L(a) to E₀. T is self-adjoint and has 1 as a cyclic vector. So the commutant of T consists of polynomials in T (or a). By the spectral theorem for self-adjoint operators,

${\displaystyle \displaystyle {T=\sum \lambda _{i}P_{i}}}$

where the P_i are orthogonal projections on E₀ with sum I and the λ_i's are the distinct real eigenvalues of T. Since the P_i's commute with T and are self-adjoint, they are given by multiplication elements e_i of R[a] and thus form a partition of 1. Uniqueness follows because if f_i is a partition of 1 and a = Σ μ_i f_i, then with p(t)=Π (t - μ_j) and p_i = p/(t − μ_i), f_i = p_i(a)/p_i(μ_i). So the f_i's are polynomials in a and uniqueness follows from uniqueness of the spectral decomposition of T.

The spectral theorem implies that the rank is independent of the Jordan frame. For a Jordan frame with k minimal idempotents can be used to construct an element a with k distinct eigenvalues. As above the minimal polynomial p of a has degree k and R[a] has dimension k. Its dimension is also the largest k such that F_k(a) ≠ 0 where F_k(a) is the determinant of a Gram matrix:

${\displaystyle \displaystyle {F_{k}(a)=\det _{0\leq m,n<k}(a^{m},a^{n}).}}$

So the rank r is the largest integer k for which F_k is not identically zero on E. In this case, as a non-vanishing polynomial, F_r is non-zero on an open dense subset of E. the regular elements. Any other a is a limit of regular elements a⁽ⁿ⁾. Since the operator norm of L(x) gives an equivalent norm on E, a standard compactness argument shows that, passing to a subsequence if necessary, the spectral idempotents of the a⁽ⁿ⁾ and their corresponding eigenvalues are convergent. The limit of Jordan frames is a Jordan frame, since a limit of non-zero idempotents yields a non-zero idempotent by continuity of the operator norm. It follows that every Jordan frame is made up of r minimal idempotents.

If e and f are orthogonal idempotents, the spectral theorem shows that e and f are polynomials in a = e − f, so that L(e) and L(f) commute. This can be seen directly from the polarized Jordan identity which implies L(e)L(f) = 2 L(e)L(f)L(e). Commutativity follows by taking adjoints.

Spectral decomposition for an idempotent

If e is a non-zero idempotent then the eigenvalues of L(e) can only be 0, 1/2 and 1, since taking a = b = e in the polarized Jordan identity yields

${\displaystyle \displaystyle {2L(e)^{3}-3L(e)^{2}+L(e)=0.}}$

In particular the operator norm of L(e) is 1 and its trace is strictly positive.

There is a corresponding orthogonal eigenspace decomposition of E

${\displaystyle \displaystyle {E=E_{0}(e)\oplus E_{1/2}(e)\oplus E_{1}(e),}}$

where, for a in E, E_λ(a) denotes the λ-eigenspace of L(a). In this decomposition E₁(e) and E₀(e) are Jordan algebras with identity elements e and 1 − e. Their sum E₁(e) ⊕ E₀(e) is a direct sum of Jordan algebras in that any product between them is zero. It is the centralizer subalgebra of e and consists of all a such that L(a) commutes with L(e). The subspace E_1/2(e) is a module for the centralizer of e, the centralizer module, and the product of any two elements in it lies in the centralizer subalgebra. On the other hand, if

${\displaystyle \displaystyle {U=8L(e)^{2}-8L(e)+I,}}$

then U is self-adjoint equal to 1 on the centralizer algebra and −1 on the centralizer module. So U² = I and the properties above show that

${\displaystyle \displaystyle {\sigma (x)=Ux}}$

defines an involutive Jordan algebra automorphism σ of E.

In fact the Jordan algebra and module properties follow by replacing a and b in the polarized Jordan identity by e and a. If ea = 0, this gives L(e)L(a) = 2L(e)L(a)L(e). Taking adjoints it follows that L(a) commutes with L(e). Similarly if (1 − e)a = 0, L(a) commutes with I − L(e) and hence L(e). This implies the Jordan algebra and module properties. To check that a product of elements in the module lies in the algebra, it is enough to check this for squares: but if L(e)a = 1/2 a, then ea = 1/2 a, so L(a)² + L(a²)L(e) = 2L(a)L(e)L(a) + L(a²e). Taking adjoints it follows that L(a²) commutes with L(e), which implies the property for squares.

Trace form

The trace form is defined by

${\displaystyle \displaystyle {\tau (a,b)=\mathrm {Tr} \,L(ab).}}$

It is an inner product since, for non-zero a = Σ λ_i e_i,

${\displaystyle \displaystyle {\tau (a,a)=\sum \lambda _{i}^{2}\mathrm {Tr} \,L(e_{i})>0.}}$

The polarized Jordan identity can be polarized again by replacing a by a + tc and taking the coefficient of t. A further anyisymmetrization in a and c yields:

${\displaystyle \displaystyle {L(a(bc)-(ab)c)=[[L(a),L(b)],L(c)].}}$

Applying the trace to both sides

${\displaystyle \displaystyle {\tau (a,bc)=\tau (ba,c),}}$

so that L(b) is self-adjoint for the trace form.

Central decomposition

If E is a Euclidean Jordan algebra an ideal F in E is a linear subspace closed under multiplication by elements of E, i.e. F is invariant under the operators L(a) for a in E. If P is the orthogonal projection onto F it commutes with the operators L(a), In particular F^⊥ = (I − P)E is also an ideal and E = F ⊕ F^⊥. Furthermore, if e = P(1), then P = L(e). In fact for a in E

${\displaystyle \displaystyle {ea=ae=L(a)P(1)=P(L(a)1)=P(a),}}$

so that ea = a for a in F and 0 for a in F^⊥. In particular e and 1 − e are orthogonal idempotents with L(e) = P and L(1 − e) = I − P. e and 1 − e are the identities in the Euclidean Jordan algebras F and F^⊥. The idempotent e is central in E, where the center of E is defined to be the set of all z such that L(z) commutes with L(a) for all a. It forms a commutative associative subalgebra.

Continuing in this way E can be written as a direct sum of minimal ideals

${\displaystyle \displaystyle {E=\oplus E_{i}.}}$

If P_i is the projection onto E_i and e_i = P_i(1) then P_i = L(e_i). The e_i's are orthogonal with sum 1 and are the identities in E_i. Minimality forces E_i to be simple, i.e. to have no non-trivial ideals. For since L(e_i) commutes with all L(a)'s, any ideal F ⊂ E_i would be invariant under E since F = e_iF. Such a decomposition into a direct sum of simple Euclidean algebras is unique. If E = ⊕ F_j is another decomposition, then F_j=⊕ e_iF_j. By minimality only one of the terms here is non-zero so equals F_j. By minimality the corresponding E_i equals F_j, proving uniqueness.

In this way the classification of Euclidean Jordan algebras is reduced to that of simple ones. For a simple algebra E all inner products for which the operators L(a) are self adjoint are proportional. Indeed, any other product has the form (Ta, b) for some positive self-adjoint operator commuting with the L(a)'s. Any non-zero eigenspace of T is an ideal in A and therefore by simplicity T must act on the whole of E as a positive scalar.

List of all simple Euclidean Jordan algebras

• Let H_n(R) be the space of real symmetric n by n matrices with inner product (a,b) = Tr ab and Jordan product a ∘ b = 1/2(ab + ba). Then H_n(R) is a simple Euclidean Jordan algebra of rank n for n ≥ 3.
• Let H_n(C) be the space of complex self-adjoint n by n matrices with inner product (a,b) = Re Tr ab* and Jordan product a ∘ b = 1/2(ab + ba). Then H_n(C) is a simple Euclidean Jordan algebra of rank n for n ≥ 3.
• Let H_n(H) be the space of self-adjoint n by n matrices with entries in the quaternions, inner product (a,b) = Re Tr ab* and Jordan product a ∘ b = 1/2(ab + ba). Then H_n(H) is a simple Euclidean Jordan algebra of rank n for n ≥ 3.
• Let V be a finite dimensional real inner product space and set E = V ⊕ R with inner product (u⊕λ,v⊕μ) =(u,v) + λμ and product (u⊕λ)∘(v⊕μ)=( μu + λv) ⊕ [(u,v) + λμ]. This is a Euclidean Jordan algebra of rank 2, called a spin factor.
• The above examples in fact give all the simple Euclidean Jordan algebras, except for one exceptional case H₃(O), the self-adjoint matrices over the octonions or Cayley numbers, another rank 3 simple Euclidean Jordan algebra of dimension 27 (see below).

The Jordan algebras H₂(R), H₂(C), H₂(H) and H₂(O) are isomorphic to spin factors V ⊕ R where V has dimension 2, 3, 5 and 9, respectively: that is, one more than the dimension of the relevant division algebra.

Peirce decomposition

Let E be a simple Euclidean Jordan algebra with inner product given by the trace form τ(a)= Tr L(a). The proof that E has the above form rests on constructing an analogue of matrix units for a Jordan frame in E. The following properties of idempotents hold in E.

• An idempotent e is minimal in E if and only if E₁(e) has dimension one (so equals Re). Moreover E_1/2(e) ≠ (0). In fact the spectral projections of any element of E₁(e) lie in E so if non-zero must equal e. If the 1/2 eigenspace vanished then E₁(e) = Re would be an ideal.
• If e and f are non-orthogonal minimal idempotents, then there is a period 2 automorphism σ of E such that σe=f, so that e and f have the same trace.
• If e and f are orthogonal minimal idempotents then E_1/2(e) ∩ E_1/2(f) ≠ (0). Moreover, there is a period 2 automorphism σ of E such that σe=f, so that e and f have the same trace, and for any a in this intersection, a² = 1/2 τ(e) |a|² (e + f).
• All minimal idempotents in E are in the same orbit of the automorphism group so have the same trace τ₀.
• If e, f, g are three minimal orthogonal idempotents, then for a in E_1/2(e) ∩ E_1/2(f) and b in E_1/2(f) ∩ E_1/2(g), L(a)² b = 1/8 τ₀ |a|² b and |ab|² = 1/8 τ₀ |a|²|b|². Moreover, E_1/2(e) ∩ E_1/2(f) ∩ E_1/2(g) = (0).
• If e₁, ..., e_r and f₁, ..., f_r are Jordan frames in E, then there is an automorphism α such that αe_i = f_i.
• If (e_i) is a Jordan frame and E_ii = E₁(e_i) and E_ij = E_1/2(e_i) ∩ E_1/2(e_j), then E is the orthogonal direct sum the E_ii's and E_ij's. Since E is simple, the E_ii's are one-dimensional and the subspaces E_ij are all non-zero for i ≠ j.
• If a = Σ α_i e_i for some Jordan frame (e_i), then L(a) acts as α_i on E_ii and (α_i + α_i)/2 on E_ij.

Reduction to Euclidean Hurwitz algebras

Let E be a simple Euclidean Jordan algebra. From the properties of the Peirce decomposition it follows that:

• If E has rank 2, then it has the form V ⊕ R for some inner product space V with Jordan product as described above.
• If E has rank r > 2, then there is a non-associative unital algebra A, associative if r > 3, equipped with an inner product satisfying (ab,ab)= (a,a)(b,b) and such that E = H_r(A). (Conjugation in A is defined by a* = −a + 2(a,1)1.)

Such an algebra A is called a Euclidean Hurwitz algebra. In A if λ(a)b = ab and ρ(a)b = ba, then:

• the involution is an antiautomorphism, i.e. (a b)*=b* a*
• a a* = ‖ a ‖² 1 = a* a
• λ(a*) = λ(a)*, ρ(a*) = ρ(a)*, so that the involution on the algebra corresponds to taking adjoints
• Re(a b) = Re(b a) if Re x = (x + x*)/2 = (x, 1)1
• Re(a b) c = Re a(b c)
• λ(a²) = λ(a)², ρ(a²) = ρ(a)², so that A is an alternative algebra.

By Hurwitz's theorem A must be isomorphic to R, C, H or O. The first three are associative division algebras. The octonions do not form an associative algebra, so H_r(O) can only give a Jordan algebra for r = 3. Because A is associative when A = R, C or H, it is immediate that H_r(A) is a Jordan algebra for r ≥ 3. A separate argument, given originally by Albert (1934), is required to show that H₃(O) with Jordan product a∘b = 1/2(ab + ba) satisfies the Jordan identity [L(a),L(a²)] = 0. There is a later more direct proof using the Freudenthal diagonalization theorem due to Freudenthal (1951): he proved that given any matrix in the algebra H_r(A) there is an algebra automorphism carrying the matrix onto a diagonal matrix with real entries; it is then straightforward to check that [L(a),L(b)] = 0 for real diagonal matrices.^[4]

Exceptional and special Euclidean Jordan algebras

The exceptional Euclidean Jordan algebra E= H₃(O) is called the Albert algebra. The Cohn–Shirshov theorem implies that it cannot be generated by two elements (and the identity). This can be seen directly. For by Freudenthal's diagonalization theorem one element X can be taken to be a diagonal matrix with real entries and the other Y to be orthogonal to the Jordan subalgebra generated by X. If all the diagonal entries of X are distinct, the Jordan subalgebra generated by X and Y is generated by the diagonal matrices and three elements

${\displaystyle \displaystyle {Y_{1}={\begin{pmatrix}0&0&0\\0&0&y_{1}\\0&y_{1}^{*}&0\end{pmatrix}},\,\,\,Y_{2}={\begin{pmatrix}0&0&y_{2}^{*}\\0&0&0\\y_{2}&0&0\end{pmatrix}},\,\,\,Y_{3}={\begin{pmatrix}0&y_{3}&0\\y_{3}^{*}&0&0\\0&0&0\end{pmatrix}}.}}$

It is straightforward to verify that the real linear span of the diagonal matrices, these matrices and similar matrices with real entries form a unital Jordan subalgebra. If the diagonal entries of X are not distinct, X can be taken to be the primitive idempotent e₁ with diagonal entries 1, 0 and 0. The analysis in Springer & Veldkamp (2000) then shows that the unital Jordan subalgebra generated by X and Y is proper. Indeed, if 1 − e₁ is the sum of two primitive idempotents in the subalgebra, then, after applying an automorphism of E if necessary, the subalgebra will be generated by the diagonal matrices and a matrix orthogonal to the diagonal matrices. By the previous argument it will be proper. If 1 - e₁ is a primitive idempotent, the subalgebra must be proper, by the properties of the rank in E.

A Euclidean algebra is said to be special if its central decomposition contains no copies of the Albert algebra. Since the Albert algebra cannot be generated by two elements, it follows that a Euclidean Jordan algebra generated by two elements is special. This is the Shirshov–Cohn theorem for Euclidean Jordan algebras.^[5]

The classification shows that each non-exceptional simple Euclidean Jordan algebra is a subalgebra of some H_n(R). The same is therefore true of any special algebra.

On the other hand, as Albert (1934) showed, the Albert algebra H₃(O) cannot be realized as a subalgebra of H_n(R) for any n.^[6]

Indeed, let π is a real-linear map of E = H₃(O) into the self-adjoint operators on V = Rⁿ with π(ab) = 1/2(π(a)π(b) + π(b)π(a)) and π(1) = I. If e₁, e₂, e₃ are the diagonal minimal idempotents then P_i = π(e_i are mutually orthogonal projections on V onto orthogonal subspaces V_i. If i ≠ j, the elements e_ij of E with 1 in the (i,j) and (j,i) entries and 0 elsewhere satisfy e_ij² = e_i + e_j. Moreover, e_ije_jk = 1/2 e_ik if i, j and k are distinct. The operators T_ij are zero on V_k (k ≠ i, j) and restrict to involutions on V_i ⊕ V_j interchanging V_i and V_j. Letting P_ij = P_i T_ij P_j and setting P_ii = P_i, the (P_ij) form a system of matrix units on V, i.e. P_ij* = P_ji, Σ P_ii = I and P_ijP_km = δ_jk P_im. Let E_i and E_ij be the subspaces of the Peirce decomposition of E. For x in O, set π_ij = P_ij π(xe_ij), regarded as an operator on V_i. This does not depend on j and for x, y in O

${\displaystyle \displaystyle {\pi _{ij}(xy)=\pi _{ij}(x)\pi _{ij}(y),\,\,\,\pi _{ij}(1)=I.}}$

Since every x in O has a right inverse y with xy = 1, the map π_ij is injective. On the other hand, it is an algebra homomorphism from the nonassociative algebra O into the associative algebra End V_i, a contradiction.^[7]

Definition

When (e_i) is a partition of 1 in a Euclidean Jordan algebra E, the self-adjoint operators L(e_i) commute and there is a decomposition into simultaneous eigenspaces. If a = Σ λ_i e_i the eigenvalues of L(a) have the form Σ ε_i λ_i is 0, 1/2 or 1. The e_i themselves give the eigenvalues λ_i. In particular an element a has non-negative spectrum if and only if L(a) has non-negative spectrum. Moreover, a has positive spectrum if and only if L(a) has positive spectrum. For if a has positive spectrum, a - ε1 has non-negative spectrum for some ε > 0.

The positive cone C in E is defined to be the set of elements a such that a has positive spectrum. This condition is equivalent to the operator L(a) being a positive self-adjoint operator on E.

• C is a convex cone in E because positivity of a self-adjoint operator T— the property that its eigenvalues be strictly positive—is equivalent to (Tv,v) > 0 for all v ≠ 0.
• C is an open because the positive matrices are open in the self-adjoint matrices and L is a continuous map: in fact, if the lowest eigenvalue of T is ε > 0, then T + S is positive whenever ||S|| < ε.
• The closure of C consists of all a such that L(a) is non-negative or equivalently a has non-negative spectrum. From the elementary properties of convex cones, C is the interior of its closure and is a proper cone. The elements in the closure of C are precisely the square of elements in E.
• C is self-dual. In fact the elements of the closure of C are just set of all squares x² in E, the dual cone is given by all a such that (a,x²) > 0. On the other hand, (a,x²) = (L(a)x,x), so this is equivalent to the positivity of L(a).^[8]

Quadratic representation

To show that the positive cone C is homogeneous, i.e. has a transitive group of automorphisms, a generalization of the quadratic action of self-adjoint matrices on themselves given by X ↦ YXY has to be defined. If Y is invertible and self-adjoint, this map is invertible and carries positive operators onto positive operators.

For a in E, define an endomorphism of E, called the quadratic representation, by^[9]

${\displaystyle \displaystyle {Q(a)=2L(a)^{2}-L\left(a^{2}\right).}}$

Note that for self-adjoint matrices L(X)Y = 1/2(XY + YX), so that Q(X)Y = XYX.

An element a in E is called invertible if it is invertible in R[a]. If b denotes the inverse, then the spectral decomposition of a shows that L(a) and L(b) commute.

In fact a is invertible if and only if Q(a) is invertible. In that case

Indeed, if Q(a) is invertible it carries R[a] onto itself. On the other hand, Q(a)1 = a², so

${\displaystyle \displaystyle {\left(Q(a)^{-1}a\right)a=aQ(a)^{-1}a=L(a)Q(a)^{-1}a=Q(a)^{-1}a^{2}=1.}}$

Taking b = a⁻¹ in the polarized Jordan identity, yields

${\displaystyle \displaystyle {Q(a)L\left(a^{-1}\right)=L(a).}}$

Replacing a by its inverse, the relation follows if L(a) and L(a⁻¹) are invertible. If not it holds for a + ε1 with ε arbitrarily small and hence also in the limit.

These identities are easy to prove in a finite-dimensional (Euclidean) Jordan algebra (see below) or in a special Jordan algebra, i.e. the Jordan algebra defined by a unital associative algebra.^[10] They are valid in any Jordan algebra. This was conjectured by Jacobson and proved in Macdonald (1960): Macdonald showed that if a polynomial identity in three variables, linear in the third, is valid in any special Jordan algebra, then it holds in all Jordan algebras.^[11]

In fact for c in A and F(a) a function on A with values in End A, let D_cF(a) be the derivative at t = 0 of F(a + tc). Then

${\displaystyle \displaystyle {c=D_{c}\left(Q(a)a^{-1}\right)=2\left[(L(a)L(c)+L(c)L(a)-L(ac))a^{-1}\right]+Q(a)D_{c}\left(a^{-1}\right)=2c+Q(a)D_{c}\left(a^{-1}\right).}}$

The expression in square brackets simplifies to c because L(a) commutes with L(a⁻¹).

Thus

Applying D_c to L(a⁻¹)Q(a) = L(a) and acting on b = c⁻¹ yields

${\displaystyle \displaystyle {(Q(a)b)\left(Q\left(a^{-1}\right)b^{-1}\right)=1.}}$

On the other hand, L(Q(a)b) is invertible on an open dense set where Q(a)b must also be invertible with

${\displaystyle \displaystyle {(Q(a)b)^{-1}=Q\left(a^{-1}\right)b^{-1}.}}$

Taking the derivative D_c in the variable b in the expression above gives

${\displaystyle \displaystyle {-Q(Q(a)b)^{-1}Q(a)c=-Q(a)^{-1}Q(b)^{-1}c.}}$

This yields the fundamental identity for a dense set of invertible elements, so it follows in general by continuity. The fundamental identity implies that c = Q(a)b is invertible if a and b are invertible and gives a formula for the inverse of Q(c). Applying it to c gives the inverse identity in full generality.

Finally it can be verified immediately from the definitions that, if u = 1 − 2e for some idempotent e, then Q(u) is the period 2 automorphism constructed above for the centralizer algebra and module of e.

Homogeneity of positive cone

The proof of this relies on elementary continuity properties of eigenvalues of self-adjoint operators.^[12]

Let T(t) (α ≤ t ≤ β) be a continuous family of self-adjoint operators on E with T(α) positive and T(β) having a negative eigenvalue. Set S(t)= –T(t) + M with M > 0 chosen so large that S(t) is positive for all t. The operator norm ||S(t)|| is continuous. It is less than M for t = α and greater than M for t = β. So for some α < s < β, ||S(s)|| = M and there is a vector v ≠ 0 such that S(s)v = Mv. In particular T(s)v = 0, so that T(s) is not invertible.

Suppose that x = Q(a)b does not lie in C. Let b(t) = (1 − t) + tb with 0 ≤ t ≤ 1. By convexity b(t) lies in C. Let x(t) = Q(a)b(t) and X(t) = L(x(t)). If X(t) is invertible for all t with 0 ≤ t ≤ 1, the eigenvalue argument gives a contradiction since it is positive at t = 0 and has negative eigenvalues at t = 1. So X(s) has a zero eigenvalue for some s with 0 < s ≤ 1: X(s)w = 0 with w ≠ 0. By the properties of the quadratic representation, x(t) is invertible for all t. Let Y(t) = L(x(t)²). This is a positive operator since x(t)² lies in C. Let T(t) = Q(x(t)), an invertible self-adjoint operator by the invertibility of x(t). On the other hand, T(t) = 2X(t)² - Y(t). So (T(s)w,w) < 0 since Y(s) is positive and X(s)w = 0. In particular T(s) has some negative eigenvalues. On the other hand, the operator T(0) = Q(a²) = Q(a)² is positive. By the eigenvalue argument, T(t) has eigenvalue 0 for some t with 0 < t < s, a contradiction.

It follows that the linear operators Q(a) with a invertible, and their inverses, take the cone C onto itself. Indeed, the inverse of Q(a) is just Q(a⁻¹). Since Q(a)1 = a², there is thus a transitive group of symmetries:

Construction

Let C be a symmetric cone in the Euclidean space E. As above, Aut C denotes the closed subgroup of GL(E) taking C (or equivalently its closure) onto itself. Let G = Aut₀ C be its identity component. K = G ∩ O(E). It is a maximal compact subgroup of G and the stabilizer of a point e in C. It is connected. The group G is invariant under taking adjoints. Let σg =(g*)⁻¹, period 2 automorphism. Thus K is the fixed point subgroup of σ. Let ${\displaystyle {\mathfrak {g}}}$ be the Lie algebra of G. Thus σ induces an involution of ${\displaystyle {\mathfrak {g}}}$ and hence a ±1 eigenspace decomposition

${\displaystyle \displaystyle {{\mathfrak {g}}={\mathfrak {k}}\oplus {\mathfrak {p}},}}$

where ${\displaystyle {\mathfrak {k}}}$, the +1 eigenspace, is the Lie algebra of K and ${\displaystyle {\mathfrak {p}}}$ is the −1 eigenspace. Thus ${\displaystyle {\mathfrak {p}}}$⋅e is an affine subspace of dimension dim ${\displaystyle {\mathfrak {p}}}$. Since C = G/K is an open subspace of E, it follows that dim E = dim ${\displaystyle {\mathfrak {p}}}$ and hence ${\displaystyle {\mathfrak {p}}}$⋅e = E. For a in E let L(a) be the unique element of ${\displaystyle {\mathfrak {p}}}$ such that L(a)e = a. Define a ∘ b = L(a)b. Then E with its Euclidean structure and this bilinear product is a Euclidean Jordan algebra with identity 1 = e. The convex cone coincides C with the positive cone of E.^[13]

Since the elements of ${\displaystyle {\mathfrak {p}}}$ are self-adjoint, L(a)* = L(a). The product is commutative since [${\displaystyle {\mathfrak {p}}}$, ${\displaystyle {\mathfrak {p}}}$] ⊆ ${\displaystyle {\mathfrak {k}}}$ annihilates e, so that ab = L(a)L(b)e = L(b)L(a)e = ba. It remains to check the Jordan identity [L(a),L(a²)] = 0.

The associator is given by [a,b,c] = [L(a),L(c)]b. Since [L(a),L(c)] lies in ${\displaystyle {\mathfrak {k}}}$ it follows that [[L(a),L(c)],L(b)] = L([a,b,c]). Making both sides act on c yields

${\displaystyle \displaystyle {[a,b^{2},c]=2[a,b,c]b.}}$

On the other hand,

${\displaystyle \displaystyle {([b^{2},a,b],c)=(b^{2}(ba)-b(b^{2}a),c)=-(b^{2},[a,b,c])}}$

and likewise

${\displaystyle \displaystyle {([b^{2},a,b],c)=(b,[a,b^{2},c]).}}$

Combining these expressions gives

${\displaystyle \displaystyle {([b^{2},a,b],c)=0,}}$

which implies the Jordan identity.

Finally the positive cone of E coincides with C. This depends on the fact that in any Euclidean Jordan algebra E

${\displaystyle \displaystyle {Q(e^{a})=e^{2L(a)}.}}$

In fact Q(e^a) is a positive operator, Q(e^ta) is a one-parameter group of positive operators: this follows by continuity for rational t, where it is a consequence of the behaviour of powers So it has the form exp tX for some self-adjoint operator X. Taking the derivative at 0 gives X = 2L(a).

Hence the positive cone is given by all elements

${\displaystyle \displaystyle {e^{2a}=Q(e^{a})1=e^{2L(a)}1=e^{X}\cdot 1,}}$

with X in ${\displaystyle {\mathfrak {p}}}$. Thus the positive cone of E lies inside C. Since both are self-dual, they must coincide.

Automorphism groups and trace form

Let C be the positive cone in a simple Euclidean Jordan algebra E. Aut C is the closed subgroup of GL(E) taking C (or its closure) onto itself. Let G = Aut₀ C be the identity component of Aut C and let K be the closed subgroup of G fixing 1. From the group theoretic properties of cones, K is a connected compact subgroup of G and equals the identity component of the compact Lie group Aut E. Let ${\displaystyle {\mathfrak {g}}}$ and ${\displaystyle {\mathfrak {k}}}$ be the Lie algebras of G and K. G is closed under taking adjoints and K is the fixed point subgroup of the period 2 automorphism σ(g) = (g*)⁻¹. Thus K = G ∩ SO(E). Let ${\displaystyle {\mathfrak {p}}}$ be the −1 eigenspace of σ.

• ${\displaystyle {\mathfrak {k}}}$ consists of derivations of E that are skew-adjoint for the inner product defined by the trace form.
• [[L(a),L(c)],L(b)] = L([a,b,c]).
• If a and b are in E, then D = [L(a),L(b)] is a derivation of E, so lies in ${\displaystyle {\mathfrak {k}}}$. These derivations span ${\displaystyle {\mathfrak {k}}}$.
• If a is in C, then Q(a) lies in G.
• C is the connected component of the open set of invertible elements of E containing 1. It consists of exponentials of elements of E and the exponential map gives a diffeomorphism of E onto C.
• The map a ↦ L(a) gives an isomorphism of E onto ${\displaystyle {\mathfrak {p}}}$ and e^L(a) = Q(e^a/2). This space of such exponentials coincides with P the positive self-adjoint elements in G.
• For g in G and a in E, Q(g(a)) = g Q(a) g*.

Cartan decomposition

• G = P ⋅ K = K ⋅ P and the decomposition g = pk corresponds to the polar decomposition in GL(E).
• If (e_i) is a Jordan frame in E, then the subspace ${\displaystyle {\mathfrak {a}}}$ of ${\displaystyle {\mathfrak {p}}}$ spanned by L(e_i) is maximal Abelian in ${\displaystyle {\mathfrak {p}}}$. A = exp ${\displaystyle {\mathfrak {a}}}$ is the Abelian subgroup of operators Q(a) where a = Σ λ_i e_i with λ_i > 0. A is closed in P and hence G. If b =Σ μ_i e_i with μ_i > 0, then Q(ab)=Q(a)Q(b).
• ${\displaystyle {\mathfrak {p}}}$ and P are the union of the K translates of ${\displaystyle {\mathfrak {a}}}$ and A.

Iwasawa decomposition for cone

If E has Peirce decomposition relative to the Jordan frame (e_i)

${\displaystyle \displaystyle {E=\bigoplus _{i\leq j}E_{ij},}}$

then ${\displaystyle {\mathfrak {a}}}$ is diagonalized by this decomposition with L(a) acting as (α_i + α_j)/2 on E_ij, where a = Σ α_i e_i.

Define the closed subgroup S of G by

${\displaystyle \displaystyle {S=\{g\in G|gE_{ij}\subseteq \bigoplus _{(p,q)\geq (i,j)}E_{pq}\},}}$

where the ordering on pairs p ≤ q is lexicographic. S contains the group A, since it acts as scalars on E_ij. If N is the closed subgroup of S such that nx = x modulo ⊕_{(p,q) > (i,j)} E_pq, then S = AN = NA, a semidirect product with A normalizing N. Moreover, G has the following Iwasawa decomposition:

${\displaystyle \displaystyle {G=KAN.}}$

For i ≠ j let

${\displaystyle \displaystyle {{\mathfrak {g}}_{ij}=\{X\in {\mathfrak {g}}:[L(a),X]={1 \over 2}(\alpha _{i}-\alpha _{j})X,\,\,\,\mathrm {for} \,\,\,a=\sum \alpha _{i}e_{i}\}.}}$

Then the Lie algebra of N is

${\displaystyle \displaystyle {{\mathfrak {n}}=\bigoplus _{i<j}{\mathfrak {g}}_{ij},\,\,\,\,{\mathfrak {g}}_{ij}=\{L(a)+2[L(a),L(e_{i})]:a\in E_{ij}\}.}}$

Taking ordered orthonormal bases of the E_ij gives a basis of E, using the lexicographic order on pairs (i,j). The group N is lower unitriangular and its Lie algebra lower triangular. In particular the exponential map is a polynomial mapping of ${\displaystyle {\mathfrak {n}}}$ onto N, with polynomial inverse given by the logarithm.

Definition of complexification

Let E be a Euclidean Jordan algebra. The complexification E_C = E ⊕ iE has a natural conjugation operation (a + ib)* = a − ib and a natural complex inner product and norm. The Jordan product on E extends bilinearly to E_C, so that (a + ib)(c + id) = (ac − bd) + i(ad + bc). If multiplication is defined by L(a)b = ab then the Jordan axiom

${\displaystyle \displaystyle {[L(a),L(a^{2})]=0}}$

still holds by analytic continuation. Indeed, the identity above holds when a is replaced by a + tb for t real; and since the left side is then a polynomial with values in End E_C vanishing for real t, it vanishes also t complex. Analytic continuation also shows that all for the formulas involving power-associativity for a single element a in E, including recursion formulas for L(a^m), also hold in E_C. Since for b in E, L(b) is still self-adjoint on E_C, the adjoint relation L(a*) = L(a)* holds for a in E_C. Similarly the symmetric bilinear form β(a,b) = (a,b*) satisfies β(ab,c) = β(b,ac). If the inner product comes from the trace form, then β(a,b) = Tr L(ab).

For a in E_C, the quadratic representation is defined as before by Q(a)=2L(a)² − L(a²). By analytic continuation the fundamental identity still holds:

${\displaystyle \displaystyle {Q(Q(a)b)=Q(a)Q(b)Q(a),\,\,\,Q(a^{m})=Q(a)^{m}\,\,(m\geq 0).}}$

An element a in E is called invertible if it is invertible in C[a]. Power associativity shows that L(a) and L(a⁻¹) commute. Moreover, a⁻¹ is invertible with inverse a.

As in E, a is invertible if and only if Q(a) is invertible. In that case

${\displaystyle \displaystyle {Q(a)^{-1}a=a^{-1},\,\,\,Q(a^{-1})=Q(a)^{-1}.}}$

Indeed, as for E, if Q(a) is invertible it carries C[a] onto itself, while Q(a)1 = a², so

${\displaystyle \displaystyle {(Q(a)^{-1}a)a=aQ(a)^{-1}a=L(a)Q(a)^{-1}a=Q(a)^{-1}a^{2}=1,}}$

so a is invertible. Conversely if a is invertible, taking b = a⁻² in the fundamental identity shows that Q(a) is invertible. Replacing a by a⁻¹ and b by a then shows that its inverse is Q(a⁻¹). Finally if a and b are invertible then so is c = Q(a)b and it satisfies the inverse identity:

${\displaystyle \displaystyle {(Q(a)b)^{-1}=Q(a^{-1})b^{-1}.}}$

Invertibility of c follows from the fundamental formula which gives Q(c) = Q(a)Q(b)Q(a). Hence

${\displaystyle \displaystyle {c^{-1}=Q(c)^{-1}c=Q(a)^{-1}Q(b)^{-1}b=Q(a)^{-1}b^{-1}.}}$

The formula

${\displaystyle \displaystyle {Q(e^{a})=e^{2L(a)}}}$

also follows by analytic continuation.

Complexification of automorphism group

Aut E_C is the complexification of the compact Lie group Aut E in GL(E_C). This follows because the Lie algebras of Aut E_C and Aut E consist of derivations of the complex and real Jordan algebras E_C and E. Under the isomorphism identifying End E_C with the complexification of End E, the complex derivations is identified with the complexification of the real derivations.^[14]

Structure groups

The Jordan operator L(a) are symmetric with respect to the trace form, so that L(a)^t = L(a) for a in E_C. The automorphism groups of E and E_C consist of invertible real and complex linear operators g such that L(ga) = gL(a)g⁻¹ and g1 = 1. Aut E_C is the complexification of Aut E. Since an automorphism g preserves the trace form, g⁻¹ = g^t.

The structure groups of E and E_C consist of invertible real and complex linear operators g such that

${\displaystyle \displaystyle {Q(ga)=gQ(a)g^{t}.}}$

They form groups Γ(E) and Γ(E_C) with Γ(E) ⊂ Γ(E_C).

• The structure group is closed under taking transposes g ↦ g^t and adjoints g ↦ g*.
• The structure group contains the automorphism group. The automorphism group can be identified with the stabilizer of 1 in the structure group.
• If a is invertible, Q(a) lies in the structure group.
• If g is in the structure group and a is invertible, ga is also invertible with (ga)⁻¹ = (g^t)⁻¹a⁻¹.
• If E is simple, Γ(E) = Aut C × {±1}, Γ(E) ∩ O(E) = Aut E × {±1} and the identity component of Γ(E) acts transitively on C.
• Γ(E_C) is the complexification of Γ(E), which has Lie algebra ${\displaystyle {\mathfrak {k}}\oplus {\mathfrak {p}}}$.
• The structure group Γ(E_C) acts transitively on the set of invertible elements in E_C.
• Every g in Γ(E_C) has the form g = h Q(a) with h an automorphism and a invertible.

The unitary structure group Γ_u(E_C) is the subgroup of Γ(E_C) consisting of unitary operators, so that Γ_u(E_C) = Γ(E_C) ∩ U(E_C).

• The stabilizer of 1 in Γ_u(E_C) is Aut E.
• Every g in Γ_u(E_C) has the form g = h Q(u) with h in Aut E and u invertible in E_C with u* = u⁻¹.
• Γ(E_C) is the complexification of Γ_u(E_C), which has Lie algebra ${\displaystyle {\mathfrak {k}}\oplus i{\mathfrak {p}}}$.
• The set S of invertible elements u such that u* = u⁻¹ can be characterized equivalently either as those u for which L(u) is a normal operator with uu* = 1 or as those u of the form exp ia for some a in E. In particular S is connected.
• The identity component of Γ_u(E_C) acts transitively on S
• g in GL(E_C) is in the unitary structure group if and only if gS = S
• Given a Jordan frame (e_i) and v in E_C, there is an operator u in the identity component of Γ_u(E_C) such that uv = Σ α_i e_i with α_i ≥ 0. If v is invertible, then α_i > 0.

Given a frame (e_i) in a Euclidean Jordan algebra E, the restricted Weyl group can be identified with the group of operators on ⊕ R e_i arising from elements in the identity component of Γ_u(E_C) that leave ⊕ R e_i invariant.

Spectral norm

Let E be a Euclidean Jordan algebra with the inner product given by the trace form. Let (e_i) be a fixed Jordan frame in E. For given a in E_C choose u in Γ_u(E_C) such that ua = Σ α_i e_i with α_i ≥ 0. Then the spectral norm ||a|| = max α_i is independent of all choices. It is a norm on E_C with

${\displaystyle \displaystyle {\|a^{*}\|=\|a\|,\,\,\,\|\{a,a^{*},a\}\|=\|a\|^{3}.}}$

In addition ||a||² is given by the operator norm of Q(a) on the inner product space E_C. The fundamental identity for the quadratic representation implies that ||Q(a)b|| ≤ ||a||²||b||. The spectral norm of an element a is defined in terms of C[a] so depends only on a and not the particular Euclidean Jordan algebra in which it is calculated.^[15]

The compact set S is the set of extreme points of the closed unit ball ||x|| ≤ 1. Each u in S has norm one. Moreover, if u = e^ia and v = e^ib, then ||uv|| ≤ 1. Indeed, by the Cohn–Shirshov theorem the unital Jordan subalgebra of E generated by a and b is special. The inequality is easy to establish in non-exceptional simple Euclidean Jordan algebras, since each such Jordan algebra and its complexification can be realized as a subalgebra of some H_n(R) and its complexification H_n(C) ⊂ M_n(C). The spectral norm in H_n(C) is the usual operator norm. In that case, for unitary matrices U and V in M_n(C), clearly ||1/2(UV + VU)|| ≤ 1. The inequality therefore follows in any special Euclidean Jordan algebra and hence in general.^[16]

On the other hand, by the Krein–Milman theorem, the closed unit ball is the (closed) convex span of S.^[17] It follows that ||L(u)|| = 1, in the operator norm corresponding to either the inner product norm or spectral norm. Hence ||L(a)|| ≤ ||a|| for all a, so that the spectral norm satisfies

${\displaystyle \displaystyle {\|ab\|\leq \|a\|\cdot \|b\|.}}$

It follows that E_C is a Jordan C* algebra.^[18]

Complex simple Jordan algebras

The complexification of a simple Euclidean Jordan algebra is a simple complex Jordan algebra which is also separable, i.e. its trace form is non-degenerate. Conversely, using the existence of a real form of the Lie algebra of the structure group, it can be shown that every complex separable simple Jordan algebra is the complexification of a simple Euclidean Jordan algebra.^[19]

To verify that the complexification of a simple Euclidean Jordan algebra E has no ideals, note that if F is an ideal in E^C then so too is F^⊥, the orthogonal complement for the trace norm. As in the real case, J = F^⊥ ∩ F must equal (0). For the associativity property of the trace form shows that F^⊥ is an ideal and that ab = 0 if a and b lie in J. Hence J is an ideal. But if z is in J, L(z) takes E_C into J and J into (0). Hence Tr L(z) = 0. Since J is an ideal and the trace form degenerate, this forces z = 0. It follows that E_C = F ⊕ F^⊥. If P is the corresponding projection onto F, it commutes with the operators L(a) and F^⊥ = (I − P)E_C. is also an ideal and E = F ⊕ F^⊥. Furthermore, if e = P(1), then P = L(e). In fact for a in E

${\displaystyle \displaystyle {ea=ae=L(a)P(1)=P(L(a)1)=P(a),}}$

so that ea = a for a in F and 0 for a in F^⊥. In particular e and 1 − e are orthogonal central idempotents with L(e) = P and L(1 − e) = I − P.

So simplicity follows from the fact that the center of E_C is the complexification of the center of E.

Definitions

Let E be a Euclidean Jordan algebra with complexification A = E_C = E + iE.

The unit ball or disk D in A is just the convex bounded open set of elements a such the ||a|| < 1, i.e. the unit ball for the spectral norm.

The tube domain T in A is the unbounded convex open set T = E + iC, where C is the open positive cone in E.

Möbius transformations

The group SL(2,C) acts by Möbius transformations on the Riemann sphere C ∪ {∞}, the one-point compactification of C. If g in SL(2,C) is given by the matrix

${\displaystyle \displaystyle {g={\begin{pmatrix}\alpha &\beta \\\gamma &\delta \end{pmatrix}},}}$

then

${\displaystyle \displaystyle {g(z)=(\alpha z+\beta )(\gamma z+\delta )^{-1}.}}$

Similarly the group SL(2,R) acts by Möbius transformations on the circle R ∪ {∞}, the one-point compactification of R.

Let k = R or C. Then SL(2,k) is generated by the three subgroups of lower and upper unitriangular matrices, L and U', and the diagonal matrices D. It is also generated by the lower (or upper) unitriangular matrices, the diagonal matrices and the matrix

${\displaystyle \displaystyle {J={\begin{pmatrix}0&1\\-1&0\end{pmatrix}}.}}$

The matrix J corresponds to the Möbius transformation j(z) = −z⁻¹ and can be written

${\displaystyle \displaystyle {J={\begin{pmatrix}1&0\\-1&1\end{pmatrix}}{\begin{pmatrix}1&1\\0&1\end{pmatrix}}{\begin{pmatrix}1&0\\-1&1\end{pmatrix}}.}}$

The Möbius transformations fixing ∞ are just the upper triangular matrices B = UD = DU. If g does not fix ∞, it sends ∞ to a finite point a. But then g can be composed with an upper unitriangular matrix to send a to 0 and then with J to send 0 to infinity. This argument gives one of the simplest examples of the Bruhat decomposition:

${\displaystyle \displaystyle {\mathbf {SL} (2,k)=\mathbf {B} \cup \mathbf {B} \cdot J\cdot \mathbf {B} ,}}$

the double coset decomposition of SL(2,k). In fact the union is disjoint and can be written more precisely as

${\displaystyle \displaystyle {\mathbf {SL} (2,k)=\mathbf {B} \cup \mathbf {B} \cdot J\cdot \mathbf {U} ,}}$

where the product occurring in the second term is direct.

Now let

${\displaystyle \displaystyle {T(\beta )={\begin{pmatrix}1&\beta \\0&1\end{pmatrix}}.}}$

Then

${\displaystyle \displaystyle {{\begin{pmatrix}\alpha &0\\0&\alpha ^{-1}\end{pmatrix}}=JT(\alpha ^{-1})JT(\alpha )JT(\alpha ^{-1}).}}$

It follows SL(2,k) is generated by the group of operators T(β) and J subject to the following relations:

• β ↦ T(β) is an additive homomorphism
• α ↦ D(α) = JT(α⁻¹)JT(α)JT(α⁻¹) is a multiplicative homomorphism
• D(−1) = J
• D(α)T(β)D(α)⁻¹ = T(α²β)
• JD(α)J⁻¹ = D(α)⁻¹

The last relation follows from the definition of D(α). The generator and relations above is fact gives a presentation of SL(2,k). Indeed, consider the free group Φ generated by J and T(β) with J of order 4 and its square central. This consists of all products T(β₁)JT(β₂)JT(β₃)J ... T(β_m)J for m ≥ 0. There is a natural homomorphism of Φ onto SL(2,k). Its kernel contain the normal subgroup Δ generated by the relations above. So there is a natural homomorphism of Φ/Δ onto SL(2,k). To show that it is injective it suffices to show that the Bruhat decomposition also holds in Φ/Δ. It is enough to prove the first version, since the more precise version follows from the commutation relations between J and D(α). The set B ∪ B J B is invariant under inversion, contains operators T(β) and J, so it is enough to show it is invariant under multiplication. By construction it is invariant under multiplication by B. It is invariant under multiplication by J because of the defining equation for D(α).^[22]

In particular the center of SL(2,k) consists of the scalar matrices ±I and it is the only non-trivial normal subgroup of SL(2,k), so that PSL(2,k) = SL(2,k)/{±I} is simple.^[23] In fact if K is a normal subgroup, then the Bruhat decomposition implies that B is a maximal subgroup, so that either K is contained in B or KB = SL(2,k). In the first case K fixes one point and hence every point of k ∪ {∞}, so lies in the center. In the second case, the commutator subgroup of SL(2,k) is the whole group, since it is the group generated by lower and upper unitriangular matrices and the fourth relation shows that all such matrices are commutators since [T(β),D(α)] = T(β − α²β). Writing J = kb with k in K and b in B, it follows that L = k U k⁻¹. Since U and L generate the whole group, SL(2,k) = KU. But then SL(2,k)/K ≅ U/U ∩ K. The right hand side here is Abelian while the left hand side is its own commutator subgroup. Hence this must be the trivial group and K = SL(2,k).

Given an element a in the complex Jordan algebra A = E_C, the unital Jordan subalgebra C[a] is associative and commutative. Multiplication by a defines an operator on C[a] which has a spectrum, namely its set of complex eigenvalues. If p(t) is a complex polynomial, then p(a) is defined in C[a]. It is invertible in A if and only if it is invertible in C[a], which happen precisely when p does not vanish on the spectrum of a. This permits rational functions of a to be defined whenever the function is defined on the spectrum of a. If F and G are rational functions with G and F∘G defined on a, then F is defined on G(a) and F(G(a)) = (F∘G)(a). This applies in particular to complex Möbius transformations which can be defined by g(a) = (αa + β1)(γa + δ1)⁻¹. They leave C[a] invariant and, when defined, the group composition law holds. (In the next section complex Möbius transformations will be defined on the compactification of A.)^[24]

Given a primitive idempotent e in E with Peirce decomposition

${\displaystyle \displaystyle {E=E_{1}(e)\oplus E_{1/2}(e)\oplus E_{0}(e),\,\,\,\,A=A_{1}(e)\oplus A_{1/2}(e)\oplus A_{0}(e).}}$

the action of SL(2,C) by Möbius transformations on E₁(e) = C e can be extended to an action on A so that the action leaves invariant the components A_i(e) and in particular acts trivially on E₀(e).^[25] If P₀ is the projection onto A₀(e), the action is given be the formula

${\displaystyle \displaystyle {g(ze\oplus x_{1/2}\oplus x_{0})={\alpha z+\beta \over \gamma z+\delta }\cdot e\oplus (\gamma z+\delta )^{-1}x_{1/2}\oplus x_{0}-(\gamma z+\delta )^{-1}P_{0}(x_{1/2}^{2}).}}$

For a Jordan frame of primitive idempotents e₁, ..., e_m, the actions of SL(2,C) associated with different e_i commute, thus giving an action of SL(2,C)^m. The diagonal copy of SL(2,C) gives again the action by Möbius transformations on A.

Cayley transform

The Möbius transformation defined by

${\displaystyle \displaystyle {C(z)=i{1+z \over 1-z}=-i+{2i \over 1-z}}}$

is called the Cayley transform. Its inverse is given by

${\displaystyle \displaystyle {P(w)={w-i \over w+i}=1-{2i \over w+i}.}}$

The inverse Cayley transform carries the real line onto the circle with the point 1 omitted. It carries the upper halfplane onto the unit disk and the lower halfplane onto the complement of the closed unit disk. In operator theory the mapping T ↦ P(T) takes self-adjoint operators T onto unitary operators U not containing 1 in their spectrum. For matrices this follows because unitary and self-adjoint matrices can be diagonalized and their eigenvalues lie on the unit circle or real line. In this finite-dimensional setting the Cayley transform and its inverse establish a bijection between the matrices of operator norm less than one and operators with imaginary part a positive operator. This is the special case for A = M_n(C) of the Jordan algebraic result, explained below, which asserts that the Cayley transform and its inverse establish a bijection between the bounded domain D and the tube domain T.

In the case of matrices, the bijection follows from resolvant formulas.^[26] In fact if the imaginary part of T is positive, then T + iI is invertible since

${\displaystyle \displaystyle {\|(T+iI)x\|^{2}=\|(T-iI)x\|^{2}+4(\mathrm {Im} (T)x,x).}}$

In particular, setting y = (T + iI)x,

${\displaystyle \displaystyle {\|y\|^{2}=\|P(T)y\|^{2}+4(\mathrm {Im} (T)x,x).}}$

Equivalently

${\displaystyle \displaystyle {I-P(T)^{*}P(T)=4(T^{*}-iI)^{-1}[\mathrm {Im} \,T](T+iI)^{-1}}}$

is a positive operator, so that ||P(T)|| < 1. Conversely if ||U|| < 1 then I − U is invertible and

${\displaystyle \displaystyle {\mathrm {Im} \,C(U)=(2i)^{-1}[C(U)-C(U)^{*}]=(1-U^{*})^{-1}[I-U^{*}U](I-U)^{-1}.}}$

Since the Cayley transform and its inverse commute with the transpose, they also establish a bijection for symmetric matrices. This corresponds to the Jordan algebra of symmetric complex matrices, the complexification of H_n(R).

In A = E_C the above resolvant identities take the following form:^[27]

${\displaystyle \displaystyle {Q(1-u^{*})Q(C(u)+C(u^{*}))Q(1-u)=-4B(u^{*},u)}}$

and equivalently

${\displaystyle \displaystyle {4Q(\mathrm {Im} \,a)=Q(a^{*}-i)B(P(a)^{*},P(a))Q(a+i),}}$

where the Bergman operator B(x,y) is defined by B(x,y) = I − 2R(x,y) + Q(x)Q(y) with R(x,y) = [L(x),L(y)] + L(xy). The inverses here are well defined. In fact in one direction 1 − u is invertible for ||u|| < 1: this follows either using the fact that the norm satisfies ||ab|| ≤ ||a|| ||b||; or using the resolvant identity and the invertibility of B(u*,u) (see below). In the other direction if the imaginary part of a is in C then the imaginary part of L(a) is positive definite so that a is invertible. This argument can be applied to a + i, so it also invertible.

To establish the correspondence, it is enough to check it when E is simple. In that case it follows from the connectivity of T and D and because:

The first criterion follows from the fact that the eigenvalues of Q(x) are exactly λ_iλ_j if the eigenvalues of x are λ_i. So the λ_i are either all positive or all negative. The second criterion follows from the fact that if a = u Σ α_i e_i = ux with α_i ≥ 0 and u in Γ_u(E_C), then B(a*,a) = u*Q(1 − x²)u has eigenvalues (1 − α_i²)(1 − α_j²). So the α_i are either all less than one or all greater than one.

The resolvant identity is a consequence of the following identity for a and b invertible

${\displaystyle \displaystyle {Q(a)Q(a^{-1}+b^{-1})Q(b)=Q(a+b).}}$

In fact in this case the relations for a quadratic Jordan algebra imply

${\displaystyle \displaystyle {R(a,b)=2Q(a)Q(a^{-1},b)=2Q(a,b^{-1})Q(b)}}$

so that

${\displaystyle \displaystyle {B(a,b)=Q(a)Q(a^{-1}-b)=Q(b^{-1}-a)Q(b).}}$

The equality of the last two terms implies the identity, replacing b by −b⁻¹.

Now set a = 1 − x and b = 1 − y. The resolvant identity is a special case of the more following more general identity:

${\displaystyle \displaystyle {Q(1-x)Q(C(x)+C(y))Q(1-y)=-4B(x,y).}}$

In fact

${\displaystyle \displaystyle {C(x)+C(y)=-2i(1-a^{-1}-b^{-1}),}}$

so the identity is equivalent to

${\displaystyle \displaystyle {Q(a)Q(1-a^{-1}-b^{-1})Q(b)=B(1-a,1-b).}}$

Using the identity above together with Q(c)L(c⁻¹) = L(c), the left hand side equals Q(a)Q(b) + Q(a + b) − 2L(a)Q(b) − 2Q(a)L(b). The right hand side equals 2L(a)L(b) + 2L(b)L(a) − 2L(ab) − 2L(a)Q(b) − 2Q(a)L(b) + Q(a)Q(b) + Q(a) + Q(b). These are equal because of the formula 1/2[Q(a + b) − Q(a) − Q(b)] = L(a)L(b) + L(b)L(a) − L(ab).

Automorphism group of bounded domain

If a lies in the bounded domain D, then a − 1 is invertible. Since D is invariant under multiplication by scalars of modulus ≤ 1, it follows that a − λ is invertible for |λ| ≥ 1. Hence for ||a|| ≤ 1, a − λ is invertible for |λ| > 1. It follows that the Möbius transformation ga is defined for ||a|| ≤ 1 and g in SU(1,1). Where defined it is injective. It is holomorphic on D. By the maximum modulus principle, to show that g maps D onto D it suffices to show it maps S onto itself. For in that case g and its inverse preserve D so must be surjective. If u = e^ix with x = Σ ξ_ie_i in E, then gu lies in ⊕ C e_i. This is a commutative associative algebra and the spectral norm is the supremum norm. Since u = Σ ς_ie_i with |ς_i| = 1, it follows that gu = Σ g(ς_i)e_i where |g(ς_i)| = 1. So gu lies in S.

This is a direct consequence of the definition of the spectral norm.

This is already known for the Möbius transformations, i.e. the diagonal in SU(1,1)^m. It follows for diagonal matrices in a fixed component in SU(1,1)^m because they correspond to transformations in the unitary structure group. Conjugating by a Möbius transformation is equivalent to conjugation by a matrix in that component. Since the only non-trivial normal subgroup of SU(1,1) is its center, every matrix in a fixed component carries D onto itself.

Given an element in D an transformation in the identity component of the unitary structure group carries it in an element in ⊕ C e_i with supremum norm less than 1. An transformation in SU(1,1)^m the carries it onto zero. Thus there is a transitive group of biholomorphic transformations of D. The symmetry z ↦ −z is a biholomorphic Möbius transformation fixing only 0.

If f is a biholomorphic self-mapping of D with f(0) = 0 and derivative I at 0, then f must be the identity.^[28] If not, f has Taylor series expansion f(z) = z + f_k + f_{k + 1}(z) + ⋅⋅⋅ with f_i homogeneous of degree iand f_k ≠ 0. But then fⁿ(z) = z + n f_k(z). Let ψ be a functional in A* of norm one. Then for fixed z in D, the holomorphic functions of a complex variable w given by h_n(w) = ψ(fⁿ(wz)) must have modulus less than 1 for |w| < 1. By Cauchy's inequality, the coefficients of w^k must be uniformly bounded independent of n, which is not possible if f_k ≠ 0.

If g is a biholomorphic mapping of D onto itself just fixing 0 then if h(z) = e^iα z, the mapping f = g ∘ h ∘ g⁻¹ ∘ h^−α fixes 0 and has derivative I there. It is therefore the identity map. So g(e^iα z) = e^iαg(z) for any α. This implies g is a linear mapping. Since it maps D onto itself it maps the closure onto itself. In particular it must map the Shilov boundary S onto itself. This forces g to be in the unitary structure group.

The orbit of 0 under A_D is the set of all points Σ α_i e_i with −1 < α_i < 1. The orbit of these points under the unitary structure group is the whole of D. The Cartan decomposition follows because K_D is the stabilizer of 0 in G_D.