Mutation (Jordan algebra)

Let A be a unital Jordan algebra over a field k of characteristic ≠ 2.[1] For a in A define the Jordan multiplication operator on A by

${\displaystyle \displaystyle {L(a)b=ab}}$

and the quadratic representation Q(a) by

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

It satisfies

${\displaystyle Q(1)=I.\,}$

the fundamental identity

${\displaystyle \displaystyle {Q(Q(a)b)=Q(a)Q(b)Q(a)}}$

the commutation or homotopy identity

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

where

${\displaystyle R(a,b)c=2Q(a,c)b,\,\,\,Q(x,y)={\frac {1}{2}}(Q(x+y)-Q(x)-Q(y)).}$

In particular if a or b is invertible then

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

It follows that A with the operations Q and R and the identity element defines a quadratic Jordan algebra, where a quadratic Jordan algebra consists of a vector space A with a distinguished element 1 and a quadratic map of A into endomorphisms of A, a ↦ Q(a), satisfying the conditions:

• Q(1) = id
• Q(Q(a)b) = Q(a)Q(b)Q(a) ("fundamental identity")
• Q(a)R(b,a) = R(a,b)Q(a) ("commutation or homotopy identity"), where R(a,b)c = (Q(a + c) − Q(a) − Q(c))b

The Jordan triple product is defined by

${\displaystyle \{a,b,c\}=(ab)c+(cb)a-(ac)b,\,}$

so that

${\displaystyle Q(a)b=\{a,b,a\},\,\,\,Q(a,c)b=\{a,b,c\},\,\,\,R(a,b)c=\{a,b,c\}.\,}$

There are also the formulas

${\displaystyle Q(a,b)=L(a)L(b)+L(b)L(a)-L(ab),\,\,\,R(a,b)=[L(a),L(b)]+L(ab).\,}$

For y in A the mutation A^y is defined to the vector space A with multiplication

${\displaystyle a\circ b=\{a,y,b\}.\,}$

If Q(y) is invertible, the mutual is called a proper mutation or isotope.

Let A be a quadratic Jordan algebra over a field k of characteristic ≠ 2. Following Jacobson (1969), a linear Jordan algebra structure can be associated with A such that, if L(a) is Jordan multiplication, then the quadratic structure is given by Q(a) = 2L(a)² − L(a²).

Firstly the axiom Q(a)R(b,a) = R(a,b)Q(a) can be strengthened to

${\displaystyle \displaystyle {Q(a)R(b,a)=R(a,b)Q(a)=2Q(Q(a)b,a).}}$

Indeed, applied to c, the first two terms give

${\displaystyle \displaystyle {2Q(a)Q(b,c)a=2Q(Q(a)c,a)b.}}$

Switching b and c then gives

${\displaystyle \displaystyle {Q(a)R(b,a)c=2Q(Q(a)b,a)c.}}$

Now let

${\displaystyle \displaystyle {L(a)={\frac {1}{2}}R(a,1).}}$

Replacing b by a and a by 1 in the identity above gives

${\displaystyle \displaystyle {R(a,1)=R(1,a)=2Q(a,1).}}$

In particular

${\displaystyle \displaystyle {L(a)=Q(a,1),\,\,\,L(1)=Q(1,1)=I.}}$

The Jordan product is given by

${\displaystyle \displaystyle {a\circ b=L(a)b={\frac {1}{2}}R(a,1)b=Q(a,b)1,}}$

so that

${\displaystyle \displaystyle {a\circ b=b\circ a.}}$

The formula above shows that 1 is an identity. Defining a² by a∘a = Q(a)1, the only remaining condition to be verified is the Jordan identity

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

In the fundamental identity

${\displaystyle \displaystyle {Q(Q(a)b)=Q(a)Q(b)Q(a),}}$

Replace a by a + t1, set b = 1 and compare the coefficients of t² on both sides:

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

Setting b = 1 in the second axiom gives

${\displaystyle \displaystyle {Q(a)L(a)=L(a)Q(a),}}$

and therefore L(a) must commute with L(a²).

Let A be a unital Jordan algebra over a field k of characteristic ≠ 2. An element a in a unital Jordan algebra A is said to be invertible if there is an element b such that ab = 1 and a²b = a.[2]

Properties.[3]

If ab = 1 and a²b = a, then Q(a)b = 2a(ab) − (a²)b = a. The Jordan identity [L(x),L(x²)] = 0 can be polarized by replacing x by x + ty and taking the coefficient of t. This gives

${\displaystyle \displaystyle {[L(x^{2}),L(y)]+2[L(xy),L(x)]=0.}}$

Taking x = a or b and y = b or a shows that L(a²) commutes with L(b) and L(b²) commutes with L(a). Hence (b²)(a²) = 1. Applying L(b) gives b²a = b. Hence Q(a)b² = 1. Conversely if Q(a)b = a and Q(a)b² = 1, then the second relation gives Q(a)Q(b)² Q(a) = I. So both Q(a) and Q(b) are invertible. The first gives Q(a)Q(b)Q(a) = Q(a) so that Q(a) and Q(b) are each other's inverses. Since L(b) commutes with Q(b) it commutes with its inverse Q(a). Similarly L(a) commutes with Q(b). So (a²)b = L(b)a² = Q(a)b = a and ab = L(b)Q(a)b= Q(a)Q(b)1= 1.

Indeed, if a is invertible then the above implies Q(a) is invertible with inverse Q(b). Any inverse b satisfies Q(a)b = a, so b = Q(a)⁻¹a. Conversely if Q(a) is invertible let b = Q(a)⁻¹a. Then Q(a)b = a. The fundamental identity then implies that Q(b) and Q(a) are each other's inverses so that Q(a)b² = Q(a)Q(b)1=1.

This follows from the formula a⁻¹ = Q(a)⁻¹a.

Suppose that Q(a)c = 1. Then by the fundamental identity Q(a) is invertible, so a is invertible.

This is an immediate consequence of the fundamental identity and the fact that STS is invertible if and only S and T are invertible.

In the commutation identity Q(a)R(b,a) = Q(Q(a)b,a), set b = c² with c = a⁻¹. Then Q(a)b = 1 and Q(1,a) = L(a). Since L(a) commutes with L(c²), R(b,a) = L(c) = L(a⁻¹).

If L(a) and L(b) commute, then ba = 1 implies b(a²) = a. Conversely suppose that a is invertible with inverse b. Then ab = 1. Morevoer L(b) commutes with Q(b) and hence its inverse Q(a). So it commutes with L(a) = Q(a)L(b).

The algebra k[a] is commutative and associative, so if b is an inverse there ab =1 and a²b = a. Conversely Q(a) leaves k[a] invariant. So if it is bijective on A it is bijective there. Thus a⁻¹ = Q(a)⁻¹a lies in k[a].

In fact [4] multiplication in the algebra A^y is given by

${\displaystyle \displaystyle {a\circ b=\{a,y,b\},}}$

so by definition is commutative. It follows that

${\displaystyle \displaystyle {a\circ b=L_{y}(a)b,}}$

with

${\displaystyle \displaystyle {L_{y}(a)=[L(a),L(y)]+L(ay).}}$

If e satisfies a ∘ e = a, then taking a = 1 gives

${\displaystyle \displaystyle {ye=1.}}$

Taking a = e gives

${\displaystyle \displaystyle {e(ya)=y(ea)}}$

so that L(y) and L(e) commute. Hence y is invertible and e = y⁻¹.

Now for y invertible set

${\displaystyle \displaystyle {Q_{y}(a)=Q(a)Q(y),\,\,\,R_{y}(a,b)=R(a,Q(y)b).}}$

Then

${\displaystyle \displaystyle {Q_{y}(e)=Q_{y}(y^{-1})=Q(y^{-1})Q(y)=I.}}$

Moreover,

${\displaystyle \displaystyle {Q_{y}(a)Q_{y}(b)Q_{y}(a)=Q(a)Q(y)Q(b)Q(y)Q(a)Q(y)=Q(a)Q(Q(y)b)Q(a)Q(y)=Q(Q(a)Q(y)b)Q(y)=Q_{y}(Q_{y}(a)b).}}$

Finally

${\displaystyle \displaystyle {Q(y)R(c,Q(y)d)Q(y)^{-1}=R(Q(y)c,d),}}$

since

${\displaystyle \displaystyle {Q(y)R(c,Q(y)d)x=2Q(y)Q(c,x)Q(y)d=2Q(Q(y)c,Q(y)x)d=R(Q(y)c,d)Q(y)x.}}$

Hence

${\displaystyle \displaystyle {Q_{y}(a)R_{y}(b,a)=Q(a)Q(y)R(b,Q(y)a)=Q(y^{-1})Q(Q(y)a)R(b,Q(y)a)=Q(y)^{-1}R(Q(y)a,b)Q(Q(y)a)=R_{y}(a,b)Q_{y}(a).}}$

Thus (A,Q^y,y⁻¹) is a unital quadratic Jordan algebra. It therefore corresponds to a linear Jordan algebra with the associated Jordan multiplication operator M(a) given by

${\displaystyle \displaystyle {M(a)b={\frac {1}{2}}R_{y}(a,e)b={\frac {1}{2}}R(a,Q(y)e)b={\frac {1}{2}}R(a,y)b=\{a,y,b\}=L_{y}(a)b.}}$

This shows that the operators L_y(a) satisfy the Jordan identity so that the proper mutation or isotope A^y is a unital Jordan algebra. The correspondence with quadratic Jordan algebras shows that its quadratic representation is given by Q_y.

The definition of mutation also applies to non-invertible elements y. If A is finite-dimensional over R or C, invertible elements a in A are dense, since invertibility is equivalent to the condition that det Q(a) ≠ 0. So by continuity the Jordan identity for proper mutations implies the Jordan identity for arbitrary mutations. In general the Jordan identity can be deduced from Macdonald's theorem for Jordan algebras because it involves only two elements of the Jordan algebra. Alternatively, the Jordan identity can be deduced by realizing the mutation inside a unital quadratic algebra.[5]

For a in A define a quadratic structure on A₁ = A ⊕ k by

${\displaystyle \displaystyle {Q_{1}(a\oplus \alpha 1)(b\oplus \beta 1)=\alpha ^{2}\beta 1\oplus [\alpha ^{2}a+\alpha ^{2}b+2\alpha \beta a+\alpha \{a,y,b\}+\beta Q(a)y+Q(a)Q(y)b].}}$

It can then be verified that (A₁, Q₁, 1) is a unital quadratic Jordan algebra. The unital Jordan algebra to which it corresponds has A^y as an ideal, so that in particular A^y satisfies the Jordan identity. The identities for a unital quadratic Jordan algebra follow from the following compatibility properties of the quadratic map Q_y(a) = Q(a)Q(y) and the squaring map S_y(a) = Q(a)y:

• R_y(a,a) = L_y(S_y(a)).
• [Q_y(a),L_y(a)] = 0.
• Q_y(a)S_y(a) = S_y(S_y(a)).
• Q_y∘ S_y = S_y ∘ Q_y.
• Q_y(a) Q_y(b) S_y(a) = S_y(Q_y(a)b).
• Q_y(Q_y(a)b) = Q_y(a) Q_y(b) Q_y(a).

Let A be a unital Jordan algebra. If a, b and a – b are invertible, then Hua's identity holds:[6]

In particular if x and 1 – x are invertible, then so too is 1 – x⁻¹ with

${\displaystyle \displaystyle {(1-x)^{-1}+(1-x^{-1})^{-1}=1.}}$

To prove the identity for x, set y = (1 – x)⁻¹. Then L(y) = Q(1 – x)⁻¹L(1 – x). Thus L(y) commutes with L(x) and Q(x). Since Q(y) = Q(1 – x)⁻¹, it also commutes with L(x) and Q(x). Since L(x⁻¹) = Q(x)⁻¹L(x), L(y) also commutes with L(x⁻¹) and Q(x⁻¹).

It follows that (x⁻¹ – 1)xy =(1 – x) y = 1. Moreover, y – 1 = xy since (1 – x)y = 1. So L(xy) commutes with L(x) and hence L(x⁻¹ – 1). Thus 1 – x⁻¹ has inverse 1 – y.

Now let A^a be the mutation of A defined by a. The identity element of A^a is a⁻¹. Moreover, an invertible element c in A is also invertible in A^a with inverse Q(a)⁻¹ c⁻¹.

Let x = Q(a)⁻¹b in A^a. It is invertible in A, as is a⁻¹ – Q(a)⁻¹b = Q(a)⁻¹(a – b). So by the special case of Hua's identity for x in A^a

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

If A is a unital Jordan algebra, the Bergman operator is defined for a, b in A by[7]

${\displaystyle \displaystyle {B(a,b)=I-R(a,b)+Q(a)Q(b).}}$

If a is invertible then

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

while if b is invertible then

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

In fact if a is invertible

Q(a)Q(a⁻¹ − b) = Q(a)[Q(a⁻¹ − 2Q(a⁻¹,b) + Q(b)]=I − 2Q(a)Q(a⁻¹,b) + Q(a)Q(b)=I − R(a,b) + Q(a)Q(b)

and similarly if b is invertible.

More generally the Bergman operator satisfies a version of the commutation or homotopy identity:

and a version of the fundamental identity:

There is also a third more technical identity:

Let A be a finite-dimensional unital Jordan algebra over a field k of characteristic ≠ 2.[8] For a pair (a,b) with a and a⁻¹ − b invertible define

In this case the Bergman operator B(a,b) = Q(a)Q(a⁻¹ − b) defines an invertible operator on A and

In fact

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

Moreover, by definition a⁻¹ − b − c is invertible if and only if (a^b)⁻¹ − c is invertible. In that case

Indeed,

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

The assumption that a be invertible can be dropped since a^b can be defined only supposing that the Bergman operator B(a,b) is invertible. The pair (a,b) is then said to be quasi-invertible. In that case a^b is defined by the formula

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

If B(a,b) is invertible, then B(a,b)c = 1 for some c. The fundamental identity implies that B(a,b)Q(c)B(b,a) = I. So by finite-dimensionality B(b,a) is invertible. Thus (a,b) is invertible if and only if (b,a) is invertible and in this case

In fact

B(a,b)(a + Q(a)b^a) = a − 2R(a,b)a + Q(a)Q(b)a + Q(a)(b − Q(b)a) = a − Q(a)b,

so the formula follows by applying B(a,b)⁻¹ to both sides.

As before (a,b+c) is quasi-invertible if and only if (a^b,c) is quasi-invertible; and in that case

${\displaystyle \displaystyle {a^{b+c}=(a^{b})^{c}.}}$

If k = R or C, this would follow by continuity from the special case where a and a⁻¹ − b were invertible. In general the proof requires four identities for the Bergman operator:

In fact applying Q to the identity B(a,b)a^b = a − Q(a)b yields

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

The first identity follows by cancelling B(a,b) and B(b,a). The second identity follows by similar cancellation in

B(a,b)Q(a^b,c)B(b,a) = Q(B(a,b)a^b,B(a,b)c) = Q(a − Q(a)b,B(a,b)c) = B(a,b)(Q(a,c) − R(c,b)Q(a)) = (Q(a,c) − Q(a)R(b,c))B(b,a).

The third identity follows by applying the second identity to an element d and then switching the roles of c and d. The fourth follows because

B(a,b)B(a^b,c) = B(a,b)(I − R(a^b,c) + Q(a^b)Q(c)) = I − R(a,b + c) + Q(a) Q(b + c) = B(a,b+c).

In fact (a,b) is quasi-invertible if and only if a is quasi-invertible in the mutation A^b. Since this mutation might not necessarily unital this means that when an identity is adjoint 1 − a becomes invertible in A^b ⊕ k1. This condition can be expressed as follows without mentioning the mutation or homotope:

In fact if (a,b) is quasi-invertible, then c = a^b satisfies the first identity by definition. The second follows because B(a,b)Q(a^b) = Q(a). Conversely the conditions state that in A^b ⊕ k1 the conditions imply that 1 + c is the inverse of 1 − a. On the other hand, ( 1 − a) ∘ x = B(a,b)x for x in A^b. Hence B(a,b) is invertible.

Let A be a finite-dimensional unital Jordan algebra over a field k of characteristic ≠ 2.[9] Two pairs (a_i,b_i) with a_i invertible are said to be equivalent if (a₁)⁻¹ − b₁ + b₂ is invertible and a₂ = (a₁)^{b₁ − b₂}.

This is an equivalence relation, since if a is invertible a⁰ = a so that a pair (a,b) is equivalent to itself. It is symmetric since from the definition a₁ = (a₂)^{b₂ − b₁}. It is transitive. For suppose that (a₃,b₃) is a third pair with (a₂)⁻¹ − b₂ + b₃ invertible and a₃ = (a₂)^{b₂ − b₃}. From the above

${\displaystyle \displaystyle {a_{1}^{-1}-b_{1}+b_{3}=(a_{1}^{-1}-b_{1}+b_{2})-b_{2}+b_{3}=a_{2}^{-1}-b_{2}+b_{3}}}$

is invertible and

${\displaystyle \displaystyle {a_{3}=a_{2}^{b_{2}-b_{3}}=(a_{1}^{b_{1}-b_{2}})^{b_{2}-b_{3}}=a_{1}^{b_{1}-b_{3}}.}}$

As for quasi-invertibility, this definition can be extended to the case where a and a⁻¹ − b are not assumed to be invertible.

Two pairs (a_i,b_i) are said to be equivalent if (a₁, b₁ − b₂) is quasi-invertible and a₂ = (a₁)^{b₁ − b₂}. When k = R or C, the fact that this more general definition also gives an equivalence relation can deduced from the invertible case by continuity. For general k, it can also be verified directly:

• The relation is reflexive since (a,0) is quasi-invertible and a⁰ = a.
• The relation is symmetric, since a₁ = (a₂)^{b₂ − b₁}.
• The relation is transitive. For suppose that (a₃,b₃) is a third pair with (a₂, b₂ − b₃) quasi-invertible and a₃ = (a₂)^{b₂ − b₃}. In this case

${\displaystyle \displaystyle {B(a_{1},b_{1}-b_{3})=B(a_{1},b_{1}-b_{2})B(a_{2},b_{2}-b_{3}),}}$

so that (a₁,b₁ − b₃) is quasi-invertible with

${\displaystyle \displaystyle {a_{3}=a_{2}^{b_{2}-b_{3}}=(a_{1}^{b_{1}-b_{2}})^{b_{2}-b_{3}}=a_{1}^{b_{1}-b_{3}}.}}$

The equivalence class of (a,b) is denoted by (a:b).

Let A be a finite-dimensional complex semisimple unital Jordan algebra. If T is an operator on A, let T^t be its transpose with respect to the trace form. Thus L(a)^t = L(a), Q(a)^t = Q(a), R(a,b)^t = R(b,a) and B(a,b)^t = B(b,a). The structure group of A consists of g in GL(A) such that

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

They form a group Γ(A). The automorphism group Aut A of A consists of invertible complex linear operators g such that L(ga) = gL(a)g⁻¹ and g1 = 1. Since an automorphism g preserves the trace form, g⁻¹ = g^t.

• 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.
• The structure group Γ(A) acts transitively on the set of invertible elements in A.
• Every g in Γ(A) has the form g = h Q(a) with h an automorphism and a invertible.

The complex Jordan algebra A is the complexification of a real Euclidean Jordan algebra E, for which the trace form defines an inner product. There is an associated involution a ↦ a* on A which gives rise to a complex inner product on A. The unitary structure group Γ_u(A) is the subgroup of Γ(A) consisting of unitary operators, so that Γ_u(A) = Γ(A) ∩ U(A). The identity component of Γ_u(A) is denoted by K. It is a connected closed subgroup of U(A).

• The stabilizer of 1 in Γ_u(A) is Aut E.
• Every g in Γ_u(A) has the form g = h Q(u) with h in Aut E and u invertible in A with u* = u⁻¹.
• Γ(A) is the complexification of Γ_u(A).
• The set S of invertible elements u in A 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(A) acts transitively on S
• Given a Jordan frame (e_i) and v in A, there is an operator u in the identity component of Γ_u(A) such that uv = Σ α_i e_i with α_i ≥ 0. If v is invertible, then α_i > 0.

The structure group Γ(A) acts naturally on X.[10] For g in Γ(A), set

${\displaystyle \displaystyle {g(a,b)=(ga,(g^{t})^{-1}b).}}$

Then (x,y) is quasi-invertible if and only if (gx,(g^t)⁻¹y) is quasi-invertible and

${\displaystyle \displaystyle {g(x^{y})=(gx)^{(g^{t})^{-1}y}.}}$

In fact the covariance relations for g with Q and the inverse imply that

${\displaystyle \displaystyle {gB(x,y)g^{-1}=B(gx,(g^{t})^{-1}y)}}$

if x is invertible and so everywhere by density. In turn this implies the relation for the quasi-inverse. If a is invertible then Q(a) lies in Γ(A) and if (a,b) is quasi-invertible B(a,b) lies in Γ(A). So both types of operators act on X.

The defining relations for the structure group show that it is a closed subgroup of ${\displaystyle {\mathfrak {g}}_{0}}$ of GL(A). Since Q(e^a) = e^2L(a), the corresponding complex Lie algebra contains the operators L(a). The commutators [L(a),L(b)] span the complex Lie algebra of derivations of A. The operators R(a,b) = [L(a),L(b)] + L(ab) span ${\displaystyle {\mathfrak {g}}_{0}}$ and satisfy R(a,b)^t = R(b,a) and [R(a,b),R(c,d)]=R(R(a,b)c,d) − R(c,R(b,a)d).

Let A be a finite-dimensional complex unital Jordan algebra which is semisimple, i.e. the trace form Tr L(ab) is non-degenerate. Let X be the quotient of A×A by the equivalence relation. Let X_b be the subset of X of classes (a:b). The map φ_b:X_b → A, (a:b) ↦ a is injective. A subset U of X is defined to be open if and only if U ∩ X_b is open for all b.

The transition maps of the atlas with charts φ_b are given by

${\displaystyle \displaystyle {\varphi _{cb}=\varphi _{c}\circ \varphi _{b}^{-1}:\varphi _{b}(X_{b}\cap X_{c})\rightarrow \varphi _{c}(X_{b}\cap X_{c}).}}$

and are injective and holomorphic since

${\displaystyle \displaystyle {\varphi _{cb}(a)=a^{b-c}}}$

with derivative

${\displaystyle \displaystyle {\varphi _{cb}^{\prime }(a)=B(a,b-c)^{-1}.}}$

This defines the structure of a complex manifold on X because φ_dc ∘ φ_cb = φ_db on φ_b(X_b ∩ X_c ∩ X_d).

Indeed, all the polynomial functions p_i(b) = det B(a_i,b_i − b) are non-trivial since p_i(b_i) = 1. Therefore, there is a b such that p_i(b) ≠ 0 for all i, which is precisely the criterion for (a_i:b_i) to lie in X_b.

Loos (1977) uses the Bergman operators to construct an explicit biholomorphism between X and a closed smooth algebraic subvariety of complex projective space.[11] This implies in particular that X is compact. There is a more direct proof of compactness using symmetry groups.

Given a Jordan frame (e_i) in E, for every a in A there is a k in U = Γ_u(A) such that a=k(Σ α_i e_i) with α_i ≥ 0 (and α_i > 0 if a is invertible). In fact, if (a,b) is in X then it is equivalent to k(c,d) with c and d in the unital Jordan subalgebra A_e = ⊕ Ce_i, which is the complexification of E_e = ⊕ Re_i. Let Z be the complex manifold constructed for A_e. Because A_e is a direct sum of copies of C, Z is just a product of Riemann spheres, one for each e_i. In particular it is compact. There is a natural map of Z into X which is continuous. Let Y be the image of Z. It is compact and therefore coincides with the closure of Y₀ = A_e ⊂ A = X₀. The set U⋅Y is the continuous image of the compact set U × Y. It is therefore compact. On the other hand, U⋅Y₀ = X₀, so it contains a dense subset of X and must therefore coincide with X. So X is compact.

The above argument shows that every (a,b) in X is equivalent to k(c,d) with c and d in A_e and k in Γ_u(A). The mapping of Z into X is in fact an embedding. This is a consequence of (x,y) being quasi-invertible in A_e if and only if it is quasi-invertible in A. Indeed, if B(x,y) is injective on A, its restriction to A_e is also injective. Conversely, the two equations for the quasi-inverse in A_e imply that it is also a quasi-inverse in A.

Let A be a finite-dimensional complex semisimple unital Jordan algebra. The group SL(2,C) acts by Möbius transformation 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

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

There is a generalization of this action of SL(2,C) to A and its compactification X. In order to define this action, note that SL(2,C) is generated by the three subgroups of lower and upper unitriangular matrices and the diagonal matrices. 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. If g does not fix ∞, it sends ∞ to a finite point a. But then g can be composed with an upper unitriangular to send a to 0 and then with J to send 0 to infinity.

For an element a of A, the action of g in SL(2,C) is defined by the same formula

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

This defines an element of C[a] provided that γa + δ1 is invertible in A. The action is thus defined everywhere on A if g is upper triangular. On the other hand, the action on X is simple to define for lower triangular matrices.[12]

• For diagonal matrices g with diagonal entries α and α⁻¹, g(a,b) = (α²a, α⁻²b) is a well-defined holomorphic action on A² which passes to the quotient X. On X₀ = A it agrees with the Möbius action.
• For lower unitriangular matrices, with off-diagonal parameter γ, define g(a,b) = (a,b − γ1). Again this is holomorphic on A² and passes to the quotient X. When b = 0 and γ ≠ 0,

${\displaystyle \displaystyle {g(a:0)=(a:-\gamma )=(a^{-\gamma }:0)=(a(\gamma a+1)^{-1}:0)}}$

if γa + 1 is invertible, so this is an extension of the Möbius action.

• For upper unitriangular matrices, with off-diagonal parameter β, the action on X₀ = (A:0) is defined by g(a,0) = (a + β1). Loos (1977) showed that this defined a complex one-parameter flow on A. The corresponding holomorphic complex vector field extended to X, so that the action on the compact complex manifold X could be defined by the associated complex flow. A simpler method is to note that the operator J can be implemented directly using its intertwining relations with the unitary structure group.

In fact on the invertible elements in A, the operator j(a) = −a⁻¹ satisfies j(ga) = (g^t)⁻¹j(a). To define a biholomorphism j on X such that j ∘ g = (g^t)⁻¹ ∘ j, it is enough to define these for (a:b) in some suitable orbit of Γ(A) or Γ_u(A). On the other hand, as indicated above, given a Jordan frame (e_i) in E, for every a in A there is a k in U = Γ_u(A) such that a=k(Σ α_i e_i) with α_i ≥ 0.

The computation of j in the associative commutative algebra A_e is straightforward since it is a direct product. For c = Σ α_i e_i and d = Σ β_i e_i, the Bergman operator on A_e has determinant det B(c,d) = Π(1 − α_iβ_i)². In particular det B(c,d − λ) ≠ 0 for some λ ≠ 0. So that (c,d) is equivalent to (x,λ). Let μ = −λ⁻¹. On A, for a dense set of a, the pair (a,λ) is equivalent to (b,0) with b invertible. Then (−b⁻¹,0) is equivalent to (μ − μ²a,μ). Since a ↦ μ − μ²a is holomorphic it follows that j has a unique continuous extension to X such that j ∘ g = (g^t)⁻¹ ∘ j for g in Γ(A), the extension is holomorphic and for λ ≠ 0, μ = −λ⁻¹

The holomorphic transformations corresponding to upper unitriangular matrices can be defined using the fact that they are the conjugates by J of lower unitriangular matrices, for which the action is already known. A direct algebraic construction is given in Dineen, Mackey & Mellon (1999).

This action of SL(2,C) is compatible with inclusions. More generally if e₁, ..., e_m is a Jordan frame, there is an action of SL(2,C)^m on A_e which extends to A. If c = Σ γ_ie_i and b = Σ β_ie_i, then S(c) and T(b) give the action of the product of the lower and upper unitriangular matrices. If a = Σ α_ie_i is invertible, the corresponding product of diagonal matrices act as W = Q(a).[13] In particular the diagonal matrices give an action of (C*)^m and T^m.

Let A be a finite-dimensional complex semisimple unital Jordan algebra. There is a transitive holomorphic action of a complex matrix group G on the compact complex manifold X. Koecher (1967) described G analogously to SL(2,C) in terms of generators and relations. G acts on the corresponding finite-dimensional Lie algebra of holomorphic vector fields restricted to X₀ = A, so that G is realized as a closed matrix group. It is the complexification of a compact Lie group without center, so a semisimple algebraic group. The identity component H of the compact group acts transitively on X, so that X can be identified as a Hermitian symmetric space of compact type.[14]

The group G is generated by three types of holomorphic transformation on X:

• Operators W corresponding to elements W in Γ(A) given by W(a,b) = (Wa, (W^t)⁻¹b). These were already described above. On X₀ = A, they are given by a ↦ Wa.
• Operators S_c defined by S_c(a,b) = (a,b + c). These are the analogue of lower unitriangular matrices and form a subgroup isomorphic to the additive group of A, with the given parametrization. Again these act holomorphically on A² and the action passes to the quotient X. On A the action is given by a ↦ a^c if (a,c) is quasi-invertible.
• The transformation j corresponding to J in SL(2,C). It was constructed above as part of the action of PSL(2,C) = SL(2,C)/{±I} on X. On invertible elements in A it is given by a ↦ −a⁻¹.

The operators W normalize the group of operators S_c. Similarly the operator j normalizes the structure group, j ∘ W = (W^t)⁻¹ ∘ j. The operators T_c = j ∘ S_−c ∘ j also form a group of holomorphic transformations isomorphic to the additive group of A. They generalize the upper unitriangular subgroup of SL(2,C). This group is normalized by the operators W of the structure group. The operator T_c acts on A as a ↦ a + c. If c is a scalar the operators S_c and T_c coincide with the operators corresponding to lower and upper unitriangular matrices in SL(2,C). Accordingly, there is a relation j = S₁ ∘ T₁ ∘ S₁ and PSL(2,C) is a subgroup of G. Loos (1977) defines the operators T_c in terms of the flow associated to a holomorphic vector field on X, while Dineen, Mackey & Mellon (1999) give a direct algebraic description.

Indeed, S_bT_a(0:0) = (a:b).

Let G₋₁ and G₊₁ be the complex Abelian groups formed by the symmetries T_c and S_c respectively. Let G₀ = Γ(A).

The two expressions for G are equivalent as follows by conjugating by j.

For a invertible, Hua's identity can be rewritten

${\displaystyle \displaystyle {Q(a)=T_{a}\circ j\circ T_{a^{-1}}\circ j\circ T_{a}\circ j.}}$

Moreover, j = S₁ ∘ T₁ ∘ S₁ and S_c = j ∘ T_−c ∘ j.[15]

The convariance relations show that the elements of G fall into sets G₀G₁, G₀G₁jG₁, G₀G₁jG₁jG₁, G₀G₁jG₁jG₁jG₁. ... The first expression for G follows once it is established that no new elements appear in the fourth or subsequent sets. For this it suffices to show that[16]

j ∘ G₁ ∘j ∘ G₁ ∘j ⊆ G₀ G₁ ∘j ∘ G₁ ∘ j ∘ G₁.

For then if there are three or more occurrences of j, the number can be recursively reduced to two. Given a, b in A, pick λ ≠ 0 so that c = a − λ and d = b − λ⁻¹ are invertible. Then

${\displaystyle \displaystyle {jT_{a}jT_{b}j=jT_{c}T_{\lambda }jT_{\lambda ^{-1}}T_{d}\circ j=\lambda ^{2}jT_{c}jT_{-\lambda }jT_{d}\ j=\lambda ^{2}T_{-c^{-1}}jQ(c^{-1})T_{-c^{-1}-\lambda -d^{-1}}jQ(d^{-1})jT_{-d^{-1}},}}$

which lies in G₀G₁ ∘ j ∘ G₁ ∘ j ∘ G₁.

It suffices to check that if S_aT_b(0:0) = (0:0), then b = 0. If so (b:0) = (0: −a) = (0:0), so b = 0.

For a invertible, Hua's identity can be rewritten

${\displaystyle \displaystyle {Q(a)=T_{a}\circ j\circ T_{a^{-1}}\circ j\circ T_{a}\circ j.}}$

Since j = S₁ ∘ T₁ ∘ S₁, the operators Q(a) belong to the group generated by G_±1.[17]

For quasi-invertible pairs (a,b), there are the "exchange relations"[18]

This identity shows that B(a,b) is in the group generated by G_±1. Taking inverses, it is equivalent to the identity T_aS_b = S_b^aB(a,b)T_a^b.

To prove the exchange relations, it suffices to check that it valid when applied to points the dense set of points (c:0) in X for which (a+c,b) is quasi-invertible. It then reduces to the identity:

In fact, if (a,b) is quasi-invertible, then (a + c,b) is quasi-invertible if and only if (c,b^a) is quasi-invertible. This follows because (x,y) is quasi-invertible if and only if (y,x) is. Moreover, the above formula holds in this case.

For the proof, two more identities are required:

${\displaystyle \displaystyle {B(c+b,a)=B(c,a^{b})B(b,a)}}$

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

The first follows from a previous identity by applying the transpose. For the second, because of the transpose, it suffices to prove the first equality. Setting c = b − Q(b)a in the identity B(a,b)R(a^b,c) = R(a,c) − Q(a)Q(b,c) yields

B(a,b)R(a^b,b − Q(b)c) = B(a,b)R(a,b),

so the identity follows by cancelling B(a,b).

To prove the formula, the relations (a + c)^b = B(a,c)⁻¹(a + c − Q(a + c)b) and a^b + B(a,b)⁻¹c^(ba) = B(a + c,b)⁻¹(B(c,b^a) (a − Q(a)b) + c − Q(c)b^a) show that it is enough to prove that

a + c − Q(a + c)b = B(c,b^a) (a − Q(a)b) + c − Q(c)b^a.

Indeed, B(c,b^a) (a − Q(a)b) + c − Q(c)b^a = a + c − Q(a)b + 2R(c,b^a)(a − Q(a)b) − Q(c)[ b^a − Q(b^a)(a − Q(a)b)]. On the other hand, 2R(c,b^a)(a − Q(a)b) = 2R(c,a − Q(a)b)b^a = R(a,b)c = 2Q(a,c)b and b^a − Q(b^a)(a − Q(a)b) = b^a − Q(b)B(a,b)⁻¹(a − Q(a)b) = b^a − Q(b)a^b = b. So B(c,b^a) (a − Q(a)b) + c − Q(c)b^a = a + c − Q(a)b − 2Q(a,c)b − Q(c)b = a + c − Q(a + c)b.

Now set Ω = G₊₁G₀G₋₁. Then the exchange relations imply that S_b T_a lies in Ω if and only if (a,b) is quasi-invertible; and that g lies in Ω if and only if g(0:0) is in X₀.[19]

In fact if S_b T_a lies in Ω, then (a,b) is equivalent to (x,0), so it a quasi-invertible pair; the converse follows from the exchange relations. Clearly Ω(0:0) = G₁(0:0) = X₀. The converse follows from G = G₋₁G₁ G₀G₋₁ and the criterion for S_b T_a to lie in Ω.

The compact complex manifold X is modelled on the space A. The derivatives of the transition maps describe the tangent bundle through holomorphic transition functions F_bc:X_b ∩ X_c → GL(A). These are given by F_bc(a,b) = B(a,b − c), so the structure group of the corresponding principal fiber bundle reduces to Γ(A), the structure group of A.[20] The corresponding holomorphic vector bundle with fibre A is the tangent bundle of the complex manifold X. Its holomorphic sections are just holomorphic vector fields on X. They can be determined directly using the fact that they must be invariant under the natural adjoint action of the known holomorphic symmetries of X. They form a finite-dimensional complex semisimple Lie algebra. The restriction of these vector fields to X₀ can be described explicitly. A direct consequence of this description is that the Lie algebra is three-graded and that the group of holomorphic symmetries of X, described by generators and relations in Koecher (1967) and Loos (1979), is a complex linear semisimple algebraic group that coincides with the group of biholomorphisms of X.

The Lie algebras of the three subgroups of holomorphic automorphisms of X give rise to linear spaces of holomorphic vector fields on X and hence X₀ = A.

• The structure group Γ(A) has Lie algebra ${\displaystyle {\mathfrak {g}}_{0}}$ spanned by the operators R(x,y). These define a complex Lie algebra of linear vector fields a ↦ R(x,y)a on A.
• The translation operators act on A as T_c(a) = a + c. The corresponding one-parameter subgroups are given by T_tc and correspond to the constant vector fields a ↦ c. These give an Abelian Lie algebra ${\displaystyle {\mathfrak {g}}_{-1}}$ of vector fields on A.
• The operators S_c defined on X by S_c(a,b) = (a,b − c). The corresponding one-parameter groups S_tc define quadratic vector fields a ↦ Q(a)c on A. These give an Abelian Lie algebra ${\displaystyle {\mathfrak {g}}_{1}}$ of vector fields on A.

Let

${\displaystyle \displaystyle {{\mathfrak {g}}={\mathfrak {g}}_{-1}\oplus {\mathfrak {g}}_{0}\oplus {\mathfrak {g}}_{1}.}}$

Then, defining ${\displaystyle {\mathfrak {g}}_{i}=(0)}$ for i ≠ −1, 0, 1, ${\displaystyle {\mathfrak {g}}}$ forms a complex Lie algebra with

${\displaystyle \displaystyle {[{\mathfrak {g}}_{p},{\mathfrak {g}}_{q}]\subseteq {\mathfrak {g}}_{p+q}}.}$

This gives the structure of a 3-graded Lie algebra. For elements (a,T,b) in ${\displaystyle {\mathfrak {g}}}$, the Lie bracket is given by

${\displaystyle \displaystyle {[(a_{1},T_{1},b_{1}),(a_{2},T_{2},b_{2})]=(T_{1}a_{2}-T_{2}a_{1},[T_{1},T_{2}]+R(a_{1},b_{2})-R(a_{2},b_{1}),T_{2}^{t}b_{1}-T_{1}^{t}b_{2})}}$