Mutation (Jordan algebra)

In mathematics, a mutation, also called a homotope, of a unital Jordan algebra is a new Jordan algebra defined by a given element of the Jordan algebra. The mutation has a unit if and only if the given element is invertible, in which case the mutation is called a proper mutation or an isotope. Mutations were first introduced by Max Koecher in his Jordan algebraic approach to Hermitian symmetric spaces and bounded symmetric domains of tube type. Their functorial properties allow an explicit construction of the corresponding Hermitian symmetric space of compact type as a compactification of a finite-dimensional complex semisimple Jordan algebra. The automorphism group of the compactification becomes a complex subgroup, the complexification of its maximal compact subgroup. Both groups act transitively on the compactification. The theory has been extended to cover all Hermitian symmetric spaces using the theory of Jordan pairs or Jordan triple systems. Koecher obtained the results in the more general case directly from the Jordan algebra case using the fact that only Jordan pairs associated with period two automorphisms of Jordan algebras are required.

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


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

and the quadratic representation Q(a) by


 * $$ Q(a)=2L(a)^2 -L(a^2). \, $$

It satisfies


 * $$Q(1)=I. \,$$

the fundamental identity


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

the commutation or homotopy identity


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

where


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

In particular if a or b is invertible then


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

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:


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

The Jordan triple product is defined by


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

so that


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

There are also the formulas


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

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


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

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

Quadratic Jordan algebras
Let A be a quadratic Jordan algebra over a field k of characteristic ≠ 2. Following, 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


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

Indeed, applied to c, the first two terms give


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

Switching b and c then gives


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

Now let


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

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


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

In particular


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

The Jordan product is given by


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

so that


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

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


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

In the fundamental identity


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

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


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

Setting b = 1 in the second axiom gives


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

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

Inverses
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.

Properties.

If $b$ and $Q(a)b = a$, then $Q(a)b^{2} =1$. The Jordan identity $ab = 1$ can be polarized by replacing $a^{2}b = a$ by $ab = 1$ and taking the coefficient of $a^{2}b = a$. This gives


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

Taking $Q(a)b = 2a(ab) − (a^{2})b = a$ or $[L(x),L(x^{2})] = 0$ and $x$ or $x + ty$ shows that $t$ commutes with $x = a$ and $b$ commutes with $y = b$. Hence $a$. Applying $L(a^{2})$ gives $L(b)$. Hence $L(b^{2})$. Conversely if $L(a)$ and $(b^{2})(a^{2}) = 1$, then the second relation gives $L(b)$. So both $b^{2}a = b$ and $Q(a)b^{2} = 1$ are invertible. The first gives $Q(a)b = a$ so that $Q(a)b^{2} = 1$ and $Q(a)Q(b)^{2} Q(a) = I$ are each other's inverses. Since $Q(a)$ commutes with $Q(b)$ it commutes with its inverse $Q(a)Q(b)Q(a) = Q(a)$. Similarly $Q(a)$ commutes with $Q(b)$. So $L(b)$ and $Q(b)$.

Indeed, if $Q(a)$ is invertible then the above implies $L(a)$ is invertible with inverse $Q(b)$. Any inverse b satisfies $(a^{2})b = L(b)a^{2} = Q(a)b = a$, so $ab = L(b)Q(a)b= Q(a)Q(b)1= 1$. Conversely if $a$ is invertible let $Q(a)$. Then $A$. The fundamental identity then implies that $a^{−1} = Q(a)^{−1}a$ and $Q(a)^{−1} = Q(a^{−1})$ are each other's inverses so that $a$.

This follows from the formula $Q(a)$.

Suppose that $Q(b)$. Then by the fundamental identity $Q(a)b = a$ is invertible, so $b = Q(a)^{−1}a$ is invertible.

This is an immediate consequence of the fundamental identity and the fact that $Q(a)$ is invertible if and only $b = Q(a)^{−1}a$ and $Q(a)b = a$ are invertible.

In the commutation identity $Q(b)$, set $Q(a)$ with $Q(a)b^{2} = Q(a)Q(b)1=1$. Then $a$ and $a^{−1}$. Since $a^{−1} = Q(a)^{−1}a$ commutes with $a$, $Q(a)$.

If $Q(a)c = 1$ and $Q(a)$ commute, then $a$ implies $Q(a)b$. Conversely suppose that $a$ is invertible with inverse $b$. Then $(Q(a)b)^{−1} = Q(a^{−1})b^{−1}$. Morevoer $STS$ commutes with $S$ and hence its inverse $T$. So it commutes with $a$.

The algebra $Q(a)L(a^{−1}) = L(a)$ is commutative and associative, so if $Q(a)R(b,a) = Q(Q(a)b,a)$ is an inverse there $b = c^{2}$ and $c = a^{−1}$. Conversely $Q(a)b = 1$ leaves $Q(1,a) = L(a)$ invariant. So if it is bijective on $L(a)$ it is bijective there. Thus $L(c^{2})$ lies in $R(b,a) = L(c) = L(a^{−1})$.

Elementary properties of proper mutations
In fact multiplication in the algebra Ay is given by


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

so by definition is commutative. It follows that


 * $$\displaystyle{a\circ b = L_y(a)b,}$$

with


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

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


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

Taking a = e gives


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

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

Now for y invertible set


 * $$\displaystyle{Q_y(a)=Q(a)Q(y),\,\,\, R_y(a,b)=R(a,Q(y)b).}$$

Then


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

Moreover,


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

Finally


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

since


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

Hence


 * $$\displaystyle{Q_y(a)R_y(b,a)=Q(a)Q(y)R(b,Q(y)a)=Q(y^{-1}) Q(Q(y)a)R(b,Q(y)a)= Q(y)^{-1} R(Q(y)a,b)Q(Q(y)a)=R_y(a,b) Q_y(a).}$$

Thus $b$ is a unital quadratic Jordan algebra. It therefore corresponds to a linear Jordan algebra with the associated Jordan multiplication operator M(a) given by


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

This shows that the operators $ab = 1$ satisfy the Jordan identity so that the proper mutation or isotope $[L(a),L(b)] = 0$ is a unital Jordan algebra. The correspondence with quadratic Jordan algebras shows that its quadratic representation is given by $a$.

Nonunital mutations
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.

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

$$\displaystyle{Q_1(a\oplus \alpha 1)(b\oplus \beta 1)= \alpha^2\beta 1\oplus [\alpha^2 a + \alpha^2 b + 2\alpha\beta a + \alpha \{a,y, b\} + \beta Q(a)y + Q(a)Q(y)b].}$$

It can then be verified that $b$ is a unital quadratic Jordan algebra. The unital Jordan algebra to which it corresponds has Ay as an ideal, so that in particular Ay satisfies the Jordan identity. The identities for a unital quadratic Jordan algebra follow from the following compatibility properties of the quadratic map $b = a^{−1}$ and the squaring map $L(a)$:



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

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


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

To prove the identity for x, set $L(b)$. Then $ba = 1$. Thus $b(a^{2}) = a$ commutes with $a$ and $b$. Since $ab = 1$, it also commutes with $L(b)$ and $Q(b)$. Since $Q(a)$, $L(a) = Q(a)L(b)$ also commutes with $A$ and $k$.

It follows that $a$. Moreover, $k[a]$ since $a^{−1}$. So $k[a]$ commutes with $k[a]$ and hence $b$. Thus $ab =1$ has inverse $a^{2}b = a$.

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

Let $A^{y}$ in $y$. It is invertible in A, as is $y^{−1}$. So by the special case of Hua's identity for x in $A^{y}$


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

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


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

If a is invertible then


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

while if b is invertible then


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

In fact if a is invertible



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:

Quasi-invertibility
Let A be a finite-dimensional unital Jordan algebra over a field k of characteristic ≠ 2. For a pair $y$ with $A^{y}$ and $Q_{y}(x) = Q(x)Q(y)$ invertible define

In this case the Bergman operator $a ∘ e = a$ defines an invertible operator on A and

In fact


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

Moreover, by definition $(A,Q^{y},y^{−1})$ is invertible if and only if $L_{y}(a)$ is invertible. In that case

Indeed,


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

The assumption that $A^{y}$ be invertible can be dropped since $Q_{y}$ can be defined only supposing that the Bergman operator $(A_{1}, Q_{1}, 1)$ is invertible. The pair $Q_{y}(a) = Q(a)Q(y)$ is then said to be quasi-invertible. In that case $S_{y}(a) = Q(a)y$ is defined by the formula


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

If $R_{y}(a,a) = L_{y}(S_{y}(a)).$ is invertible, then $[Q_{y}(a),L_{y}(a)] = 0.$ for some $Q_{y}(a)S_{y}(a) = S_{y}(S_{y}(a)).$. The fundamental identity implies that $Q_{y}∘ S_{y} = S_{y} ∘ Q_{y}.$. So by finite-dimensionality $Q_{y}(a) Q_{y}(b) S_{y}(a) = S_{y}(Q_{y}(a)b).$ is invertible. Thus $Q_{y}(Q_{y}(a)b) = Q_{y}(a) Q_{y}(b) Q_{y}(a)$ is invertible if and only if $y = (1 – x)^{−1}$ is invertible and in this case

In fact



so the formula follows by applying $L(y) = Q(1 – x)^{−1}L(1 – x)$ to both sides. As before $L(y)$ is quasi-invertible if and only if $L(x)$ is quasi-invertible; and in that case


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

If k = R or C, this would follow by continuity from the special case where $Q(x)$ and $Q(y) = Q(1 – x)^{−1}$ were invertible. In general the proof requires four identities for the Bergman operator:

In fact applying $L(x)$ to the identity $Q(x)$ yields


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

The first identity follows by cancelling $L(x^{−1}) = Q(x)^{−1}L(x)$ and $L(y)$. The second identity follows by similar cancellation in



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



In fact $L(x^{−1})$ is quasi-invertible if and only if $Q(x^{−1})$ is quasi-invertible in the mutation $(x^{−1} – 1)xy =(1 – x) y = 1$. Since this mutation might not necessarily unital this means that when an identity is adjoint $y – 1 = xy$ becomes invertible in $(1 – x)y = 1$. This condition can be expressed as follows without mentioning the mutation or homotope:

In fact if $L(xy)$ is quasi-invertible, then $L(x)$ satisfies the first identity by definition. The second follows because $L(x^{−1} – 1)$. Conversely the conditions state that in $1 – x^{−1}$ the conditions imply that $1 – y$ is the inverse of $A^{a}$. On the other hand, $A^{a}$ for $a^{−1}$ in $A^{a}$. Hence $Q(a)^{−1} c^{−1}$ is invertible.

Equivalence relation
Let A be a finite-dimensional unital Jordan algebra over a field k of characteristic ≠ 2. Two pairs $x = Q(a)^{−1}b$ with $A^{a}$ invertible are said to be equivalent if $a^{−1} – Q(a)^{−1}b = Q(a)^{−1}(a – b)$ is invertible and $A^{a}$.

This is an equivalence relation, since if $Q(a)Q(a^{−1} − b) = Q(a)[Q(a^{−1} − 2Q(a^{−1},b) + Q(b)]=I − 2Q(a)Q(a^{−1},b) + Q(a)Q(b)=I − R(a,b) + Q(a)Q(b)$ is invertible $(a,b)$ so that a pair $a$ is equivalent to itself. It is symmetric since from the definition $a^{−1} − b$. It is transitive. For suppose that $B(a,b) = Q(a)Q(a^{−1} − b)$ is a third pair with $a^{−1} − b − c$ invertible and $(a^{b})^{−1} − c$. From the above


 * $$\displaystyle{a_1^{-1} - b_1 + b_3= (a_1^{-1} - b_1 +b_2) -b_2 + b_3=a_2^{-1} -b_2+b_3}$$

is invertible and


 * $$\displaystyle{a_3=a_2^{b_2-b_3}=(a_1^{b_1-b_2})^{b_2-b_3}=a_1^{b_1-b_3}.}$$

As for quasi-invertibility, this definition can be extended to the case where $a$ and $a^{b}$ are not assumed to be invertible.

Two pairs $B(a,b)$ are said to be equivalent if $(a,b)$ is quasi-invertible and $a^{b}$. 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 $B(a,b)$ is quasi-invertible and $B(a,b)c = 1$.
 * The relation is symmetric, since $c$.
 * The relation is transitive. For suppose that $B(a,b)Q(c)B(b,a) = I$ is a third pair with $B(b,a)$ quasi-invertible and $(a,b)$. In this case


 * $$\displaystyle{B(a_1,b_1-b_3)=B(a_1,b_1-b_2)B(a_2,b_2-b_3),}$$


 * so that $(b,a)$ is quasi-invertible with


 * $$\displaystyle{a_3=a_2^{b_2-b_3}=(a_1^{b_1-b_2})^{b_2-b_3}=a_1^{b_1-b_3}.}$$

The equivalence class of $B(a,b)(a + Q(a)b^{a}) = a − 2R(a,b)a + Q(a)Q(b)a + Q(a)(b − Q(b)a) = a − Q(a)b,$ is denoted by $B(a,b)^{−1}$.

Structure groups
Let $(a,b+c)$ be a finite-dimensional complex semisimple unital Jordan algebra. If $(a^{b},c)$ is an operator on $a$, let $a^{−1} − b$ be its transpose with respect to the trace form. Thus $Q$, $B(a,b)a^{b} = a − Q(a)b$, $B(a,b)$ and $B(b,a)$. The structure group of A consists of g in $B(a,b)Q(a^{b},c)B(b,a) = Q(B(a,b)a^{b},B(a,b)c) = Q(a − Q(a)b,B(a,b)c) = B(a,b)(Q(a,c) − R(c,b)Q(a)) = (Q(a,c) − Q(a)R(b,c))B(b,a)$ such that


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

They form a group $B(a,b)B(a^{b},c) = B(a,b)(I − R(a^{b},c) + Q(a^{b})Q(c)) = I − R(a,b + c) + Q(a) Q(b + c) = B(a,b+c)$. 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 = gt.


 * The structure group is closed under taking transposes g ↦ gt 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 = (gt)−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 $(a,b)$ on $a$ which gives rise to a complex inner product on $A^{b}$. The unitary structure group Γu(A) is the subgroup of Γ(A) consisting of unitary operators, so that $1 − a$. The identity component of $A^{b} ⊕ k1$ is denoted by $(a,b)$. It is a connected closed subgroup of $c$.


 * The stabilizer of 1 in Γu(A) is Aut E.
 * Every g in Γu(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 Γu(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 Γu(A) acts transitively on S
 * Given a Jordan frame (ei) and v in A, there is an operator u in the identity component of Γu(A) such that uv = Σ αi ei with αi ≥ 0. If v is invertible, then αi > 0.

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


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

Then $B(a,b)c = a − Q(a)b$ is quasi-invertible if and only if $B(a,b)Q(c)b = Q(a)b$ is quasi-invertible and


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

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


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

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 $$\mathfrak{g}_0$$ of $c = a^{b}$. Since $(a,b)$, the corresponding complex Lie algebra contains the operators $c = a^{b}$. The commutators $B(a,b)Q(a^{b}) = Q(a)$ span the complex Lie algebra of derivations of $A^{b} ⊕ k1$. The operators $1 + c$ span $$\mathfrak{g}_0$$ and satisfy $1 − a$ and $( 1 − a) ∘ x = B(a,b)x$.

Geometric properties of quotient space
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 $A^{b}$ by the equivalence relation. Let $B(a,b)$ be the subset of X of classes $(a_{i},b_{i})$. The map $a_{i}$, $(a_{1})^{−1} − b_{1} + b_{2}$ is injective. A subset $a_{2} = (a_{1})^{b_{1} − b_{2}}|undefined$ of $a$ is defined to be open if and only if $a^{0} = a$ is open for all $(a,b)$.

The transition maps of the atlas with charts $a_{1} = (a_{2})^{b_{2} − b_{1}}|undefined$ are given by


 * $$\displaystyle{\varphi_{cb}=\varphi_c\circ\varphi_b^{-1}:\varphi_b(X_b\cap X_c)\rightarrow \varphi_c(X_b\cap X_c).}$$

and are injective and holomorphic since


 * $$\displaystyle{\varphi_{cb}(a)=a^{b-c}}$$

with derivative


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

This defines the structure of a complex manifold on X because $(a_{3},b_{3})$ on $(a_{2})^{−1} − b_{2} + b_{3}$.

Indeed, all the polynomial functions $a_{3} = (a_{2})^{b_{2} − b_{3}}|undefined$ are non-trivial since $a$. Therefore, there is a $a^{−1} − b$ such that $(a_{i},b_{i})$ for all i, which is precisely the criterion for $(a_{1}, b_{1} − b_{2})$ to lie in $a_{2} = (a_{1})^{b_{1} − b_{2}}|undefined$.

uses the Bergman operators to construct an explicit biholomorphism between X and a closed smooth algebraic subvariety of complex projective space. This implies in particular that $(a,0)$ is compact. There is a more direct proof of compactness using symmetry groups.

Given a Jordan frame (ei) in E, for every a in A there is a k in U = Γu(A) such that $a^{0} = a$ with $a_{1} = (a_{2})^{b_{2} − b_{1}}|undefined$ (and $(a_{3},b_{3})$ 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 $(a_{2}, b_{2} − b_{3})$, which is the complexification of $a_{3} = (a_{2})^{b_{2} − b_{3}}|undefined$. Let $(a_{1},b_{1} − b_{3})$ be the complex manifold constructed for $(a,b)$. Because $(a:b)$ is a direct sum of copies of C, Z is just a product of Riemann spheres, one for each $A$. 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 = Ae ⊂ 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 $T$ and k in $A$. The mapping of Z into X is in fact an embedding. This is a consequence of $T^{t}$ being quasi-invertible in $L(a)^{t} = L(a)$ if and only if it is quasi-invertible in $Q(a)^{t} = Q(a)$. Indeed, if $R(a,b)^{t} = R(b,a)$ is injective on A, its restriction to $B(a,b)^{t} = B(b,a)$ is also injective. Conversely, the two equations for the quasi-inverse in $GL(A)$ imply that it is also a quasi-inverse in $Γ(A)$.

Möbius transformations
Let $a ↦ a*$ 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


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

then


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

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


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

The matrix J corresponds to the Möbius transformation $A$ and can be written


 * $$\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 $Γ_{u}(A) = Γ(A) ∩ U(A)$, the action of $Γ_{u}(A)$ in SL(2,C) is defined by the same formula


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

This defines an element of $K$ provided that $U(A)$ is invertible in $(x,y)$. The action is thus defined everywhere on $(gx,(g^{t})^{−1}y)$ if g is upper triangular. On the other hand, the action on X is simple to define for lower triangular matrices.


 * For diagonal matrices g with diagonal entries $GL(A)$ and $Q(e^{a}) = e^{2L(a)}$, $L(a)$ is a well-defined holomorphic action on $[L(a),L(b)]$ which passes to the quotient X. On $A$ it agrees with the Möbius action.
 * For lower unitriangular matrices, with off-diagonal parameter γ, define $R(a,b) = [L(a),L(b)] + L(ab)$. Again this is holomorphic on $R(a,b)^{t} = R(b,a)$ and passes to the quotient X. When $[R(a,b),R(c,d)]=R(R(a,b)c,d) − R(c,R(b,a)d)$ and $X$,


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


 * if $A×A$ is invertible, so this is an extension of the Möbius action.


 * For upper unitriangular matrices, with off-diagonal parameter β, the action on $X_{b}$ is defined by $(a:b)$. showed that this defined a complex one-parameter flow on $φ_{b}:X_{b} → A$. The corresponding holomorphic complex vector field extended to $(a:b) ↦ a$, so that the action on the compact complex manifold $U$ could be defined by the associated complex flow. A simpler method is to note that the operator $X$ can be implemented directly using its intertwining relations with the unitary structure group.

In fact on the invertible elements in A, the operator $U ∩ X_{b}$ satisfies $b$. To define a biholomorphism j on X such that $φ_{b}$, it is enough to define these for $φ_{dc} ∘ φ_{cb} = φ_{db}$ in some suitable orbit of Γ(A) or Γu(A). On the other hand, as indicated above, given a Jordan frame (ei) in E, for every a in A there is a k in U = Γu(A) such that $φ_{b}(X_{b} ∩ X_{c} ∩ X_{d})$ with $(a_{i}:b_{i})$.

The computation of $X$ in the associative commutative algebra $X_{b}$ is straightforward since it is a direct product. For $p_{i}(b) = det B(a_{i},b_{i} − b)$ and $p_{i}(b_{i}) = 1$, the Bergman operator on $b$ has determinant $p_{i}(b) ≠ 0$. In particular $(a_{i}:b_{i})$ for some λ ≠ 0. So that $X_{b}$ is equivalent to $X$. Let $a=k(Σ α_{i} e_{i})$. On $α_{i} ≥ 0$, for a dense set of $α_{i} > 0$, the pair $A_{e} = ⊕ Ce_{i}$ is equivalent to $E_{e} = ⊕ Re_{i}$ with b invertible. Then $Z$ is equivalent to $A_{e}$. Since $A_{e}$ is holomorphic it follows that j has a unique continuous extension to X such that $e_{i}$ for $A_{e}$ in $Γ_{u}(A)$, the extension is holomorphic and for $(x,y)$, $A_{e}$

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.

This action of $A$ is compatible with inclusions. More generally if $B(x,y)$ is a Jordan frame, there is an action of $A_{e}$ on Ae which extends to A. If $A_{e}$ and $A$, then $A$ and $j(z) = −z^{−1}$ give the action of the product of the lower and upper unitriangular matrices. If $a$ is invertible, the corresponding product of diagonal matrices act as $A$. In particular the diagonal matrices give an action of $g$ and $C[a]$.

Holomorphic symmetry group
Let $γa + δ1$ be a finite-dimensional complex semisimple unital Jordan algebra. There is a transitive holomorphic action of a complex matrix group $A$ on the compact complex manifold $A$. described $α$ analogously to $α^{−1}$ in terms of generators and relations. $g(a,b) = (α^{2}a, α^{−2}b)$ acts on the corresponding finite-dimensional Lie algebra of holomorphic vector fields restricted to $A^{2}$, so that $X_{0} = A$ 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 $g(a,b) = (a,b − γ1)$ of the compact group acts transitively on $A^{2}$, so that $b = 0$ can be identified as a Hermitian symmetric space of compact type.

The group G is generated by three types of holomorphic transformation on $γ ≠ 0$:


 * Operators W corresponding to elements W in $γa + 1$ given by $X_{0} = (A:0)$. These were already described above. On $g(a,0) = (a + β1)$, they are given by $A$.
 * Operators Sc defined by $X$. These are the analogue of lower unitriangular matrices and form a subgroup isomorphic to the additive group of $X$, with the given parametrization. Again these act holomorphically on $J$ and the action passes to the quotient X. On $j(a) = −a^{−1}$ the action is given by $j(ga) = (g^{t})^{−1}j(a)$ if $j ∘ g = (g^{t})^{−1} ∘ j$ is quasi-invertible.
 * The transformation $(a:b)$ corresponding to $a=k(Σ α_{i} e_{i})$ in $α_{i} ≥ 0$. It was constructed above as part of the action of $j$ on $A_{e}$. On invertible elements in $c = Σ α_{i} e_{i}$ it is given by $d = Σ β_{i} e_{i}$.

The operators $A_{e}$ normalize the group of operators $det B(c,d) = Π(1 − α_{i}β_{i})^{2}$. Similarly the operator $det B(c,d − λ) ≠ 0$ normalizes the structure group, $(c,d)$. The operators $(x,λ)$ also form a group of holomorphic transformations isomorphic to the additive group of $μ = −λ^{−1}$. They generalize the upper unitriangular subgroup of $A$. This group is normalized by the operators W of the structure group. The operator $a$ acts on $(a,λ)$ as $(b,0)$. If $(−b^{−1},0)$ is a scalar the operators $(μ − μ^{2}a,μ)$ and $a ↦ μ − μ^{2}a$ coincide with the operators corresponding to lower and upper unitriangular matrices in $j ∘ g = (g^{t})^{−1} ∘ j$. Accordingly, there is a relation $g$ and $Γ(A)$ is a subgroup of G. defines the operators $λ ≠ 0$ in terms of the flow associated to a holomorphic vector field on $μ = −λ^{−1}$, while give a direct algebraic description.

Indeed, $SL(2,C)$.

Let $e_{1}, ..., e_{m}$ and $SL(2,C)^{m}$ be the complex Abelian groups formed by the symmetries $c = Σ γ_{i}e_{i}$ and $b = Σ β_{i}e_{i}$ respectively. Let $S(c)$.

The two expressions for $T(b)$ are equivalent as follows by conjugating by $a = Σ α_{i}e_{i}$.

For $W = Q(a)$ invertible, Hua's identity can be rewritten


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

Moreover, $(C*)^{m}$ and $T^{m}$.

The convariance relations show that the elements of $A$ fall into sets $G$, $X$, $G$, $SL(2,C)$. ... 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



For then if there are three or more occurrences of $X_{0} = A$, the number can be recursively reduced to two. Given $G$ in $H$, pick $X$ so that $X$ and $X$ are invertible. Then


 * $$\displaystyle{j T_a j T_b  j =  j  T_c  T_\lambda  j  T_{\lambda^{-1}}  T_d \circ j = \lambda^2 j T_c  j  T_{-\lambda}  j T_d \ j = \lambda^2 T_{-c^{-1}} j Q(c^{-1})T_{-c^{-1} -\lambda - d^{-1}} j Q(d^{-1})jT_{-d^{-1}},}$$

which lies in $Γ(A)$.

It suffices to check that if $W(a,b) = (Wa, (W^{t})^{−1}b)$, then $X_{0} = A$. If so $a ↦ Wa$, so $S_{c}(a,b) = (a,b + c)$.

Exchange relations
For $A$ invertible, Hua's identity can be rewritten


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

Since $A^{2}$, the operators $A$ belong to the group generated by $a ↦ a^{c}$.

For quasi-invertible pairs $(a,c)$, there are the "exchange relations"

This identity shows that $j$ is in the group generated by $J$. Taking inverses, it is equivalent to the identity $SL(2,C)$.

To prove the exchange relations, it suffices to check that it valid when applied to points the dense set of points $PSL(2,C) = SL(2,C)/{±I}$ in $X$ for which $A$ is quasi-invertible. It then reduces to the identity:

In fact, if $a ↦ −a^{−1}$ is quasi-invertible, then $W$ is quasi-invertible if and only if $S_{c}$ is quasi-invertible. This follows because $j$ is quasi-invertible if and only if $j ∘ W = (W^{t})^{−1} ∘ j$ is. Moreover, the above formula holds in this case.

For the proof, two more identities are required:


 * $$\displaystyle{B(c+b,a)=B(c,a^b)B(b,a)}$$
 * $$\displaystyle{R(a,b)=R(a^b,b-Q(b)a)=R(a-Q(a)b,b^a)}$$

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 $T_{c} = j ∘ S_{−c} ∘ j$ in the identity $A$ yields



so the identity follows by cancelling $SL(2,C)$.

To prove the formula, the relations $T_{c}$ and $A$ show that it is enough to prove that



Indeed, $a ↦ a + c$. On the other hand, $c$ and $S_{c}$. So $T_{c}$.

Now set $SL(2,C)$. Then the exchange relations imply that $j = S_{1} ∘ T_{1} ∘ S_{1}$ lies in $PSL(2,C)$ if and only if $T_{c}$ is quasi-invertible; and that $X$ lies in $G$ if and only if $X$ is in $S_{b}T_{a}(0:0) = (a:b)$.

In fact if $G_{−1}$ lies in $G_{+1}$, then $T_{c}$ is equivalent to $S_{c}$, so it a quasi-invertible pair; the converse follows from the exchange relations. Clearly $G_{0} = Γ(A)$. The converse follows from $G$ and the criterion for $j$ to lie in $a$.

Lie algebra of holomorphic vector fields
The compact complex manifold $j = S_{1} ∘ T_{1} ∘ S_{1}$ is modelled on the space $S_{c} = j ∘ T_{−c} ∘ j$. The derivatives of the transition maps describe the tangent bundle through holomorphic transition functions $G$. These are given by $G_{0}G_{1}$, so the structure group of the corresponding principal fiber bundle reduces to $G_{0}G_{1}jG_{1}$, the structure group of $G_{0}G_{1}jG_{1}jG_{1}$. The corresponding holomorphic vector bundle with fibre $G_{0}G_{1}jG_{1}jG_{1}jG_{1}$ is the tangent bundle of the complex manifold $G$. 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  and, 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 $j ∘ G_{1} ∘j ∘ G_{1} ∘j ⊆ G_{0} G_{1} ∘j ∘ G_{1} ∘ j ∘ G_{1}$ give rise to linear spaces of holomorphic vector fields on $j$ and hence $a, b$.


 * The structure group $A$ has Lie algebra $$\mathfrak{g}_0$$ spanned by the operators $λ ≠ 0$. These define a complex Lie algebra of linear vector fields $c = a − λ$ on $d = b − λ^{−1}$.
 * The translation operators act on $G_{0}G_{1} ∘ j ∘ G_{1} ∘ j ∘ G_{1}$ as $(0:0)$. The corresponding one-parameter subgroups are given by $G$ and correspond to the constant vector fields $G_{0}G_{−1}$. These give an Abelian Lie algebra $$\mathfrak{g}_{-1}$$ of vector fields on $S_{a}T_{b}(0:0) = (0:0)$.
 * The operators $b = 0$ defined on $(b:0) = (0: −a) = (0:0)$ by $b = 0$. The corresponding one-parameter groups $G$ define quadratic vector fields $G_{±1}$ on $a$. These give an Abelian Lie algebra $$\mathfrak{g}_{1}$$ of vector fields on $j = S_{1} ∘ T_{1} ∘ S_{1}$.

Let


 * $$\displaystyle{\mathfrak{g}=\mathfrak{g}_{-1}\oplus\mathfrak{g}_0\oplus\mathfrak{g}_1.}$$

Then, defining $$\mathfrak{g}_i = (0)$$ for $Q(a)$, $$\mathfrak{g}$$ forms a complex Lie algebra with


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

This gives the structure of a 3-graded Lie algebra. For elements $G_{±1}$ in $$\mathfrak{g}$$, the Lie bracket is given by


 * $$\displaystyle{[(a_1,T_1,b_1),(a_2,T_2,b_2)]=(T_1a_2-T_2a_1,[T_1,T_2]+R(a_1,b_2)-R(a_2,b_1),T_2^tb_1-T_1^tb_2)}$$

The group $(a,b)$ of Möbius transformations of X normalizes the Lie algebra $$\mathfrak{g}$$. The transformation $S_{b}T_{a} = T_{a^{b}}B(a,b)^{−1}S_{b^{a}}|undefined$ corresponding to the Weyl group element $B(a,b)$ induces the involutive automorphism $G_{±1}$ given by


 * $$\displaystyle{\sigma(a,T,b)=(b,-T^t,a).}$$

More generally the action of a Möbius transformation


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

can be described explicitly. In terms of generators diagonal matrices act as


 * $$\displaystyle{\begin{pmatrix}\alpha & 0 \\ 0 & \alpha^{-1}\end{pmatrix}(a,T,b)=(\alpha^2 a,T,\alpha^{-2}b),}$$

upper unitriangular matrices act as


 * $$\displaystyle{\begin{pmatrix}1 & \beta \\ 0 & 1\end{pmatrix}(a,T,b)=(a +\beta T(1) -\beta^2 b,T - \beta L(a),b),}$$

and lower unitriangular matrices act as


 * $$\displaystyle{\begin{pmatrix}1 & 0 \\ \gamma & 1\end{pmatrix}(a,T,b)=(a,T-\gamma L(b),b-\gamma T^t(1) -\gamma^2 a).}$$

This can be written uniformly in matrix notation as


 * $$\displaystyle{\begin{pmatrix}g(T) & g(a) \\ g(b) & g(T)^t\end{pmatrix}= g \begin{pmatrix}T & a \\ b & T^t\end{pmatrix} g^{-1}.}$$

In particular the grading corresponds to the action of the diagonal subgroup of $T_{a}S_{b} = S_{b^{a}}B(a,b)T_{a^{b}}|undefined$, even with |α| = 1, so a copy of T.

The Killing form is given by


 * $$\displaystyle{\mathbf((a_1,T_1,b_1),(a_2,T_2,b_2))= (a_1,b_2) + (b_1,a_2) + \beta(T_1,T_2),}$$

where $(c:0)$ is the symmetric bilinear form defined by


 * $$\displaystyle{\beta(R(a,b),R(c,d))=(R(a,b)c,d)=(R(c,d)a,b),}$$

with the bilinear form $X$ corresponding to the trace form: $(a+c,b)$.

More generally the generators of the group $(a,b)$ act by automorphisms on $$\mathfrak{g}$$ as


 * $$\displaystyle{W(a,T,b)=(Wa,WTW^{-1},(W^t)^{-1}b),}$$
 * $$\displaystyle{J(a,T,b)=(-b,-T^t,-a),}$$
 * $$\displaystyle{T_x(a,T,b)=(a+Tx -Q(x)b, T-R(x,b),b),}$$
 * $$\displaystyle{S_y(a,T,b)=(a,T-R(a,y),b-T^ty-Q(y)a).}$$

The nondegeneracy of the Killing form is immediate from the explicit formula. By Cartan's criterion, $$\mathfrak{g}$$ is semisimple. In the next section the group $(a + c,b)$ is realized as the complexification of a connected compact Lie group $(c,b^{a})$ with trivial center, so semisimple. This gives a direct means to verify semisimplicity. The group H also acts transitively on X.

To prove that $$\mathfrak{g}$$ exhausts the holomorphic vector fields on $(x,y)$, note the group T acts on holomorphic vector fields. The restriction of such a vector field to $(y,x)$ gives a holomorphic map of A into A. The power series expansion around 0 is a convergent sum of homogeneous parts of degree $c = b − Q(b)a$. The action of $B(a,b)R(a^{b},c) = R(a,c) − Q(a)Q(b,c)$ scales the part of degree $B(a,b)R(a^{b},b − Q(b)c) = B(a,b)R(a,b),$ by $B(a,b)$. By taking Fourier coefficients with respect to T, the part of degree m is also a holomorphic vector field. Since conjugation by $(a + c)^{b} = B(a,c)^{−1}(a + c − Q(a + c)b)$ gives the inverse on $a^{b} + B(a,b)^{−1}c^{(b^{a})} = B(a + c,b)^{−1}(B(c,b^{a}) (a − Q(a)b) + c − Q(c)b^{a})$, it follows that the only possible degrees are 0, 1 and 2. Degree 0 is accounted for by the constant fields. Since conjugation by $a + c − Q(a + c)b = B(c,b^{a}) (a − Q(a)b) + c − Q(c)b^{a}.$ interchanges degree 0 and degree 2, it follows that $$\mathfrak{g}_{\pm 1}$$ account for all these holomorphic vector fields. Any further holomorphic vector field would have to appear in degree 1 and so would have the form $B(c,b^{a}) (a − Q(a)b) + c − Q(c)b^{a} = a + c − Q(a)b + 2R(c,b^{a})(a − Q(a)b) − Q(c)[ b^{a} − Q(b^{a})(a − Q(a)b)]$ for some $2R(c,b^{a})(a − Q(a)b) = 2R(c,a − Q(a)b)b^{a} = R(a,b)c = 2Q(a,c)b$ in $b^{a} − Q(b^{a})(a − Q(a)b) = b^{a} − Q(b)B(a,b)^{−1}(a − Q(a)b) = b^{a} − Q(b)a^{b} = b$. Conjugation by J would give another such map N. Moreover, $B(c,b^{a}) (a − Q(a)b) + c − Q(c)b^{a} = a + c − Q(a)b − 2Q(a,c)b − Q(c)b = a + c − Q(a + c)b$. But then


 * $$\displaystyle{e^{tM}(0,0,b)=Je^{tN}J(0,0,b)=Je^{tN}(b,0,0)=(0,0,e^{tN}b).}$$

Set $Ω = G_{+1}G_{0}G_{−1}$ and $S_{b} T_{a}$. Then


 * $$\displaystyle{Q(U_ta)b = U_tQ(a)V_{-t}b.}$$

It follows that $Ω$ lies in $(a,b)$ for all $g$ and hence that $Ω$ lies in $$\mathfrak{g}_0$$. So $$\mathfrak{g}$$ is exactly the space of holomorphic vector fields on X.

Compact real form
Suppose $g(0:0)$ acts trivially on $$\mathfrak{g}$$. Then $X_{0}$ must leave the subalgebra $S_{b} T_{a}$ invariant. Hence so must $Ω$. This forces $(a,b)$, so that $(x,0)$. But then $Ω(0:0) = G_{1}(0:0) = X_{0}$ must leave the subalgebra $G = G_{−1}G_{1} G_{0}G_{−1}$ invariant, so that $S_{b} T_{a}$ and $Ω$. If $X$ acts trivially, $A$.

The group $F_{bc}:X_{b} ∩ X_{c} → GL(A)$ can thus be identified with its image in GL $$\mathfrak{g}$$.

Let $F_{bc}(a,b) = B(a,b − c)$ be the complexification of a Euclidean Jordan algebra $Γ(A)$. For $A$, set $A$. The trace form on $X$ defines a complex inner product on $X$ and hence an adjoint operation. The unitary structure group $X$ consists of those $X_{0} = A$ in $Γ(A)$ that are in $R(x,y)$, i.e. satisfy $a ↦ R(x,y)a$. It is a closed subgroup of U(A). Its Lie algebra consists of the skew-adjoint elements in $$\mathfrak{g}_0$$. Define a conjugate linear involution $A$ on $$\mathfrak{g}$$ by


 * $$\displaystyle{\theta(a,T,b)=(b^*,-T^*, a^*).}$$

This is a period 2 conjugate-linear automorphism of the Lie algebra. It induces an automorphism of $A$, which on the generators is given by


 * $$\displaystyle{\theta(S_a)=T_{a^*},\,\,\, \theta(j)=j,\,\,\, \theta(T_b)=S_{b^*},\,\,\, \theta(W)=(W^*)^{-1}.}$$

Let $T_{c}(a) = a + c$ be the fixed point subgroup of $T_{tc}$ in $a ↦ c$. Let $$\mathfrak{h}$$ be the fixed point subalgebra of $A$ in $$\mathfrak{g}$$. Define a sesquilinear form on $$\mathfrak{g}$$ by $S_{c}$. This defines a complex inner product on $$\mathfrak{g}$$ which restricts to a real inner product on $$\mathfrak{h}$$. Both are preserved by $X$. Let $S_{c}(a,b) = (a,b − c)$ be the identity component of $S_{tc}$. It lies in $a ↦ Q(a)c$. Let $A$ be the diagonal torus associated with a Jordan frame in E. The action of $A$ is compatible with $i ≠ −1, 0, 1$ which sends a unimodular matrix $$\begin{pmatrix} \alpha & \beta\\ \gamma & \delta\end{pmatrix}$$ to $$\begin{pmatrix} \overline{\delta} & -\overline{\gamma}\\ -\overline{\beta} & \overline{\alpha}\end{pmatrix}$$. In particular this gives a homomorphism of $(a,T,b)$ into $PSL(2,C)$.

Now every matrix $j(z) = −z^{−1}$ in $J$ can be written as a product


 * $$\displaystyle{M = \begin{pmatrix} \zeta_1 & 0\\ 0 & \zeta_1^{-1}\end{pmatrix}

\begin{pmatrix} \cos \varphi & \sin \varphi\\ -\sin \varphi & \cos \varphi\end{pmatrix} \begin{pmatrix} \zeta_2 & 0\\ 0 & \zeta_2^{-1}\end{pmatrix}.}$$

The factor in the middle gives another maximal torus in $σ$ obtained by conjugating by $SL(2,C)$. If $β(T_{1},T_{2})$ with |αi| = 1, then $(a,b)$ gives the action of the diagonal torus $(a,b) = Tr L(ab)$ and corresponds to an element of  $G$. The element $G$ lies in $H$ and its image is a Möbius transformation $X$ lying in $X$. Thus $X_{0} = A$ is another torus in $m ≥ 0$ and $T$ coincides with the image of $m$.

Since $α^{2m − 2}$ for the compact complex manifold corresponding to $J$, if follows that $T$, where $J$ is the image of $a ↦ Ma$. On the other hand, $M$, so that $End A$. On the other hand, the stabilizer of $e^{tM}(a,0,0)= (e^{tM}a,0,0)$ in $U_{t} = e^{tM}$ is $V_{t} = e^{tB}$, since the fixed point subgroup of $U_{t}$ under $Γ(A)$ is $t$. Hence $M$. In particular H is compact and connected since both K and S are. Because it is a closed subgroup of U $$\mathfrak{g}$$, it is a Lie group. It contains K and hence its Lie algebra contains the operators $g = WT_{x}S_{y} T_{z}$ with $S_{y} T_{z}$. It contains the image of $(0,0,A)$ and hence the elements $S_{y}$ with $y = 0$ in $g = WT_{x + z}$. Since $T_{x+z}$ and $(A,0,0)$, it follows that the Lie algebra $$\mathfrak{h}_1$$ of $x + z = 0$ contains $g = W$ for all $W$ in $W = I$. Thus it contains $$\mathfrak{h}$$.

They are equal because all skew-adjoint derivations of $$\mathfrak{h}$$ are inner. In fact, since $G$ normalizes $$\mathfrak{h}$$ and the action by conjugation is faithful, the map of $$\mathfrak{h}_1$$ into the Lie algebra $$\mathfrak{d}$$ of derivations of $$\mathfrak{h}$$ is faithful. In particular $$\mathfrak{h}$$ has trivial center. To show that $$\mathfrak{h}$$ equals $$\mathfrak{h}_1$$, it suffices to show that $$\mathfrak{d}$$ coincides with $$\mathfrak{h}$$. Derivations on $$\mathfrak{h}$$ are skew-adjoint for the inner product given by minus the Killing form. Take the invariant inner product on $$\mathfrak{d}$$ given by $A = E + iE$. Since $$\mathfrak{h}$$ is invariant under $$\mathfrak{d}$$ so is its orthogonal complement. They are both ideals in $$\mathfrak{d}$$, so the Lie bracket between them must vanish. But then any derivation in the orthogonal complement would have 0 Lie bracket with $$\mathfrak{h}$$, so must be zero. Hence $$\mathfrak{h}$$ is the Lie algebra of $E$. (This also follows from a dimension count since $a = x + iy$.)

The formulas above for the action of $a* = x − iy$ and $E$ show that the image of $A$ is closed in GL $$\mathfrak{g}$$. Since $Γ_{u}(A)$ acts transitively on $g$ and the stabilizer of $Γ(A)$ in $U(A)$ is $gg*=g*g = I$, it follows that $θ$. The compactness of $G$ and closedness of $H$ implies that $θ$ is closed in GL $$\mathfrak{g}$$.

$G$ is a closed subgroup of GL  $$\mathfrak{g}$$ so a real Lie group. Since it contains $θ$ with $(a,b) = −B(a,θ(b))$ or $H$, its Lie algebra contains $$\mathfrak{g}$$. Since $$\mathfrak{g}$$ is the complexification of $$\mathfrak{h}$$, like $$\mathfrak{h}$$ all its derivations are inner and it has trivial center. Since the Lie algebra of $K$ normalizes $$\mathfrak{g}$$ and o is the only element centralizing $$\mathfrak{g}$$, as in the compact case the Lie algebra of $Γ_{u}(A)$ must be $$\mathfrak{g}$$. (This can also be seen by a dimension count since $H$.) Since it is a complex subspace, $K_{e} = T^{m}$ is a complex Lie group. It is connected because it is the continuous image of the connected set $SL(2,C)^{m}$. Since $$\mathfrak{g}$$ is the complexification of $$\mathfrak{h}$$, $θ$ is the complexification of $SU(2)^{m}$.

Noncompact real form
For $H$ in $M$ the spectral norm ||a|| is defined to be $SU(2)$ if $SU(2)$ with $J$ and $a = Σ α_{i}e_{i}$ in $Q(a)$. It is independent of choices and defines a norm on $T = T^{m}$. Let $K ⊆ H$ be the set of $J$ with ||a|| < 1 and let $SU(2)^{m}$ be the identity component of the closed subgroup of G carrying $j$ onto itself. It is generated by $H$, the Möbius transformations in $S = j ∘ T ∘ j$ and the image of $H$ corresponding to a Jordan frame. Let τ be the conjugate-linear period 2 automorphism of $$\mathfrak{g}$$ defined by


 * $$\displaystyle{\tau(a,T,b)=(-a^*,-T^*,-b^*).}$$

Let $$\mathfrak{h}^*$$ be the fixed point algebra of τ. It is the Lie algebra of $T ∘ S ∘ T$. It induces a period 2 automorphism of $SU(2)^{m}$ with fixed point subgroup $(0:0)$. The group $K$ acts transitively on $H = KSK$. The stabilizer of 0 is $H$.

The noncompact real semisimple Lie group $Z = SU(2)^{m}(0:0)$ acts on X with an open orbit $A^{e}$. As with the action of $Y = T S (0:0)$ on the Riemann sphere, it has only finitely many orbits. This orbit structure can be explicitly described when the Jordan algebra $Y$ is simple. Let $Z$ be the subset of $X = KY$ consisting of elements $X = KTS(0:0) = KS(0:0)$ with exactly $(0:0)$ of the αi less than one and exactly $H$ of them greater than one. Thus $K$. These sets are the intersections of the orbits $G_{0}G_{−1}$ of $θ$ with $K$. The orbits with $H = KSK$ are open. There is a unique compact orbit $(0,T,0)$. It is the Shilov boundary S of D consisting of elements $T* = −T$ with $SU(2)^{m}$ in $(a,0,a*)$, the underlying Euclidean Jordan algebra. $a$ is in the closure of $A_{e}$ if and only if $A = KA_{e}$ and $(k^{t})^{−1}(a*) = (ka)*$. In particular $H$ is in the closure of every orbit.

Jordan algebras with involution
The preceding theory describes irreducible Hermitian symmetric spaces of tube type in terms of unital Jordan algebras. In general Hermitian symmetric spaces are described  by a systematic extension of the above theory to Jordan pairs. In the development of, however, irreducible Hermitian symmetric spaces not of tube type are described in terms of period two automorphisms of simple Euclidean Jordan algebras. In fact any period 2 automorphism defines a Jordan pair: the general results of on Jordan pairs can be specialized to that setting.

Let τ be a period two automorphism of a simple Euclidean Jordan algebra E with complexification A. There are corresponding decompositions E = E+ ⊕ E− and A = A+ ⊕ A− into ±1 eigenspaces of τ. Let $(a,0,a*)$. τ is assumed to satisfy the additional condition that the trace form on $a$ defines an inner product. For $A$ in $H$, define $−Tr D_{1}D_{2}$ to be the restriction of $H$ to V. For a pair $dim X = dim H − dim K$ in $G$, define $W$ and $S_{y}$ to be the restriction of $G_{0}G_{−1}$ and $H$ to $X$. Then $(0:0)$ is simple if and only if the only subspaces invariant under all the operators $G$ and $G_{0}G_{−1}$ are $G = HG_{0}G_{−1}$ and $H$.

The conditions for quasi-invertibility in $G_{0}G_{−1}$ show that $G$ is invertible if and only if $G$ is invertible. The quasi-inverse $G$ is the same whether computed in $G_{i}$ or $i = 0$. A space of equivalence classes $±1$ can be defined on pairs $G$. It is a closed subspace of $G$, so compact. It also has the structure of a complex manifold, modelled on $dim X = dim G − dim G_{0}G_{−1}$. The structure group $G$ can be defined in terms of $H × G_{0}G_{−1}$ and it has as a subgroup the unitary structure group $G$ with identity component $H$. The group $a$ is the identity component of the fixed point subgroup of τ in $A$. Let $max α_{i}$ be the group of biholomorphisms of $a = u Σ α_{i}e_{i}$ generated by $α_{i} ≥ 0$ in $u$, the identity component of $K$, and the Abelian groups $A$ consisting of the $D$ and $a$ consisting of the $H*$ with $D$ and $K$ in $PSU(1,1)$. It acts transitively on $SU(1,1)^{m}$ with stabilizer $H*$ and $G$. The Lie algebra $$\mathfrak{g}_\tau$$ of holomorphic vector fields on $H*$ is a 3-graded Lie algebra,


 * $$\displaystyle{\mathfrak{g}_\tau = \mathfrak{g}_{\tau,+1} \oplus \mathfrak{g}_{\tau,0} \oplus

\mathfrak{g}_{\tau,-1}.}$$

Restricted to $H*$ the components are generated as before by the constant functions into $D$, by the operators $K$ and by the operators $H*$. The Lie brackets are given by exactly the same formula as before.

The spectral decomposition in $D$ and $SU(1,1)$ is accomplished using tripotents, i.e. elements $A$ such that $X_{0}(r,s)$. In this case $A$ is an idempotent in $a = u Σ α_{i}a_{i}$. There is a Pierce decomposition $r$ into eigenspaces of $s$. The operators $0 ≤ r + s ≤ m$ and $X(r,s)$ commute, so $H*$ leaves the eigenspaces above invariant. In fact $X_{0}$ acts as 0 on $r + s = m$, as 1/4 on $X(0,0)$ and 1 on $e^{ix}$. This induces a Pierce decomposition $x$. The subspace $E$ becomes a Euclidean Jordan algebra with unit $X(p,q)$ under the mutation Jordan product $X(r,s)$.

Two tripotents $p ≤ r$ and $q ≤ s$ are said to be orthogonal if all the operators $S$ when a and b are powers of $V ≡ A_{τ} = A_{−}$ and $V$ and if the corresponding idempotents $a$ and $V$ are orthogonal. In this case $Q_{τ}(a)$ and $Q(a)$ generate a commutative associative algebra and $(a,b)$, since $V^{2}$. Let $B_{τ}(a,b)$ be in $R_{τ}(a,b)$. Let $B(a,b)$ be the finite-dimensional real subspace spanned by odd powers of $R(a,b)$. The commuting self-adjoint operators $V$ with $V$ odd powers of $Q_{τ}(a)$ act on $R_{τ}(a,b)$, so can be simultaneously diagonalized by an orthonormal basis $(0)$. Since $V$ is a positive multiple of $A$, rescaling if necessary, $B_{τ}(a,b)$ can be chosen to be a tripotent. They form an orthogonal family by construction. Since $B(a,b)$ is in $a^{b}$, it can be written $A$ with $V$ real. These are called the eigenvalues of $X_{τ}$ (with respect to τ). Any other tripotent $V^{2}$ in $X$ has the form $V$ with $Γ(V)$, so the $Q_{τ}$ are up to sign the minimal tripotents in $Γ_{u}(V) = Γ(V) ∩ U(V)$.

A maximal family of orthogonal tripotents in $K_{τ}$ is called a Jordan frame. The tripotents are necessarily minimal. All Jordan frames have the same number of elements, called the rank of $K_{τ}$. Any two frames are related by an element in the subgroup of the structure group of $K$ preserving the trace form. For a given Jordan frame $G_{τ}$, any element $X_{τ}$ in $W$ can be written in the form $G_{τ,0}$ with $Γ(V)$ and $G_{τ,−1}$ an operator in $S_{a}$. The spectral norm of $G_{τ,+1}$ is defined by ||a|| = sup αi and is independent of choices. Its square equals the operator norm of $T_{b}$. Thus $a$ becomes a complex normed space with open unit ball $b$.

Note that for $V$ in $X_{τ}$, the operator $G_{τ,0}G_{τ,−1}$ is self-adjoint so that the norm ||$G_{τ} = G_{τ,0}G_{τ,−1}G_{τ,+1}G_{τ,−1}$|| = ||$X_{τ}$||n. Since $V$, it follows that ||$V$|| = ||$R_{τ}(a,b)$||n. In particular the spectral norm of $Q_{τ}(a)$ in $E_{τ}$ is the square root of the spectral norm of $V$. It follows that the spectral norm of $e$ is the same whether calculated in $e^{3} = e$ or $f = e^{2}$. Since $E_{+}$ preserves both norms, the spectral norm on $E = E_{0}(f) ⊕ E_(f) ⊕ E_{1}(f)$ is obtained by restricting the spectral norm on $L(f)$.

For a Jordan frame $L(e)$, let $L(f)$. There is an action of $L(e)$ on $L(e)^{2}$ which extends to V. If $E_{0}(f)$ and $E_(f)$, then $E_{1}(f)$ and $E_{τ} = E_{τ,0}(f) ⊕ E_{τ,1⁄2}(f) ⊕ E_{τ,1}(f)$ give the action of the product of the lower and upper unitriangular matrices. If $E_{τ,1}(f)$ with $f$, then the corresponding product of diagonal matrices act as $x ∘ y = {x,e,y}$, where $e_{1}$. In particular the diagonal matrices give an action of $e_{2}$ and $[L(a),L(b)] = 0$.

As in the case without an automorphism τ, there is an automorphism θ of $e_{1}$. The same arguments show that the fixed point subgroup $e_{2}$ is generated by $f_{1}$ and the image of $f_{2}$. It is a compact connected Lie group. It acts transitively on $e_{1}$; the stabilizer of $e_{2}$ is $e_{1}e_{2} = 0$. Thus $(e_{1}e_{2},e_{1}e_{2}) =(f_{1},f_{2}) =0$, a Hermitian symmetric space of compact type.

Let $a$ be the identity component of the closed subgroup of $E_{τ}$ carrying $F$ onto itself. It is generated by $a$ and the image of $L(x)L(y)$ corresponding to a Jordan frame. Let ρ be the conjugate-linear period 2 automorphism of $$\mathfrak{g}_\tau$$ defined by


 * $$\displaystyle{\rho(a,T,b)=(-a^*,-T^*,-b^*).}$$

Let $$\mathfrak{h}^*_\tau$$ be the fixed point algebra of ρ. It is the Lie algebra of $x, y$. It induces a period 2 automorphism of $a$ with fixed point subgroup $F$. The group $e_{i}$ acts transitively on $(e_{i})^{3}$. The stabilizer of 0 is $e_{i}$. $e_{i}$ is the Hermitian symmetric space of noncompact type dual to $a$.

The Hermitian symmetric space of non-compact type have an unbounded realization, analogous the upper half-plane in C. Möbius transformations in $F$ corresponding to the Cayley transform and its inverse give biholomorphisms of the Riemann sphere exchanging the unit disk and the upper halfplane. When the Hermitian symmetric space is of tube type the same Möbius transformations map the disk $a = Σ α_{i} e_{i}$ in $α_{i}$ onto the tube domain $a$ were $e$ is the open self-dual convex cone of squares in the Euclidean Jordan algebra $F$.

For Hermitian symmetric space not of tube type there is no action of $a = Σ ε_{i} e_{i}$ on X, so no analogous Cayley transform. A partial Cayley transform can be defined in that case for any given maximal tripotent $ε_{i} = 0, ±1$ in $e_{i}$. It takes the disk $F$ in $E_{τ}$ onto a Siegel domain of the second kind.

In this case $E_{τ}$ is a Euclidean Jordan algebra and there is symmetric $E_{τ}$-valued bilinear form on $(e_{i})$ such that the corresponding quadratic form $a$ takes values in its positive cone $V$. The Siegel domain consists of pairs $a = u Σ α_{i} e_{i}$ such that $α_{i} ≥ 0$ lies in $u$. The quadratic form $K_{τ}$ on $a$ and the squaring operation on $Q_{τ}(a)$ are given by $V$. The positive cone $D_{τ}$ corresponds to $x$ with $E$ invertible.

Examples
For simple Euclidean Jordan algebras $Q(x)$ with complexication $Q(x)^{n}$, the Hermitian symmetric spaces of compact type $Q(x)$ can be described explicitly as follows, using Cartan's classification.

Type In. A is the Jordan algebra of n × n complex matrices $Q(x)^{n} = Q(x^{n})$ with the operator Jordan product $x^{n}$. It is the complexification of $x$, the Euclidean Jordan algebra of self-adjoint n × n complex matrices. In this case $x = Σ α_{i} e_{i}$ acting on $A$ with $$g=\begin{pmatrix}a & b \\ c & d\end{pmatrix}$$ acting as $x^{2} = Σ (α_{i})^{2} f_{i}$. Indeed, this can be verified directly for diagonal, upper and lower unitriangular matrices which correspond to the operators $x$, $A$ and $A_{τ}$. The subset $K_{τ}$ corresponds to the matrices $A_{τ}$ with $A$ invertible. In fact consider the space of linear maps from $e_{1}, ..., e_{m}$ to $V_{e} = ⊕ C e_{i}$. It is described by a pair ($SL(2,C)^{m}$|$V_{e}$) with $c = Σ γ_{i}e_{i}$ in $b = Σ β_{i}e_{i}$. This is a module for $S(c)$ acting on the target space. There is also an action of $T(b)$ induced by the action on the source space. The space of injective maps $a = Σ α_{i}e_{i}$ is invariant and $α_{i} ≠ 0$ acts freely on it. The quotient is the Grassmannian $W = B_{τ}(a, e − a)$ consisting of n-dimensional subspaces of $e = Σ e_{i}$. Define a map of $(C*)^{m}$ into $T^{m}$ by sending $G_{τ}$ to the injective map ($H_{τ}$|$K_{τ}$). This map induces an isomorphism of $SU(2)^{m}$ onto $X_{τ}$.

In fact let $(0:0)$ be an n-dimensional subspace of $K_{τ}$. If it is in general position, i.e. it and its orthogonal complement have trivial intersection with $X_{τ} = H_{τ}/K_{τ}$ and $H_{τ}*$, it is the graph of an invertible operator $G_{τ}$. So the image corresponds to ($D_{τ}$|$K_{τ}$) with $SU(1,1)^{m}$ and $H_{τ}*$.

At the other extreme, $G$ and its orthogonal complement $H_{τ}*$ can be written as orthogonal sums $H_{τ}*$, $D_{τ}$, where $K_{τ}*$ and $H_{τ}*/K_{τ}$ are the intersections with $H_{τ}/K_{τ}$ and $PSL(2,C)$ and $D$ with $A$. Then $T = E + iC$ and $C$. Moreover, $E$ and $PSL(2,C)$. The subspace $e = Σ ε_{i} e_{i}$ corresponds to the pair ($E_{τ}$|$D_{τ}$), where $A_{τ} = A_{τ,1}(f) ⊕ A_{τ,1⁄2}(f)$ is the orthogonal projection of $E_{τ,1}(f)$ onto $E_{τ,1}(f)$. So $E_{τ,1⁄2}(f)$ and $q$.

The general case is a direct sum of these two cases. $C_{τ}$ can be written as an orthogonal sum $(x + iy,u + iv)$ where $y − q(u) − q(v)$ and $C_{τ}$ are the intersections with $q$ and $E_{τ,1⁄2}(f)$ and $E_{τ,1}(f)$ is their orthogonal complement in $x ↦ Q_{τ}(x)e$. Similarly the orthogonal complement $C_{τ}$ of $x$ can be written $Q_{τ}(x)$. Thus $E$ and $A$, where $X$ are orthogonal complements. The direct sum $M_{n}(C)$ is of the second kind and its orthogonal complement of the first.

Maps $x ∘ y = 1⁄2(xy + yx)$ in the structure group correspond to $E = H_{n}(C)$ in $G = PSL(2n,C)$, with $A$. The corresponding map on $g(z) = (az + b)(cz + d)^{−1}$ sends ($W$|$S_{c}$) to ($T_{b}$|$Ω$). Similarly the map corresponding to $g$ sends ($d$|$C^{n}$) to ($C^{2n} = C^{n} ⊕ C^{n}$|$T_{1}$), the map corresponding to $T_{2}$ sends ($T_{i}$|$M_{n}(C)$) to ($GL(2n,C)$|$GL(n,C)$) and the map corresponding to $U$ sends ($GL(n,C)$|$M$) to ($C^{2n}$|$A^{2}$). It follows that the map corresponding to $M$ sends ($(a,b)$|$a$) to ($I − b^{t}a$|$X$). On the other hand, if $M$ is invertible, ($V$|$C^{n} ⊕ C^{n}$) is equivalent to ($C^{n} ⊕ (0)$|$(0) ⊕ C^{n}$), whence the formula for the fractional linear transformation.

Type IIIn. A is the Jordan algebra of n × n symmetric complex matrices $T$ with the operator Jordan product $a$. It is the complexification of $I − b^{t}a$, the Euclidean Jordan algebra of n × n symmetric real matrices. On $a = I$, define a nondegenerate alternating bilinear form by $b^{t} = I − T$. In matrix notation if $$J=\begin{pmatrix} 0 & I \\ -I & 0\end{pmatrix}$$,


 * $$\displaystyle{\omega(z_1,z_2)=zJz^t.}$$

Let $V$ denote the complex symplectic group, the subgroup of $U$ preserving ω. It consists of $V = V_{1} ⊕ V_{2}$ such that $U = U_{1} ⊕ U_{2}$ and is closed under $V_{1}$. If $$g=\begin{pmatrix}a & b \\ c & d\end{pmatrix}$$ belongs to $U_{1}$ then


 * $$\displaystyle{g^{-1}=\begin{pmatrix}d^t & -c^t \\ -b^t & a^t\end{pmatrix}.}$$

It has center $C^{n} ⊕ (0)$. In this case $V_{2}$ acting on $U_{2}$ as $(0) ⊕ C^{n}$. Indeed, this can be verified directly for diagonal, upper and lower unitriangular matrices which correspond to the operators $dim V_{1} = dim U_{2}$, $dim V_{2} = dim U_{1}$ and $C^{n} ⊕ (0) = V_{1} ⊕ U_{1}$. The subset $(0) ⊕ C^{n} = V_{2} ⊕ U_{2}$ corresponds to the matrices $V$ with $e$ invertible. In fact consider the space of linear maps from $I − e$ to $e$. It is described by a pair ($C^{n} ⊕ (0)$|$V_{1}$) with $a = e$ in $b = I$. This is a module for $V$ acting on the target space. There is also an action of $V = V_{0} ⊕ V_{1} ⊕ V_{2}$ induced by the action on the source space. The space of injective maps $V_{1}$ with isotropic image, i.e. ω vanishes on the image, is invariant. Moreover, $V_{2}$ acts freely on it. The quotient is the symplectic Grassmannian $C^{n} ⊕ (0)$ consisting of n-dimensional Lagrangian subspaces of $(0) ⊕ C^{n}$. Define a map of $V_{0}$ into $V$ by sending $U$ to the injective map ($V$|$U = U_{0} ⊕ U_{1} ⊕ U_{2}$). This map induces an isomorphism of $C^{n} ⊕ (0) = V_{1} ⊕ U_{1} ⊕ W_{1}$ onto $(0) ⊕ C^{n} = V_{2} ⊕ U_{2} ⊕ W_{2}$.

In fact let $W_{i}$ be an n-dimensional Lagrangian subspace of $(V_{1} ⊕ U_{1}) ⊕ (V_{2} ⊕ U_{2}) ⊆ C^{n} ⊕ C^{n}$. Let $W$ be a Lagrangian subspace complementing $h$. If they are in general position, i.e. they have trivial intersection with $GL(n,C)$ and $W(a) = hah^{t}$, than $M$ is the graph of an invertible operator $x$ with $y$. So the image corresponds to ($hx$|$(h^{t})^{−1}y$) with $S_{c}$ and $x$.

At the other extreme, $y$ and $x$ can be written as direct sums $y + c$, $T_{b}$, where $x$ and $y$ are the intersections with $x + b$ and $y$ and $J$ with $x$. Then $y$ and $y$. Moreover, $−x$ and $g$. The subspace $x$ corresponds to the pair ($y$|$ax + by$), where $cx + dy$ is the projection of $y$ onto $x$. Note that the pair ($y$, $xy^{−1}$) is in duality with respect to ω and the identification between them induces the canonical symmetric bilinear form on $I$. In particular V1 is identified with U2 and V2 with U1. Moreover, they are V1 and U1 are orthogonal with respect to the symmetric bilinear form on ($S_{n}(C)$. Hence the idempotent $x ∘ y = 1⁄2(xy + yx)$ satisfies $E = H_{n}(R)$. So $C^{2n} = C^{n} ⊕ C^{n}$ and $ω(x_{1} ⊕ y_{1}, x_{2} ⊕ y_{2}) = x_{1} • y_{2} − y_{1} • x_{2}$ lie in $Sp(2n,C)$ and $GL(2n,C)$ is the image of ($g$|$gJg^{t} = J$).

The general case is a direct sum of these two cases. $g ↦ g^{t}$ can be written as a direct sum $Sp(2n,C)$ where ${±I}$ and $G = Sp(2n,C)/{±I}$ are the intersections with $A$ and $g(z) = (az + b)(cz + d)^{−1}$ and $W$ is a complement in $S_{c}$. Similarly $T_{b}$ can be written $Ω$. Thus $g$ and $d$, where $C^{n}$ are complements. The direct sum $C^{2n} = C^{n} ⊕ C^{n}$ is of the second kind. It has a complement of the first kind.

Maps $T_{1}$ in the structure group correspond to $T_{2}$ in $T_{i}$, with $M_{n}(C)$. The corresponding map on $Sp(2n,C)$ sends ($GL(n,C)$|$U$) to ($GL(n,C)$|$M$). Similarly the map corresponding to $C^{2n}$ sends ($A^{2}$|$M$) to ($(a,b)$|$a$), the map corresponding to $I − ba$ sends ($X$|$M$) to ($V$|$C^{n} ⊕ C^{n}$) and the map corresponding to $U$ sends ($V$|$C^{n} ⊕ (0)$) to ($(0) ⊕ C^{n}$|$V$). It follows that the map corresponding to $T$ sends ($T^{t} = T$|$a$) to ($I − ba$|$a = I$). On the other hand, if $b = I − T$ is invertible, ($V$|$U$) is equivalent to ($V = V_{1} ⊕ V_{2}$|$U = U_{1} ⊕ U_{2}$), whence the formula for the fractional linear transformation.

Type II2n. A is the Jordan algebra of 2n × 2n skew-symmetric complex matrices $V_{1}$ and Jordan product $U_{1}$ where the unit is given by $$J=\begin{pmatrix}0 & I \\ -I & 0\end{pmatrix}$$. It is the complexification of $C^{n} ⊕ (0)$, the Euclidean Jordan algebra of self-adjoint n × n matrices with entries in the quaternions. This is discussed in and.

Type IVn. A is the Jordan algebra $V_{2}$ with Jordan product $U_{2}$. It is the complexication of the rank 2 Euclidean Jordan algebra defined by the same formulas but with real coefficients. This is discussed in.

Type VI. The complexified Albert algebra. This is discussed in, and.

The Hermitian symmetric spaces of compact type $(0) ⊕ C^{n}$ for simple Euclidean Jordan algebras $dim V_{1} = dim U_{2}$ with period two automorphism can be described explicitly as follows, using Cartan's classification.

Type Ip,q. Let F be the space of q by p matrices over R with p ≠ q. This corresponds to the automorphism of E = Hp + q(R) given by conjugating by the diagonal matrix with p diagonal entries equal to 1 and q to −1. Without loss of generality $dim V_{2} = dim U_{1}$ can be taken greater than $C^{n} ⊕ (0) = V_{1} ⊕ U_{1}$. The structure is given by the triple product $(0) ⊕ C^{n} = V_{2} ⊕ U_{2}$. The space X can be identified with the Grassmannian of $V$-dimensional subspace of $e$. This has a natural embedding in $I − e$ by adding 0's in the last $e$ coordinates. Since any $C^{n} ⊕ (0)$-dimensional subspace of $V_{1}$ can be represented in the form [$C^{n} ⊕ (0)$|$(0) ⊕ C^{n}$], the same is true for subspaces lying in $C^{n}$. The last $C^{n} ⊕ (0)$ rows of $e$ must vanish and the mapping does not change if the last $e^{t} = e$ rows of $a = e$ are set equal to zero. So a similar representation holds for mappings, but now with q by p matrices. Exactly as when $b = I$, it follows that there is an action of $A$ by fractional linear transformations.

Type IIn F is the space of real skew-symmetric m by m matrices. After removing a factor of $\sqrt{-1}$, this corresponds to the period 2 automorphism given by complex conjugation on E = Hn(C).

Type V. F is the direct sum of two copies of the Cayley numbers, regarded as 1 by 2 matrices. This corresponds to the canonical period 2 automorphism defined by any minimal idempotent in E = H3(O).