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Real Forms and Real Semisimple Lie Algebras — Examples

1 · Prerequisites

2 · Summary

These examples accompany real-forms-and-real-semisimple-lie-algebras. They compute the compact and split real forms of sl2(C), the Cartan involution and Cartan decomposition of sln(R), polar and Iwasawa decompositions, the compact and split Cartan subalgebras of sl2(R), restricted roots of sln(R), a nonreduced BC restricted root system, Vogan diagrams for the real forms of sl3(C), a complex simple algebra viewed as a real one, two nonconjugate real Cartan subalgebras, two real forms with the same complexification but different Killing-form signatures, and hyperbolic space as SO(n,1)/SO(n).

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

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Compact and split real forms of sl two c

Example

Assume ACω. Let s=sl2(C) be the complex special linear Lie algebra with its standard basis e,f,h (The special linear Lie algebra sl_2). The conjugate-transpose map σc(X)=X and entrywise complex conjugation σs(X)=X are conjugate-linear involutive automorphisms of s, and their fixed algebras

sσc=su(2),sσs=sl2(R)

are real forms of s. Moreover su(2) is a compact real form of s and sl2(R) is a split real form of s.

Facts & Assumptions

Given: ACω and s=sl2(C) with the basis e=(0100),f=(0010),h=(1001), so that [h,e]=2e, [h,f]=2f, [e,f]=h, and the two maps σc(X)=X and σs(X)=X.

[A1]

ACω is countable choice; it is used only through the matrix Lie-group examples cited in [L1] and [L2].

[L1]

s is the Lie algebra of traceless complex 2×2 matrices with bracket [A,B]=ABBA, with basis e,f,h and the displayed relations, and the real traceless matrices form the real Lie subalgebra sl2(R) (The special linear Lie algebra sl_2, General and special linear Lie groups).

[L2]

su(2)={XM2(C):X+X=0, trX=0} is a real Lie subalgebra of s with the same bracket (Unitary and special unitary Lie groups).

[L3]

If σ is a conjugate-linear involutive automorphism of a finite-dimensional complex Lie algebra g, then its fixed locus gσ is a real form of g (Real forms correspond to conjugate-linear involutions, Real form of a complex Lie algebra).

[L4]

The Killing form of sl2 satisfies B(h,h)=8, B(e,f)=B(f,e)=4, and all other pairings of basis vectors zero; equivalently B(X,Y)=4tr(XY) for all X,Y (Killing form of sl_2, Killing form).

[L5]

When the ambient complex Lie algebra is finite-dimensional and semisimple, a real form g0 is a compact real form exactly when B(X,X)<0 for every nonzero Xg0; it is a split real form exactly when it contains a Cartan subalgebra h0 such that every adH, Hh0, is diagonalizable over R (Compact real form of a complex semisimple Lie algebra, Split real form, Cartan subalgebra).

[L6]

The conjugate transpose X=XT is additive, and (XY)=YX (The transpose AT of a matrix).

Proof technique: direct matrix computation.

1.1 Both maps σc and σs are real-linear, additive, involutive and conjugate-linear, and they preserve brackets. By [L6] the conjugate transpose reverses products, σc(AB)=(AB)=BA, while σc(A)σc(B)=AB; taking the difference gives σc[A,B]=(ABBA)=ABBA=[σcA,σcB]. Entrywise conjugation is multiplicative, AB=AB, so σs[A,B]=[σsA,σsB]. [given, L6, algebra]

1.2 The fixed algebra of σc inside s is su(2): a traceless X satisfies X=X exactly when X is skew-Hermitian, so sσc={XM2(C):X+X=0,trX=0}, which is su(2) by [L2]. [given, L2, algebra]

1.3 The fixed algebra of σs is sl2(R): a matrix is fixed by entrywise conjugation exactly when it is real, and a real traceless matrix lies in sl2(R) by [L1]. [given, L1, algebra]

1.4 The Killing form of s is B(X,Y)=4tr(XY). Both sides are symmetric bilinear, so it suffices to compare them on basis pairs: 4tr(ef)=4=B(e,f), 4tr(h2)=8=B(h,h), and 4tr(e2)=4tr(f2)=4tr(eh)=4tr(fh)=0, matching the vanishing pairings of [L4]. In the basis (e,f,h) its Gram matrix is (040400008), whose determinant is 1280; hence B is nondegenerate and s is semisimple by Cartan's criterion over the characteristic-zero field C. Thus the ambient hypothesis in [L5] has been established before either definition is invoked. [L4, Cartan's semisimplicity criterion, algebra]

1.5 The line Rh is a Cartan subalgebra of sl2(R): it is abelian, hence nilpotent, and its normalizer is itself because [h,X]=0 for X=(abca) forces 2b=0 and 2c=0, so XRh. Since [h,h]=0, [h,e]=2e and [h,f]=2f by the given relations, adh is diagonal on the basis (h,e,f) with real eigenvalues 0,2,2, so sl2(R) is a split real form by [L5]. [given, L1, L5, algebra]

2.1 Both fixed loci are real forms. Every Zs decomposes as Z=X+iY with X=12(Z+Z) and Y=12i(ZZ) real traceless, so sl2(R) spans s over C and has real dimension 3=dimCs; alternatively this is the general conclusion of [L3] applied to the involution σs of step 1.1. Likewise every Zs is Z=X+iY with X=12(ZZ) and Y=12i(Z+Z) both skew-Hermitian and traceless, so su(2) spans s over C and has real dimension 3=dimCs; again [L3] gives the same conclusion from σc. [L3, step 1.1, step 1.2, step 1.3, algebra]

2.2 For nonzero Xsu(2) one has X=X, so step 1.4 gives B(X,X)=4tr(X2)=4tr(XX)=4i,jXij2<0, because a nonzero matrix has a nonzero entry. Hence B is negative definite on su(2), and su(2) is a compact real form by [L5]. [step 1.4, L5, algebra]

3.1 The two forms are genuinely different: B(h,h)=8>0 on sl2(R) while B(ef,ef)=B(e,e)2B(e,f)+B(f,f)=8<0, so the Killing form of sl2(R) is indefinite, as a noncompact real form must be, whereas the form on su(2) is definite by step 2.2. All computations are finite; ACω enters only through [L1] and [L2]. [A1, step 1.4, step 2.2, step 1.5, algebra] ∎

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Cartan involution and k plus p for sl n r

Example

Let n2 and let g0=sln(R) be the real Lie algebra of real traceless n×n matrices (General and special linear Lie groups). Then

θ(X)=XT,k0={X:XT+X=0}=so(n),p0={Xsln(R):XT=X}

is a Cartan involution of g0 together with its Cartan decomposition: k0 is the special orthogonal Lie algebra and p0 is the space of symmetric traceless matrices (Cartan involution of a real semisimple Lie algebra, Cartan decomposition of a real semisimple Lie algebra).

Facts & Assumptions

Given: An integer n2, the real Lie algebra g0=sln(R) of real traceless matrices, the map θ(X)=XT, and the Killing form B of g0.

[L1]

sln(R) is a real Lie subalgebra of Mn(R) under [X,Y]=XYYX, with tr(XY)=tr(YX) and trX=0 for every X (General and special linear Lie groups).

[L2]

Transposition is additive, involutive and reverses products: (XY)T=YTXT (The transpose AT of a matrix).

[L3]

The Killing form of sln is B(X,Y)=2ntr(XY) for n2, so it is nondegenerate on sln(R), and sln(R) is therefore semisimple (Classical simple Lie algebras and their Killing forms, Killing form, Cartan's semisimplicity criterion).

[L4]

The orthogonal Lie algebra is so(n)={XMn(R):XT+X=0} (Orthogonal and special orthogonal Lie groups).

[L5]

A Cartan involution of a real semisimple Lie algebra g0 is an involutive automorphism θ with Bθ(X,Y)=B(X,θY) positive definite, and its Cartan decomposition is the decomposition into the +1 and 1 eigenspaces; then [k0,k0]k0, [k0,p0]p0, [p0,p0]k0, with B negative definite on k0 and positive definite on p0 (Cartan involution of a real semisimple Lie algebra, Cartan decomposition of a real semisimple Lie algebra, Bracket relations and Killing signs in a Cartan decomposition).

Proof technique: direct matrix computation.

1.1 The map θ is an involutive automorphism of g0: by [L2], θ2X=XTT=X; tr(XT)=trX makes θ preserve tracelessness; and θ[X,Y]=(XYYX)T=(YTXTXTYT)=[θX,θY] by [L2]. [given, L1, L2, algebra]

1.2 The Killing form satisfies B(X,Y)=2ntr(XY) on all of g0 by [L3]. [L3]

2.1 The fixed space of θ is k0=so(n): θX=X means XT=X, that is XT+X=0, which is the defining condition of so(n) by [L4]; such an X automatically has trX=0, so no tracelessness is lost. [step 1.1, L1, L4, algebra]

2.2 The anti-fixed space of θ is the space of symmetric traceless matrices: θX=X means XT=X, that is XT=X, and membership in g0 adds trX=0. [step 1.1, L1, algebra]

2.3 The form Bθ(X,Y)=B(X,θY) is positive definite: by steps 1.2 and 2.2, Bθ(X,Y)=2ntr(XYT)=2ntr(XYT)=2ni,j=1nXijYij, and tr(XXT)=i,jXij2>0 for X0. Hence θ is a Cartan involution of the semisimple algebra g0 of [L3]. [step 1.1, step 1.2, L3, L5, algebra]

3.1 The eigenspace decomposition g0=k0p0 holds with k0 as in step 2.1 and p0 as in step 2.2, since every X is 12(XθX)+12(X+θX) and the two summands are respectively symmetric and skew-symmetric. [step 2.1, step 2.2, algebra]

3.2 The bracket relations follow directly from transposition: for skew X,Y one has [X,Y]T=[X,Y], for skew X and symmetric Y one has [X,Y]T=[X,Y], and for symmetric X,Y one has [X,Y]T=[X,Y]; hence [k0,k0]k0, [k0,p0]p0 and [p0,p0]k0, in agreement with [L5]. [step 2.1, step 2.2, L5, algebra]

3.3 The Killing signs also follow from the computations: for Xk0 we have B(X,X)=2ntr(X2)=2ntr(XXT)<0 for X0, and for Yp0 we have B(Y,Y)=2ntr(Y2)=2ntr(YYT)>0 for Y0, so B is negative definite on k0 and positive definite on p0. [step 1.2, step 2.1, step 2.2, algebra]

4.1 Endpoints and scope: for n=1 the algebra sl1(R)=0 is semisimple (its Killing form is nondegenerate vacuously), and the same construction degenerates to the zero Cartan decomposition. The hypothesis n2 isolates the nonzero classical case covered by [L3]. For n=2 one has dimk0=1 and dimp0=2, so both summands are nonzero. The computations are finite, use no choice principle, and the displayed identification of k0 with so(n) is an equality of matrix sets, not merely an isomorphism. [given, step 2.1, step 2.2, L3, algebra] ∎

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Polar cartan decomposition of sl n r

Example

Assume the Axiom of Choice. Let n2 and let g0=sln(R)=k0p0 be the Cartan decomposition θ(X)=XT of Cartan involution and k plus p for sl n r, so that k0=so(n) and p0={Xsln(R):XT=X} (Cartan decomposition of a real semisimple Lie algebra). Then every gSLn(R) has a unique factorization

g=kexpX,kSO(n),Xp0,

with exp the matrix exponential; equivalently, the global Cartan decomposition of SLn(R) is its polar factorization, and the multiplication map SO(n)×p0SLn(R) is a diffeomorphism (Global Cartan decomposition for a connected finite center semisimple Lie group).

Facts & Assumptions

Given: The Axiom of Choice; an integer n2; the Lie group SLn(R) with Lie algebra sln(R); the Cartan decomposition g0=k0p0 with k0=so(n) and p0 the symmetric traceless matrices of Cartan involution and k plus p for sl n r; and an element gSLn(R).

[A1]

The Axiom of Choice is The Axiom of Choice; it enters only through the global Cartan decomposition of [L5] and the smooth structure of SLn(R).

[L1]

SLn(R)={A:detA=1} is an embedded Lie subgroup of GLn(R) with Lie algebra sln(R), and SO(n) is the determinant-one subgroup of O(n) with Lie algebra so(n) (General and special linear Lie groups, Orthogonal and special orthogonal Lie groups).

[L2]

Every endomorphism T of a finite-dimensional real inner product space has a polar decomposition T=SU with U=TT non-negative and S an isometry on the orthogonal complement of kerT; the factor U is unique and S is unique when T is invertible (Every endomorphism has a polar decomposition T = SU with U non-negative and S an isometry on the orthogonal complement of ker T, and S is unique exactly when T is invertible).

[L3]

A self-adjoint endomorphism of a finite-dimensional real inner product space has an orthonormal eigenbasis with real eigenvalues (Real spectral theorem: a self-adjoint endomorphism of a finite-dimensional real inner product space has an orthonormal eigenbasis).

[L4]

For a matrix Lie group the Lie-group exponential is the matrix exponential eX=k0Xk/k! (Matrix exponential as the Lie-group exponential).

[L5]

If G is a connected real semisimple Lie group with finite center and K=GΘ for a global Cartan involution with differential θ that fixes Z(G) pointwise, then K×p0G, (k,X)kexpX, is a diffeomorphism (Global Cartan decomposition for a connected finite center semisimple Lie group, Cartan decomposition of a real semisimple Lie algebra).

[L7]

The real algebra sln(R) is semisimple and θ(X)=XT is its Cartan involution with symmetric traceless anti-fixed space (Cartan involution and k plus p for sl n r, Cartan decomposition of a real semisimple Lie algebra).

Verification

technique · direct matrix computation
1.1

Put P:=gTg, the non-negative square root of the self-adjoint positive definite matrix gTg. By [L3] the matrix P is self-adjoint with an orthonormal eigenbasis and positive eigenvalues λ1,,λn>0, because g is invertible and gTgv,v=gv2>0 for v0.

givenL2L3algebra
1.2

The group SO(n) is path connected. For a unit vector v, a rotation of the plane spanned by v,e1 sends v to e1 and is joined to the identity by varying its angle. If v=e1, use a rotation through π in the (e1,e2) plane; if v=e1, use the identity. For RSO(n) apply this to its first column; after this rotation the matrix is diag(1,R) with RSO(n1). Induction, ending with SO(1)={1}, expresses every R as a product of rotations, each with a path to the identity.

L1algebra
2.1

The determinant of P is 1: det(P)2=det(P2)=det(gTg)=det(gT)det(g)=(detg)2=1 by [L6] and detg=1, while detP=λ1λn>0 by step 1.1, so detP=1.

step 1.1L6algebra
2.2

Define X:=logP by the spectral decomposition P=Qdiag(λ1,,λn)QT with Q orthogonal and X:=Qdiag(logλ1,,logλn)QT. Then X is self-adjoint, and Xk=Qdiag((logλ1)k,,(logλn)k)QT for every k0, so the matrix exponential of [L4] gives expX=Qdiag(λ1,,λn)QT=P.

step 1.1L3L4algebra
3.1

Define k:=gP1, which is well defined because P is invertible with positive eigenvalues. Then kTk=P1gTgP1=P1P2P1=I and detk=detgdetP1=1 by [L6], so kSO(n) by [L1].

step 1.1step 2.1L1L6algebra
3.2

The trace of X vanishes: trX=i=1nlogλi=log(λ1λn)=logdetP=0 by step 2.1, so Xp0 by [L7].

L7step 2.1step 2.2algebra
4.1

Existence: steps 3.1, 2.2 and 3.2 give g=kP=kexpX with kSO(n) and Xp0.

step 3.1step 2.2step 3.2
5.1

Define Θ(g)=(gT)1 on G=SLn(R). It is a smooth involutive automorphism, differentiates to XT, and has fixed group SO(n). To compute the center, a central matrix commutes with I+tEijG for every ij and real t, hence with every Eij. Comparing entries of zEij=Eijz forces all off-diagonal entries of z to vanish and its diagonal entries to be equal. Thus z=λI with real λn=1, so the center consists of I and, only when n is even, I. It is finite and fixed pointwise by Θ. Semisimplicity and the Cartan differential are [L7]. Finally, G is path connected without assuming the global theorem: by step 4.1, g=kexpX, and tkexp(tX) joins k to g inside G, since diagonalization gives detexp(tX)=exp(ttrX)=1. Step 1.2 joins I to k. Every hypothesis of [L5] is therefore established.

L1L3L4L5L7step 1.2step 4.1algebra
5.2

Uniqueness: suppose g=kexpX=kexpX with k,kSO(n) and X,Xp0. Both expX and expX are self-adjoint positive definite, since the eigenvalues of X and X are real by self-adjointness and exponentiate to positive numbers, and both k,k are orthogonal. By the uniqueness clause of [L2] for the invertible element g the non-negative factor is unique, so expX=expX=P and k=k; then X=X because the self-adjoint logarithm is unique: both X and X commute with P=expX=expX, hence preserve each eigenspace of P, and on the eigenspace for the eigenvalue λ>0 the equality eX=eX=λI forces every eigenvalue of the self-adjoint operator XEλ to lie in logλ+2πiZ, hence to equal the real number logλ.

step 3.1step 2.2step 4.1L2L3algebra
6.1

Consequently the map SO(n)×p0SLn(R), (k,X)kexpX, is a bijection by steps 4.1 and 5.2 and a diffeomorphism by [L5] applied as in step 5.1; its inverse is g(gP1,logP) with P=gTg.

step 4.1step 5.2step 5.1L5L2
7.1

The argument includes repeated eigenvalues, since the logarithm is scalar on each positive eigenspace. The zero logarithm gives the orthogonal elements of G. Although n2 is the stated scope, at n=1 both factors and the group are singletons. More generally the orthogonal polar factor is special orthogonal whenever detg>0, since detP=detg; the stronger condition detg=1 additionally makes trlogP=0. AC covers the cited Lie-group and global Cartan interfaces; no arbitrary eigenbasis selection over an indexed family is needed in the finite matrix calculations.

A1L6step 6.1algebra
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Compact and split cartan subalgebras of sl two r

Example

Let g0=sl2(R) with the Cartan involution θ(X)=XT (Cartan involution and k plus p for sl n r) and g0=k0p0 its Cartan decomposition, so that k0=so(2) and p0 is the space of symmetric traceless matrices. Put

K0=(0110),H=(1001).

Then h0:=RK0 and h0:=RH are θ-stable Cartan subalgebras of g0, and h0=t0a0 with t0=h0, a0=0 is the compact (maximally compact) one, while h0 has t0=0, a0=h0 and is the split (maximally noncompact) one, in the sense of Theta-stable Cartan subalgebras and their compact and split parts.

Facts & Assumptions

Given: g0=sl2(R) with the basis e=(0100), f=(0010), H=(1001) satisfying [H,e]=2e, [H,f]=2f, [e,f]=H, and the matrices K0=fe=(0110) and θ.

[L1]

sl2(R) consists of the real traceless 2×2 matrices, with basis e,f,H and the displayed bracket relations (The special linear Lie algebra sl_2, General and special linear Lie groups).

[L2]

The Cartan involution θ(X)=XT of sl2(R) has k0={X:XT+X=0}=so(2) and p0={Xsl2(R):XT=X} (Cartan involution and k plus p for sl n r, Cartan involution of a real semisimple Lie algebra, Cartan decomposition of a real semisimple Lie algebra, Orthogonal and special orthogonal Lie groups).

[L3]

A Cartan subalgebra of a Lie algebra is a nilpotent self-normalizing subalgebra, and a θ-stable Cartan subalgebra h0 decomposes as h0=t0a0 with t0=h0k0 the compact part and a0=h0p0 the split part; it is maximally compact when dimt0 is maximal and maximally noncompact when dima0 is maximal (Cartan subalgebra, Theta-stable Cartan subalgebras and their compact and split parts).

Proof technique: direct matrix computation.

1.1 The elements K0 and H lie where claimed: K0T=K0, so θK0=K0 and K0k0, while HT=H and trH=0, so θH=H and Hp0. In particular both lines RK0 and RH are θ-stable. [given, L2, algebra]

1.2 The normalizer of RK0 in g0 is RK0: for X=(abca) one computes [K0,X]=(bc2a2ab+c). For this matrix to equal rK0 its diagonal entries force c=b, while its two off-diagonal entries give 2a=r and 2a=r, hence a=r=0. Thus XRK0, and conversely every such X normalizes the line. Hence RK0 is abelian, nilpotent and self-normalizing, so it is a Cartan subalgebra by [L3]. [given, L1, L3, algebra]

1.3 The normalizer of RH in g0 is RH: for X=(abca) one computes [H,X]=(02b2c0). If this equals rH, comparison of diagonal and off-diagonal entries gives r=0 and b=c=0, so X=diag(a,a)RH; conversely every such X normalizes the line. Hence RH is abelian, nilpotent and self-normalizing, so it is a Cartan subalgebra by [L3]. [given, L1, L3, algebra]

1.4 The adjoint spectra distinguish the two: by [L1], [K0,H]=[fe,H]=[f,H][e,H]=2f+2e=2(e+f) and [K0,e+f]=[f,e]+[f,f][e,e][e,f]=[f,e][e,f]=2H, so adK0 is zero on K0 and has the matrix (0220) in the basis (H,e+f) of p0, with eigenvalues 0,±2i; whereas adH is diagonal on e,f,H with real eigenvalues 2,2,0. [given, L1, algebra]

2.1 Both Cartan subalgebras are θ-stable by step 1.1, so the decomposition of [L3] applies. For h0=RK0 one has h0k0=RK0 and h0p0=0, so t0=h0 and a0=0: every element of h0 is compact, and the compact dimension 1 equals dimk0, so h0 is maximally compact. For h0=RH one has h0k0=0 and h0p0=RH, so t0=0 and a0=h0. Its noncompact dimension is maximal: if a θ-stable Cartan subalgebra q had dim(qp0)=2, then p0=RHR(e+f) would lie in q. Closure under brackets and [H,e+f]=2(ef)=2K0 would then put K0 in q, hence q=g0. But g0 is not nilpotent (indeed [g0,g0]=g0 from the displayed basis relations), whereas every Cartan subalgebra is nilpotent by [L3]. Thus every split part has dimension at most 1, and RH attains that bound. [step 1.1, step 1.2, step 1.3, L1, L2, L3, algebra]

3.1 Consistency of brackets and signs with the general theory: [h0,h0]=0 lies in both t0 and a0 cases because the Cartan subalgebras are abelian, and the compact line lies in k0 where the Killing form is negative definite while the split line lies in p0 where it is positive definite, so the Killing form is negative definite on t0=h0 in the first case and positive definite on a0=h0 in the second. Both subalgebras are one-dimensional, and the two eigenvalue computations of step 1.4 match the compact/purely imaginary and split/real terminology. [step 1.1, step 1.4, step 2.1, L2, algebra]

4.1 Endpoints and scope: the algebra is three-dimensional and both Cartan subalgebras are one-dimensional, which equals its rank; h0h0, and the compact one is the compact line RK0=iRhB in the notation of the source and the split one is the diagonal line RH. No choice principle enters, and the computations are finite. [given, step 1.2, step 1.3, step 2.1, algebra] ∎

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Iwasawa decomposition of sl two r

Example

Assume the Axiom of Choice. In SL2(R) put

K=SO(2)={(cosθsinθsinθcosθ)},A={(a00a1):a>0},N={(1x01):xR}.

Then every gSL2(R) has a unique factorization g=kan with kK, aA and nN, and the multiplication map K×A×NSL2(R) is a diffeomorphism: this is the Iwasawa decomposition of SL2(R) (Global iwasawa decomposition).

Facts & Assumptions

Given: The Axiom of Choice; the group SL2(R) of real 2×2 matrices of determinant 1; the Cartan involution θ(X)=XT with k0=so(2) and p0 the symmetric traceless matrices; and a matrix g=(prqs)SL2(R).

[A1]

The Axiom of Choice is The Axiom of Choice; it enters only through the global Iwasawa theorem of [L4] and the smooth structure of the group.

[L1]

SL2(R) is an embedded Lie group with Lie algebra sl2(R) and SO(2) is a closed connected subgroup with Lie algebra so(2)=R(0110) (General and special linear Lie groups, Orthogonal and special orthogonal Lie groups).

[L2]

The Cartan decomposition sl2(R)=k0p0 has k0=so(2) and p0={X:XT=X, trX=0}; p0 is spanned by H=(1001) and σ=(0110), and [H,σ]=(0220)0, so RH is a maximal abelian subspace of p0 (Cartan involution and k plus p for sl n r, Compact and split cartan subalgebras of sl two r, Cartan decomposition of a real semisimple Lie algebra, Theta-stable Cartan subalgebras and their compact and split parts).

[L3]

The matrix exponential is the Lie-group exponential of a matrix group, and exp(X)=I+X whenever X2=0 (Matrix exponential as the Lie-group exponential, Exponential map of a Lie group).

[L4]

In the global Cartan setup of a connected real semisimple Lie group G with finite center, with A=exp(a) and N the connected subgroup with Lie algebra the sum n of the positive restricted root spaces, the multiplication map K×A×NG is a diffeomorphism (Global iwasawa decomposition).

Proof technique: direct matrix computation.

1.1 Write E=(0100) and H=(1001). Then [H,E]=2E, so n=RE is the positive restricted root space for the functional 2f1 with f1(H)=1 on the maximal abelian a=RH of [L2], and sl2(R)=k0an is the Iwasawa decomposition on the Lie-algebra level, since the three spaces have dimensions 1,1,1 and the sum is direct. [given, L2, algebra]

1.2 The subgroups generated by a and n are the sets displayed: exp(tH)=(et00et), so exp(a)=A with a=et>0, and E2=0 gives exp(xE)=I+xE=(1x01) by [L3], so exp(n)=N. [given, L3, algebra]

2.1 Existence of the factorization. Since g is invertible, its first column (p,q) is nonzero, so a:=p2+q2>0 is well defined. Put k:=1a(pqqp) and x:=pr+qsa2; then kTk=I and detk=p2+q2a2=1, so kSO(2), and a direct multiplication gives k1g=1a(pqqp)(prqs)=(apr+qsa0psqra)=(aax0a1), using psqr=1 and a2=p2+q2. Hence g=k(a00a1)(1x01) with all three factors in K,A,N. [given, step 1.2, algebra]

3.1 Uniqueness. Suppose g=k1a1n1=k2a2n2 with kiK, ai=diag(ai,ai1) and niN. Applying both sides to the first standard basis vector and using nie1=e1 gives k1a1e1=k2a2e1, so taking norms and using that k1,k2 are orthogonal yields a1=a2>0; hence k1e1=k2e1 and therefore k1=k2, because a rotation of R2 fixing e1 is the identity. Then a1n1=a2n2 forces n1=n2, so all three factors are unique. [given, step 2.1, algebra]

4.1 Define Θ(g)=(gT)1 on SL2(R). The identities Θ(gh)=Θ(g)Θ(h) and Θ2=id make it an involutive Lie-group automorphism; its differential is dΘI(X)=XT=θ(X), and its fixed group is {g:gTg=I, detg=1}=SO(2)=K. It fixes the center {±I} pointwise. Thus the global Cartan setup required by [L4] holds for G=SL2(R): its Lie algebra is semisimple, its center is finite, it is connected, and by steps 1.1 and 1.2 the data k0,a,n are exactly those of the displayed K,A,N. Therefore K×A×NSL2(R), (k,a,n)kan, is a diffeomorphism, and by step 2.1 it is the unique factorization of each element. [step 1.1, step 1.2, step 2.1, step 3.1, L4, A1, algebra]

5.1 Endpoints and scope: for g=I the factorization is k=a=n=I with a=1 and x=0, which is the endpoint t=0 of the positive parameter domain a>0; for a0+ no element of A is lost because A is defined by positivity of a. The choice principle is inherited only from [L4], while the matrix computations use none. The diagonal factor is exactly exp(a) and the unipotent factor is exactly exp(n), so the group-level statement matches the Lie-algebra-level Iwasawa decomposition of step 1.1. [given, step 1.1, step 1.2, step 4.1, algebra] ∎

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Restricted roots of sl n r

Example

Let n2 and let g0=sln(R) with the Cartan involution θ(X)=XT, so that p0={Xsln(R):XT=X} (Cartan involution and k plus p for sl n r). Let

a={diag(h1,,hn):h1++hn=0}

be the space of real diagonal traceless matrices, and let fia be the coordinate functional fi(H)=hi. Then a is a maximal abelian subspace of p0 and the restricted roots of (g0,a) are exactly the functionals fifj with ij, each of them with one-dimensional restricted root space

g0fifj=REij,

where Eij is the matrix unit; in particular the restricted root system is of type An1 and is reduced (Restricted root and restricted root space, Maximal split abelian subspace and real rank).

Facts & Assumptions

Given: An integer n2, the real Lie algebra g0=sln(R) of real traceless matrices, the Cartan involution θ(X)=XT with p0 the symmetric traceless matrices, the diagonal subspace a, and the matrix units Eij, ij.

[L1]

sln(R) is a real Lie algebra under [X,Y]=XYYX, with trX=0 for every element and [H,Eij]=(hihj)Eij for H=diag(h1,,hn) (General and special linear Lie groups, Lie algebras over a field).

[L2]

a consists of diagonal symmetric traceless matrices, hence is a subspace of p0; the Cartan decomposition of g0 is g0=k0p0 with k0=so(n) (Cartan involution and k plus p for sl n r, Cartan decomposition of a real semisimple Lie algebra).

[L3]

A restricted root of (g0,a) for a maximal abelian ap0 is a nonzero real functional λ on a whose restricted root space g0λ={Xg0:[H,X]=λ(H)X for all Ha} is nonzero, and its multiplicity is dimRg0λ; a maximal abelian subspace of p0 has dimension equal to the real rank (Restricted root and restricted root space, Maximal split abelian subspace and real rank).

[L4]

In the complex analogue, the diagonal traceless subalgebra of sln(C) is a Cartan subalgebra with roots εiεj and one-dimensional root spaces CEij (Diagonal Cartan subalgebra and roots of sl_n).

Proof technique: direct matrix computation.

1.1 The subspace a is a maximal abelian subspace of p0. It is abelian because its elements are diagonal, and it lies in p0 by [L2]. Conversely, let Xp0 satisfy [H,X]=0 for every Ha. Choosing H=diag(h1,,hn) with pairwise distinct entries hi (possible with ihi=0 in dimension n2), the identity [H,X]=i,j(hihj)XijEij of [L1] shows (hihj)Xij=0 for all i,j, so Xij=0 whenever ij: the centralizer of a in p0 is a itself, which is therefore maximal abelian. [given, L1, L2, algebra]

1.2 Every functional fifj with ij is a restricted root with REijg0fifj: for H=diag(h)a, [L1] gives [H,Eij]=(hihj)Eij=(fifj)(H)Eij, and Eij is a nonzero real matrix of trace zero, while fifj0 since ij. [given, L1, L3, algebra]

2.1 There are no further restricted roots. Let X=i,jxijEijg0 and suppose [H,X]=λ(H)X for every Ha. Comparing the (i,j)-entry using [L1] gives xij(fifj)(H)=xijλ(H)for every Ha. Thus, if an off-diagonal coefficient xij is nonzero, then λ(H)=(fifj)(H) for every H, so λ=fifj as functionals. If λ0, choose H with λ(H)0; the diagonal-entry equations then force every xii=0. Distinct functionals fifj have disjoint eigenspaces, so step 1.2 now gives g0fifj=REij. For λ=0, choose one diagonal Ha with pairwise distinct entries. If Xg00 then [H,X]=0, so the same entry computation forces every off-diagonal coefficient of X to vanish; as X is traceless, it is a diagonal traceless matrix and hence belongs to a. The reverse inclusion is immediate because diagonal matrices commute, so g00=a. Hence the nonzero restricted roots are exactly the fifj, each with multiplicity one. [step 1.1, step 1.2, L1, L3, algebra]

3.1 The decomposition is consistent dimensionally: dima=n1 and there are n(n1) roots each of multiplicity 1, so dimg0=(n1)+n(n1)=n21, which is the dimension of sln(R); the restricted root system {fifj:ij} is the standard realization of An1 and is reduced, since for every root fifj its double 2(fifj) is not of the form fkfl. [step 1.1, step 2.1, L3, algebra]

3.2 The computation matches the complex root computation of [L4]: the functionals fifj are the restrictions to the real diagonal traceless subspace of the root functionals εiεj of the complexification, and the real root space REij is the real form of CEij fixed by complex conjugation, which is why each multiplicity is 1. [step 2.1, L4, algebra]

4.1 Endpoints and scope: for n=2 there is a single pair of opposite roots ±(f1f2) with one-dimensional spaces RE12 and RE21, and dima=1; the case n=1 is excluded because sl1(R)=0 has no nonzero diagonal traceless element. The computation uses no choice principle, and the diagonal element with pairwise distinct entries exists by an explicit choice of coordinates. [given, step 2.1, step 3.1, algebra] ∎

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A nonreduced bc root system from a real form

Example

Assume the Axiom of Choice. Fix integers 1p<q and put m=p+q. Let

g0=su(p,q)={XMm(C):XI+IX=0, trX=0},I=diag(Ip,Iq),

written in block form as X=(ACCE) with Au(p), Eu(q), trA+trE=0 and CCp×q, with Cartan involution θ(X)=X and Cartan decomposition g0=k0p0, k0={X:C=0}, p0={X:A=E=0}. Let

a={HD:D=diag(d1,,dp)Mp(R)},HD=(0[D  0][D  0]0),

where [D  0] is the p×q matrix whose first p columns are D, and let fi(HD)=di. Then a is a maximal abelian subspace of p0 and the restricted root system of (g0,a) is

Σ={±fi±fj: ij}{±fi}{±2fi},

the classical nonreduced system of type BCp, with multiplicities 2 for ±fi±fj (ij), 2(qp) for ±fi and 1 for ±2fi (Restricted root and restricted root space, Maximal split abelian subspace and real rank).

Facts & Assumptions

Given: AC; integers 1p<q, r=qp>0, m=p+q; the displayed trace-zero matrix algebra and the matrices HD. All vector spaces and dimensions below are real unless explicitly described as complex.

[A1]

AC is The Axiom of Choice. It is retained as a standing assumption of the example; the finite matrix argument below needs no additional choices and does not invoke a general classification theorem.

[L1]

On slm(C), the Killing form is 2mtr(XY) for m2 (Classical simple Lie algebras and their Killing forms, special-linear formula). A Killing form is the trace of the product of adjoint maps, and nondegeneracy is equivalent to semisimplicity in characteristic zero (Killing form, Cartan's semisimplicity criterion).

[L2]

A Cartan involution is an involutive automorphism with Bθ(X,Y)=B(X,θY) positive definite; its fixed and anti-fixed spaces give the Cartan decomposition (Cartan involution of a real semisimple Lie algebra, Cartan decomposition of a real semisimple Lie algebra).

[L3]

The restricted root spaces are the simultaneous real adjoint eigenspaces for nonzero real functionals on a maximal abelian subspace of p0; multiplicity means real dimension. The dimension of that maximal split subspace is the real rank (Restricted root and restricted root space, Maximal split abelian subspace and real rank).

[L4]

Reducedness means that a root line meets the root set in exactly the two signs of that root (Reduced crystallographic Euclidean root system). Here BCp denotes the standard set {±ei,±2ei,±ei±ej:i<j}; its reflection and integrality properties will be checked directly.

Verification

technique · direct matrix computation and simultaneous weights
1.1

On slm(C) define σ(Z)=IZI. It is a conjugate-linear involutive Lie automorphism: adjoint reverses products, so the minus sign preserves the commutator, and I2=1. Its fixed space is exactly the trace-zero algebra in the statement. Every Z decomposes uniquely as X+iY, where X=(Z+σZ)/2 and Y=(ZσZ)/(2i) are fixed by σ. Thus this fixed real algebra has complexification slm(C) and real dimension m21. A real basis of it is a complex basis of the complexification; the adjoint matrices of real elements in that basis have the same real and complex traces. Consequently its real Killing form is the restriction B(X,Y)=2mtr(XY) by [L1]. This establishes the real-form assertion rather than attributing it to the compact unitary-group example.

L1algebra
2.1

Solving XI+IX=0 gives the stated skew-Hermitian blocks A,E and the off-diagonal pair C,C, with the single imaginary trace constraint. The map θ(X)=X preserves this algebra, squares to the identity, and preserves brackets by the same adjoint calculation as in step 1.1. Moreover Bθ(X,Y)=2mtr(XY) on this real space, and Bθ(X,X)=2mi,jXij2>0 for X0. The form is real by step 1.1 and symmetric by conjugate symmetry of the displayed trace. Hence B is nondegenerate: if B(X,Y)=0 for all Y, take Y=θX. By [L1] the algebra is semisimple, and by [L2] θ is a Cartan involution with exactly the displayed k0,p0.

L1L2step 1.1algebra
2.2

Simultaneously diagonalize the matrices HD on Cm using the basis vi+=ei+ep+i, vi=eiep+i for 1ip, and zk=e2p+k for 1kr. These have respective weights fi,fi,0. Their independence follows separately on each two-dimensional plane and on the remaining coordinates. In the corresponding matrix-unit basis of End(Cm), the operator taking a basis vector of weight ν to one of weight μ has adjoint weight μν. These units form a simultaneous eigenbasis. Every nonzero-weight unit is traceless, while the zero-weight space in slm is the trace-zero part of its zero-weight endomorphism space.

step 1.1algebra
3.1

The HD commute, since both products have diagonal blocks DD and diag(DD,0r). To compute their centralizer in p0, put C=[C1 C2]. Vanishing of [HD,Y] for all real diagonal D gives DC1=C1D, DC1=C1D, and DC2=0. Taking D=1 in the last equation gives C2=0. In the first equation the (i,j) entry reads di(C1)ji=(C1)ijdj. Independent di,dj force off-diagonal entries to vanish, and the diagonal entries are real. Conversely every real diagonal C1 satisfies all equations. Thus this centralizer is exactly a, proving maximality: any abelian subspace containing it lies in that centralizer. Its dimension is p, so the real rank is p.

L3step 2.1algebra
3.2

Counting the units in step 2.2 gives the complete nonzero weight list and complex dimensions. For distinct i,j, the weight fifj has the two ordered pairs (fi,fj) and (fj,fi); fi+fj has (fi,fj) and (fj,fi). Reversing pairs gives the negatives, each also of dimension two. Weight fi has r pairs (fi,0) and r pairs (0,fi), giving dimension 2r; its negative has the same dimension. Weight 2fi has only the pair (fi,fi), giving dimension one, and similarly for its negative. No other differences occur. The zero-weight endomorphisms have dimension 2p+r2, from the 2p separate nonzero-weight lines and the full endomorphisms of the r-dimensional zero space; trace zero imposes one independent condition, giving 2p+r21.

step 2.2algebra
4.1

These complex dimensions equal the required real multiplicities. Indeed σ fixes every HD and commutes with their adjoint action on a weight space of real weight λ. That complex weight space is therefore σ-stable. Its real fixed space is precisely g0λ, and every vector decomposes as X+iY with both X,Y in that fixed space by the formulas of step 1.1. A real basis of the fixed space is a complex basis of the weight space, so the dimensions agree. This applies also to weight zero, and proves a complete real simultaneous decomposition without dividing by λ(HD), which may vanish at particular D.

L3step 1.1step 3.1step 3.2algebra
4.2

For clarity, the zero space in the original blocks has C1 real diagonal, C2=0, E12=E21=0, and A=E11 diagonal and purely imaginary, while E22 is an arbitrary skew-Hermitian r-by-r matrix subject to 2trA+trE22=0. The C equations follow as in step 3.1; the other equations are DE11=AD, DE12=0 and E21D=0 for every D, which give exactly these conditions. The C1 part is a of dimension p; the other part is Zk0(a) of dimension p+r21. In particular the zero space contains every HD, as it must.

step 2.1step 3.1algebra
5.1

The real dimensions sum to (2p+r21)+4p(p1)+4pr+2p=(2p+r)21=m21, agreeing with step 1.1. Also B(HD,HD)=4mididi, so the dual inner product gives the fi equal lengths and mutual orthogonality. Reflections in fi or 2fi negate one coordinate and reflections in fi±fj are signed coordinate swaps; all preserve the displayed set. For denominator root fi, 2fi, or fi±fj, the Cartan integer is respectively 2βi, βi, or βi±βj in these coordinates, always integral. The set is finite and spans, and it is exactly the standard BCp set in [L4]. It is nonreduced because both fi and 2fi occur with positive multiplicities.

L4step 1.1step 3.2step 4.1step 4.2algebra
6.1

At p=1<q the mixed-root family is empty and the roots are ±f1,±2f1 of multiplicities 2(q1) and one. The zero space has dimension (q1)2+1, so the same count gives q2+2q. The hypotheses exclude p=0 and p=q; in particular qp>0 guarantees that the short roots counted above actually occur. This proves all assertions, retaining the standing AC assumption [A1] but using only finite matrix calculations.

A1step 3.1step 4.1step 5.1algebra
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Vogan diagrams for real forms of sl three c

Example

Assume AC. For g=sl3(C), the following three real forms are pairwise non-isomorphic and have the indicated Vogan classes: su(3):identity involution, no painting; su(2,1):identity involution, one painted vertex; sl3(R):the vertex interchange of A2, no painting. For the standard diagonal Cartan and simple roots α1=ε1ε2, α2=ε2ε3, the second form paints α2, equivalently α1 by diagram isomorphism. The split form uses a different, maximally compact real Cartan as constructed below. Diagram equivalence is Vogan diagram.

Facts & Assumptions

Given: The three real matrix algebras su(3)={X:trX=0, X+X=0}, su(2,1)={X:trX=0, XI+IX=0} with I=diag(1,1,1), and sl3(R). Real-form status is proved below.

[A1]

We assume The Axiom of Choice for the Cartan and diagram classification interfaces.

[L1]

The complex Killing form is B(X,Y)=6tr(XY). The diagonal traceless algebra is a Cartan, with roots εiεj and root vectors Eij; sl3(C) is simple (Classical simple Lie algebras and their Killing forms, Diagonal Cartan subalgebra and roots of sl_n, Classical types correspond to sl, so and sp).

[L2]

Fixed algebras of conjugate-linear bracket-preserving involutions are real forms, and semisimplicity is reflected by complexification (Real forms correspond to conjugate-linear involutions, Complexification preserves semisimplicity).

[L3]

A Cartan involution has positive form Bθ=B(,θ); its compact and split parts are its plus and minus eigenspaces. A theta-stable Cartan with no real roots is maximally compact (Cartan involution of a real semisimple Lie algebra, Theta-stable Cartan subalgebras and their compact and split parts, Cayley transforms connect theta-stable Cartans in the classification).

[L4]

Vogan diagrams use maximally compact Cartans and compact-first positive systems; painting applies only to fixed simple root spaces in the complexified eigenspaces. Equivalence is generated by diagram isomorphisms and painted reflections (Vogan diagram). The equivalence class is an isomorphism invariant, and equivalent diagrams classify isomorphic real forms (Vogan diagram for a fixed Cartan involution is well defined up to equivalence, Classification of real forms by Vogan diagrams).

Proof

technique · direct
1.1

On sl3(C) the maps XX, XIXI and XX are conjugate-linear involutions preserving brackets: conjugate transpose reverses products and its extra minus sign restores the commutator. Their fixed algebras are respectively the three displayed algebras, so [L2] proves they are real forms and semisimple. On the first algebra θ=1 is a Cartan involution since B(X,X)=6tr(XX)>0 for nonzero skew-Hermitian X. On the second θX=IXI=X, which is a real automorphism squaring to one and has Bθ(X,X)=6tr(XX)>0. On the third θX=XT has Bθ(X,X)=6tr(XXT)>0. The real diagonal traceless Cartan has real adjoint eigenvalues, so this third real form is split.

L1L2L3algebra
2.1

In both unitary forms take the diagonal traceless skew-Hermitian Cartan. Its complexification is the diagonal Cartan of [L1]; it is abelian and its real normalizer lies in the real part of the complex normalizer, hence equals itself. It is fixed pointwise by θ, so every root is imaginary and there are no real roots; it is maximally compact by [L3]. For su(3) the complex extension of θ is the identity, so both simple roots are compact and neither is painted. For su(2,1) the complex extension is XIXI, with θEij=IiiIjjEij. Thus CE12 is in k and CE23 is in p, giving precisely one painted vertex α2. These statements concern complexified root spaces, not membership of Eij in real eigenspaces. Interchanging the labels of the two vertices gives the equivalent painting at α1.

L1L3L4step 1.1algebra
2.2

In sl3(R) put H1=(010100000) and H2=diag(1,1,2). They commute, with θH1=H1 and θH2=H2. For Q=diag((11ii),1), direct multiplication gives Q1H1Q=diag(i,i,0) and Q1H2Q=H2. Their complex span is therefore conjugate to the full diagonal Cartan. Its real part h0=RH1RH2 is abelian and self-normalizing by complexification, hence a theta-stable real Cartan. In these diagonal coordinates put β1=ε1ε2 and β2=ε2ε3. On H=xH1+yH2, the positive-root values in the standard ordering are 2ix, ix+3y, and ix+3y. None vanishes identically on RH1, so no root is real and this Cartan is maximally compact by [L3].

L1L3step 1.1algebra
3.1

The root action is θβ1=β1, θβ2=(β1+β2) and θ(β1+β2)=β2. Choose positive roots {β1,β2,β1+β2}, all positive on iH1it0, so the system is compact-first compatible. Its simple roots are δ1=β2, δ2=β1+β2, with δ1+δ2=β1. The involution exchanges δ1 and δ2. Thus this is an A2 diagram with vertex interchange and no fixed vertex to paint. In particular the orbit containing β2 is {β2,(β1+β2)}, not the pair of positive roots in the incompatible standard ordering.

L1L4step 2.2algebra
4.1

These three diagrams represent distinct equivalence classes, not merely different drawings. The empty painting with identity involution admits no painted reflection and stays empty under isomorphisms. A painted reflection retains its reflected vertex as painted, so cannot take a nonempty painting to the empty painting. Such a reflection commutes with the root involution, and relabeling conjugates its vertex permutation, so cannot turn the identity permutation into the interchange. Thus all three classes are different under the exact moves in [L4]. Isomorphic real forms would have equivalent diagrams by [L4], proving the pairwise non-isomorphism. The identity-involution empty-painting form is compact by step 1.1; in the intermediate form, E13+E31 is a real element with positive Killing square, so it is noncompact. All computations are in rank two, and AC is inherited only through the stated classification and Cartan interfaces.

A1L4step 1.1step 2.1step 3.1algebra
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Complex simple lie algebra viewed as a real simple algebra

Example

Let s be a finite-dimensional complex simple Lie algebra and let sR be the same real vector space with the bracket restricted to real scalars, regarded as a real Lie algebra. Then sR is a simple real Lie algebra, and its complexification is C-isomorphic to ss, where s is s with the conjugate complex structure, with the canonical conjugation of the real form interchanging the two factors. This is the complex-as-real case of the dichotomy of Complexification dichotomy for a real simple lie algebra.

Facts & Assumptions

Given: A finite-dimensional complex simple Lie algebra s with multiplication by i written J, and the real Lie algebra sR obtained by restricting scalars.

[L1]

sR is a real Lie algebra whose bracket is the restriction of the bracket of s; J is R-linear with J2=id and J[X,Y]=[JX,Y]=[X,JY], and the complexification (sR)C carries the bracket extending the one of sR (Complexification of a real Lie algebra).

[L2]

The Killing form of sR is BR(X,Y)=2ReBs(X,Y), and a finite-dimensional real Lie algebra is semisimple exactly when its Killing form is nondegenerate (Killing form, Cartan's semisimplicity criterion).

[L3]

Every ideal of a finite-dimensional semisimple Lie algebra is a direct sum of simple ideals with an ideal complement, hence is itself semisimple; a semisimple Lie algebra equals its own derived algebra (Ideals and quotients of semisimple Lie algebras, Semisimple Lie algebras are centerless and perfect).

[L4]

The complexification (sR)C carries the canonical conjugation σ(X+iY)=XiY, whose fixed locus is the embedded copy of sR, and complexification preserves semisimplicity (Complexification has a canonical conjugation with fixed algebra g zero, Complexification of a real Lie algebra, Complexification preserves semisimplicity).

Proof technique: direct computation with ideals and with the explicit isomorphism.

1.1 The real Lie algebra sR is semisimple: its Killing form is BR(X,Y)=2ReBs(X,Y) because the adjoint operators of sR are the C-linear operators adX viewed over R and the real trace of a complex-linear operator is twice the real part of its complex trace, so if BR(X,Y)=0 for all Y, then replacing Y by JY and using Bs(X,JY)=iBs(X,Y) gives ImBs(X,Y)=0 as well, hence Bs(X,)=0 and X=0; thus BR is nondegenerate and [L2] applies. [given, L1, L2, algebra]

2.1 For every ideal asR one has a=[a,sR]: by [L3] applied to the semisimple algebra sR of step 1.1, a is semisimple and satisfies a=[a,a], so a=[a,a][a,sR]a. [step 1.1, L3, algebra]

3.1 Every ideal is J-stable: if Xa and X=j[Xj,Yj] with Xja and YjsR as in step 2.1, then JX=jJ[Xj,Yj]=j[Xj,JYj][a,sR]=a by [L1]. Hence a real ideal of sR is a complex subspace and a complex ideal of s. [step 2.1, L1, algebra]

4.1 Consequently sR is simple over R: since s is complex simple, a complex ideal is 0 or s, so every ideal of sR is 0 or sR; the algebra is nonabelian because s is nonabelian, so it is simple. [step 3.1, algebra]

5.1 Define s to be the real space s with the complex structure J; it is a complex Lie algebra with the same bracket. Then the map L:(sR)Css, L(X+iY)=(X+JY, XJY), is a C-linear isomorphism of complex vector spaces: it is additive and R-bilinear in the obvious way, its inverse is (U,V)12(U+V)+i12J1(UV), and L(i(X+iY))=L(Y+iX)=(Y+JX, YJX)=(J(X+JY), J(XJY)), which is i times L(X+iY) in the complex structure (J,J) of ss. Dimension counts agree: both sides have complex dimension 2dimCs. [given, step 4.1, algebra]

6.1 The map L preserves brackets: for X,Y,X,YsR one has [L(X+iY),L(X+iY)]=([X+JY,X+JY], [XJY,XJY]) and L([X+iY,X+iY])=L([X,X][Y,Y]+i([X,Y]+[Y,X]))=([X,X][Y,Y]+J([X,Y]+[Y,X]), [X,X][Y,Y]J([X,Y]+[Y,X])), and the two expressions agree because [JY,JY]=J2[Y,Y]=[Y,Y] and [JY,X]=J[Y,X] by [L1]. Hence L is an isomorphism of complex Lie algebras. [step 5.1, L1, algebra]

7.1 The canonical conjugation of (sR)C, namely σ(X+iY)=XiY, corresponds under L to the swap of the two factors: L(σ(X+iY))=L(XiY)=(XJY, X+JY), which is the interchange of the entries of L(X+iY)=(X+JY,XJY); its fixed locus is the image of sR under the embedding, in agreement with [L4]. [step 5.1, step 6.1, L4, algebra]

8.1 Combining the steps: sR is a simple real Lie algebra by step 4.1, and its complexification is isomorphic to ss by steps 5.1 and 6.1, with the canonical conjugation of the real form acting as the swap of the two factors by step 7.1. This realizes the complex-as-real alternative of Complexification dichotomy for a real simple lie algebra directly, from the explicit isomorphism L and without using any supplementary clause of that theorem: a complex simple algebra regarded as real has a complexification that is a direct sum of two simple ideals interchanged by conjugation, and the example supplies the isomorphism and the swap. [step 4.1, step 5.1, step 6.1, step 7.1]

9.1 Endpoints and scope: s is nonabelian by hypothesis, so sR has nonzero bracket and simplicity is not vacuous; for s=sl2(C) the real dimension dimRsR=6 equals dimC(ss) computed over C as 3+3, in agreement with step 5.1; the zero algebra is excluded because it is not simple, and every step is a finite computation, so the argument uses no choice principle. [given, step 5.1, step 7.1, algebra] ∎

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Two nonconjugate real cartan subalgebras

Statement refuted

Any two Cartan subalgebras of the real Lie algebra sl2(R) are conjugate by an inner automorphism of sl2(R).

Facts & Assumptions

Given: The real Lie algebra g0=sl2(R) with the matrices H=(1001), K0=(0110), e=(0100), f=(0010) and the relations [H,e]=2e, [H,f]=2f, [e,f]=H.

[L1]

sl2(R) is the real Lie algebra of real traceless matrices; its inner automorphisms are the maps Adg(X)=gXg1 for gSL2(R), and more generally every automorphism φ satisfies φ(adH)=adφ(H)φ (General and special linear Lie groups, The special linear Lie algebra sl_2).

[L2]

A Cartan subalgebra is a nilpotent self-normalizing subalgebra (Cartan subalgebra).

[L3]

The lines RK0 and RH are θ-stable Cartan subalgebras of sl2(R): RK0 is the compact one and RH the split one (Compact and split cartan subalgebras of sl two r, Cartan involution and k plus p for sl n r).

[L4]

The two Cartan subalgebras RH and RK0 of sl2(R) are not conjugate by any real inner automorphism (Real Cartan subalgebras need not be conjugate).

Proof technique: direct computation of adjoint spectra.

1.1 The subspaces RK0 and RH are Cartan subalgebras of sl2(R) by [L3], and they are distinct, because K0 is skew-symmetric while H is symmetric and diagonal. [given, L3, algebra]

1.2 The adjoint operator of K0 has spectrum {0,2i,2i}: from the relations one computes [K0,H]=[fe,H]=2f+2e=2(e+f) and [K0,e+f]=[f,e][e,f]=HH=2H, while [K0,K0]=0; hence in the basis (H,e+f,K0) of sl2(R) the operator adK0 has the block matrix (020200000), whose characteristic polynomial is t(t2+4). [given, algebra]

1.3 The adjoint operator of H has spectrum {0,2,2}: by the given relations adH is diagonal in the basis (H,e,f) with eigenvalues 0,2,2, so its characteristic polynomial is t(t2)(t+2). [given, algebra]

2.1 No automorphism of sl2(R) carries RH onto RK0: if φ were such an automorphism with φ(H)=cK0 for some c0, then by [L1] the operators adφ(H) and adH would be conjugate, hence would have the same characteristic polynomial; but step 1.3 gives t(t2)(t+2) for adH and step 1.2 gives t(t2+c24) for adcK0, and no nonzero c makes these polynomials equal (the first has three distinct real roots, the second has a nonzero purely imaginary pair). [step 1.2, step 1.3, L1, algebra]

3.1 Consequently the two Cartan subalgebras RH and RK0 are not conjugate by any automorphism, and in particular not by an inner automorphism; since they are distinct Cartan subalgebras of sl2(R) by step 1.1, they refute the displayed statement, and they are exactly a witness pair for the general phenomenon of [L4]. [step 1.1, step 2.1, L4]

4.1 Scope: the invariant that separates the two lines is the isomorphism type of adH as a real operator, equivalently the position of the line inside k0 or p0: the compact line consists of elements whose adjoint operators have purely imaginary nonzero spectrum, the split line of elements with real nonzero spectrum. The computation is finite, uses no choice principle, and shows that the failure of conjugacy is detected already at the level of all automorphisms, not merely inner ones. [step 1.2, step 1.3, step 2.1, algebra] ∎

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Same complexification with different killing form signatures

Statement refuted

Real forms of one complex semisimple Lie algebra have congruent Killing forms; equivalently, the inertia of the Killing form of a real semisimple Lie algebra is determined by its complexification.

Facts & Assumptions

Given: The two real Lie algebras su(2) and sl2(R), both real forms of s=sl2(C), and the basis e,f,h of s with the Killing form B.

[L1]

su(2) and sl2(R) are real forms of s=sl2(C); su(2) is a compact real form and sl2(R) is a split real form (Compact and split real forms of sl two c, Compact real form of a complex semisimple Lie algebra, Split real form).

[L2]

The Killing form of sl2 satisfies B(h,h)=8, B(e,f)=B(f,e)=4, and all other pairings of the basis e,f,h vanish; equivalently B(X,Y)=4tr(XY) (Killing form of sl_2, Killing form).

Proof technique: direct computation of the two Killing forms.

1.1 The two algebras have the same complexification: by [L1] both su(2) and sl2(R) are real forms of s=sl2(C), so their complexifications are both isomorphic to s. [L1]

1.2 The Killing form of su(2) has inertia (0,3,0): for Xsu(2) one has X=X, and [L2] gives B(X,X)=4tr(X2)=4tr(XX)=4i,jXij2, which is negative for every nonzero X and zero only at X=0; hence B is negative definite on the three-dimensional space su(2), with no positive and no null directions. [L2, algebra]

1.3 The Killing form of sl2(R) has inertia (2,1,0): in the basis (h,e,f) the Gram matrix of B is (800004040) by [L2], whose characteristic polynomial is (8λ)(λ216), so the eigenvalues are 8, 4 and 4; a symmetric matrix is diagonalized by an orthogonal change of basis, so the form has two positive and one negative square and is nondegenerate. [L2, algebra]

2.1 The two forms are not congruent: their inertias (0,3,0) and (2,1,0) differ, and by [L3] congruent forms of the same dimension have equal inertia. [step 1.2, step 1.3, L3]

3.1 No Lie-algebra isomorphism can exist between them: if φ ⁣:sl2(R)su(2) were an isomorphism, then adφ(X)=φadXφ1 would give Bsu(2)(φX,φY)=tr(adφXadφY)=tr(adXadY)=Bsl2(R)(X,Y), so the two Killing forms would be congruent via the invertible matrix of φ, contradicting step 2.1. [step 2.1, L2, algebra]

4.1 Consequently the complexification does not determine the inertia of the Killing form: the real forms su(2) and sl2(R) of the same complex algebra sl2(C) carry Killing forms of inertia (0,3,0) and (2,1,0) and are not isomorphic. The compactness of su(2) corresponds exactly to the vanishing of the positive part of the inertia, while the split form has a positive-definite subspace of dimension 2. [step 1.1, step 1.2, step 1.3, step 3.1, L1]

5.1 Endpoints and scope: both algebras are three-dimensional and nondegenerate, so the nullity is 0 in both cases and the difference is entirely in the signature; the computation is finite, uses the explicit basis of sl2 only, and needs no choice principle. [step 1.2, step 1.3, algebra] ∎

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Hyperbolic space as so zero n one mod so n

Example

Assume the Axiom of Choice and let n2. Write v,w=i=1nviwivn+1wn+1 for the Lorentz form on Rn+1, let

Hn={vRn+1:v,v=1, vn+1>0}

be the upper sheet of the hyperboloid, and let G=SO0(n,1) be the identity component of the group of J-preserving matrices, J=diag(In,1). Then K=SO(n) is a maximal compact subgroup of G and the orbit map induces a diffeomorphism G/KHn; under it the Cartan metric of Riemannian symmetric pair of noncompact type is a G-invariant Riemannian metric on real hyperbolic n-space of constant sectional curvature 1/(2(n1)); equivalently the Cartan metric is 2(n1) times the standard normalization of curvature 1, namely the metric (2(n1))1Bθ (Cartan decomposition gives the invariant metric and curvature of G mod K, Cartan decomposition identifies p with the noncompact symmetric space).

Facts & Assumptions

Given: The Axiom of Choice; an integer n2; the Lorentz form , with matrix J; the groups O(n,1)={AGLn+1(R):ATJA=J}, SO(n,1)=O(n,1)SLn+1(R) and its identity component G=SO0(n,1); the hyperboloid Hn and its point e0=(0,,0,1).

[A1]

The Axiom of Choice is The Axiom of Choice; it enters through the closed-subgroup theorem, the quotient-manifold structure and the global Cartan decomposition used below.

[L1]

Every closed subgroup of a finite-dimensional real Lie group is an embedded Lie subgroup; GLn+1(R) is a Lie group with Lie algebra Mn+1(R); and SO(n)={R:RTR=I, detR=1} is a closed subgroup with Lie algebra so(n)={X:XT+X=0} (Cartan closed subgroup theorem, General and special linear Lie groups, Orthogonal and special orthogonal Lie groups). Closed and bounded subsets of a finite-dimensional real matrix space are compact by Heine-Borel in Rn: with the Euclidean metric a subset of Rn is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line.

[L2]

A regular level set of a smooth map is an embedded submanifold whose tangent space at a point is the kernel of the differential (A regular level set is an embedded submanifold, The tangent space of a regular level set is the kernel).

[L3]

The Killing form is B(X,Y)=tr(adXadY), and a finite-dimensional characteristic-zero Lie algebra is semisimple exactly when its Killing form is nondegenerate (Killing form, Cartan's semisimplicity criterion).

[L4]

For a Riemannian symmetric pair (G,K) of noncompact type with Cartan decomposition g0=k0p0, the form Bθ defines a G-invariant Riemannian metric on G/K with value Bθ at the origin, the curvature at the origin is R(X,Y)Z=[[X,Y],Z] for X,Y,Zp0, and the sectional curvature of a plane with basis X,Yp0 is Rm(X,Y,Y,X)/(Bθ(X,X)Bθ(Y,Y)Bθ(X,Y)2); moreover G/K is diffeomorphic to p0 by Xexp(X)K (Riemannian symmetric pair of noncompact type, Cartan decomposition gives the invariant metric and curvature of G mod K, Cartan decomposition identifies p with the noncompact symmetric space, Sectional curvature).

[L5]

G/K carries the unique smooth structure making the quotient map a submersion and the left G-action smooth, and the orbit map G/KHn, gKge0, is smooth and G-equivariant (Homogeneous spaces of Lie groups, Quotient manifold by a closed Lie subgroup).

[L6]

The matrix exponential is the Lie-group exponential of a matrix group, and the exponential map carries a neighborhood of 0 diffeomorphically onto a neighborhood of the identity (Matrix exponential as the Lie-group exponential, The exponential map is a local diffeomorphism at zero). Consequently the subgroup H generated by exp(g) is the identity component: H contains an open identity neighborhood and is therefore an open subgroup, while every path texp(tX) lies in the identity component, so HG0; the cosets of H make both H and its complement open in the connected group G0, forcing H=G0.

[L7]

For a connected real semisimple Lie group with finite center and a global Cartan involution, its fixed subgroup is maximal compact (Maximal compact subgroups exist and are conjugate in a connected finite center semisimple Lie group).

Verification

technique · direct matrix computation
1.1

The group O(n,1) is closed in GLn+1(R), so [L1] makes it an embedded Lie subgroup. Differentiating ATJA=J gives XTJ+JX=0; conversely this condition implies exp(tX)TJexp(tX)=J by differentiation in t, so it characterizes the Lie algebra. It consists of (AbbT0) with AT=A. Determinant has values ±1 on O(n,1), hence equals 1 on its identity component G. This component has the same Lie algebra. Write Xb=(0bbT0) and Kij=EijEji for i<jn.

L1L6algebra
1.2

The level function F(v)=v,v has differential dFv(w)=2v,w, nonzero at every F(v)=1. Thus [L2] gives tangent space v and dimension n. The upper sheet is the graph v=(u,1+u2), hence connected. Its Lorentz tangent metric is positive: if w=(a,b)v, then b=ua/1+u2 and w,wa2/(1+u2)>0 for w0. Every gG preserves this sheet, since the sign of the last coordinate of gv cannot change continuously on connected G.

L2algebra
1.3

The group SO(n) is compact, being closed and bounded in matrix space and hence compact by Heine--Borel in [L1], and is path connected: plane rotations can carry any unit first column to the first coordinate vector; after doing so the remaining block is in SO(n1), and induction ends with SO(1)={1}. Each plane rotation has a path to the identity through its angle. Therefore K={diag(R,1):RSO(n)} lies in G.

L1algebra
2.1

The basis Kij,Xi:=Xei satisfies [Kij,Xl]=δjlXiδilXj and [Xi,Xj]=Kij for i<j. For fixed Kij, its adjoint square is I on each two-dimensional span of the rotations joining i,j to a third spatial index, and on span(Xi,Xj); it vanishes on the remaining basis vectors. Its trace is 2(n2)2=2(n1). For fixed Xi, its adjoint square is +I on each span of Xj and the rotation joining i,j (ji), and zero on the rest, giving trace 2(n1). Mixed Killing pairings of different basis vectors vanish: conjugation by diag(ϵ1,,ϵn,1), ϵi=±1, is a Lie-algebra automorphism, preserves the adjoint trace, and acts with distinct sign characters ϵiϵj on Kij and ϵi on Xi. A sign choice therefore negates any mixed pairing while preserving it. This proves on the whole basis, hence bilinearly, B(X,Y)=(n1)tr(XY). It is nondegenerate for every n2, including n=3, so [L3] proves semisimplicity. The involution θX=XT has eigenspaces k=span(Kij) and p={Xb}, with Bθ(X,X)=(n1)tr(XXT)>0 for X0. Since [p,p]=k, no proper ideal contains p, so the noncompact-type criterion of [L4] is satisfied.

L3L4step 1.1algebra
2.2

The action is transitive: for v=(u,1+u2) with u0, put a=u/u and t0 with sinht=u. The matrix exponential gives exp(tXa)e0=(sinhta,cosht)=v and belongs to G; v=e0 uses the identity. The stabilizer of e0 consists exactly of diag(R,1) with RSO(n), since it preserves e0 and determinant one; these matrices are in G by step 1.3. Thus it is K.

L6step 1.1step 1.2step 1.3algebra
3.1

The smooth orbit map factors through the quotient submersion to a smooth bijection Φ:G/KHn by [L5] and step 2.2. Its derivative at eK, using g/kp, is Xb(b,0), an isomorphism. Equivariance makes the derivative an isomorphism everywhere, so the inverse function theorem gives a local diffeomorphism everywhere; a bijective local diffeomorphism has a smooth inverse.

L5step 2.1step 2.2algebra
3.2

The center of G is trivial. If z is central, ze0 is fixed by K; the only spatial vector fixed by all spatial rotations for n2 is zero. Since ze0Hn, it equals e0, so z=diag(R,1)K. Commuting with every exp(tXb) and differentiating forces Rb=b for every b, hence z=I. The group automorphism Θ(g)=(gT)1=JgJ preserves G, is involutive and differentiates to θ. Its fixed elements lie in both O(n+1) and O(n,1), hence commute with J and have block form diag(R,c); the upper-sheet condition gives c=1 and determinant one gives RSO(n). Thus GΘ=K. Together with step 2.1 this verifies all hypotheses of the symmetric-pair interface [L4].

L4step 2.1step 2.2algebra
4.1

Step 3.2 proves that G is connected semisimple with finite center, that Θ is a global Cartan involution, and that GΘ=K. Therefore [L7] applies directly and makes K maximal compact.

L7step 3.2
4.2

The metric of [L4] is now applicable by step 3.2. At the origin Bθ(Xa,Xb)=B(Xa,Xb)=2(n1)ab by step 2.1. Under dΦ, the Lorentz metric is ab. Both metrics are G-invariant, so the Cartan metric is 2(n1) times the Lorentz metric everywhere. For independent a,b let D=a2b2(ab)2>0. The bracket is [Xa,Xb]=diag(abTbaT,0), whose squared Bθ-norm is 2(n1)D, while the Gram determinant of Xa,Xb is 4(n1)2D. The sectional formula in [L4] therefore gives 1/(2(n1)) at the origin and, by transitivity, everywhere.

L4step 2.1step 3.1step 3.2algebra
5.1

Scaling a metric by a constant c>0 preserves its Levi-Civita connection and its curvature operator of type (1,3): the same connection remains torsion free and metric compatible. The sectional numerator scales by c and its Gram denominator by c2. Thus (2(n1))1Bθ, the Lorentz metric from step 4.2, has curvature 1. At n=2 the Cartan curvature is 1/2; at n=3 the direct trace proof remains valid. Rank n=1 is excluded because the algebra is abelian with zero Killing form and there are no tangent two-planes. AC covers the Lie-group, quotient, maximal-compact and symmetric-space interfaces; the finite matrix computations require no further choice.

A1L4step 4.2algebra

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