Decomposition of Algebras - Mathematical Physics

Algebras

3.5 Decomposition of Algebras

• Let₁and₂be antiderivations with respect to the same involutionω such thatω_i= −_iωfori=1,2. Then₁₂+₂₁is a derivation.

• A particular example of the second case is whenis an antiderivation with respect to an involution ωsuch that ω= −ω. Then² is a derivation.

it is easily seen that

A²=

⎛

⎜⎜

⎝

0 0 a₁₂a₂₃ a₁₂a₂₄+a₁₃a₃₄

0 0 0 a23a34

0 0 0 0

⎞

⎟⎟

⎠,

A³=

⎛

⎜⎜

⎝

0 0 0 a₁₂a₂₃a₃₄

0 0 0 0

⎞

⎟⎟

⎠,

and

A⁴=

⎛

⎜⎜

⎝

0 0 0 0

⎞

⎟⎟

⎠.

Thus, the strictly upper triangular 4×4 matrices are nilpotent of index 4. In fact, one can show that thesubalgebraof the strictly upper triangular 4×4 matrices has index 4.

The reader can convince him/herself that strictly upper triangularn×n matrices are nilpotent of indexn, and that the subalgebra of the strictly upper triangularn×nmatrices is nilpotent of indexn.

3.5.1 The Radical

Nilpotent subalgebras play a fundamental role in the classification of algebras. It is remarkable that all the left, right, and two-sided nilpotent ideals of an algebra are contained is a single nilpotent ideal, which we shall explore now.

Lemma 3.5.5 LetLandMbe two nilpotent left(right)ideals of the alge- braA.Letλandμbe the indices ofLandM,respectively.ThenL+Mis a left(right)ideal ofAof index at mostλ+μ−1.

Proof We prove the Lemma for left ideals. Clearly,L+Mis a left ideal.

Any element ofL+Mraised to thekth power can be written as a linear combination of elements of the forma₁a₂. . .a_kwitha_i belonging to either LorM. Suppose thatl terms of this product are inLandmterms inM. Letj be the largest integer such thata_j ∈L. Starting witha_j move to the left until you reach another element ofL, saya_r. All the termsa_r₊₁toa_j₋₁ are inM. SinceLis a left ideal,

ar+1. . .aj−1

∈A

aj≡a^′_j∈L.

This contracts the producta_ra_r+1. . .a_j−1a_j toa_ra^′_jwith both factors inL. Continuing this process, we obtain

a1a2. . .ak=b1b2. . .blc, bi∈L,c∈M. Similarly,

a1a2. . .ak=c1c2. . .cmb, b∈L, ci ∈M.

Sincek=l+m, ifk=μ+λ−1, then(μ−m)+(λ−l)=1. This shows that ifm < μ, thenl≥λand ifl < λ, thenm≥μ. In either case,a1. . .a_k=0, by one of the last two equations above. Hence,L+Mis nilpotent with an index of at mostμ+λ−1. The proof for the right ideals is identical to this

proof.

Lemma 3.5.6 LetLbe a nilpotent left ideal of the algebraA.Then the sum I=L+LAis a nilpotent two-sided ideal.

Proof SinceLis a left ideal,AL⊆L. Therefore, AI=AL+ALA⊆L+LA=I, showing thatIis a left ideal. On the other hand,

IA=LA+LAA⊆LA+LA=LA⊂I showing thatIis a right ideal.

Now consider a product ofkelements ofLA:

l₁a₁l₂a₂. . .l_ka_k=l₁l^′₂l^′₃. . .l^′_ka_k, l_j∈L, a_j∈A

wherel^′_i≡ai−1li ∈L. This shows that ifkis equal to the index ofL, then the product is zero and hence,LAis nilpotent. Note that since some of the a’s may be inL, the index ofLAis at most equal to the index ofL. Invoking

Lemma3.5.5completes the proof.

The preceding two lemmas were introduced for the following:

Theorem 3.5.7 There exists a unique nilpotent ideal inAwhich contains every nilpotent left,right,and two-sided ideal ofA.

Proof Let Nbe a nilpotent ideal of maximum dimension. Let Mbe any nilpotent ideal. By Lemma3.5.5,N+Mis both a left and a right nilpotent ideal, hence, a nilpotent ideal. By assumptionN+M⊂N, and therefore, M⊂N, proving thatNcontains all ideals. If there were another maximal idealN^′, thenN^′⊂NandN⊂N^′, implying thatN^′=N, and that Nis unique.

IfLis a left nilpotent ideal, then by Lemma3.5.6,L⊂I=L+LA⊂N, becauseIis an ideal. Thus,Ncontains all the nilpotent left ideals. Similarly,

Ncontains all the nilpotent right ideals.

Definition 3.5.8 The unique maximal ideal of an algebra A guar- anteed by Theorem3.5.7is called theradicalof Aand denoted by Rad(A).

We have seen that a nilpotent algebra cannot contain an idempotent. In radical of an algebra

fact, the reverse implication is also true. To show that, we need the following Lemma 3.5.9 Suppose that A contains an element a such that Aa^k = Aa^k⁻¹for some positive integerk.ThenAcontains an idempotent.

Proof LetB≡Aa^k⁻¹. ThenBis a left ideal ofAsatisfyingBa=B. Mul- tiplying both sides bya, we see that

Ba²=Ba=B, Ba³=Ba=B,

andBa^k=B. Buta^k∈BbecauseB≡Aa^k⁻¹. Thus, withb=a^k, we get Bb=B. This means that there must exist an elementP∈Bsuch thatPb= b, or(P²−P)b=0. By Problem3.32,P²=P. Hence,B, and thereforeA

has an idempotent.

Proposition 3.5.10 An algebra is nilpotent if and only if it contains no idempotent.

Proof The “only if” part was shown after Definition3.5.1. We now show that ifAhas no idempotent, then it must be nilpotent. To begin, we note that in general,Aa⊆A, and therefore,Aa^k⊆Aa^k⁻¹for allk. IfAhas no idempotent, then the equality is ruled out by Lemma3.5.9. Hence,Aa^k⊂ Aa^k⁻¹. This being true for allk, we have

A⊃Aa⊃Aa²⊃ · · · ⊃Aa^k⊃ · · ·.

SinceAhas a finite dimension, there must exist an integerrsuch thatAa^r = {0}for alla∈A. In particular,a^r⁺¹=0for alla∈A. This shows thatAis

nil, and by Theorem3.5.3, nilpotent.

LetPbe an idempotent ofA. ConsiderL(P), the left annihilator ofP(see Example3.2.2), and note that(a−aP)∈L(P)for anya∈A. Furthermore, ifa∈PL(P), thena=Pxfor somex∈L(P). Thus,ahas the property that Pa=aandaP=0.

Similarly, considerR(P), the right annihilator of P, and note that (a− Pa)∈R(P)for anya∈A. Furthermore, ifa∈R(P)P, thena=xPfor some x∈R(P). Thus,ahas the property thataP=aandPa=0.

LetI(P)=L(P)∩R(P). Then, clearlyI(P)is a two-sided ideal consist- ing of elementsa∈Asuch thataP=Pa=0. To these, we add the subalge- braPAP, whose elementsacan be shown to have the propertyPa=aP=a.

We thus have

PAP= {a∈A|Pa=aP=a}, PL(P)= {a∈A|Pa=a, aP=0}, R(P)P= {a∈A|aP=a, Pa=0},

I(P)= {a∈A|aP=Pa=0},

(3.13)

and the following

Theorem 3.5.11 LetAbe any algebra with an idempotentP.Then we have

thePeirce decompositionofA: Peirce decomposition

A=PAP⊕^VPL(P)⊕^VR(P)P⊕^V I(P),

where⊕V indicates a vector space direct sum,and each factor is a subal- gebra.

Proof By Eq. (3.13), each summand is actually an algebra. Furthermore, it is not hard to show that the only vector common to any two of the summands is the zero vector. Thus the sum is indeed a direct sum of subspaces. Next note that for anya∈A,

a=PaP+P(a−aP)

∈L(P)

+(a−Pa)

∈R(P)

P+(a−Pa−aP+PaP)

∈I(P)

Problem3.33provides the details of the proof.

Definition 3.5.12 An elementa∈Aisorthogonalto an idempotentPif principal idempotent and elements orthogonal to an idempotent aP=Pa=0. ThusI(P)houses such elements. An idempotentPis called

principalifI(P)contains no idempotent.

Let P0 be an idempotent. If it is not principal, then I(P0)contains an idempotentq. LetP1=P0+q. Then using the fact thatP0q=qP0=0, we can show thatP₁is an idempotent and that

P1P0=P0P1=P0 and P1q=qP1=q. (3.14) If x∈I(P1), then xP1=P1x=0, and the first equation in (3.14) gives xP0=P0x=0, i.e., x∈I(P0), demonstrating that I(P1)⊆I(P0). Since q∈I(P0), butq∈I(P1),I(P1)is a proper subset ofI(P0). IfI(P1)is not principal, thenI(P₁)contains an idempotentr. LetP₂=P₁+r. ThenP₂is an idempotent and, as before,I(P2)is a proper subset ofI(P1). We continue this process and obtain

I(P₀)⊃I(P₁)⊃I(P₂)⊃ · · · ⊃I(P_k)⊃ · · ·.

However, we cannot continue this chain indefinitely, becauseI(P0)has finite dimension. This means that there is a positive integernsuch thatI(Pn)has no idempotent, i.e.,P_nis principal. We have just proved

Proposition 3.5.13 Every algebra that is not nilpotent has a principal idempotent.

Definition 3.5.14 An idempotent is primitiveif it is not the sum of two primitive idempotent

orthogonal idempotents.

Proposition 3.5.15 Pis primitive if and only if it is the only idempotent of PAP.

Proof Suppose thatPis not primitive. Then there are orthogonal idempo- tentsP1andP2such thatP=P1+P2. It is easy to show thatPPi=PiP=Pi

fori=1,2. Hence, by the first equation in (3.13),Pi ∈PAP, andPis not the only idempotent ofPAP.

Conversely, suppose thatPis not the only idempotent inPAP, so that PAP contains another idempotent, say P^′. Then by the first equation in (3.13),PP^′=P^′P=P^′. This shows that

P−P^′ P^′=P^′

P−P^′

=0 and P−P^′

P=P P−P^′

=P−P^′, i.e., that (P−P^′)∈PAP and it is orthogonal to P^′. Furthermore, P= (P−P^′)+P^′, i.e.,Pis the sum of two primitive idempotents, and thus not

primitive.

LetPbe an idempotent that is not primitive. WriteP=P₁+Q, withP₁ andQorthogonal. If either of the two, say Q, is not primitive, write it as Q=P2+P3, withP2andP3orthogonal. By Problem3.34, the set{Pi}³i=1

are mutually orthogonal idempotents andP=P1+P2+P3. We can continue this process until allP_is are primitive. Therefore, we have

Theorem 3.5.16 Every idempotent of an algebraAcan be expressed as the sum of a finite number of mutually orthogonal primitive idempotents.

3.5.2 Semi-simple Algebras

Algebras which have no nilpotent ideals play an important role in the classification of algebras.

semi-simple algebras

Definition 3.5.17 An algebra whose radical is zero is calledsemi-

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