lie algebras and their root systems · sophus lie memorial conference, oslo, august, 1992, oslo:...
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Lie algebras and their root systemsA case study in the classification of Lie algebras
Joshua Ruiter
Calvin College Mathematics ColloquiumApril 28, 2016
Joshua Ruiter Lie algebras and their root systems
![Page 2: Lie algebras and their root systems · Sophus Lie Memorial Conference, Oslo, August, 1992, Oslo: Scandinavian University Press, pp. 3–21. Joshua Ruiter Lie algebras and their root](https://reader034.vdocuments.us/reader034/viewer/2022042710/5f5a3cad1f053c6c8050df17/html5/thumbnails/2.jpg)
Historical background
Source for historical information: Helgason, Sigurdur (1994), "Sophus Lie, the Mathematician," Proceedings of TheSophus Lie Memorial Conference, Oslo, August, 1992, Oslo: Scandinavian University Press, pp. 3–21.
Joshua Ruiter Lie algebras and their root systems
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The matrix algebra sl(3,C)
sl(3,C) is the set of 3× 3 matrices with complex entries andtrace zero.
sl(3,C) =
a b c
d e fg h i
: a + e + i = 0
This is a vector space, because:
The zero matrix has trace zero.A scalar multiple of a traceless matrix is traceless.The sum of two traceless matrices is traceless.
Joshua Ruiter Lie algebras and their root systems
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The matrix algebra sl(3,C)
sl(3,C) is the set of 3× 3 matrices with complex entries andtrace zero.
sl(3,C) =
a b c
d e fg h i
: a + e + i = 0
This is a vector space, because:
The zero matrix has trace zero.A scalar multiple of a traceless matrix is traceless.The sum of two traceless matrices is traceless.
Joshua Ruiter Lie algebras and their root systems
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Basis matrices
DefinitionThe matrix eij is a matrix with a one in the ij th place and zeroeselsewhere.
Usual basis for sl(3,C):
{e11 − e22,e22 − e33} ∪ {eij : i 6= j}
Direct sum expression for sl(3,C):
sl(3,C) = span{e11 − e22,e22 − e33} ⊕⊕i 6=j
span{eij}
Joshua Ruiter Lie algebras and their root systems
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Basis matrices
DefinitionThe matrix eij is a matrix with a one in the ij th place and zeroeselsewhere.
Usual basis for sl(3,C):
{e11 − e22,e22 − e33} ∪ {eij : i 6= j}
Direct sum expression for sl(3,C):
sl(3,C) = span{e11 − e22,e22 − e33} ⊕⊕i 6=j
span{eij}
Joshua Ruiter Lie algebras and their root systems
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Basis matrices
DefinitionThe matrix eij is a matrix with a one in the ij th place and zeroeselsewhere.
Usual basis for sl(3,C):
{e11 − e22,e22 − e33} ∪ {eij : i 6= j}
Direct sum expression for sl(3,C):
sl(3,C) = span{e11 − e22,e22 − e33} ⊕⊕i 6=j
span{eij}
Joshua Ruiter Lie algebras and their root systems
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Is it a ring?
We just determined that sl(3,C) is closed under matrix addition.Is it closed under matrix multiplication?−2 0 0
0 1 00 0 1
1 0 00 −2 00 0 1
=
−2 0 00 −2 00 0 1
Joshua Ruiter Lie algebras and their root systems
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Is it a ring?
We just determined that sl(3,C) is closed under matrix addition.Is it closed under matrix multiplication?−2 0 0
0 1 00 0 1
1 0 00 −2 00 0 1
=
−2 0 00 −2 00 0 1
Joshua Ruiter Lie algebras and their root systems
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A new binary operation
DefinitionLet x , y be matrices. The bracket of x and y is[A,B] := AB − BA.
Proposition
Let A,B be matrices. Then [A,B] has trace zero.
Proof.From linear algebra, we know that tr(AB) = tr(BA) and tr islinear. Hence tr[A,B] = tr(AB − BA) = tr(AB)− tr(BA) = 0.
Corollary
A,B ∈ sl(3,C) =⇒ [A,B] ∈ sl(3,C)
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A new binary operation
DefinitionLet x , y be matrices. The bracket of x and y is[A,B] := AB − BA.
Proposition
Let A,B be matrices. Then [A,B] has trace zero.
Proof.From linear algebra, we know that tr(AB) = tr(BA) and tr islinear. Hence tr[A,B] = tr(AB − BA) = tr(AB)− tr(BA) = 0.
Corollary
A,B ∈ sl(3,C) =⇒ [A,B] ∈ sl(3,C)
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A new binary operation
DefinitionLet x , y be matrices. The bracket of x and y is[A,B] := AB − BA.
Proposition
Let A,B be matrices. Then [A,B] has trace zero.
Proof.From linear algebra, we know that tr(AB) = tr(BA) and tr islinear. Hence tr[A,B] = tr(AB − BA) = tr(AB)− tr(BA) = 0.
Corollary
A,B ∈ sl(3,C) =⇒ [A,B] ∈ sl(3,C)
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A new binary operation
DefinitionLet x , y be matrices. The bracket of x and y is[A,B] := AB − BA.
Proposition
Let A,B be matrices. Then [A,B] has trace zero.
Proof.From linear algebra, we know that tr(AB) = tr(BA) and tr islinear. Hence tr[A,B] = tr(AB − BA) = tr(AB)− tr(BA) = 0.
Corollary
A,B ∈ sl(3,C) =⇒ [A,B] ∈ sl(3,C)
Joshua Ruiter Lie algebras and their root systems
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Properties of bracket (1)
Proposition (Bracket is alternating)
Let A be a matrix. Then [A,A] = 0.
Proof.
[A,A] = A2 − A2 = 0
Proposition (Bracket is antisymmetric)
Let A,B be matrices. Then [A,B] = −[B,A].
Proof.[A,B] = AB − BA = −BA + AB = −(BA− AB) = −[B,A]
Corollary
Let A,B ∈ sl(3,C). Then [A,B] = [B,A] ⇐⇒ [A,B] = 0.
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Properties of bracket (1)
Proposition (Bracket is alternating)
Let A be a matrix. Then [A,A] = 0.
Proof.
[A,A] = A2 − A2 = 0
Proposition (Bracket is antisymmetric)
Let A,B be matrices. Then [A,B] = −[B,A].
Proof.[A,B] = AB − BA = −BA + AB = −(BA− AB) = −[B,A]
Corollary
Let A,B ∈ sl(3,C). Then [A,B] = [B,A] ⇐⇒ [A,B] = 0.
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Properties of bracket (1)
Proposition (Bracket is alternating)
Let A be a matrix. Then [A,A] = 0.
Proof.
[A,A] = A2 − A2 = 0
Proposition (Bracket is antisymmetric)
Let A,B be matrices. Then [A,B] = −[B,A].
Proof.[A,B] = AB − BA = −BA + AB = −(BA− AB) = −[B,A]
Corollary
Let A,B ∈ sl(3,C). Then [A,B] = [B,A] ⇐⇒ [A,B] = 0.
Joshua Ruiter Lie algebras and their root systems
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Properties of bracket (1)
Proposition (Bracket is alternating)
Let A be a matrix. Then [A,A] = 0.
Proof.
[A,A] = A2 − A2 = 0
Proposition (Bracket is antisymmetric)
Let A,B be matrices. Then [A,B] = −[B,A].
Proof.[A,B] = AB − BA = −BA + AB = −(BA− AB) = −[B,A]
Corollary
Let A,B ∈ sl(3,C). Then [A,B] = [B,A] ⇐⇒ [A,B] = 0.
Joshua Ruiter Lie algebras and their root systems
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Properties of bracket (1)
Proposition (Bracket is alternating)
Let A be a matrix. Then [A,A] = 0.
Proof.
[A,A] = A2 − A2 = 0
Proposition (Bracket is antisymmetric)
Let A,B be matrices. Then [A,B] = −[B,A].
Proof.[A,B] = AB − BA = −BA + AB = −(BA− AB) = −[B,A]
Corollary
Let A,B ∈ sl(3,C). Then [A,B] = [B,A] ⇐⇒ [A,B] = 0.
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Properties of bracket (2)
Proposition (Bracket is bilinear)Let A,B,C be matrices and let λ be a scalar. Then
[A + B,C] = [A,C] + [B,C]
[C,A + B] = [C,A] + [C,B]
[λA,C] = [A, λC] = λ[A,C]
Proof.Apply the fact that matrix multiplication distributes over matrixaddition and commutes with scalar multiplication.
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Properties of bracket (2)
Proposition (Bracket is bilinear)Let A,B,C be matrices and let λ be a scalar. Then
[A + B,C] = [A,C] + [B,C]
[C,A + B] = [C,A] + [C,B]
[λA,C] = [A, λC] = λ[A,C]
Proof.Apply the fact that matrix multiplication distributes over matrixaddition and commutes with scalar multiplication.
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Properties of bracket (3)
Proposition (Jacobi identity)Let A,B,C be matrices. Then[A, [B,C]] + [B, [C,A]] + [C, [A,B]] = 0.
Proof.Remember that matrix multiplication distributes over matrixaddition. Expand all the terms out, get a lot of terms like ABC,and see that they all cancel in pairs.
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Properties of bracket (3)
Proposition (Jacobi identity)Let A,B,C be matrices. Then[A, [B,C]] + [B, [C,A]] + [C, [A,B]] = 0.
Proof.Remember that matrix multiplication distributes over matrixaddition. Expand all the terms out, get a lot of terms like ABC,and see that they all cancel in pairs.
Joshua Ruiter Lie algebras and their root systems
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Brackets of basis matrices
DefinitionThe Kronecker delta function is the function
δij =
{1 i = j0 i 6= j
Lemma
[eij ,ekl ] = δjkeil − δilekj
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Brackets of basis matrices
DefinitionThe Kronecker delta function is the function
δij =
{1 i = j0 i 6= j
Lemma
[eij ,ekl ] = δjkeil − δilekj
Joshua Ruiter Lie algebras and their root systems
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sl(3,C) is Lie algebra
DefinitionA Lie algebra is a vector space L over a field F with a bilinearform [, ] : L× L→ L that satisfies [x , x ] = 0 and[x , [y , z]] + [y , [z, x ]] + [z, [x , y ]] = 0 for x , y , z ∈ L.
As a consequence of this definition, the bracket must beantisymmetric.
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Brief excursion: connection to Lie groups
DefinitionSL(3,C) = {A ∈ M(3,C) : det A = 1}
α(t) =
1 0 00 t2 + 1 t3 + 2t0 t t2 + 1
α(0) =
1 0 00 1 00 0 1
det(α(t)) = 1 =⇒ α(t) ∈ SL(3,C)
α′(t) =
0 0 00 2t 3t2 + 20 1 2t
α′(0) =
0 0 00 0 20 1 0
tr(α′(0)) = 0 =⇒ α′(0) ∈ sl(3,C)
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Brief excursion: connection to Lie groups
DefinitionSL(3,C) = {A ∈ M(3,C) : det A = 1}
α(t) =
1 0 00 t2 + 1 t3 + 2t0 t t2 + 1
α(0) =
1 0 00 1 00 0 1
det(α(t)) = 1 =⇒ α(t) ∈ SL(3,C)
α′(t) =
0 0 00 2t 3t2 + 20 1 2t
α′(0) =
0 0 00 0 20 1 0
tr(α′(0)) = 0 =⇒ α′(0) ∈ sl(3,C)
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Brief excursion: connection to Lie groups
DefinitionSL(3,C) = {A ∈ M(3,C) : det A = 1}
α(t) =
1 0 00 t2 + 1 t3 + 2t0 t t2 + 1
α(0) =
1 0 00 1 00 0 1
det(α(t)) = 1 =⇒ α(t) ∈ SL(3,C)
α′(t) =
0 0 00 2t 3t2 + 20 1 2t
α′(0) =
0 0 00 0 20 1 0
tr(α′(0)) = 0 =⇒ α′(0) ∈ sl(3,C)
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Brief excursion: connection to Lie groups
DefinitionSL(3,C) = {A ∈ M(3,C) : det A = 1}
α(t) =
1 0 00 t2 + 1 t3 + 2t0 t t2 + 1
α(0) =
1 0 00 1 00 0 1
det(α(t)) = 1 =⇒ α(t) ∈ SL(3,C)
α′(t) =
0 0 00 2t 3t2 + 20 1 2t
α′(0) =
0 0 00 0 20 1 0
tr(α′(0)) = 0 =⇒ α′(0) ∈ sl(3,C)
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Brief excursion: connection to Lie groups
DefinitionSL(3,C) = {A ∈ M(3,C) : det A = 1}
α(t) =
1 0 00 t2 + 1 t3 + 2t0 t t2 + 1
α(0) =
1 0 00 1 00 0 1
det(α(t)) = 1 =⇒ α(t) ∈ SL(3,C)
α′(t) =
0 0 00 2t 3t2 + 20 1 2t
α′(0) =
0 0 00 0 20 1 0
tr(α′(0)) = 0 =⇒ α′(0) ∈ sl(3,C)
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Linear maps and ad x
DefinitionLet V ,W be vector spaces. A linear map is a functionφ : V →W such that φ(av1 + v2) = aφ(v1) + φ(v2).
For each x ∈ sl(3,C), we can associate to x a map from L→ Cby y → [x , y ]. This map associated to x is called ad x .
Proposition
For x ∈ sl(3,C), ad x is linear.
Proof.Follows from linearity of the bracket in the 2nd entry.
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Linear maps and ad x
DefinitionLet V ,W be vector spaces. A linear map is a functionφ : V →W such that φ(av1 + v2) = aφ(v1) + φ(v2).
For each x ∈ sl(3,C), we can associate to x a map from L→ Cby y → [x , y ]. This map associated to x is called ad x .
Proposition
For x ∈ sl(3,C), ad x is linear.
Proof.Follows from linearity of the bracket in the 2nd entry.
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Linear maps and ad x
DefinitionLet V ,W be vector spaces. A linear map is a functionφ : V →W such that φ(av1 + v2) = aφ(v1) + φ(v2).
For each x ∈ sl(3,C), we can associate to x a map from L→ Cby y → [x , y ]. This map associated to x is called ad x .
Proposition
For x ∈ sl(3,C), ad x is linear.
Proof.Follows from linearity of the bracket in the 2nd entry.
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Linear maps and ad x
DefinitionLet V ,W be vector spaces. A linear map is a functionφ : V →W such that φ(av1 + v2) = aφ(v1) + φ(v2).
For each x ∈ sl(3,C), we can associate to x a map from L→ Cby y → [x , y ]. This map associated to x is called ad x .
Proposition
For x ∈ sl(3,C), ad x is linear.
Proof.Follows from linearity of the bracket in the 2nd entry.
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The subalgebra of diagonal matrices
H =
d1 0 0
0 d2 00 0 d3
: d1 + d2 + d3 = 0
⊂ sl(3,C)
Proposition
Let A,B ∈ H. Then [A,B] = 0, hence [A,B] ∈ H.
Proof.A,B are diagonal, so AB = BA, so [A,B] = AB − BA = 0.
DefinitionA subalgebra of a Lie algebra L is a vector subspace H that isclosed under the bracket (x , y ∈ H =⇒ [x , y ] ∈ H).
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The subalgebra of diagonal matrices
H =
d1 0 0
0 d2 00 0 d3
: d1 + d2 + d3 = 0
⊂ sl(3,C)
Proposition
Let A,B ∈ H. Then [A,B] = 0, hence [A,B] ∈ H.
Proof.A,B are diagonal, so AB = BA, so [A,B] = AB − BA = 0.
DefinitionA subalgebra of a Lie algebra L is a vector subspace H that isclosed under the bracket (x , y ∈ H =⇒ [x , y ] ∈ H).
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The subalgebra of diagonal matrices
H =
d1 0 0
0 d2 00 0 d3
: d1 + d2 + d3 = 0
⊂ sl(3,C)
Proposition
Let A,B ∈ H. Then [A,B] = 0, hence [A,B] ∈ H.
Proof.A,B are diagonal, so AB = BA, so [A,B] = AB − BA = 0.
DefinitionA subalgebra of a Lie algebra L is a vector subspace H that isclosed under the bracket (x , y ∈ H =⇒ [x , y ] ∈ H).
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The subalgebra of diagonal matrices
H =
d1 0 0
0 d2 00 0 d3
: d1 + d2 + d3 = 0
⊂ sl(3,C)
Proposition
Let A,B ∈ H. Then [A,B] = 0, hence [A,B] ∈ H.
Proof.A,B are diagonal, so AB = BA, so [A,B] = AB − BA = 0.
DefinitionA subalgebra of a Lie algebra L is a vector subspace H that isclosed under the bracket (x , y ∈ H =⇒ [x , y ] ∈ H).
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Ideals: sl(3,C) is simple
DefinitionAn ideal of a Lie algebra L is a vector subspace I such that forx ∈ L,a ∈ I, [x ,a] ∈ I.
Note that every ideal is a subalgebra, but not every subalgebrais an ideal.
Proposition
sl(3,C) is simple, that is, it has no nonzero proper ideals.
DefinitionA Lie algebra L is semisimple if it can be written as a directsum of simple Lie algebras.
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Ideals: sl(3,C) is simple
DefinitionAn ideal of a Lie algebra L is a vector subspace I such that forx ∈ L,a ∈ I, [x ,a] ∈ I.
Note that every ideal is a subalgebra, but not every subalgebrais an ideal.
Proposition
sl(3,C) is simple, that is, it has no nonzero proper ideals.
DefinitionA Lie algebra L is semisimple if it can be written as a directsum of simple Lie algebras.
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Ideals: sl(3,C) is simple
DefinitionAn ideal of a Lie algebra L is a vector subspace I such that forx ∈ L,a ∈ I, [x ,a] ∈ I.
Note that every ideal is a subalgebra, but not every subalgebrais an ideal.
Proposition
sl(3,C) is simple, that is, it has no nonzero proper ideals.
DefinitionA Lie algebra L is semisimple if it can be written as a directsum of simple Lie algebras.
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Sketch of proof that sl(n,C) is simple
Suppose I is a nonzero proper ideal with v 6= 0, v ∈ I. Let
v =∑i 6=j
cijeij +n∑
i=1
dieii
where cij ,di ∈ C. Then
[ekl , [ekl , v ]] = −2c lkekl
[ekl , v ] = (d l − dk )ekl
From this it follows that eij ∈ I for some i 6= j . One can thenshow that if an ideal of sl(n,C) contains some eij , thenI = sl(n,C).
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Sketch of proof that sl(n,C) is simple
Suppose I is a nonzero proper ideal with v 6= 0, v ∈ I. Let
v =∑i 6=j
cijeij +n∑
i=1
dieii
where cij ,di ∈ C. Then
[ekl , [ekl , v ]] = −2c lkekl
[ekl , v ] = (d l − dk )ekl
From this it follows that eij ∈ I for some i 6= j . One can thenshow that if an ideal of sl(n,C) contains some eij , thenI = sl(n,C).
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Sketch of proof that sl(n,C) is simple
Suppose I is a nonzero proper ideal with v 6= 0, v ∈ I. Let
v =∑i 6=j
cijeij +n∑
i=1
dieii
where cij ,di ∈ C. Then
[ekl , [ekl , v ]] = −2c lkekl
[ekl , v ] = (d l − dk )ekl
From this it follows that eij ∈ I for some i 6= j . One can thenshow that if an ideal of sl(n,C) contains some eij , thenI = sl(n,C).
Joshua Ruiter Lie algebras and their root systems
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Sketch of proof that sl(n,C) is simple
Suppose I is a nonzero proper ideal with v 6= 0, v ∈ I. Let
v =∑i 6=j
cijeij +n∑
i=1
dieii
where cij ,di ∈ C. Then
[ekl , [ekl , v ]] = −2c lkekl
[ekl , v ] = (d l − dk )ekl
From this it follows that eij ∈ I for some i 6= j . One can thenshow that if an ideal of sl(n,C) contains some eij , thenI = sl(n,C).
Joshua Ruiter Lie algebras and their root systems
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Sketch of proof that sl(n,C) is simple
Suppose I is a nonzero proper ideal with v 6= 0, v ∈ I. Let
v =∑i 6=j
cijeij +n∑
i=1
dieii
where cij ,di ∈ C. Then
[ekl , [ekl , v ]] = −2c lkekl
[ekl , v ] = (d l − dk )ekl
From this it follows that eij ∈ I for some i 6= j . One can thenshow that if an ideal of sl(n,C) contains some eij , thenI = sl(n,C).
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Diagonal subalgebra, continued
Let h be the diagonal matrix with entries d1,d2,d3. Then
ad h(eij) = [h,eij ] = heij − eijh = dieij − djeij = (di − dj)eij
DefinitionA linear map φ : V → V is diagonalizable if there is a basis ofV consisting of eigenvectors for φ.
Specifically, ad h is diagonalizable.
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Diagonal subalgebra, continued
Let h be the diagonal matrix with entries d1,d2,d3. Then
ad h(eij) = [h,eij ] = heij − eijh = dieij − djeij = (di − dj)eij
DefinitionA linear map φ : V → V is diagonalizable if there is a basis ofV consisting of eigenvectors for φ.
Specifically, ad h is diagonalizable.
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Simultaneous diagonalization
Proposition (Lemma 16.7)Let x1, . . . xk : V → V be diagonalizable linear transformations.There is a basis of V the simultaneously diagonalizes all xi ifand only if each pair xi , xj commutes.
Application: If h1,h2, . . .hk are diagonal elements of sl(3,C),then ad h1,ad h2, . . .ad hk are all simultaneously diagonalizedin the usual basis, so they all pairwise commute.
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Roots (1)
Let h = diag(d1,d2,d3), and let Lij = span{eij}. Based on ourcomputations so far, we have
H ⊂ {x ∈ sl(3,C) : ad h(x) = 0, for all h ∈ H}Lij ⊂ {x ∈ sl(3,C) : ad h(x) = (di − dj)x , for all h ∈ H}
Actually, we can replace ⊂ with =, this takes a bit more work.Define εi : H → C by εi(diag(d1,d2,d3)) = di . Then we canrewrite this as
Lij = {x ∈ sl(3,C) : ad h(x) = (εi − εj)(h)x , for all h ∈ H}
This makes Lij a root space with associated root (εi − εj).
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Roots (1)
Let h = diag(d1,d2,d3), and let Lij = span{eij}. Based on ourcomputations so far, we have
H ⊂ {x ∈ sl(3,C) : ad h(x) = 0, for all h ∈ H}Lij ⊂ {x ∈ sl(3,C) : ad h(x) = (di − dj)x , for all h ∈ H}
Actually, we can replace ⊂ with =, this takes a bit more work.Define εi : H → C by εi(diag(d1,d2,d3)) = di . Then we canrewrite this as
Lij = {x ∈ sl(3,C) : ad h(x) = (εi − εj)(h)x , for all h ∈ H}
This makes Lij a root space with associated root (εi − εj).
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Roots (1)
Let h = diag(d1,d2,d3), and let Lij = span{eij}. Based on ourcomputations so far, we have
H ⊂ {x ∈ sl(3,C) : ad h(x) = 0, for all h ∈ H}Lij ⊂ {x ∈ sl(3,C) : ad h(x) = (di − dj)x , for all h ∈ H}
Actually, we can replace ⊂ with =, this takes a bit more work.Define εi : H → C by εi(diag(d1,d2,d3)) = di . Then we canrewrite this as
Lij = {x ∈ sl(3,C) : ad h(x) = (εi − εj)(h)x , for all h ∈ H}
This makes Lij a root space with associated root (εi − εj).
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Roots (1)
Let h = diag(d1,d2,d3), and let Lij = span{eij}. Based on ourcomputations so far, we have
H ⊂ {x ∈ sl(3,C) : ad h(x) = 0, for all h ∈ H}Lij ⊂ {x ∈ sl(3,C) : ad h(x) = (di − dj)x , for all h ∈ H}
Actually, we can replace ⊂ with =, this takes a bit more work.Define εi : H → C by εi(diag(d1,d2,d3)) = di . Then we canrewrite this as
Lij = {x ∈ sl(3,C) : ad h(x) = (εi − εj)(h)x , for all h ∈ H}
This makes Lij a root space with associated root (εi − εj).
Joshua Ruiter Lie algebras and their root systems
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Roots (1)
Let h = diag(d1,d2,d3), and let Lij = span{eij}. Based on ourcomputations so far, we have
H ⊂ {x ∈ sl(3,C) : ad h(x) = 0, for all h ∈ H}Lij ⊂ {x ∈ sl(3,C) : ad h(x) = (di − dj)x , for all h ∈ H}
Actually, we can replace ⊂ with =, this takes a bit more work.Define εi : H → C by εi(diag(d1,d2,d3)) = di . Then we canrewrite this as
Lij = {x ∈ sl(3,C) : ad h(x) = (εi − εj)(h)x , for all h ∈ H}
This makes Lij a root space with associated root (εi − εj).
Joshua Ruiter Lie algebras and their root systems
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Roots (1)
Let h = diag(d1,d2,d3), and let Lij = span{eij}. Based on ourcomputations so far, we have
H ⊂ {x ∈ sl(3,C) : ad h(x) = 0, for all h ∈ H}Lij ⊂ {x ∈ sl(3,C) : ad h(x) = (di − dj)x , for all h ∈ H}
Actually, we can replace ⊂ with =, this takes a bit more work.Define εi : H → C by εi(diag(d1,d2,d3)) = di . Then we canrewrite this as
Lij = {x ∈ sl(3,C) : ad h(x) = (εi − εj)(h)x , for all h ∈ H}
This makes Lij a root space with associated root (εi − εj).
Joshua Ruiter Lie algebras and their root systems
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Roots (2)
DefinitionA root is a linear map α : H → C such that
{x ∈ sl(3,C) : ad h(x) = α(h)x for all h ∈ H}
is a nonzero subspace of sl(3,C).
DefinitionA root space is a nonzero subspace of sl(3,C) of the form
{x ∈ sl(3,C) : ad h(x) = α(h)x for all h ∈ H}
where α : H → C is a linear map.
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Roots (2)
DefinitionA root is a linear map α : H → C such that
{x ∈ sl(3,C) : ad h(x) = α(h)x for all h ∈ H}
is a nonzero subspace of sl(3,C).
DefinitionA root space is a nonzero subspace of sl(3,C) of the form
{x ∈ sl(3,C) : ad h(x) = α(h)x for all h ∈ H}
where α : H → C is a linear map.
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Root space decomposition
Proposition
Φ = {εi − εj : i 6= j} is the entire set of roots for sl(3,C).
Recall:
sl(3,C) = span{e11 − e22,e22 − e33} ⊕⊕i 6=j
span{eij}
Now that we have the language of roots, we can write this as
sl(3,C) = H ⊕⊕i 6=j
Lij = H ⊕⊕α∈Φ
Lα
This is called the root space decomposition.
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Root space decomposition
Proposition
Φ = {εi − εj : i 6= j} is the entire set of roots for sl(3,C).
Recall:
sl(3,C) = span{e11 − e22,e22 − e33} ⊕⊕i 6=j
span{eij}
Now that we have the language of roots, we can write this as
sl(3,C) = H ⊕⊕i 6=j
Lij = H ⊕⊕α∈Φ
Lα
This is called the root space decomposition.
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Root space decomposition
Proposition
Φ = {εi − εj : i 6= j} is the entire set of roots for sl(3,C).
Recall:
sl(3,C) = span{e11 − e22,e22 − e33} ⊕⊕i 6=j
span{eij}
Now that we have the language of roots, we can write this as
sl(3,C) = H ⊕⊕i 6=j
Lij = H ⊕⊕α∈Φ
Lα
This is called the root space decomposition.
Joshua Ruiter Lie algebras and their root systems
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Root systems
DefinitionA root system is a subset R of a real inner-product space Esatisfying:(R1) R is finite, it spans E , and it does not contain 0.(R2) If α ∈ R, then the only scalar multiples of α in R are ±α.(R3) If α ∈ R, then the reflection sα permutes R.(R4) If α, β ∈ R, then 2(α, β)/(β, β) ∈ Z.
PropositionLet L be a complex semisimple Lie algebra, and let Φ be a setof roots of L. Then Φ is a root system.
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Root systems
DefinitionA root system is a subset R of a real inner-product space Esatisfying:(R1) R is finite, it spans E , and it does not contain 0.(R2) If α ∈ R, then the only scalar multiples of α in R are ±α.(R3) If α ∈ R, then the reflection sα permutes R.(R4) If α, β ∈ R, then 2(α, β)/(β, β) ∈ Z.
PropositionLet L be a complex semisimple Lie algebra, and let Φ be a setof roots of L. Then Φ is a root system.
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Root diagram for sl(3,C)
α
α + ββ
−α
−(α + β) −β
α = ε1 − ε2 β = ε2 − ε3
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Classification by root systems
Theorem (Cartan)Up to isomorphism, there is just one complex semisimple Liealgebra for each root system.
Theorem (Cartan)With five exceptions, every finite-dimensional simple Liealgebra over C is isomorphic to one of the classical Lie algebras
sl(n,C) so(n,C) sp(2n,C)
The five exceptional Lie algebras are known as e6,e7,e8, f4,and g2.
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Classification by root systems
Theorem (Cartan)Up to isomorphism, there is just one complex semisimple Liealgebra for each root system.
Theorem (Cartan)With five exceptions, every finite-dimensional simple Liealgebra over C is isomorphic to one of the classical Lie algebras
sl(n,C) so(n,C) sp(2n,C)
The five exceptional Lie algebras are known as e6,e7,e8, f4,and g2.
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Joshua Ruiter Lie algebras and their root systems