Normal forms of lone symmetries Let � 0 − 1 � 0 1 � � ε = , ω = i 1 0 1 0 det ε = det ω = 1 We’ll see: Depending on classes, N := dim V = n , 2 n , 4 n , ( n = 1 , 2 , . . . ). We’ll construct adapted orthonormal bases V = ( v j ) k j = 1 = ( v 1 , . . . , v k ) of invariant subspaces of V of dimension k = 1 , 2 , 4; generated by arbitrary v 1 , ( � v 1 � = 1). Exhaust the complement. Left action: UV = ( Uv 1 , . . . , Uv k ) ( U : V → V map) i v i M ij ) k Right action: VM = ( � j = 1 ( M : matrix of order k ) ◮ Θ ( α = + 1) has k = 1: v 1 such that Θ v 1 = v 1 ,
Normal forms of lone symmetries Let � 0 − 1 � 0 1 � � ε = , ω = i 1 0 1 0 det ε = det ω = 1 We’ll see: Depending on classes, N := dim V = n , 2 n , 4 n , ( n = 1 , 2 , . . . ). We’ll construct adapted orthonormal bases V = ( v j ) k j = 1 = ( v 1 , . . . , v k ) of invariant subspaces of V of dimension k = 1 , 2 , 4; generated by arbitrary v 1 , ( � v 1 � = 1). Exhaust the complement. Left action: UV = ( Uv 1 , . . . , Uv k ) ( U : V → V map) i v i M ij ) k Right action: VM = ( � j = 1 ( M : matrix of order k ) ◮ Θ ( α = + 1) has k = 1: v 1 such that Θ v 1 = v 1 , i.e. Θ V = V
Normal forms of lone symmetries Let � 0 − 1 � 0 1 � � ε = , ω = i 1 0 1 0 det ε = det ω = 1 We’ll see: Depending on classes, N := dim V = n , 2 n , 4 n , ( n = 1 , 2 , . . . ). We’ll construct adapted orthonormal bases V = ( v j ) k j = 1 = ( v 1 , . . . , v k ) of invariant subspaces of V of dimension k = 1 , 2 , 4; generated by arbitrary v 1 , ( � v 1 � = 1). Exhaust the complement. Left action: UV = ( Uv 1 , . . . , Uv k ) ( U : V → V map) i v i M ij ) k Right action: VM = ( � j = 1 ( M : matrix of order k ) ◮ Θ ( α = + 1) has k = 1: v 1 such that Θ v 1 = v 1 , i.e. Θ V = V ◮ Θ ( α = − 1) has k = 2:
Normal forms of lone symmetries Let � 0 − 1 � 0 1 � � ε = , ω = i 1 0 1 0 det ε = det ω = 1 We’ll see: Depending on classes, N := dim V = n , 2 n , 4 n , ( n = 1 , 2 , . . . ). We’ll construct adapted orthonormal bases V = ( v j ) k j = 1 = ( v 1 , . . . , v k ) of invariant subspaces of V of dimension k = 1 , 2 , 4; generated by arbitrary v 1 , ( � v 1 � = 1). Exhaust the complement. Left action: UV = ( Uv 1 , . . . , Uv k ) ( U : V → V map) i v i M ij ) k Right action: VM = ( � j = 1 ( M : matrix of order k ) ◮ Θ ( α = + 1) has k = 1: v 1 such that Θ v 1 = v 1 , i.e. Θ V = V ◮ Θ ( α = − 1) has k = 2: v 2 := Θ v 1 , Θ v 2 = − v 1 ; then Θ V = (Θ v 1 , Θ v 2 ) = ( v 2 , − v 1 ) = V ε
Normal forms of lone symmetries (cont.) ◮ Σ ( β = + 1) has k = 2:
Normal forms of lone symmetries (cont.) ◮ Σ ( β = + 1) has k = 2: v 1 ∈ V + , v 2 := − i Σ v 1 ∈ V − ; then Σ V = ( i v 2 , i v 1 ) = V ω
Normal forms of lone symmetries (cont.) ◮ Σ ( β = + 1) has k = 2: v 1 ∈ V + , v 2 := − i Σ v 1 ∈ V − ; then Σ V = ( i v 2 , i v 1 ) = V ω ◮ Σ ( β = − 1) has k = 2:
Normal forms of lone symmetries (cont.) ◮ Σ ( β = + 1) has k = 2: v 1 ∈ V + , v 2 := − i Σ v 1 ∈ V − ; then Σ V = ( i v 2 , i v 1 ) = V ω ◮ Σ ( β = − 1) has k = 2: v 1 ∈ V + , v 2 := Σ v 1 ∈ V − ; then Σ V = ( v 2 , − v 1 ) = V ε
Normal forms of lone symmetries (cont.) ◮ Σ ( β = + 1) has k = 2: v 1 ∈ V + , v 2 := − i Σ v 1 ∈ V − ; then Σ V = ( i v 2 , i v 1 ) = V ω ◮ Σ ( β = − 1) has k = 2: v 1 ∈ V + , v 2 := Σ v 1 ∈ V − ; then Σ V = ( v 2 , − v 1 ) = V ε ◮ Π ( γ = − 1 w.l.o.g.) has k = 2:
Normal forms of lone symmetries (cont.) ◮ Σ ( β = + 1) has k = 2: v 1 ∈ V + , v 2 := − i Σ v 1 ∈ V − ; then Σ V = ( i v 2 , i v 1 ) = V ω ◮ Σ ( β = − 1) has k = 2: v 1 ∈ V + , v 2 := Σ v 1 ∈ V − ; then Σ V = ( v 2 , − v 1 ) = V ε ◮ Π ( γ = − 1 w.l.o.g.) has k = 2: v 1 ∈ V + , v 2 := − i Π v 1 ∈ V − ; then Π V = ( i v 2 , i v 1 ) = V ω
Normal forms of combined symmetries Θ , Σ ◮ α = + 1 , β = ± 1 has k = 2: � ω ( β = + 1 ) Θ V = V , Σ V = V ε ( β = − 1 )
Normal forms of combined symmetries Θ , Σ ◮ α = + 1 , β = ± 1 has k = 2: � ω ( β = + 1 ) Θ V = V , Σ V = V ε ( β = − 1 ) ◮ α = − 1 , β = ± 1 has k = 4: � ε 0 � � 0 β 1 2 � Θ V = V , Σ V = V 0 ε 1 2 0
The identification V ∼ = C N Let K : C N → C N be the standard complex conjugation
The identification V ∼ = C N Let K : C N → C N be the standard complex conjugation Claim: Lone symmetries U = Θ , Σ ( U 2 = γ = ± 1; γ = α or β , hence 2 + 2 cases) on V can be brought to the following form on C N ◮ γ = + 1 ( N = n , 2 n ): U = K ◮ γ = − 1 ( N = 2 n ): U = ε K
The identification V ∼ = C N Let K : C N → C N be the standard complex conjugation Claim: Lone symmetries U = Θ , Σ ( U 2 = γ = ± 1; γ = α or β , hence 2 + 2 cases) on V can be brought to the following form on C N ◮ γ = + 1 ( N = n , 2 n ): U = K ◮ γ = − 1 ( N = 2 n ): U = ε K Remarks. 1) Note that Σ qualifies as Θ (see earlier remark)
The identification V ∼ = C N Let K : C N → C N be the standard complex conjugation Claim: Lone symmetries U = Θ , Σ ( U 2 = γ = ± 1; γ = α or β , hence 2 + 2 cases) on V can be brought to the following form on C N ◮ γ = + 1 ( N = n , 2 n ): U = K ◮ γ = − 1 ( N = 2 n ): U = ε K Remarks. 1) Note that Σ qualifies as Θ (see earlier remark) 2) For U = Σ the split V = V + ⊕ V − is compatibly realized as V ± = { ( v , ± i v ) | v ∈ C n } ⊂ C n ⊕ C n = C N , (or flipped)
The identification V ∼ = C N Let K : C N → C N be the standard complex conjugation Claim: Lone symmetries U = Θ , Σ ( U 2 = γ = ± 1; γ = α or β , hence 2 + 2 cases) on V can be brought to the following form on C N ◮ γ = + 1 ( N = n , 2 n ): U = K ◮ γ = − 1 ( N = 2 n ): U = ε K Remarks. 1) Note that Σ qualifies as Θ (see earlier remark) 2) For U = Σ the split V = V + ⊕ V − is compatibly realized as V ± = { ( v , ± i v ) | v ∈ C n } ⊂ C n ⊕ C n = C N , (or flipped) 3) Proof: ◮ In the cases α = ± 1, the claim is seen by mapping an adapted basis V to the standard basis of C N
The identification V ∼ = C N Let K : C N → C N be the standard complex conjugation Claim: Lone symmetries U = Θ , Σ ( U 2 = γ = ± 1; γ = α or β , hence 2 + 2 cases) on V can be brought to the following form on C N ◮ γ = + 1 ( N = n , 2 n ): U = K ◮ γ = − 1 ( N = 2 n ): U = ε K Remarks. 1) Note that Σ qualifies as Θ (see earlier remark) 2) For U = Σ the split V = V + ⊕ V − is compatibly realized as V ± = { ( v , ± i v ) | v ∈ C n } ⊂ C n ⊕ C n = C N , (or flipped) 3) Proof: ◮ In the cases α = ± 1, the claim is seen by mapping an adapted basis V to the standard basis of C N � 1 ◮ In the cases β = ± 1 the mapping is (e.g.) v 1 �→ � , � 1 i � v 2 �→ − i β , compatibly with the stated V ± (unflipped) − i
The identification V ∼ = C N Let K : C N → C N be the standard complex conjugation Claim: Lone symmetries U = Θ , Σ ( U 2 = γ = ± 1; γ = α or β , hence 2 + 2 cases) on V can be brought to the following form on C N ◮ γ = + 1 ( N = n , 2 n ): U = K ◮ γ = − 1 ( N = 2 n ): U = ε K Remarks. 1) Note that Σ qualifies as Θ (see earlier remark) 2) For U = Σ the split V = V + ⊕ V − is compatibly realized as V ± = { ( v , ± i v ) | v ∈ C n } ⊂ C n ⊕ C n = C N , (or flipped) 3) Proof: ◮ In the cases α = ± 1, the claim is seen by mapping an adapted basis V to the standard basis of C N � 1 ◮ In the cases β = ± 1 the mapping is (e.g.) v 1 �→ � , � 1 i � v 2 �→ − i β , compatibly with the stated V ± (unflipped) − i 4) This foreshadows: The lone symmetry Θ will not contribute to G (to be checked).
The classification task in general Any (symmetry equipped) vector space V can be identified with C N (symmetry equipped as explained).
The classification task in general Any (symmetry equipped) vector space V can be identified with C N (symmetry equipped as explained). However this does not imply that any two of them are homotopic, as the identifications might be non homotopic.
The classification task in general Any (symmetry equipped) vector space V can be identified with C N (symmetry equipped as explained). However this does not imply that any two of them are homotopic, as the identifications might be non homotopic. Viewed through C N , the setting and the task become ◮ Bases (unrestricted) are matrices V ∈ U ( N ) (unitary group as a set)
The classification task in general Any (symmetry equipped) vector space V can be identified with C N (symmetry equipped as explained). However this does not imply that any two of them are homotopic, as the identifications might be non homotopic. Viewed through C N , the setting and the task become ◮ Bases (unrestricted) are matrices V ∈ U ( N ) (unitary group as a set) ◮ Change of basis (in general) is by action of T ∈ U ( N ) (unitary group as a group): V �→ VT
The classification task in general Any (symmetry equipped) vector space V can be identified with C N (symmetry equipped as explained). However this does not imply that any two of them are homotopic, as the identifications might be non homotopic. Viewed through C N , the setting and the task become ◮ Bases (unrestricted) are matrices V ∈ U ( N ) (unitary group as a set) ◮ Change of basis (in general) is by action of T ∈ U ( N ) (unitary group as a group): V �→ VT ◮ Adapted bases form the set B = { V ∈ U ( N ) | symmetry constraint }
The classification task in general Any (symmetry equipped) vector space V can be identified with C N (symmetry equipped as explained). However this does not imply that any two of them are homotopic, as the identifications might be non homotopic. Viewed through C N , the setting and the task become ◮ Bases (unrestricted) are matrices V ∈ U ( N ) (unitary group as a set) ◮ Change of basis (in general) is by action of T ∈ U ( N ) (unitary group as a group): V �→ VT ◮ Adapted bases form the set B = { V ∈ U ( N ) | symmetry constraint } ◮ Change of adapted basis is by action of T = { T ∈ U ( N ) | symmetry constraint preserving }
The classification task in general Any (symmetry equipped) vector space V can be identified with C N (symmetry equipped as explained). However this does not imply that any two of them are homotopic, as the identifications might be non homotopic. Viewed through C N , the setting and the task become ◮ Bases (unrestricted) are matrices V ∈ U ( N ) (unitary group as a set) ◮ Change of basis (in general) is by action of T ∈ U ( N ) (unitary group as a group): V �→ VT ◮ Adapted bases form the set B = { V ∈ U ( N ) | symmetry constraint } ◮ Change of adapted basis is by action of T = { T ∈ U ( N ) | symmetry constraint preserving } ◮ Classification: Right cosets B / T , connected components thereof.
Class A: no symmetry B = U ( N ) , T = U ( N ) , hence B / T trivial.
Class A: no symmetry B = U ( N ) , T = U ( N ) , hence B / T trivial. Only obstruction is N = dim V . So index group is G = Z
Class A: no symmetry B = U ( N ) , T = U ( N ) , hence B / T trivial. Only obstruction is N = dim V . So index group is G = Z Remarks. 1) The index trivializes for Σ , Π by dim V + − dim V − = 0. Do new indices appear? 2) The index survives for just Θ (classes AI, AII). Does the group become larger? (Likely not by earlier remark)
Class AI: Θ with α = + 1 ◮ Basis (normal form): V = Θ V
Class AI: Θ with α = + 1 ◮ Basis (normal form): V = Θ V ◮ Concretely: Θ = K , so V = ¯ V
Class AI: Θ with α = + 1 ◮ Basis (normal form): V = Θ V ◮ Concretely: Θ = K , so V = ¯ V ◮ So: B = O ( N ) . Index Z 2 by det V = ± 1?
Class AI: Θ with α = + 1 ◮ Basis (normal form): V = Θ V ◮ Concretely: Θ = K , so V = ¯ V ◮ So: B = O ( N ) . Index Z 2 by det V = ± 1? ◮ Change of basis V → VT .
Class AI: Θ with α = + 1 ◮ Basis (normal form): V = Θ V ◮ Concretely: Θ = K , so V = ¯ V ◮ So: B = O ( N ) . Index Z 2 by det V = ± 1? ◮ Change of basis V → VT . “Is vs. Ought”: VT = Θ( VT ) = (Θ V )¯ VT = (Θ V ) T , T
Class AI: Θ with α = + 1 ◮ Basis (normal form): V = Θ V ◮ Concretely: Θ = K , so V = ¯ V ◮ So: B = O ( N ) . Index Z 2 by det V = ± 1? ◮ Change of basis V → VT . “Is vs. Ought”: VT = Θ( VT ) = (Θ V )¯ VT = (Θ V ) T , T ◮ So T = ¯ T ,
Class AI: Θ with α = + 1 ◮ Basis (normal form): V = Θ V ◮ Concretely: Θ = K , so V = ¯ V ◮ So: B = O ( N ) . Index Z 2 by det V = ± 1? ◮ Change of basis V → VT . “Is vs. Ought”: VT = Θ( VT ) = (Θ V )¯ VT = (Θ V ) T , T ◮ So T = ¯ T , T = O ( N )
Class AI: Θ with α = + 1 ◮ Basis (normal form): V = Θ V ◮ Concretely: Θ = K , so V = ¯ V ◮ So: B = O ( N ) . Index Z 2 by det V = ± 1? ◮ Change of basis V → VT . “Is vs. Ought”: VT = Θ( VT ) = (Θ V )¯ VT = (Θ V ) T , T ◮ So T = ¯ T , T = O ( N ) and B / T is trivial
Class AI: Θ with α = + 1 ◮ Basis (normal form): V = Θ V ◮ Concretely: Θ = K , so V = ¯ V ◮ So: B = O ( N ) . Index Z 2 by det V = ± 1? ◮ Change of basis V → VT . “Is vs. Ought”: VT = Θ( VT ) = (Θ V )¯ VT = (Θ V ) T , T ◮ So T = ¯ T , T = O ( N ) and B / T is trivial ◮ Index group remains G = Z
Class AII: Θ with α = − 1 ◮ Basis (normal form): Θ V = V ε (or V = − Θ V ε )
Class AII: Θ with α = − 1 ◮ Basis (normal form): Θ V = V ε (or V = − Θ V ε ) ◮ Concretely: Θ = ε K , so ε ¯ V = V ε
Class AII: Θ with α = − 1 ◮ Basis (normal form): Θ V = V ε (or V = − Θ V ε ) ◮ Concretely: Θ = ε K , so ε ¯ V = V ε ◮ So B = { V | V unitary , ε ¯ V = V ε }
Class AII: Θ with α = − 1 ◮ Basis (normal form): Θ V = V ε (or V = − Θ V ε ) ◮ Concretely: Θ = ε K , so ε ¯ V = V ε ◮ So B = { V | V unitary , ε ¯ V = V ε } ◮ Change of basis V → VT .
Class AII: Θ with α = − 1 ◮ Basis (normal form): Θ V = V ε (or V = − Θ V ε ) ◮ Concretely: Θ = ε K , so ε ¯ V = V ε ◮ So B = { V | V unitary , ε ¯ V = V ε } ◮ Change of basis V → VT . “Is vs. Ought”: VT = − Θ( VT ) ε = − (Θ V )¯ VT = − (Θ V ε ) T , T ε
Class AII: Θ with α = − 1 ◮ Basis (normal form): Θ V = V ε (or V = − Θ V ε ) ◮ Concretely: Θ = ε K , so ε ¯ V = V ε ◮ So B = { V | V unitary , ε ¯ V = V ε } ◮ Change of basis V → VT . “Is vs. Ought”: VT = − Θ( VT ) ε = − (Θ V )¯ VT = − (Θ V ε ) T , T ε ◮ So ε T = ¯ T ε ,
Class AII: Θ with α = − 1 ◮ Basis (normal form): Θ V = V ε (or V = − Θ V ε ) ◮ Concretely: Θ = ε K , so ε ¯ V = V ε ◮ So B = { V | V unitary , ε ¯ V = V ε } ◮ Change of basis V → VT . “Is vs. Ought”: VT = − Θ( VT ) ε = − (Θ V )¯ VT = − (Θ V ε ) T , T ε ◮ So ε T = ¯ T ε , T = B and B / T is trivial
Class AII: Θ with α = − 1 ◮ Basis (normal form): Θ V = V ε (or V = − Θ V ε ) ◮ Concretely: Θ = ε K , so ε ¯ V = V ε ◮ So B = { V | V unitary , ε ¯ V = V ε } ◮ Change of basis V → VT . “Is vs. Ought”: VT = − Θ( VT ) ε = − (Θ V )¯ VT = − (Θ V ε ) T , T ε ◮ So ε T = ¯ T ε , T = B and B / T is trivial ◮ Index group remains G = Z
Class D: Σ with β = + 1 ◮ Basis (normal form): Σ V = V ω (or V = − (Σ V ) ω )
Class D: Σ with β = + 1 ◮ Basis (normal form): Σ V = V ω (or V = − (Σ V ) ω ) ◮ Concretely: Σ = K , so ¯ V = V ω
Class D: Σ with β = + 1 ◮ Basis (normal form): Σ V = V ω (or V = − (Σ V ) ω ) ◮ Concretely: Σ = K , so ¯ V = V ω ◮ So B = { V | V unitary , ¯ V = V ω }
Class D: Σ with β = + 1 ◮ Basis (normal form): Σ V = V ω (or V = − (Σ V ) ω ) ◮ Concretely: Σ = K , so ¯ V = V ω ◮ So B = { V | V unitary , ¯ V = V ω } ◮ det V = det V , hence det V = ± 1. Index Z 2 ?
Class D: Σ with β = + 1 ◮ Basis (normal form): Σ V = V ω (or V = − (Σ V ) ω ) ◮ Concretely: Σ = K , so ¯ V = V ω ◮ So B = { V | V unitary , ¯ V = V ω } ◮ det V = det V , hence det V = ± 1. Index Z 2 ? ◮ Change of basis V → VT , T = diag( T + , T − ) .
Class D: Σ with β = + 1 ◮ Basis (normal form): Σ V = V ω (or V = − (Σ V ) ω ) ◮ Concretely: Σ = K , so ¯ V = V ω ◮ So B = { V | V unitary , ¯ V = V ω } ◮ det V = det V , hence det V = ± 1. Index Z 2 ? ◮ Change of basis V → VT , T = diag( T + , T − ) . “Is vs. Ought”: ω T = ¯ T ω , i.e. T − = T +
Class D: Σ with β = + 1 ◮ Basis (normal form): Σ V = V ω (or V = − (Σ V ) ω ) ◮ Concretely: Σ = K , so ¯ V = V ω ◮ So B = { V | V unitary , ¯ V = V ω } ◮ det V = det V , hence det V = ± 1. Index Z 2 ? ◮ Change of basis V → VT , T = diag( T + , T − ) . “Is vs. Ought”: ω T = ¯ T ω , i.e. T − = T + ◮ So T = U ( n )
Class D: Σ with β = + 1 ◮ Basis (normal form): Σ V = V ω (or V = − (Σ V ) ω ) ◮ Concretely: Σ = K , so ¯ V = V ω ◮ So B = { V | V unitary , ¯ V = V ω } ◮ det V = det V , hence det V = ± 1. Index Z 2 ? ◮ Change of basis V → VT , T = diag( T + , T − ) . “Is vs. Ought”: ω T = ¯ T ω , i.e. T − = T + ◮ So T = U ( n ) ◮ Index group is G = Z 2
Class C: Σ with β = − 1 ◮ Basis (normal form): Σ V = V ε ◮ Concretely: Σ = ε K , so ε ¯ V = V ε
Class C: Σ with β = − 1 ◮ Basis (normal form): Σ V = V ε ◮ Concretely: Σ = ε K , so ε ¯ V = V ε ◮ So B = { V | V unitary , V ε V T = ε }
Class C: Σ with β = − 1 ◮ Basis (normal form): Σ V = V ε ◮ Concretely: Σ = ε K , so ε ¯ V = V ε ◮ So B = { V | V unitary , V ε V T = ε } ◮ V symplectic, so det V = + 1.
Class C: Σ with β = − 1 ◮ Basis (normal form): Σ V = V ε ◮ Concretely: Σ = ε K , so ε ¯ V = V ε ◮ So B = { V | V unitary , V ε V T = ε } ◮ V symplectic, so det V = + 1. (Use pf( ABA T ) = det A · pf ( B ) )
Class C: Σ with β = − 1 ◮ Basis (normal form): Σ V = V ε ◮ Concretely: Σ = ε K , so ε ¯ V = V ε ◮ So B = { V | V unitary , V ε V T = ε } ◮ V symplectic, so det V = + 1. (Use pf( ABA T ) = det A · pf ( B ) ) ◮ Change of basis V → VT , T = diag( T + , T − ) .
Class C: Σ with β = − 1 ◮ Basis (normal form): Σ V = V ε ◮ Concretely: Σ = ε K , so ε ¯ V = V ε ◮ So B = { V | V unitary , V ε V T = ε } ◮ V symplectic, so det V = + 1. (Use pf( ABA T ) = det A · pf ( B ) ) ◮ Change of basis V → VT , T = diag( T + , T − ) . “Is vs. Ought”: ε T = T ε , i.e. T − = T +
Class C: Σ with β = − 1 ◮ Basis (normal form): Σ V = V ε ◮ Concretely: Σ = ε K , so ε ¯ V = V ε ◮ So B = { V | V unitary , V ε V T = ε } ◮ V symplectic, so det V = + 1. (Use pf( ABA T ) = det A · pf ( B ) ) ◮ Change of basis V → VT , T = diag( T + , T − ) . “Is vs. Ought”: ε T = T ε , i.e. T − = T + ◮ So T = U ( n )
Class C: Σ with β = − 1 ◮ Basis (normal form): Σ V = V ε ◮ Concretely: Σ = ε K , so ε ¯ V = V ε ◮ So B = { V | V unitary , V ε V T = ε } ◮ V symplectic, so det V = + 1. (Use pf( ABA T ) = det A · pf ( B ) ) ◮ Change of basis V → VT , T = diag( T + , T − ) . “Is vs. Ought”: ε T = T ε , i.e. T − = T + ◮ So T = U ( n ) ◮ Index group is G = 0.
Classes involving Π (AIII, BDI, CI, DIII, CII) An alternate procedure is available. Recall bases adapted to Θ or Σ (but not to Π = ΘΣ ).
Classes involving Π (AIII, BDI, CI, DIII, CII) An alternate procedure is available. Recall bases adapted to Θ or Σ (but not to Π = ΘΣ ). Change to a basis adapted to Π (or i Π ): � 1 0 � Π = , ( 1 = 1 n , 1 2 n ) 0 − 1 Let K be complex conjugation in that basis (thus real).
Classes involving Π (AIII, BDI, CI, DIII, CII) An alternate procedure is available. Recall bases adapted to Θ or Σ (but not to Π = ΘΣ ). Change to a basis adapted to Π (or i Π ): � 1 0 � Π = , ( 1 = 1 n , 1 2 n ) 0 − 1 Let K be complex conjugation in that basis (thus real). In presence of Θ or Σ (and hence all three with α, β = ± 1):
Classes involving Π (AIII, BDI, CI, DIII, CII) An alternate procedure is available. Recall bases adapted to Θ or Σ (but not to Π = ΘΣ ). Change to a basis adapted to Π (or i Π ): � 1 0 � Π = , ( 1 = 1 n , 1 2 n ) 0 − 1 Let K be complex conjugation in that basis (thus real). In presence of Θ or Σ (and hence all three with α, β = ± 1): ◮ α = + 1 ( N = 2 n )
Classes involving Π (AIII, BDI, CI, DIII, CII) An alternate procedure is available. Recall bases adapted to Θ or Σ (but not to Π = ΘΣ ). Change to a basis adapted to Π (or i Π ): � 1 0 � Π = , ( 1 = 1 n , 1 2 n ) 0 − 1 Let K be complex conjugation in that basis (thus real). In presence of Θ or Σ (and hence all three with α, β = ± 1): ◮ α = + 1 ( N = 2 n ) � 1 0 ◮ β = + 1: Θ = � K 0 1
Classes involving Π (AIII, BDI, CI, DIII, CII) An alternate procedure is available. Recall bases adapted to Θ or Σ (but not to Π = ΘΣ ). Change to a basis adapted to Π (or i Π ): � 1 0 � Π = , ( 1 = 1 n , 1 2 n ) 0 − 1 Let K be complex conjugation in that basis (thus real). In presence of Θ or Σ (and hence all three with α, β = ± 1): ◮ α = + 1 ( N = 2 n ) � 1 0 ◮ β = + 1: Θ = � K 0 1 � 0 1 ◮ β = − 1: Θ = � K 1 0
Classes involving Π (AIII, BDI, CI, DIII, CII) An alternate procedure is available. Recall bases adapted to Θ or Σ (but not to Π = ΘΣ ). Change to a basis adapted to Π (or i Π ): � 1 0 � Π = , ( 1 = 1 n , 1 2 n ) 0 − 1 Let K be complex conjugation in that basis (thus real). In presence of Θ or Σ (and hence all three with α, β = ± 1): ◮ α = + 1 ( N = 2 n ) � 1 0 ◮ β = + 1: Θ = � K 0 1 � 0 1 ◮ β = − 1: Θ = � K 1 0 ◮ α = − 1 ( N = 4 n )
Classes involving Π (AIII, BDI, CI, DIII, CII) An alternate procedure is available. Recall bases adapted to Θ or Σ (but not to Π = ΘΣ ). Change to a basis adapted to Π (or i Π ): � 1 0 � Π = , ( 1 = 1 n , 1 2 n ) 0 − 1 Let K be complex conjugation in that basis (thus real). In presence of Θ or Σ (and hence all three with α, β = ± 1): ◮ α = + 1 ( N = 2 n ) � 1 0 ◮ β = + 1: Θ = � K 0 1 � 0 1 ◮ β = − 1: Θ = � K 1 0 ◮ α = − 1 ( N = 4 n ) � 0 ε ◮ β = + 1: Θ = � K ε 0
Classes involving Π (AIII, BDI, CI, DIII, CII) An alternate procedure is available. Recall bases adapted to Θ or Σ (but not to Π = ΘΣ ). Change to a basis adapted to Π (or i Π ): � 1 0 � Π = , ( 1 = 1 n , 1 2 n ) 0 − 1 Let K be complex conjugation in that basis (thus real). In presence of Θ or Σ (and hence all three with α, β = ± 1): ◮ α = + 1 ( N = 2 n ) � 1 0 ◮ β = + 1: Θ = � K 0 1 � 0 1 ◮ β = − 1: Θ = � K 1 0 ◮ α = − 1 ( N = 4 n ) � 0 ε ◮ β = + 1: Θ = � K ε 0 � ε 0 ◮ β = − 1: Θ = � K 0 ε ( ε composite as a rule)
The classification task involving Π Setting: Vector space V = V + ⊕ V − ,
The classification task involving Π Setting: Vector space V = V + ⊕ V − , split encoded in projections P ± onto V ± (hence Π P ± = P ∓ Π ),
The classification task involving Π Setting: Vector space V = V + ⊕ V − , split encoded in projections P ± onto V ± (hence Π P ± = P ∓ Π ), or in ”flattened Hamiltonian” H = H ∗ = P + − P − , H 2 = 1 hence with { H , Π } = 0
The classification task involving Π Setting: Vector space V = V + ⊕ V − , split encoded in projections P ± onto V ± (hence Π P ± = P ∓ Π ), or in ”flattened Hamiltonian” H = H ∗ = P + − P − , H 2 = 1 hence with { H , Π } = 0 � 1 0 � Equivalently (recall Π = in adapted basis) 0 − 1 � 0 U ∗ � H = , U ∈ U ( n ) , U ( 2 n ) U 0
The classification task involving Π Setting: Vector space V = V + ⊕ V − , split encoded in projections P ± onto V ± (hence Π P ± = P ∓ Π ), or in ”flattened Hamiltonian” H = H ∗ = P + − P − , H 2 = 1 hence with { H , Π } = 0 � 1 0 � Equivalently (recall Π = in adapted basis) 0 − 1 � 0 U ∗ � H = , U ∈ U ( n ) , U ( 2 n ) U 0 If further symmetries: [ H , Θ] = 0.
Aside: Normal form of matrices ◮ (Takagi) Every complex symmetric matrix A = A T is of the form A = UNU T with U unitary and N = N T diagonal.
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