Introduction E9F5FC

Questions FFFFC0

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Bott Correspondence

Slides

• 9+1, 8+2, 7+3, 6+4
• two kinds of Clifford algebra
  • squaring products of Clifford generators
• R,C,H structure
  • {$a^2=\theta$}, {$a^4=1$}
  • 2x2 matrices and the operators
  • three group extensions: recurring activity and structure
• matrix expression of {$J_1, J_2, J_3\cdots$}

Readings

• Stone, Chiu, Roy. Symmetries, Dimensions, and Topological Insulators: the mechanism behind the face of the Bott clock.
• Trimble. The Super Brauer Group and Super Division Algebras.
• Moore. Quantum Symmetries and K-Theory.
• Moore. Quantum Symmetries and Compatible Hamiltonians
• Moore. Linear Algebra User’s Manual.
• Kulikauskas. What are the group extensions of {$\mathbb{Z}_2$} by {$U(1)$}?
• Eschenburg. Lecture Notes on Symmetric Spaces.

Questions

• Understand real and quaternionic structures.
• Understand the definition of {$(\phi, \chi )$}-representation.
• Understand why complex linear operator CT is matched with the commutative group extension, and why complex linear operators C, T are matched with the noncommutative group extension.
• Understand how to interpret C and iC as acting on the complex vector space.

Notation

{$ε±$} is odd and {$ε^2± = ±1$}

{$\eta=\mathbb{R}^{1|1}\;\;\;\rho(e)=\begin{pmatrix}0 & -1\\1 & 0 \\ \end{pmatrix} \;\;\;\; \tilde{\eta}=\mathbb{R}^{1|1}\;\;\;\rho(e)=\begin{pmatrix}0 & 1\\1 & 0 \\ \end{pmatrix}$}

{$\iota = \begin{pmatrix}0 & -1\\1 & 0 \\ \end{pmatrix} \;\;\;\; \varphi = \begin{pmatrix}1 & 0 \\ 0 & -1 \\ \end{pmatrix} \;\;\;\; \psi=\begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ \end{pmatrix}$}

Lie algebra decomposition {$\frak{g}=\frak{h}+\frak{m}$} where {$\frak{[h, h] ∈ h, [h, m] ∈ m, [m, m] ∈ h}$}.

{$m\in G/H$}

We need to compare two Clifford algebras.

• A Clifford algebra is generated by {$n$} mutually anticommuting linear complex structures, where {$G$} commutes with the first {$n-1$} linear complex structures and {$H$} commutes with all {$n$} linear complex structures. This pattern is centered on {$C$} and {$O(16r)$} and starts with {$O(16r)/U(8r)$}.
• A Clifford algebra gets its generators from CT groups. This pattern is centered on {$T$} and {$U(2r)/O(2r)$}. The direction (forwards or backwards) is determined by {$C$} and the unfolding is determined by {$T$}.

{$T$} defines the {$J_k$} by acting as {$I_m$}, splitting the context into zones with eigenvalues {$\pm 1$}. {$C$} sets up an isometry {$L$}

It looks like there are six perspectives which get divided up into what is accesible to the conscious (-1) and the unconscious (-1) and there are two additional perspectives for consciousness (+1). Together they make for an eightfold structure. The key structure is the threesome as it is complemented by a threesome for the unconscious.

DIII ≡ O(16r)/U(8r): The m ∈ m0 ’s are real skew symmetric matrices that anticommute with J1 . We can keep C = ϕ as a particle-hole symmetry and take T = ϕ ⊗ J1 as a time reversal that commutes with H = −iσ2 ⊗ M and squares to −I. The product of C and T is J1 , and this a linear (commutes with −iσ2 ⊗ I) “P ” type symmetry that anticommutes with H.

AII≡ U(8r)/Sp(4r): The generators m ∈ m1 are real skew matrices that commute with J1 and anticommute with J2 . They can be regarded as skew-quaternion-hermitian matrices with complex entries. We no longer need set i → −iσ2 ⊗ I as the matrices no longer have elements coupling between the artificial copies. We instead use J1 as the surrogate for “i.” Now H = J1 m is real symmetric, and commutes with T = J2 . This T acts as a time reversal operator squaring to −I.

CII≡ Sp(4r)/Sp(2r) × Sp(2r): The matrices m ∈ m2 commute with J1 and J2 but anti- commute with J3 . Again H = J1 m. We can set T = J3 as this commutes with with H and squares to −I. P = J2 J3 anticommutes with H but commutes with J1 (and so is a linear map) while C = J2 anticommutes with H, is antilinear and squares to −I.

C≡ {Sp(2r) × Sp(2r)}/Sp(2r) ≃ Sp(2r): The matrices m ∈ m3 commute with J1 , J2 , J3 , and anticommute with J4 , and we can restrict ourselves to the subspace in which K = J1 J2 J3 takes a definite value, say +1. The Hamiltonian J1 m commutes with J4 – but 15J4 does not commute with J1 J2 J3 and so is not allowed as an operator on our subspace. Indeed no product involving J4 is allowed. But C = J2 commutes with J1 J2 J3 and still anticommutes with H. Thus we still have a particle-hole symmetry squaring to −1. The old time reversal J3 now anticommutes with H and looks like another particle- hole symmetry, but is not really an independent one as in this subspace J3 = J2 J1 and J1 is simply multiplication by “i.”

CI≡ Sp(2r)/U(2r): The m ∈ m4 anticommute with J5 . Now J4 J5 commutes with J1 J2 J3 , and so is an allowed operator. It anticommutes with H = J1 m and commutes with J1 . It is therefore a “P ”-type linear map. The map T = J2 J4 J5 is antilinear (anticommutes with J1 ), commutes with H and T 2 = +I. We can take C = J2 again.

AI ≡ U(2r)/O(2r): The m ∈ m5 n anticommute with J6 , and we are to restrict ourselves to the subspace on which K = J1 J2 J3 = +1 and M = J1 J4 J5 = +1. The map T = J3 J4 J6 commutes with K and M and commutes H = J1 m. We have T 2 = +I. We could equivalently take T = J2 J4 J6 .

BDI≡ O(2r)/O(r) × O(r): The m ∈ m6 anticommute with J7 , and we are to restrict our- selves to the eigenspaces of K = J1 J2 J3 , M = J1 J4 J5 . The m also commute with the antilinear operator N = J2 J4 J6 which we can regard as our real structure ϕ. We therefore set C = ϕ = N. Now J3 J4 J7 , J3 J5 J6 , J2 J5 J7 and J1 J6 J7 all commute with K, M and N, each squaring to +I. In the restricted subspace J3 J4 J7 ∝ J2 J5 J7 ∝ J1 J6 J7 and J3 J5 J6 ∝ I. The problem here is what to take for “i,” as the current H → J1 m will take us out of the eigenspace of N. But this is the problem we started with. We need to double the space and keep “i”= J1 and the real structure ϕ = N. We are therefore retaining the R2r Hilbert space. With H = J1 m we have that T = J1 J6 J7 commutes withH and squares to +I.

D≡ {O(r) × O(r)}/O(r) ≃ O(r): Now the m ∈ m7 anticommute with J8 and, except for the factor “i”= J1 , we should stay in the space where K = J1 J2 J3 , M = J1 J4 J5 , N = J2 J4 J6 and P = J1 J6 J7 take the value +1 With H = J1 m we have that C = ϕ = N anticommutes with H, and brings us full circle.