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Notes on gapped boundaries of Abelian topological phases

2014年08月15日 留下评论 Go to comments

This is going to be a collection of various results on Abelian topological phases, with the focus being gapped boundaries. Part of the motivation is to make some connections between the existing mathematical literature on quadratic forms and the physical applications.

First, recall that a topological phase (unitary modular tensor category) can have a gapped boundary if and only if it contains a maximal Lagrangian subalgebra. It is also known that all such topological phases must be the Drinfeld centers of unitary fusion categories, i.e. quantum doubles. For Abelian phases (i.e. pointed categories), it means all those with gapped boundaries are essentially discrete gauge theories with an Abelian gauge group G, possibly twisted by 3-cocycles in H^3(G, U(1)). If we represent them with Abelian Chern-Simons theory, their K matrices always take the following form (up to GL equivalence):

\displaystyle \mathbf{K}=\displaystyle \begin{pmatrix} 0 & \mathbf{A} \\ \mathbf{A}^{\mathrm{T}} & \mathbf{B} \end{pmatrix}

In particular, it implies that \det \mathbf{K} is a square.

We also note that time-reversal-invariant Abelian theories must have gapped boundaries. This can be shown following Levin and Stern.

First we define the time-reversal symmetry directly in Abelian Chern-Simons theory. Let us consider a general Abelian topological phase given by a \mathbf{K} matrix with dimension 2N. The time-reversal symmetry operation \mathcal{T} acts on the gauge fields as

a_I\rightarrow \mathbf{T}_{IJ}a_J.

Also, because t\rightarrow -t, time-reversal invariance implies that

\mathbf{T}^{\mathrm{T}}\mathbf{K}\mathbf{T}=-\mathbf{K}.

For bosonic systems, we should have \mathcal{T}^2=1, which implies \mathbf{T}^2=1.

We now need to prove that there should be at least N null vectors in the theory which are mutually null. We can easily prove that \mathrm{Tr}\,(\mathbf{T})=0, and since \mathbf{T}^2=1, the eigenvalues of \mathbf{T} are \pm 1 and there must be equal number of 1 and -1. Let \mathbf{v}_i denote the eigenvectors of \mathbf{T} with +1 eigenvalue, i=1,\dots, N. We have

\mathbf{v}_i^{\mathrm{T}} \mathbf{K} \mathbf{v}_j=\mathbf{v}_i^{\mathrm{T}} \mathbf{T}^{\mathrm{T}} \mathbf{K} \mathbf{T}\mathbf{v}_j=-\mathbf{v}_i^{\mathrm{T}} \mathbf{K} \mathbf{v}_j.

Therefore \mathbf{v}_i^{\mathrm{T}} \mathbf{K} \mathbf{v}_j=0.

Levin has argued that \mathbf{v}_i can be chosen to be integer vectors. This is always possible since this eigenspace is spanned by the columns of 1 + \mathbf{T} , a matrix with integer entries. Therefore we have found a set of null vectors. QED.

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