Eigenvalue crossings in Floquet topological systems

Abstract

The topology of electrons on a lattice subject to a periodic driving is captured by the three-dimensional winding number of the propagator that describes time evolution within a cycle. This index captures the homotopy class of such a unitary map. In this paper, we provide an interpretation of this winding number in terms of local data associated with the eigenvalue crossings of such a map over a three-dimensional manifold, based on an idea from Nathan and Runder (New J Phys 17(12):125014, 2015). We show that, up to homotopy, the crossings are a finite set of points and non-degenerate. Each crossing carries a local Chern number, and the sum of these local indices coincides with the winding number. We then extend this result to fully degenerate crossings and extended submanifolds to connect with models from the physics literature. We finally classify up to homotopy the Floquet unitary maps, defined on manifolds with boundary, using the previous local indices. The results rely on a filtration of the special unitary group as well as the local data of the basic gerbe over it.

Appendices

Appendix A: Reduction in SU(N)-valued maps

Let X be a topological space, and $Y \subset X$ a subspace. For a topological group G, we denote by C((X, Y), (G, 1)) the set of continuous maps $U : X \rightarrow G$ such that $U|_Y \equiv 1$ is the constant map at the unit $1 \in G$. By the pointwise multiplication, the set gives rise to a group. A (relative) homotopy between two maps $U_0, U_1 \in C((X, Y), (G, 1))$ is a continuous map ${\tilde{U}} \in C((X \times [0, 1], Y \times [0, 1]), (G, 1))$ such that ${\tilde{U}}|_{X \times \{ i \}} = U_i$ for $i = 0, 1$. The set of homotopy classes in C((X, Y), (G, 1)) will be denoted by

$$\begin{aligned}{}[(X, Y), (G, 1)], \end{aligned}$$

which inherits a group structure from C((X, Y), (G, 1)).

Lemma A.1

Let X be a topological space, and $Y \subset X$ a subspace. There is an exact sequence of groups

$$\begin{aligned} 1 \rightarrow [(X, Y), (SU(N), 1)] \rightarrow [(X, Y), (U(N), 1)] \rightarrow [(X, Y), (U(1), 1)] \rightarrow 1. \end{aligned}$$

This admits a section to the surjection induced from $\det : U(N) \rightarrow U(1)$, so that there is an isomorphism of groups

$$\begin{aligned}{}[(X, Y), (U(N), 1)] \cong [(X, Y), (SU(N), 1)] \rtimes [(X, Y), (U(1), 1)]. \end{aligned}$$

Proof

We have the exact sequence of topological groups

$$\begin{aligned} 1 \rightarrow SU(N) \rightarrow U(N) \overset{det}{\rightarrow } U(1) \rightarrow 1, \end{aligned}$$

which admits a section $s : U(1) \rightarrow U(N)$ given by $s(u) = \mathrm {diag}(u, 1, \cdots , 1)$. Using this section, we can verify the lemma directly. For instance, the exactness at the middle term [(X, Y), (SU(N), 1)] can be verified as follows: Given a map $U \in C((X, Y), (U(N), 1))$ and a homotopy $h \in C((X \times [0, 1], Y \times [0, 1]), (U(1), 1))$ such that $h(x, 0) = \det U(x)$ and $h(x, 1) = 1$ for all $x \in X$, we define $H : X \times [0, 1] \rightarrow U(N)$ by $H(x, t) = U(x) \cdot s(h(x, t))^{-1}$. This turns out to be a homotopy $H \in C((X \times [0, 1], Y \times [0, 1]), (U(N), 1))$ such that $H(\cdot , 0) \in C((X, Y), (SU(N), 1))$ and $H(\cdot , 1) = U(\cdot )$. Hence the homotopy class of U that goes to the unit in [(X, Y), (U(1), 1)] comes from [(X, Y), (SU(N), 1)]. $\square$

To describe the obstructions for $U \in C((X, Y), (U(1), 1))$ to being homotopic to the constant map at 1, we introduce the odd-dimensional winding number as follows: It is well known that the cohomology ring $H^*(U(N); {\mathbb {Z}})$ of U(N) is isomorphic to the exterior ring

$$\begin{aligned} H^*(U(N); {\mathbb {Z}}) \cong \bigwedge (W_1, W_3, \cdots , W_{2N-1}) \end{aligned}$$

generated by $W_{2i-1} \in H^{2i-1}(U(N); {\mathbb {Z}}) \cong H^{2i-1}(U(N), 1; {\mathbb {Z}})$, ($i = 1, \cdots , N$). We then define the $(2i-1)$-dimensional winding number to be the pullback of the generator

$$\begin{aligned} W_{2i-1}(U) := U^*W_{2i-1} \in H^{2i-1}(X, Y; {\mathbb {Z}}). \end{aligned}$$

Lemma A.2

Let X be a finite CW complex, and $Y \subset X$ a subcomplex. A continuous map $u \in C((X, Y), (U(1), 1))$ is (relatively) homotopic to the constant map at 1, if and only if $W_1(u) = 0$.

Proof

The “if” part is clear. For the “only if” part (cf. [6]), a standard obstruction theory argument can be applied: Because of the assumptions about X and Y, $u \in C((X, Y), (U(1), 1))$ is relatively homotopic to the constant map at 1, if and only if so is the map ${\bar{u}} \in C((X/Y, Y/Y), (U(1), 1))$ induced from u, where X / Y is the CW complex given by collapsing Y to a point. Accordingly, we can assume $Y = \mathrm {pt}$ is a point (a 0-cell) from the beginning. For $k = 0, 1, 2, \cdots$, we denote by $X_k$ the k-skeleton of the CW complex X. Thus, $X_0$ consists of all the 0-cells, and $X_k$ is given by attaching the boundary of each k-cell $e^k$ to $X_{k-1}$.

Because U(1) is connected, there is a path between $1 \in U(1)$ and $u(e^0) \in U(1)$ for each 0-cell $e^0$. For the base 0-cell $\mathrm {pt}$, we choose the path to be the constant. Such paths together define a relative homotopy between $u|_{X_0} : X_0 \rightarrow U(1)$ and the constant map at 1. By the homotopy extension property, we can extend the relative homotopy on $X_0$ to one between $u : X \rightarrow U(1)$ and a map $u_1 : X \rightarrow U(1)$ such that $u_1|_{X_0} \equiv 1$. Now, each 1-cell $e^1$ defines a loop $u_1 : e^1/\partial e^1 \rightarrow U(1)$ based at $1 \in U(1)$. Hence its winding number defines a 1-cocycle of the cellular cochain complex $C^1(X, \mathrm {pt}; {\mathbb {Z}})$. This represents $W_1(u_1) = W_1(u) \in H^1(X, \mathrm {pt}; {\mathbb {Z}})$, in view of the case that $X = U(1)$. The assumption $W_1(U) = 0$ says that $u_1 : e^1/\partial e^1 \rightarrow U(1)$ is homotopic to the constant loop at 1. Such homotopies together define a relative homotopy between $u_1|_{X_1} : X_1 \rightarrow U(1)$ and the constant map. By the homotopy extension property, it extends to a relative homotopy between $u_1 : X \rightarrow U(1)$ and $u_2 : X \rightarrow U(1)$ such that $u_2|_{X_1} \equiv 1$. Then, each 2-cell $e^2$ defines an element $u_2 : e^2/\partial e_2 \rightarrow U(1)$. Since $\pi _2(U(1)) = 0$, each map $e^2/\partial e_2 \rightarrow U(1)$ is homotopic to the constant map, and such homotopies together constitute a homotopy from $u_2|_{X_2} : X_2 \rightarrow U(1)$ to the constant map. By the homotopy extension property, this homotopy extends one between $u_2 : X \rightarrow U(1)$ and $u_3 : X \rightarrow U(1)$ such that $u_3|_{X_2} \equiv 1$. Because $\pi _i(U(1)) = 0$ for $i \ge 2$, we can repeat the same argument to get a homotopy from $u_{i-1} : X \rightarrow U(1)$ to $u_i : X \rightarrow U(1)$ such that $u_i|_{X_{i-1}} \equiv 1$. Because X is a finite complex, this procedure terminates at a finite step, yielding a homotopy to the constant map on X. Putting all the homotopies together, we get a homotopy from u to the constant map on X. $\square$

Proposition A.3

Let X be a finite CW complex which contains only cells of dimension 3 or less, and $Y \subset X$ a subcomplex. Let $N \ge 2$. A continuous map $U \in C((X, Y), (U(N), 1))$ is (relatively) homotopic to the constant map at 1, if and only if $W_1(U) = 0$ and $W_3(U) = 0$.

Proof

The “only if” part is clear. For the “if” part, Lemma A.1 and Lemma A.2 imply that the given map U is relatively homotopic to a map in $U' \in C((X, Y), (SU(N), 1))$. We have $W_3(U') = W_3(U)$. Hence it suffices to show that $W_3(U') = 0$ implies that $U'$ is relatively homotopic to the constant map. Then its proof is essentially the same as that of Lemma A.2: We can assume that Y is a 0-cell. Since $\pi _i(SU(N)) = 0$ for $i \le 2$, the map $U'$ is relatively homotopic to $U'' : X \rightarrow SU(N)$ such that $U''|_{X_2} \equiv 1$. Then, we have $\pi _3(SU(N)) \cong {\mathbb {Z}}$, and the map $U'' : e^3/\partial e^3 \rightarrow SU(N)$ defines a cellular 3-cocycle which represents $W_3(U'') = W_3(U') = W_3(U)$. The vanishing $W_3(U) = 0$ ensures that $U''$ is homotopic to the constant map. $\square$

So far, we are in the topological setup, so that given maps and their homotopy are continuous. When the given CW complexes are smooth manifolds and given maps are smooth, then, by approximation, their (continuous) homotopy can be replaced by a smooth homotopy (through a homotopy of homotopies). In this paper, this replacement may be implicitly adapted.

As is mentioned in the introduction, when we are interested in the classification of (topological invariants of) quantum systems on two-dimensional lattices subject to a periodic driving, we would like to know the obstruction for $U \in C((T^2 \times S^1, T^2 \times \{ 0 \}), (U(N), 1))$ to being trivial. For a compact oriented two-dimensional manifold $\Sigma$ without boundary, it holds that

$$\begin{aligned} {\mathbb {Z}}\cong H^1(\Sigma \times S^1, \Sigma \times \{ 0 \}; {\mathbb {Z}}) \subset H^1(\Sigma \times S^1; {\mathbb {Z}}), \end{aligned}$$

and this subgroup is generated by the pullback of $H^1(S^1; {\mathbb {Z}}) \cong {\mathbb {Z}}$ under the projection $\Sigma \times S^1 \rightarrow S^1$. This implies that $W_1(U)$ is computed as the winding number of $\det U|_{\{ x \} \times S^1} : \{ x \} \times S^1 \rightarrow U(1)$, where $x \in \Sigma$ is any point. It also holds that

$$\begin{aligned} {\mathbb {Z}}\cong H^3(\Sigma \times S^1, \Sigma \times \{ 0 \}; {\mathbb {Z}}) \cong H^3(\Sigma \times S^1; {\mathbb {Z}}). \end{aligned}$$

Hence the relative three-dimensional winding number $W_3(U)$ agrees with the absolute three-dimensional winding number. As a matter of fact, a compact oriented manifold (without boundary) admits a CW decomposition (see [20] for example), and we can apply Proposition A.3. Then the map U is relatively homotopic to the constant map at 1, if and only if the one-dimensional winding number along a point $x \in \Sigma$ and the (absolute) three-dimensional winding number $W_3(U) \in H^3(\Sigma \times S^1; {\mathbb {Z}})$ are vanishing. Notice that the other weak invariants, i.e. the one-dimensional winding numbers along spacial directions, do not appear here. They are indeed always trivial since $U|_{\Sigma \times \{0\}} = 1$.

The one-dimensional winding number is easier to compute, and if it is non-trivial, then we can conclude that U is non-trivial. If the one-dimensional winding number is trivial, then the remaining obstruction is $W_3(U)$. Thanks to the exact sequence in Lemma A.1, we can assume in this case that U takes values in SU(N). Instead if $W_1(U) = p \ne 0$ we consider $U_p = U\cdot \mathrm {diag}({\mathrm e}^{-2 \pi {\mathrm i}p t}, 1, \ldots , 1)$ that satisfies $W_1(U_p) =0$ and $W_3(U_p) = W_3(U)$ by additivity of the winding numbers. In particular $U_p$ is homotopic to an SU(N)-valued map and shares the same value for $W_3$. This motivates us to give a local expression of the three-dimensional winding number for SU(N)-valued maps.

Appendix B: Further examples

1.1 B.1. The adjoint $S^2 \times S^1 \rightarrow SU(2)$

Let $T \subset SU(2)$ be the maximal torus consisting of diagonal matrices, which is diffeomorphic to the circle $S^1 \cong U(1)$. The quotient space SU(2) / T is readily identified with the two-dimensional sphere $S^2 = {\mathbb {C}}P^1$ by

$$\begin{aligned} \left( \begin{array}{rr} u &{}\quad -{\bar{v}} \\ v &{}\quad {\bar{u}} \end{array} \right) /T \mapsto [u : v]. \end{aligned}$$

By the adjoint action, we have a smooth surjective map

$$\begin{aligned} U&: SU(2)/T \times T \rightarrow SU(2),&(gT, h)&\mapsto ghg^{-1}, \end{aligned}$$

This map gives rise to a double covering over $SU(2) \backslash \{ \pm {\mathbb {1}} \}$, but not over the whole of SU(2). Because $SU(2)/T \times T \cong S^2 \times S^1$ is a compact oriented three-dimensional manifold without boundary, the three-dimensional winding number $W_3(U)$ makes sense. This number agrees with the mapping degree of U. It is known [2] that $W_3(U) = 2$, which we compute through our local formula.

It is easy to see the eigenvalue crossings:

$$\begin{aligned} \begin{array}{|c|c|c|c|c|c|} \hline j &{} \text{ eigenvalues } &{} \lambda _1 &{} \lambda _2 &{} \lambda _1 - 1&{} \mathrm {Cr}_j(U) \\ \hline 1 &{} 1, 1 &{} 0 &{} 0 &{} -1 &{} S^2 \times \{ 1 \} \\ \hline 2 &{} -1, -1 &{} 1/2 &{} -1/2 &{} -1/2 &{} S^2 \times \{ -1 \} \\ \hline \end{array} \end{aligned}$$

To apply Theorem 2.4 for $j = 1$, we choose a closed tubular neighborhood $N_1$ of $\mathrm {Cr}_1(U) = S^2 \times \{ 1 \}$ to be $N_1 = S^1 \times \{ {\mathrm e}^{2\pi {\mathrm i}t} |\ -1/4 \le t \le 1/4 \} \cong S^1 \times [-1/4, 1/4].$ At $([u: v], \pm 1/4)) \in \partial N_1$, the value of U is

$$\begin{aligned} U([u : v], \pm 1/4) = \left( \begin{array}{rr} u &{}\quad -{\bar{v}} \\ v &{}\quad {\bar{u}} \end{array} \right) \left( \begin{array}{cc} \pm i &{}\quad 0 \\ 0 &{}\quad \mp i \end{array} \right) \left( \begin{array}{rr} u &{}\quad -{\bar{v}} \\ v &{}\quad {\bar{u}} \end{array} \right) ^{-1}, \end{aligned}$$

so that its eigenvalues are ${\mathrm e}^{2\pi {\mathrm i}\lambda _1} = i$ and ${\mathrm e}^{2\pi {\mathrm i}\lambda _2} = -i$: $\lambda _1 = \tfrac{1}{4} \ge \lambda _2 = -\tfrac{1}{4} \ge \lambda _1-1 = -\tfrac{3}{4}$. Thus, on the connected component $S^2 \times \{ 1/4 \} \subset \partial N_1$, the eigenvector of U([u : v], 1 / 4) with eigenvalue ${\mathrm e}^{2\pi {\mathrm i}\lambda _1} = i$ is

$$\begin{aligned} \left( \begin{array}{rr} u &{}\quad -{\bar{v}} \\ v &{}\quad {\bar{u}} \end{array} \right) \left( \begin{array}{c} 1 \\ 0 \end{array} \right) = \left( \begin{array}{c} u \\ v \end{array} \right) , \end{aligned}$$

which spans the tautological line bundle on $S^2 = {\mathbb {C}}P^1$. On the other connected component $S^2 \times \{ -1/4 \} \subset \partial N_1$, the eigenvector of $U([u: v], -1/4)$ with eigenvalue ${\mathrm e}^{2\pi {\mathrm i}\lambda _1} = i$ is

$$\begin{aligned} \left( \begin{array}{rr} u &{}\quad -{\bar{v}} \\ v &{}\quad {\bar{u}} \end{array} \right) \left( \begin{array}{c} 0 \\ 1 \end{array} \right) = \left( \begin{array}{c} -{\bar{v}} \\ {\bar{u}} \end{array} \right) , \end{aligned}$$

which spans the dual of the tautological line bundle on $S^2 = {\mathbb {C}}P^1$. Taking the induced orientations on $S^2 \times \{ \pm 1/4 \}$ into account, we find that $\mathrm {Ch}(\mathrm {Cr}_1; 1) = -1 - (+1) = -2$. Hence Theorem 2.4 gives $W_3(U) = 2$, as anticipated. The application of Theorem 2.4 for $j = 2$ is similar.

1.2 B.2. The standard embedding $SU(2) \rightarrow SU(3)$

Let $U : SU(2) \rightarrow SU(3)$ be the standard embedding

$$\begin{aligned} U\left( \left( \begin{array}{rr} u &{}\quad -{\bar{v}} \\ v &{}\quad {\bar{u}} \end{array} \right) \right) = \left( \begin{array}{rrr} u &{}\quad -{\bar{v}} &{}\quad 0 \\ v &{}\quad {\bar{u}} &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 1 \end{array} \right) . \end{aligned}$$

The winding number is $W_3(U) = 1$, as can be computed directly. We here compute this number by means of the results in this paper. Using the unique expression of the eigenvalues ${\mathrm e}^{2\pi {\mathrm i}\lambda _1}, {\mathrm e}^{2\pi {\mathrm i}\lambda _2}, {\mathrm e}^{2\pi {\mathrm i}\lambda _3}$ of a matrix in SU(3) in terms of $\lambda _i \in {\mathbb {R}}$ such that $\lambda _1 + \lambda _2 + \lambda _3 = 0$ and $\lambda _1 \ge \lambda _2 \ge \lambda _3 \ge \lambda _1 - 1$, we can summarize the crossings of the eigenvalues as follows:

$$\begin{aligned} \begin{array}{|c|c|c|c|c|c|c|} \hline \text{ eigenvalues } &{} \lambda _1 &{} \lambda _2 &{} \lambda _3 &{} \lambda _1 - 1&{} \text{ subspace } \text{ in } \, SU(2) \\ \hline 1, 1, 1 &{} 0 &{} 0 &{} 0 &{} -1 &{} U^{-1}({\mathbb {1}}_{3}) = \{ {\mathbb {1}}_{2} \} \\ \hline -1, 1, -1 &{} 1/2 &{} 0 &{} -1/2 &{} -1/2 &{} \mathrm {Cr}_3(U) = \{ -{\mathbb {1}}_{2} \} \\ \hline \end{array} \end{aligned}$$

Hence we can apply Theorem 2.8:

• For $j = 1$, we have $\mathrm {Cr}_1(U) = \emptyset$ and $W_3(U) = \mathrm {Ch}(\{ {\mathbb {1}}_{2} \}; 2) + \mathrm {Ch}(\{ {\mathbb {1}}_{2} \}; 3)$.

• For $j = 2$, we have $\mathrm {Cr}_2(U) = \emptyset$ and $W_3(U) = \mathrm {Ch}(\{ {\mathbb {1}}_{2} \}; 3)$.

• For $j = 3$, we have $\mathrm {Cr}_3(U) = \{ -{\mathbb {1}}_{2} \}$ and $W_3(U) = -\mathrm {Ch}(\{-{\mathbb {1}}_{2}\}; 3)$.

To compute the local indices, we can use the three-dimensional disks $D_x$ containing $x = {\mathbb {1}}$ and $D_{x'}$ containing $x' = -{\mathbb {1}}$ in SU(2). Thus, all the relevant indices are Chern numbers of some line bundles over $\partial D_x = \partial D_{x'}$. The eigenvalues of $U \in \partial D_x = \partial D_{x'}$ are i, 1 and $-i$. Note that

$$\begin{aligned} \overbrace{\lambda _1}^{1/4}> \overbrace{\lambda _2}^{0}> \overbrace{\lambda _3}^{-1/4} > \overbrace{\lambda _1 - 1}^{-3/4}. \end{aligned}$$

For $j = 1$, the local index is the Chern number of the tensor product of the line bundles whose fibers are eigenspaces with eigenvalues ${\mathrm e}^{2\pi {\mathrm i}\lambda _2} = 1$ and ${\mathrm e}^{2\pi {\mathrm i}\lambda _3} = -i$. The Chern number of the latter line bundle is computed in Example 2.5, whereas that of the former is trivial, since, for any $y \in \partial D_x = \partial D_{x'}$, the eigenvector of $U(y) \in SU(3)$ with eigenvalue 1 is $(0,\, 0,\, 1)^t$. Therefore, we get $\mathrm {Ch}(\{ {\mathbb {1}}_{2} \}; 2)=0$ and $\mathrm {Ch}(\{ {\mathbb {1}}_{2} \}; 3) = 1$. Finally, for $j = 3$, the local index is also the Chern number of the line bundle whose fibers are eigenspaces with eigenvalues $-i$, so that $\mathrm {Ch}(\{-{\mathbb {1}}_{2}\}; 3) = -1$ by the computation in Example 2.5.

1.3 B.3. A perturbed embedding $SU(2) \rightarrow SU(3)$

Let us consider a family of embedding

$$\begin{aligned} U&: SU(2) \rightarrow SU(3),&U\left( \left( \begin{array}{rr} u &{}\quad -{\bar{v}} \\ v &{}\quad {\bar{u}} \end{array} \right) \right)&= \left( \begin{array}{ccc} u {\mathrm e}^{{\mathrm i}t} &{}\quad -{\bar{v}}{\mathrm e}^{{\mathrm i}t} &{}\quad 0 \\ v {\mathrm e}^{{\mathrm i}t} &{}\quad {\bar{u}}{\mathrm e}^{{\mathrm i}t} &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad {\mathrm e}^{-2{\mathrm i}t} \end{array} \right) \end{aligned}$$

parametrized by $t \in {\mathbb {R}}$. The three-dimensional winding number of U is $W_3(U) = 1$ for any t, since U at t is homotopic to U at $t = 0$, which is the standard embedding. The three eigenvalues of U are generally expressed as $\{ {\mathrm e}^{{\mathrm i}t}u, {\mathrm e}^{{\mathrm i}t}{\bar{u}}, {\mathrm e}^{-2{\mathrm i}t} \}$, where $u \in U(1)$ and $t \in {\mathbb {R}}$.

As a special choice, we take ${\mathrm e}^{{\mathrm i}t} = i$, so that

$$\begin{aligned} U\left( \left( \begin{array}{rr} u &{}\quad -{\bar{v}} \\ v &{}\quad {\bar{u}} \end{array} \right) \right) = \left( \begin{array}{ccc} iu &{}\quad -i{\bar{v}} &{}\quad 0 \\ iv &{}\quad i{\bar{u}} &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad -1 \end{array} \right) . \end{aligned}$$

In this case, the three eigenvalues are distinct, or two of them coincide. The following table summarizes the detail of the latter case by using the unique expression of the eigenvalues ${\mathrm e}^{2\pi {\mathrm i}\lambda _1}, {\mathrm e}^{2\pi {\mathrm i}\lambda _2}, {\mathrm e}^{2\pi {\mathrm i}\lambda _3}$ of a matrix in SU(3) in terms of $\lambda _i \in {\mathbb {R}}$ such that $\lambda _1 + \lambda _2 + \lambda _3 = 0$ and $\lambda _1 \ge \lambda _2 \ge \lambda _3 \ge \lambda _1 - 1$.

$$\begin{aligned} \begin{array}{|c|c|c|c|c|c|c|} \hline j &{} \text{ eigenvalues } &{} \lambda _1 &{} \lambda _2 &{} \lambda _3 &{} \lambda _1 - 1&{} \mathrm {Cr}_j(U) \\ \hline 1 &{} i, i, 1 &{} 1/4 &{} 1/4 &{} -1/2 &{} -3/4 &{} \mathrm {pt}\\ \hline 2 &{} -1, -i, -i &{} 1/2 &{} -1/4 &{} -1/4 &{} -1/2 &{} \mathrm {pt}\\ \hline 3 &{} -1, 1, -1 &{} 1/2 &{} 0 &{} -1/2 &{} -1/2 &{} S^2 \\ \hline \end{array} \end{aligned}$$

Note that $\mathrm {Cr}_1(U) = \{ {\mathbb {1}} \}$, $\mathrm {Cr}_2(U) = \{ -{\mathbb {1}} \}$ and

$$\begin{aligned} \mathrm {Cr}_3(U) = \left\{ \left( \begin{array}{rr} u &{}\quad -{\bar{v}} \\ v &{}\quad {\bar{u}} \end{array} \right) \in SU(2) \bigg | u + {\bar{u}} = 0 \right\} . \end{aligned}$$

Accordingly, we can apply Theorem 2.4: The application of the theorem for $j = 1, 2$ reduces to the calculations of the identity map $SU(2) \rightarrow SU(2)$ given in Example 2.5, so we omit the detail. To apply Theorem 2.4 for $j = 3$, we choose a closed tubular neighborhood N of the two-dimensional sphere $\mathrm {Cr}_3(U) \subset SU(2)$ to be

$$\begin{aligned} N = \left\{ \left( \begin{array}{rr} u &{}\quad -{\bar{v}} \\ v &{}\quad {\bar{u}} \end{array} \right) \in SU(2) \bigg | - \sqrt{2} \le u + {\bar{u}} \le \sqrt{2} \right\} . \end{aligned}$$

We can identify $S^2 \times [-1/\sqrt{2}, 1/\sqrt{2}]$ with N by

$$\begin{aligned} (X, Y, Z, t) \mapsto \left( \begin{array}{rr} t + i\sqrt{1-t^2}X &{} -\sqrt{1-t^2}(Y - iZ) \\ \sqrt{1-t^2}(Y + iZ) &{} t - i \sqrt{1 - t^2}X \end{array} \right) , \end{aligned}$$

where $S^2 = \{ (X, Y, Z) \in {\mathbb {R}}^3 |\ X^2 + Y^2 + Z^2 = 1 \}$. Thus, on the boundary $\partial N = S^2 \times \{ \pm 1/\sqrt{2} \}$, we have

$$\begin{aligned} U(X, Y, Z, \pm 1/\sqrt{2}) = \left( \begin{array}{ccc} \frac{-X \pm i}{\sqrt{2}} &{}\quad \frac{-Z - iY}{\sqrt{2}} &{}\quad 0 \\ \frac{-Z + iY}{\sqrt{2}} &{}\quad \frac{X \pm i}{\sqrt{2}} &{}\quad \\ 0 &{}\quad 0 &{}\quad -1 \end{array} \right) . \end{aligned}$$

The eigenvalues of this matrix are as follows:

$$\begin{aligned} \begin{array}{|c|c|c|c|c|c|} \hline t &{} \text{ eigenvalues } &{} \lambda _1 &{} \lambda _2 &{} \lambda _3 &{} \lambda _1 - 1 \\ \hline t = \frac{1}{\sqrt{2}} &{} \frac{-1+i}{\sqrt{2}}, \frac{1 + i}{\sqrt{2}}, -1 &{} 3/8 &{} 1/8 &{} -1/2 &{} -5/8 \\ \hline t = -\frac{1}{\sqrt{2}} &{} -1, \frac{1-i}{\sqrt{2}}, \frac{-1 - i}{\sqrt{2}} &{} 1/2 &{} -1/8 &{} -3/8 &{} -1/2 \\ \hline \end{array} \end{aligned}$$

The local index $\mathrm {Ch}(\mathrm {Cr}_3; 3)$ is the Chern number of the line bundle whose fibers are eigenspaces with eigenvalues ${\mathrm e}^{2\pi {\mathrm i}\lambda _3}$. Thus, on $S^2 \times \{ 1/\sqrt{2} \}$, we have the constant eigenvector with eigenvalue $-1$, so that the Chern number of the line bundle is trivial. On $S^2 \times \{ -1/\sqrt{2} \}$, the line bundle is non-trivial. At $g = (X, Y, Z, -1/\sqrt{2})$ with $X \ne 1$, an eigenvector $v_3^-(g)$ of U(g) with eigenvalue ${\mathrm e}^{2\pi {\mathrm i}\lambda _3} = (-1 - i)/\sqrt{2}$ is given by

$$\begin{aligned} v_3^-(g) = \left( \begin{array}{c} \frac{Z + iY}{1 - X} \\ 1 \\ 0 \end{array} \right) . \end{aligned}$$

At $g = (X, Y, Z, -1/\sqrt{2})$ with $X \ne -1$, an eigenvector $v_3^+(g)$ of U(g) with eigenvalue ${\mathrm e}^{2\pi {\mathrm i}\lambda _3} = (-1 - i)/\sqrt{2}$ is given by

$$\begin{aligned} v_3^+(g) = \left( \begin{array}{c} 1 \\ \frac{Z - iY}{1 + X} \\ 0 \end{array} \right) . \end{aligned}$$

On the circle in the sphere $S^2 \times \{ -1/\sqrt{2} \}$

$$\begin{aligned} \{ g = (X, Y, Z, -1/\sqrt{2}) |\ X = 0, Y^2 + Z^2 = 1 \} \subset S^2 \times \{ -1/\sqrt{2} \}, \end{aligned}$$

we have a U(1)-valued map $f(g) = Z - iY = -i(Y + iZ)$ which measures the discrepancy of $v_3^+(g)$ and $v_3^-(g)$ by $v_3^+(g) = f(g)v_3^-(g)$. This implies that the Chern number of the line bundle over $S^2 \times \{ -1/\sqrt{2} \}$ is 1 under a choice of an orientation. To summarize, we get $\mathrm {Ch}(\mathrm {Cr}_3; 3) = 0-1 = -1$ and $W_3(U) = 1$.

1.4 B.4. The adjoint and embedding $S^2 \times S^1 \rightarrow SU(3)$

We here consider the map $U : S^2 \times S^1 \rightarrow SU(3)$

$$\begin{aligned} U([u : v], {\mathrm e}^{{\mathrm i}t}) = \left( \begin{array}{rrr} u &{}\quad -{\bar{v}} &{}\quad 0 \\ v &{}\quad {\bar{u}} &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 1 \end{array} \right) \left( \begin{array}{ccc} {\mathrm e}^{{\mathrm i}t} &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad {\mathrm e}^{-{\mathrm i}t} &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 1 \end{array} \right) \left( \begin{array}{rrr} u &{}\quad -{\bar{v}} &{}\quad 0 \\ v &{}\quad {\bar{u}} &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 1 \end{array} \right) ^{-1} \end{aligned}$$

given by composing the adjoint map $S^2 \times S^1 \rightarrow SU(2)$ in §§B.1 and the standard embedding $SU(2) \rightarrow SU(3)$. Although $W_3(U) = 2$ is clear, we consider to apply Theorem 2.8.

The eigenvalue crossings are as follows:

$$\begin{aligned} \begin{array}{|c|c|c|c|c|c|} \hline \text{ eigenvalues } &{} \lambda _1 &{} \lambda _2 &{} \lambda _3 &{} \lambda _1 - 1 &{} \text{ spaces } \text{ in } \, S^2 \times S^1 \\ \hline 1, 1, 1 &{} 0 &{} 0 &{} 0 &{} -1 &{} U^{-1}({\mathbb {1}}) = S^2 \times \{ 1 \} \\ \hline -1, 1, -1 &{} 1/2 &{} 0 &{} -1/2 &{} -1/2 &{} \mathrm {Cr}_3(U) = S^2 \times \{ -1 \} \\ \hline \end{array} \end{aligned}$$

Note that $\mathrm {Cr}_1(U) = \emptyset$ and $\mathrm {Cr}_2(U) = \emptyset$. Theorem 2.8 produces:

• for $j = 1$, we have $W_3(U) = \mathrm {Ch}(U^{-1}({\mathbb {1}}); 2)+\mathrm {Ch}(U^{-1}({\mathbb {1}}); 3)$,

• for $j = 2$, we have $W_3(U) = \mathrm {Ch}(U^{-1}({\mathbb {1}}); 3)$,

• for $j = 3$, we have $W_3(U) = - \mathrm {Ch}(\mathrm {Cr}_3; 3)$.

It turns out that the all the calculations of the local indices reduce to those given in Example B.1. Hence we just consider the case of $j = 1$. In this case, we choose $N_1 = S^2 \times [-1/4, 1/4]$ in Example B.1 as the closed tubular neighborhood of $U^{-1}({\mathbb {1}}) = S^2 \times \{ 1 \} \subset S^2 \times S^1$. On the boundary $\partial N_1 = S^2 \times \{ \pm 1/4 \}$, the map U takes the values

$$\begin{aligned} U([u : v], {\mathrm e}^{{\mathrm i}t}) = \left( \begin{array}{rrr} u &{}\quad -{\bar{v}} &{}\quad 0 \\ v &{}\quad {\bar{u}} &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 1 \end{array} \right) \left( \begin{array}{ccc} \pm i &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad \mp i &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 1 \end{array} \right) \left( \begin{array}{rrr} u &{}\quad -{\bar{v}} &{}\quad 0 \\ v &{}\quad {\bar{u}} &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 1 \end{array} \right) ^{-1}, \end{aligned}$$

hence its eigenvalues are ${\mathrm e}^{2\pi {\mathrm i}\lambda _1} = i$, ${\mathrm e}^{2\pi {\mathrm i}\lambda _2} = 1$ and ${\mathrm e}^{2\pi {\mathrm i}\lambda _3} = -i$

$$\begin{aligned} \overbrace{\lambda _1}^{1/4} \ge \overbrace{\lambda _2}^{0} \ge \overbrace{\lambda _3}^{-1/4} \ge \overbrace{\lambda _1-1}^{-3/4}. \end{aligned}$$

On the connected component $S^2 \times \{ 1/4 \} \subset \partial N_1$, we can find the following eigenvectors with eigenvalues ${\mathrm e}^{2\pi {\mathrm i}\lambda _2} = 1$ and ${\mathrm e}^{2\pi {\mathrm i}\lambda _3} = -i$, respectively $(0,\,0,\,1)^t$ and $(-{{\bar{v}}},\,{{\bar{u}}} ,\,1)^t$. Hence the tensor product of the line bundles whose fibers are the eigenspaces with eigenvalues ${\mathrm e}^{2\pi {\mathrm i}\lambda _2} = 1$ and ${\mathrm e}^{2\pi {\mathrm i}\lambda _3} = -i$ is the dual of the tautological line bundle on $S^2 = {\mathbb {C}}P^1$. On the other connected component $S^2 \times \{ -1/4 \} \subset \partial N_1$, we have the following eigenvectors with eigenvalues ${\mathrm e}^{2\pi {\mathrm i}\lambda _2} = 1$ and ${\mathrm e}^{2\pi {\mathrm i}\lambda _3} = -i$, respectively $(0,\,0,\,1)^t$ and $(u,\,v,\,1)^t$. Then the tensor product of the line bundles whose fibers are the eigenspaces with eigenvalues ${\mathrm e}^{2\pi {\mathrm i}\lambda _2} = 1$ and ${\mathrm e}^{2\pi {\mathrm i}\lambda _3} = -i$ is the tautological line bundle on $S^2 = {\mathbb {C}}P^1$. Taking the orientation into account, we find $\mathrm {Ch}(U^{-1}({\mathbb {1}}); 2)+\mathrm {Ch}(U^{-1}({\mathbb {1}}); 3) = 2$.

1.5 B.5. Floquet map example

Let $U_i : {\mathbb {C}}P^1 \times [0, 1] \rightarrow SU(2)$ be the following maps

$$\begin{aligned} U_1([u : v], t)&= \left( \begin{array}{rr} u &{}\quad -{\bar{v}} \\ {\bar{v}} &{}\quad u \end{array} \right) \left( \begin{array}{cc} {\mathrm e}^{\pi {\mathrm i}t/2} &{}\quad 0 \\ 0 &{}\quad {\mathrm e}^{-\pi {\mathrm i}t/2} \end{array} \right) \left( \begin{array}{rr} u &{}\quad -{\bar{v}} \\ {\bar{v}} &{}\quad u \end{array} \right) ^{-1}, \\ U_2([u : v], t)&= \left( \begin{array}{rr} u &{}\quad -{\bar{v}} \\ {\bar{v}} &{}\quad u \end{array} \right) \left( \begin{array}{cc} {\mathrm e}^{-3\pi {\mathrm i}t/2} &{}\quad 0 \\ 0 &{}\quad {\mathrm e}^{3\pi {\mathrm i}t/2} \end{array} \right) \left( \begin{array}{rr} u &{}\quad -{\bar{v}} \\ {\bar{v}} &{}\quad u \end{array} \right) ^{-1}. \end{aligned}$$

Note that, as $t \in [0, 1]$ varies from $t = 0$ to $t = 1$, the element ${\mathrm e}^{\pi {\mathrm i}t/2}$ travels on U(1) anticlockwisely from $1 \in U(1)$ to $i \in U(1)$, whereas ${\mathrm e}^{-3\pi {\mathrm i}t} = {\mathrm e}^{2\pi {\mathrm i}(- 3/4t)}$ travels clockwisely from $1 \in U(1)$ to $i \in U(1)$ via $-1 \in U(1)$. They satisfy $U_i|_{{\mathbb {C}}P^1 \times \{ 0 \}} = {\mathbb {1}}$ and $U_i({\mathbb {C}}P^1 \times \{ 1 \}) \subset SU(2)_{\le 0}$. Further, $U_i({\mathbb {C}}P^1 \times (0, 1]) \subset SU(2)_{\le 1}$ and $U_1|_{{\mathbb {C}}P^1 \times \{ 1 \}} = U_2|_{{\mathbb {C}}P^2 \times \{ 1 \}}$. It is easy to see $\mathrm {Cr}_1(U_1) = {\mathbb {C}}P^1 \times \{ 0 \}$ and $\mathrm {Cr}_2(U_1) = \emptyset$ for $U_1$. We can compute the topological numbers of $U_1$ as follows:

$$\begin{aligned} {\mathcal {I}}(U_1; 1)&= -\mathrm {Ch}(\mathrm {Cr}_1; 1) = 1,&{\mathcal {I}}(U_1; 2)&= -\mathrm {Ch}(\mathrm {Cr}_2; 2) = 0. \end{aligned}$$

For $U_2$, we have $Cr_1(U_2) = {\mathbb {C}}P^1 \times \{ 0 \}$ and $\mathrm {Cr}_2(U_2) = {\mathbb {C}}P \times \{ 2/3 \}$. We can also compute

$$\begin{aligned} {\mathcal {I}}(U_2; 1)&= - \mathrm {Ch}(\mathrm {Cr}_1; 1) = -1,&{\mathcal {I}}(U_2; 2)&= - \mathrm {Ch}(\mathrm {Cr}_2; 2) = -2. \end{aligned}$$

These computations show that $U_1$ and $U_2$ are not homotopic relative to ${\mathbb {C}}P^1 \times \partial [0, 1]$. Notice that $U_1 \cup U_2 : {\mathbb {C}}P^1 \times S^1 \rightarrow SU(2)$ is just the map in Example B.1. Hence ${\mathcal {I}}(U_1; j) - {\mathcal {I}}(U_2; j) = W_3(U_1 \cup U_2) = 2$ as anticipated.

Composing the two maps $U_i : {\mathbb {C}}P^1 \times [0, 1] \rightarrow SU(2)$ and the standard embedding $SU(2) \rightarrow SU(3)$, we define $V_i : {\mathbb {C}}P^1 \times [0, 1] \rightarrow SU(3)$. In this case, $V_i^{-1}({\mathbb {1}}) = {\mathbb {C}}P^1 \times \{ 0 \}$. We have $\mathrm {Cr}_1(V_i) = \emptyset$ for $V_1$ and $V_2$. Thus, to ${\mathcal {I}}(V_i; 1)$, the contributions $\mathrm {Ch}(\mathrm {Cr}_1; 1) = 0$ from the simple crossing $\mathrm {Cr}_1(V_i)$ is trivial for both $V_1$ and $V_2$. But, the contributions $\mathrm {Ch}(V_i^{-1}({\mathbb {1}});2)$ and $\mathrm {Ch}(V_i^{-1}({\mathbb {1}});3)$ from the full crossings at ${\mathbb {C}}P^1 \times \{ 0 \}$ are non-trivial, and we get

$$\begin{aligned} {\mathcal {I}}(V_1; 1)&= - \mathrm {Ch}(\mathrm {Cr}_1; 1) + \mathrm {Ch}(V_1^{-1}({\mathbb {1}});2)+\mathrm {Ch}(V_1^{-1}({\mathbb {1}});3) = 1, \\ {\mathcal {I}}(V_2; 1)&= - \mathrm {Ch}(\mathrm {Cr}_2; 1) +\mathrm {Ch}(V_1^{-1}({\mathbb {1}});2)+ \mathrm {Ch}(V_1^{-1}({\mathbb {1}});3) = -1. \end{aligned}$$

This contribution from the full degeneracy at $t = 0$ seems not considered in [24], but the above example shows that it plays an indispensable role in the topological invariant.