- Dual Paving

For a cellular paving K with inner-oriented cells, we can compose a dual cellular paving K with outer-oriented cells. Here, we focus on ambient space of a two-dimensional plane to introduce the concept. As shown in Figure 15a, each face *fi* in K is converted to an outer-oriented point *n*ˇ*i* in K. Each edge *ej* in K is also replaced with an outer-oriented edge *e*ˇ*j* in K transverse to the original edge. The obtained edge is connected to a point in K, if the original edge is included in the original face. Furthermore, each point *nj* in K is converted to a face ˇ *fj* in K as shown in Figure 15b. The face is adjacent to an edge in K, if the original point is connected to the original edge. Under this composition, the orientation of each cell is naturally inherited from the original cell to the dual cell.

**Figure 15.** Correspondence between a cellular paving K and its dual paving K. (**a**) a face *fi* in K ↔ a point *n*ˇ*i* in K; (**b**) a point *nj* in K ↔ a face ˇ *fj* in K. In both figures, an edge *ej* in K corresponds to an edge *e*ˇ*j* in K.

Let us explicitly see this composition in the previous example, where the original cellular paving is shown again in Figure 16a. By applying the above procedure for K shown in Figure 16a, we obtain the dual paving K as shown in Figure 16b.

**Figure 16.** (**a**) cellular paving and (**b**) its dual paving.


For this dual paving, we have the following matrix representations of *∂*:

$$
\partial[\not f\_1 \not f\_2 \not f\_3] = [\not \ell\_1 \not \ell\_2 \not \ell\_3 \not \ell\_4] \begin{bmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 1 & -1 \\ -1 & 0 & 1 \end{bmatrix} \tag{5}
$$

$$
\partial \begin{bmatrix} \mathbb{K}\_1 \ \mathbb{K}\_2 \ \mathbb{K}\_3 \ \mathbb{K}\_4 \end{bmatrix} = \begin{bmatrix} \mathbb{H}\_1 \ \mathbb{H}\_2 \ \mathbb{H}\_3 \end{bmatrix} \begin{bmatrix} 1 & 1 & 0 & 1 \\ 0 & -1 & 1 & 0 \\ 1 & 0 & 1 & 1 \end{bmatrix} \tag{6}
$$

Comparing Equations (5) and (6) with Equations (1) and (4), we have

$$\left[\partial: \mathbb{C}\_1(\mathcal{K}) \to \mathbb{C}\_0(\mathcal{K})\right] = -\left[\partial: \mathbb{C}\_2(\mathcal{K}^\star) \to \mathbb{C}\_1(\mathcal{K}^\star)\right]^T,\tag{7}$$

$$\left[\partial:\mathbb{C}\_{2}(\mathcal{K})\to\mathbb{C}\_{1}(\mathcal{K})\right]=\left[\partial:\mathbb{C}\_{1}(\mathcal{K}^{\star})\to\mathbb{C}\_{0}(\mathcal{K}^{\star})\right]^{\mathrm{T}},\tag{8}$$

where the square brackets indicate matrix representation. The above relations *nj*(*∂ei*) = −*e*<sup>ˇ</sup>*<sup>i</sup>*(*<sup>∂</sup>* ˇ*fj*) and *ej*(*<sup>∂</sup> fi*) = *<sup>n</sup>*<sup>ˇ</sup>*<sup>i</sup>*(*∂e*<sup>ˇ</sup>*j*) hold from Figure 15. These relations universally hold even in a higher-dimensional space. Recalling [d : *Cp*(K) → *Cp*+<sup>1</sup>(K)] = [*∂* : *Cp*+<sup>1</sup>(K) → *Cp*(K)]T, we obtain the following commutative diagram:

$$\begin{array}{c} \mathbb{C}\_{2}(\mathbb{K}) \xrightarrow{\mathfrak{d}} \xrightarrow{\mathfrak{d}} \mathbb{C}\_{1}(\mathbb{K}) \xrightarrow{\mathfrak{d}} \xrightarrow{\mathfrak{d}} \mathbb{C}\_{0}(\mathbb{K})\\ \downarrow \star\_{2} \qquad \downarrow \star\_{1} \qquad \downarrow \star\_{0} \\ \mathbb{C}^{0}(\mathbb{K}^{\star}) \xrightarrow{\mathfrak{d}} \mathbb{C}^{1}(\mathbb{K}^{\star}) \xrightarrow{-\mathfrak{d}} \mathbb{C}^{2}(\mathbb{K}^{\star}) \end{array} \tag{9}$$

Here, we have isomorphisms *p* : *Cp*(K) → *Cm*−*<sup>p</sup>*(K) with *m* = 2, where 0 : *ni* → ˇ*f i*, 1 : *ei* → *e*ˇ*i*, and 2 : *fi* → *<sup>n</sup>*<sup>ˇ</sup>*i*. The dual counterpart of Equation (9) is given by

$$\begin{array}{ccccc}\mathcal{C}^{2}(\mathcal{K}) \xleftarrow{\mathrm{d}} & \mathcal{C}^{1}(\mathcal{K}) \xleftarrow{\mathrm{d}} & \mathcal{C}^{0}(\mathcal{K})\\\uparrow\* & \uparrow\* & \uparrow\* & \uparrow\* \\\mathcal{C}\_{0}(\mathcal{K}^{\*}) \xleftarrow{\mathrm{d}} & \mathcal{C}\_{1}(\mathcal{K}^{\*}) \xleftarrow{-\mathrm{d}} & \mathcal{C}\_{2}(\mathcal{K}^{\*}) \end{array} \tag{10}$$

with *i* = (*i*)<sup>∗</sup> using the dual map (*i*)<sup>∗</sup> of *i*. These correspondences naturally hold in higher-dimensional spaces and lead to Poincaré duality: *Hp*(*M*) ∼= *Hm*−*<sup>p</sup>*(*M*) for a compact orientable *m*-dimensional manifold *M* [63].

## *2.5. Dual Circuits*


Now, let us introduce dual circuits. Consider a circuit on a planar graph. The planar graph is seen as a cellular paving K for a two-dimensional plane (or a sphere surface). Because there is no nontrivial loop on a plane or sphere, we have the following equation:

$$\dim\left(\partial:\mathbb{C}\_2(\mathbb{X})\to\mathbb{C}\_1(\mathbb{X})\right) = \ker\left(\partial:\mathbb{C}\_1(\mathbb{X})\to\mathbb{C}\_0(\mathbb{X})\right).\tag{11}$$

Let *I* ∈ *<sup>Z</sup>*1(K) and *V* ∈ *B*<sup>1</sup>(K) be current and voltage distributions satisfying KCL and KVL, respectively. We set a reference resistance *<sup>R</sup>*ref(= 1/*G*ref) to exchange a current and voltage. Consider current *I* := *<sup>G</sup>*ref(<sup>1</sup>)−<sup>1</sup>(*V*) and voltage *V* := *R*ref 1 (*I*) distributions for a circuit on K, where 1 : *ei* → *e*ˇ*i* and (<sup>1</sup>)−<sup>1</sup> : *ei* → *e*ˇ*i* are defined with 1 := (1)<sup>∗</sup>. Note that *I* and *V* are outer-oriented, but we can obtain an inner-oriented component for a given orientation of the plane. By combining Equation (9) with Equation (11), KCL and KVL are shown to hold for *I* and *V* as explained below. Here, we check KVL for *V*. From Equation (11), a current distribution *I* ∈ *<sup>Z</sup>*1(K) can be written as *I* = *∂F* with "face" (or mesh) currents *F* ∈ *<sup>C</sup>*2(K). Then, we have *V* = *<sup>R</sup>*ref<sup>d</sup> 2 (*F*), which indicates *V* ∈ *<sup>B</sup>*<sup>1</sup>(K). A similar discussion to KCL holds for *I*.

A current and voltage relation along a circuit element located at each edge *ei* ∈ E is written by *hi*(*<sup>I</sup>*, *V*) = 0 for *I* ∈ *<sup>Z</sup>*1(K), and *V* ∈ *<sup>B</sup>*<sup>1</sup>(K). As a dual relation, we define *hi* (*I*, *V*) := *hi<sup>G</sup>*ref(1)−<sup>1</sup>(*V*), *R*ref 1 (*I*). If we assign a circuit element with *hi* (*I*, *V*) = 0 for each edge *e* ˇ *i* in K, *I* and *V* give a solution of the circuit on K. For example, consider a resistance with Ohm's law *V* = *IR* with scalar *V*, *I*, and *R*. Substituting *V* → *IR*ref, *I* → *G*ref*V*, we have *I* = *GV* with *G* = *<sup>R</sup>*/(*<sup>R</sup>*ref)2. Importantly, we obtain

$$RR^\* = (R\_{\text{ref}})^2 \tag{12}$$

for *R* = 1/*G*. The derived circuit on K is called a *dual circuit*. The dual relation is summarized in Table 1.


**Table 1.** Dual relations in electrical circuits.

For the previous example shown in Figure 17a, we construct a dual circuit as shown in Figure 17b. Here, we have *I*S = *V*S/*R*ref, *Ri* = (*<sup>R</sup>*ref)2/*Ri*, and take a specific orientation () of the plane to obtain an inner orientation of the current source. We can clearly see the duality between series and parallel connections in Figure 17. Therefore, the concept of dual circuits is considered as a generalization of series–parallel duality.

**Figure 17.** (**a**) example circuit and (**b**) its dual circuit.


Consider a one-port network *N* composed of resistors. A voltage source is attached to *N* as shown in Figure 18a. The network *N* is characterized by an equivalent resistance *R* = *V*s/*I*s, where *V*s and *I*s are the voltage and current along the source, respectively. Consider the dual counterpart as shown in Figure 18b. The dual network *N* is characterized by *R* = *<sup>V</sup>*s /*I*s with the current *I*s and voltage *V*s along the source. From the corresponding duality, we have

$$RR^\* = (R\_{\rm ref})^2,\tag{13}$$

where *R*ref is the reference resistance.

**Figure 18.** (**a**) one-port network with a voltage source and (**b**) its dual counterpart.

## *2.6. Self-Dual Circuit*

If the *self-dual* relation *N* = *N* holds for a one-port network *N*, the composite resistance satisfies *R* = *R*. In this case, we obtain *R* = *R* = *R*ref without solving circuit equations. As an example, we consider a bridge circuit shown in Figure 19a and its dual counterpart over *R*ref is given in Figure 19b. The self-dual condition is written as *R*1*R*2 = (*<sup>R</sup>*ref)2. Under the self-dual condition, the circuit behaves as an effective resistor with *R*ref.

Thus far, we only consider circuits with resistors, but the extension to alternating-current (AC) circuits is obvious. Under this extension, the real-number field (R) is replaced with the complex one (C). In AC circuits, resistances are replaced with complex impedances. Consider the example of an AC circuit shown in Figure 19c. This circuit is obtained from Figure 19a by replacing *R*1 → *Z*1 = j*ωL*, *R*2 → *Z*2 = 1/(j*ω<sup>C</sup>*), where *ω* is an angular frequency. The self-dual condition is given by *R*ref = √*L*/*C*. Under the self-duality, the circuit surprisingly behaves as a frequency-independent resistor with *R*ref, although capacitors and inductors exhibit frequency-dependent response. This result can be interpreted to mean that the frequency dependency of capacitors and inductors negate each other when the whole system is self-dual. Therefore, self-dual circuits provide a powerful way to produce constant-resistance circuits [64].

**Figure 19.** (**a**) bridge circuit and (**b**) its dual circuit; (**c**) alternating-current bridge circuit.

#### **3. Zero Backscattering from Self-Duality**

Signals can propagate in a uniform transmission line without backscattering. In this section, we associate self-duality with the zero-backscattering condition. Furthermore, we show that a large phase shift without backscattering in Huygens' metasurfaces can be understood by using a self-dual circuit model.

#### *3.1. Self-Dual Transmission Lines*

In this section, we consider signal propagation in a transmission line, which can be expressed by an LC ladder network [65]. In contrast to previous research on duality in transmission lines [66], we provide a circuit theoretical interpretation of the characteristic impedance of a transmission line in terms of self-dual response described in Section 2.
