matrix, where H ( k ) = q − 1 ∑ n $_{=0}$[ h BA n $_{c}$† $_{nB}$c$_{nA}$ + h BA ∗ n $_{c}$† $_{nB}$c$_{nA}$ + h CB n $_{c}$† $_{nC}$c$_{nB}$ + exp( − i 4 πf ) h CB ∗ n $_{$_{+1}$c}$† $_{nC}$c$_{n}$$_{+1}$$_{B}$ (7) + h AC n $_{c}$† $_{nA}$c$_{nC}$ + h AC ∗ n $_{c}$† n +1 $_{A}$c$_{nC}$ + H.c. ] , h BA n = t exp( − i k · a$_{1}$ ) exp [$_{−}$ i 8 πf ( n − 1 6 )] , h CB n = t exp( − i k · a$_{2}$ ) exp [ i 8 πf ( n + 1 12 )] , h AC n = t exp( − i k · a$_{3}$ ) . and With the above Hamiltonian (7), it is straightforward to obtain the Hofstadter butterfly-like energy spectrum of the kagom´ e lattice, which is shown in Fig. 2. From Fig. 2, one can easily observe three symmetries of this Hofstadter butterfly-like spectrum of the kagom´ e lattice [4]: (i) The spectrum at f is the same as that at f + j with j any integer; (ii) The spectrum is also unchanged on changing f to − f , because if there is an eigenstate with energy ϵ for field f , then the corresponding complex conjugate of this eigenstate is also an eigenstate with the same energy ϵ for field − f ; (iii) The spectrum is inverted when the field f changes to f + 1 / 2. Thus the highest-energy states for field near f =1 / 2 are equivalent to the lowest-energy states near zero field. To obtain the orbital magnetization of the kagom´ e lattice in an external magnetic field and with finite temperature, let us first write down the single-particle free energy as follows [16]: Here, Ω$_{n}$( k ) is the Berry curvature of electronic Bloch states defined by Ω$_{n}$( k )= i 〈∇ $_{k}$u$_{n}$$_{k}$ | × |∇ $_{k}$u$_{n}$$_{k}$ 〉 with | u$_{n}$$_{k}$ 〉 being the periodic part of Bloch wave for the n th band. Its III. THE PROPERTIES OF THE ORBITAL MAGNETIZATION F = − 1 β ∑ n ∫$_{MBZ}$ d $^{2}$k [ 1 + e $_{ℏ}$B · Ω$_{n}$( k ) ] ln[1 + e β ( µ − ϵ$_{n}$$_{k}$ $^{)}$] . (8) 6