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Note that (a) H (u) ≤ q0−1 1−q0 uq0 , (b) uG(u) ≤ H 2 (u), (c) G (u) ≥ C1 H 2(u) whenever u = , where C1 > 0 is a constant. The proofs of these inequalities are elementary. For example, for part (c), we note that, when u > , G (u) = ≥ 2q0 − 1 q0 2q0 − 1 q0 2−2q0 2q0 −2 u 2−2q0 2q0 −2 u (q0 − 1)2 2−q0 q0 −2 u q0 2q0 − 1 (H (u))2 . = q0 + Now choose ζ = (ηη)2 G(u) where η ∈ C0∞ (B) and η ∈ C0∞ (B \ {0}) are nonnegative functions. 4) is valid for any test function ζ with support in B \ {0}, we have 64 L.

33) On the other hand, the fact that < n(t), ∂δ >g(t) ˜ = 0 and < n(t), n(t) >g(t) ˜ = 1 imply that (∂n , ∂δ )+ < ∂δ , n (0) >go = 0 (∂n , ∂n ) + 2 < n (0), n(0) >go = 0 . (34) Hence, (33) becomes < D∂0α ∂β , n (0) >go = − To calculate < d t dt {D∂α ∂β }|t=0 , n(0) 1 2 n nn αβ − δ nδ αβ . (35) >go , we recall that D∂t α ∂β − D∂0α ∂β = D(0)kαβ ∂k . Hence, < d {D t ∂β }|t=0 , n(0) >go = D (0)nαβ , dt ∂α (36) 34 P. Miao where D (0)nαβ = 1 nn g { 2 o nα;β + nβ;α − αβ;n } (37) by (13). Therefore (32), (35) and (36) imply that goαβ 1 d {< D∂t α ∂β , n(t) >g(t) ˜ }|t=0 = H ( , go ) dt 2 nn + goαβ D (0)nαβ .

C2m (F ). 9) The associated topological charge is then defined as s2m = S 4m s2m (F ) = − S 4m Tr (F (m) ∧ F (m)) = −(−1)m (2π)2m (2m)! c2m . 10) We now decompose F (m) into its self-dual and anti-self-dual parts, F (m) = F + (m) + F − (m), F ± (m) = 1 (F (m) ± ∗F (m)). 4), we see that F + (m) and F − (m) are orthogonal, (F + (m), F − (m)) = 0. 10) and using the property ∗F ± (m) = ±F ± (m) and the orthogonality of F + (m) and F − (m), we obtain E = (F (m), F (m)) = (F + (m) + F − (m), F + (m) + F − (m)) = F + (m) 2 + F − (m) 2 , s2m = (F (m), ∗F (m)) = (F + (m) + F − (m), F + (m) − F − (m)) = F + (m) 2 − F − (m) 2 .