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上循环与上边缘:亲手算 H^1

条形分解把上同调化为群元素元组上的显式函数。我们写出上边缘公式,辨认交叉同态,并算出一个真实的 H^1。

标准上链复形

条形分解 使 H^n(G, A) 可由显式 上链复形 计算。一个 n-上链 就是函数 f : G^n → A(其中 C^0 = A,即常值)。上边缘 d : C^n → C^{n+1} 交替地穿插作用与各槽。真正要用的是下面两条公式。

Cochains:   C^0 = A,   C^1 = { f: G -> A },   C^2 = { f: GxG -> A }.

d^0 : A -> C^1        (d^0 a)(g)      = g.a - a
d^1 : C^1 -> C^2      (d^1 f)(g,h)    = g.f(h) - f(gh) + f(g)
d^2 : C^2 -> C^3      (d^2 f)(g,h,k)  = g.f(h,k) - f(gh,k) + f(g,hk) - f(g,h)

Definitions in degree 1:
   Z^1 = ker d^1 = { f : f(gh) = f(g) + g.f(h) }          (1-cocycles)
   B^1 = im  d^0 = { f : f(g) = g.a - a  for some fixed a }  (1-coboundaries)
   H^1(G, A) = Z^1 / B^1.

Check d^1 . d^0 = 0:  d^1(d^0 a)(g,h) = g(h.a - a) - (gh.a - a) + (g.a - a) = 0.  OK.
上链复形 C^0 → C^1 → C^2 → …,及一次上循环/上边缘群。

1-上循环即交叉同态

再读上循环条件 f(gh) = f(g) + g·f(h):这是 扭曲的莱布尼茨法则。满足它的函数恰是 交叉同态(亦称导子)。当作用平凡时它退化为 f(gh) = f(g) + f(h),即普通 同态 G → A。于是 H^1 推广了 Hom(G, A),并为非平凡作用作出修正。

上边缘 f(g) = g·a − a 是平凡的扭曲:函数仅来自挪动单个元素 a。故 H^1 = (交叉同态)/(主交叉同态) 度量交叉同态在该平凡性下的差异。两个上循环之差为上边缘时称 上同调等价——这一等价关系是整门学科的脉搏。

一个算完的 H^1

G = Z/2Z = {1, s},   A = Z[i] (Gaussian integers) as additive group,
action:  s = complex conjugation,  s.(a + bi) = a - bi.

Cyclic recipe with n = 2:   N = 1 + s,    g - 1 = s - 1.

  N(a+bi)     = (a+bi) + (a-bi) = 2a.       ker N = { a+bi : 2a = 0 } = { bi : b in Z } = iZ.
  (s-1)(a+bi) = (a-bi) - (a+bi) = -2bi.      im(s-1) = { -2bi } = 2iZ.

H^1(Z/2Z, Z[i]) = ker N / im(s-1) = iZ / 2iZ = Z/2Z.

Sanity check by hand: the cocycle f with f(s) = i satisfies f(s^2)=f(1)=0 since
  f(1) = f(s)+ s.f(s) = i + (-i) = 0.  Good. It is NOT a coboundary: i is not of the
  form (s-1).(a+bi) = -2bi (an even multiple of i). So [f] is the nonzero class.
H^1(Z/2Z, Z[i]) = Z/2Z,用循环公式 ker N / im(g−1) 算出并直接验证。

注意这份纪律:上循环是 G 上的函数,不是一个数;上同调里的相等指相差一个上边缘,绝非逐点相等。请记住——下一篇里同样的上边缘记账法将分类群扩张,而在伽罗瓦上同调中某个看似平凡的上循环,会因定理 90 而真正消失。