[五行] 末學覺得exp人很純真自然
所以末學試著證明了一下:
Let v∈G_e,
and let θ: R^1 → G be the corresponding 1-parameter subgroup of G.
Then φθ: R^1 → H is the 1-parameter subgroup of H corresponding to φ'v.
Thus, exp(φ'v) = φθ(1) = φexp(v).
The following diagram is commutative:
φ'
G_e ───→ H_e
｜ ｜
exp｜ ｜exp
｜ ｜
↓ ↓
G ───→ H
φ
i.e. exp is natrual.
Q.E.D.
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◆ From: 140.115.25.215
※ 編輯: CFE220 來自: 140.115.25.215 (01/06 05:37)
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01/07 12:19, , 1F
01/07 12:19, 1F