Let $\varphi$ be a Lie algebra homomorphism. By definition $$\varphi[X, Y] = [\varphi(X), \varphi(Y)].$$ You can then use induction to show $$\varphi(\mathrm{ad}_Y^n (X)) = \mathrm{ad}_{\varphi(Y)}^n (\varphi(X)).$$ Hence $$\varphi(e^{t \mathrm{ad}_Y (X)}) = e^{t \mathrm{ad}_{\varphi(Y)}}(\varphi(X))$$ What I don't understand is why $$\varphi(e^{\mathrm{ad}_X} e^{t\mathrm{ad}_Y} (Y)) = e^{\mathrm{ad}_{\varphi(X)}} e^{t \mathrm{ad}_{\varphi(Y)}} (\varphi(Y))?$$
2026-03-26 01:10:43.1774487443
Lie algebra homomorphism of product
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You just need to expand both exponentials. $$\varphi(\mathrm{ad}_X^n \mathrm{ad}_Y^m (Z)) = \mathrm{ad}_{\varphi(X)}^n (\varphi(\mathrm{ad}_Y^m(Z))) = \mathrm{ad}_{\varphi(X)}^n \mathrm{ad}_{\varphi(Y)}^m (\varphi(Z)).$$ Thus $$\varphi(e^{\mathrm{ad}_X} e^{t \mathrm{ad}_Y} Y) = \varphi \left( \sum_{k = 0}^{\infty} \sum_{m = 0}^{\infty} \frac{t^m}{k! m!} \mathrm{ad}_X^k \mathrm{ad}_Y^m Y \right) = \sum_{k = 0}^{\infty} \sum_{m = 0}^{\infty} \frac{t^m}{k! m!} \mathrm{ad}_{\varphi(X)}^k \mathrm{ad}_{\varphi(Y)}^m \varphi(Y).$$ Therefore $\varphi(e^{\mathrm{ad}_X} e^{t \mathrm{ad}_Y} Y) = e^{\mathrm{ad}_{\varphi(X)}} e^{t\mathrm{ad}_{\varphi(Y)}} \varphi(Y)$.