# Two Lie Group homomorphisms are equal if their induced Lie algebra homomorphisms are equal and $G$ is connected

Let $$f, g$$ be two Lie Group homomorphisms from G to H (smooth group homomorphisms).

Let the induced Lie Algebra homomorphisms $$Df(e)$$ and $$Dg(e)$$ be equal. Then if $$G$$ is connected show that $$f=g$$.

I know that the set $$S=\{x | f(x)=g(x)\}$$ is closed. If I show it is open I am done. Now I know $$e$$ is in $$S$$. How do I show a nbd is also there?

• Mm I don't think that there is an easy proof for your statement. You can see Wolfgang Ziller's notes for a proof of this fact easier than for example a classical one that is in Warner's book. Commented Feb 8, 2020 at 10:55
• Could you outline the proof briefly? Commented Feb 8, 2020 at 10:57

If $$f$$ is an homorphism of Lie group, $$f(exp(X))=exp(df_{e_G}(X))$$, there exists a neighborhood $$U$$ of $$e_G$$ such that for every $$x\in U$$, there exists $$X$$ in the Lie algebra of $$G$$ such that $$x=exp(X)$$, we deduce that $$f$$ and $$g$$ coincide in a neighborhood $$U$$ of $$e_G$$, since $$G$$ is connected, $$U$$ generates $$G$$ and $$f=g$$.
Let $$\varphi:G\to H$$ a Lie group homomorphism and $$d\varphi:\mathfrak{g}\to\mathfrak{h}$$ the induced Lie algebra homomorphism, given by $$(d\varphi X)_{e_H}=(d\varphi)_{e_G} X_{e_G}$$.
Given $$\varphi$$, let $$\Gamma_{\varphi}$$ the graph of $$\varphi$$, namely $$\Gamma_{\varphi}=\{(g,\varphi(g))\mid g\in G\}$$. You can verify that $$\Gamma_{\varphi}$$ is an abstract subgroup of $$G\times H$$ and that $$(\Gamma_{\varphi}, \iota)$$ (where $$\iota$$ is the inclusion) is an embedding of $$G\times H$$ so it is a Lie subgroup of $$G\times H$$.
What is the Lie algebra of $$\Gamma_{\varphi}$$ as a subalgebra of the Lie algebra of $$G\times H$$? Well, you can see that $$\mathrm{Lie}(\Gamma_{\varphi})\cong \Gamma_{d\varphi}$$, i.e. the graph of the homomorphism $$d\varphi$$.
We have $$\mathrm{Lie}(\Gamma_f)\cong \Gamma_{df}=\Gamma_{dg}\cong\mathrm{Lie}(\Gamma_g)$$, and if $$G$$ is connected, then $$\Gamma_f$$ and $$\Gamma_g$$ are two connected subgroups which have the same Lie algebra. By uniqueness we must have $$\Gamma_f=\Gamma_g$$, i.e. $$f(a)=g(a)\,\forall a\in G$$.