# Theorem Given two [[Lie group|Lie groups]] $G$ and $H$ and their [[Lie Algebra|Lie algebras]] $\mathfrak{g}$ and $\mathfrak{h}$. Consider a [[Continuous Map|continuous]] [[Group Homomorphism|homomorphism]] $\Phi: G \rightarrow H$, then this induces a [[Lie Algebra Homomorphism]] $\phi: \mathfrak{g} \rightarrow \mathfrak{h}$ given by: $ \phi(X) = \frac{d}{dt} \Phi(e^{tX})\big|_{t=0} \tag{1} $ or equivalently $ \Phi(e^{tX}) = e^{t \phi(X)} \tag{2} \label{eq:lie_algebra_homomorphism} $ where here $X \in \mathfrak{g}$. # Proof Heuristics Recall that the Lie algebra is just the [[tangent space]] to the Lie group [[manifold]] at the identity [[group element|element]]. What we are saying here is that, given a [[group homomorphism]] from $G$ to $H$, we also end up with a [[Lie Algebra Homomorphism]] between $\mathfrak{g}$ and $\mathfrak{h}$, i.e., between the two tangent spaces. Recall that for every tangent vector to the identity in $G$, i.e., $X \in \mathfrak{g}$, there exists a [[one-parameter subgroup]] $\gamma(t) = e^{tX}$. Then, the corresponding one-parameter subgroup in $H$ is just $(\Phi\circ\gamma)(t) = \Phi(e^{tX})$. Taking the derivative of this element and evaluating it at $t=0$ (i.e., computing the infinitesimal generator of the group) will give us $\phi(X)$. ![[Pasted image 20210707193607.png]] # Proof We have already explained the form of the proposition, but to prove that $\phi$ is indeed a Lie group homomorphism, we need to prove 3 things: 1. $\phi(sX) = s\phi(X)$ 2. $\phi(X+Y) = \phi(X) + \phi(Y)$ 3. $[\phi(X), \phi(Y)] = \phi([X, Y])$ (1) and (2) prove linearity (a requirement for [[vector space]] homomorphisms), and (3) proves closure under the Lie bracket, a requirement for [[Lie Algebra Homomorphism]]. ## Proof of $\phi(sX) = s\phi(X)$ $ \begin{align} \phi(sX) &= \frac{d}{dt}\Phi(e^{t s X}) \bigg|_{t=0}\\ &= \frac{d(st)}{dt} \frac{d}{d(st)}\Phi(e^{t s X}) \bigg|_{t=st=0}\\ &= s \frac{d}{d(t)}\Phi(e^{t X}) \bigg|_{t=0}\\ &= s \phi(X) \end{align} $ where in the second-to-last equality we renamed the dummy variable $st \rightarrow t$ ## Proof of $\phi(X+Y) = \phi(X) + \phi(Y)$ $ \begin{align} \phi(X+Y) &= \frac{d}{dt}\Phi\left(e^{t(X+Y)}\right)\bigg|_{t=0}\\ &= \frac{d}{dt}\Phi\left(\lim_{m\rightarrow\infty}\left[e^{tX/m} e^{tY/m}\right]^m\right)\bigg|_{t=0}\\ &= \frac{d}{dt}\lim_{m\rightarrow\infty}\Phi\left(\left[e^{tX/m} e^{tY/m}\right]^m\right)\bigg|_{t=0}\\ &= \frac{d}{dt}\lim_{m\rightarrow\infty}\left(\Phi(e^{tX/m}) \Phi(e^{tY/m})\right)^m\bigg|_{t=0}\\ &= \frac{d}{dt}\lim_{m\rightarrow\infty}\left(e^{t\phi(X)/m}e^{t\phi(Y)/m})\right)^m\bigg|_{t=0}\\ &= \frac{d}{dt}e^{t(\phi(X)+\phi(Y))}\bigg|_{t=0}\\ &= \phi(X) + \phi(Y) \end{align} $ In the second equality, we used the [[Lie product formula]]. In the third equality, we used the [[Continuous Map|continuity]] of $\phi$. In the fourth equality, we used the properties of [[Group Homomorphism|homomorphisms]]. In the fifth equality, we used $\eqref{eq:lie_algebra_homomorphism}$ above. In the sixth equality, we used the [[Lie product formula]] in the opposite direction. ## Proof of $[\phi(X), \phi(Y)] = \phi([X, Y])$ First, let's prove this little useful property. Given a [[linear map]] $\phi$ from a finite-[[dimension|dimensional]] [[vector space]] $V$ to a finite dimensional vector space $W$, and let $\gamma(t):\mathbb{R} \rightarrow V$ be a [[Differentiable Function|differentiable]] $V$-valued [[map]] of $t$. Observe the following: $ \begin{align} \frac{d}{dt} \phi(\gamma(t)) &= \lim_{h \rightarrow 0} \frac{\phi(\gamma(t + h)) - \phi(\gamma(t))}{h}\\ &= \lim_{h \rightarrow 0} \phi\left(\frac{\gamma(t + h)) \gamma(t)}{h}\right)\\ &= \phi\left(\lim_{h \rightarrow 0} \frac{\gamma(t + h)) \gamma(t)}{h}\right)\\ &= \phi\left(\frac{d\gamma}{dt}\right)\\ \end{align} $ In the second equality, we used the linearity of $\phi$, in the third equality, we used the continuity of $\phi$ and the fact that $V$ and $W$ are finite dimensional. so we end up with $ \frac{d}{dt} \phi(\gamma(t)) = \phi\left(\frac{d\gamma(t)}{dt}\right) \quad \forall t \label{eq:continuity_derivatives3} \tag{3} $ We will also need some identities $ e^{t A X A^{-1}} = A e^{tX} A^{-1} \tag{4} \label{eq:exp_to_exp_in_middle2} $ where $X \in \mathfrak{g}$ and $A \in G$ (recall that this makes $A X A^{-1} \in \mathfrak{g}$). Now letting $A = e^{tY}, \, Y \in \mathfrak{g}$, we have $e^{tY} X e^{-tY} \in \mathfrak{g}$, and we compute the derivative at $t = 0$: $ \begin{align} \frac{d}{dt} e^{tY}X e^{-tY} \bigg|_{t=0} = Y e^{tY}X e^{-tY}\bigg|_{t=0} - e^{tY}X e^{-tY}Y\bigg|_{t=0} = [Y, X] \tag{5} \label{eq:commutator_from_ddt2} \end{align} $ Now let's start the proof: $ \begin{align} [\phi(X), \phi(Y)] &= \frac{d}{dt} e^{t\phi(X)} \phi(Y) e^{-t\phi(X)}\\ &= \frac{d}{dt} \Phi(e^{tX}) \phi(Y) \Phi(e^{-tX})\\ &= \frac{d}{dt} \Phi(e^{tX}) \left(\frac{d}{ds} e^{s\phi(Y)})\right) \Phi(e^{-tX})\\ &= \frac{d}{dt} \frac{d}{ds} \Phi(e^{tX}) \Phi(e^{sY}) \Phi(e^{-tX})\\ &= \frac{d}{dt} \frac{d}{ds} \Phi\left(e^{tX} e^{sX} e^{-tX}\right)\\ &= \frac{d}{dt} \frac{d}{ds} \Phi\left(e^{\left[s e^{tX} Y e^{-tX}\right]}\right)\\ &= \frac{d}{dt} \phi\left(e^{tX} Y e^{-tX}\right)\\ &= \phi\left(\frac{d}{dt}e^{tX} Y e^{-tX}\right)\\ &= \phi\left([X,Y]\right) \end{align} $ In the first equality, we used Eq. $\eqref{eq:commutator_from_ddt2}$. In the second equality, we used the definition of Eq. $\eqref{eq:lie_algebra_homomorphism}$. In the third equality, we messed around with derivatives. In the fourth equality, we used $\eqref{eq:lie_algebra_homomorphism}$ again. In the fifth equality, we used the fact that $\Phi$ is a homomomorphism. In the sixth equality, we used $\eqref{eq:exp_to_exp_in_middle2}$. Finally, we used $\eqref{eq:continuity_derivatives3}$ in the seventh equality and $\eqref{eq:commutator_from_ddt2}$ in the last equality.