---- > [!definition] Definition. ([[Lie algebra of a Lie group]]) > > Recall that any [[Lie group]] $G$ [[group action|acts]] [[smooth group action|smoothly]] and [[transitive group action|transitively]] on itself by left translation: > > $\begin{align} > G &\to \text{Diff }G \\ > g & \mapsto L_{g} = g \cdot \_ \ \ > \end{align}$ > > [[related vector fields|Recall]] in general that a given [[vector field]] pushes forward under [[diffeomorphism]] to a unique vector field. A [[vector field]] $X$ on $G$ is said to be **left-(translation)-invariant** if it is invariant under left-translations, that is, if it pushes forward *to itself* under $L_{g}$ for all $g \in G$. Explicitly, [[differential of a smooth map between smooth manifolds|this means]] $\begin{align} > (L_{g})_{*}X &= X \text{ for all }g \in G, \text{ i.e., } \\ > d(L_{g})_{g'}(X_{g'})&=X_{g g'} \text{ for all } g,g' \in G. > \end{align}$ > The collection of left-invariant vector fields on $G$ forms a [[Lie subalgebra]] of the [[Lie algebra]] $\mathscr{V}(M)$[^1], denoted by $\text{Lie}(G)$ or $\mathfrak{g}$ and called the **Lie algebra of $G$**. > It follows from [[Lie's First Theorem on Differentiating Lie morphisms|Lie's First Theorem]] that $\text{Lie}(-)$ is a [[covariant functor]] from the [[category]] $\mathsf{Lie}$ of [[Lie group|Lie groups]] to the category $\mathsf{lie}$ of [[Lie algebra|Lie algebras]]. > > The evaluation map $\begin{align} > \varepsilon:\text{Lie}(G) &\to T_{e}G \\ > X &\mapsto X_{e} > \end{align}$ > is a [[natural transformation|natural]] [[isomorphism]] of [[vector space|vector spaces]], and so the space $\mathfrak{g}=T_{e}G$ is also canonically referred to as the **Lie algebra of $G$**.[^2] > [!basicproperties] > - Left-invariant vector fields are [[integral curve|complete]]. > - An LGH $F:G \to H$ pushes forward [[Lie algebra of a Lie group|left-invariant vector fields]] vector fields uniquely. > > > > [!proof]- Proof. > > Indeed, $X \in \operatorname{Lie}(G)$ is uniquely determined by $X_{e} \in T_{e}G$, which gets mapped to some $Y_{e}:=dF_{e}X_{e} \in T_{e}H$, which uniquely promotes to a left-invariant vector field $Y \in \operatorname{Lie}(H)$ given by $Y_{h}=(L_{h})_{*}Y_{e}$. We claim that $X,Y$ are [[related vector fields]]. So suppose $h=F(g)$. Then $\begin{align} > > Y_{h} &= dL_{F(g)} |_{e_{H}} Y_{e} \\ > > &= dL_{F(g)} |_{e_{H}} \, dF |_{e _{G}} X _{e} \\ > > &= (L_{F(g)} \circ F) _{*} X_{e} \\ > > &= (F \circ L_{g}) _{*} X_{e} \\ > > &= dF |_{g}\, dL_{g} |_{e} X_{e} \\ > > &= dF |_{g} X_{g} > > \end{align}$ > > as required, where in the only meaningful step we used that $F$ is a LGH to commute $L_{F(g)} \circ F=F \circ L_{g}$. > > > > > > - Similarly, an LGH $F:G\to H$ pushes forward [[integral curve|integral curves]] of LIVFs uniquely. > > > [!proof]- Proof. > > Let $X \in \operatorname{Lie}(G)$ and $\alpha_{g}(t)$ the integral curve of $X$ starting at $g$. We claim that $F \circ \alpha_{g}(t)$ is the integral curve of $Y=F_{*}X$ starting at $h=F(g)$. To show this, must check it satisfies the [[ODE]], i.e., that $\frac{d}{dt}(F \circ \alpha_{g}(t))=Y_{F\big( \alpha_{g}(t) \big)} \text{ for all } t.$ > > Indeed, using that $\dot{\alpha}_{g}(t)=X_{\alpha_{g}(t)}$ we have $\begin{align} > > \frac{d}{dt} \big( F \circ \alpha_{g} (t) \big)&= d F |_{\alpha_{g}(t)} \dot{\alpha}_{g}(t) \\ > > &= dF |_{\alpha_{g}(t)} X_{\alpha_{g}(t)} \\ > > &= Y_{F\big(\alpha_{g}(t)\big)}. > > \end{align}$ > > > > [!note] Remark. (Lie groups are parallelizable) > In the notation below, any basis [[basis]] $x_{1},\dots,x_{n}$ for $\mathfrak{g}$ gives rise to a smooth global frame $X_{1}^{x_{1}},\dots,X_{n}^{x_{n}}$, where (as below) $X_{i}^{x_{i}}(g)=(dL_{g})_{e} x_{i}$. Thus, every [[Lie group]] $G$ is [[local frame, global frame|parallelizable]]. > [!justification] Proving $\text{Lie}(G)$ is indeed a [[Lie subalgebra]] of $\mathscr{V}(M)$. > Really just need to show $[X,Y]$ is left-invariant when $X$ and $Y$ are. This follows from [[vector field|the general fact that]] $F_{*}[X,Y]=[F_{*}X, F_{*}Y]$ for $F$ a [[diffeomorphism]]. Indeed, given $g \in G$, $L_{g}$ is a [[diffeomorphism]] and so $(L_{g})_{*}[X, Y]=[(L_{g})_{*}X, (L_{g})_{*}Y]$. But $(L_{g})_{*}X$ and $(L_{g})_{*}Y$ just equal $X$ and $Y$ respectively, by left-invariance of $X$ and $Y$. ^justification > [!justification] Proving $\text{Lie}(G)=_{\varepsilon}T_{e}G$. > **Surjectivity.** Let $v \in T_{e}G$ be an arbitrary [[tangent vector to a smooth manifold|tangent vector]]. Define a [[vector field]] $X^{v}$ on $G$ by 'translating $v$ around': $X^{v}_{g}:=(dL_{g})_{e}v$. (This is smooth.) Then $\varepsilon(X^{v})=X^{v}_{e}=v$. The vector field $X^{v}$ is left-invariant because, by [[covariant functor|functoriality]] of the pushforward, $(dL_{g})_{g'}X^{v}_{g'}=(dL_{g})_{g'}(dL_{g'})_{e}v=d(L_{g} \circ L_{g'})_{e}v=d(L_{g g'})_{e} v=X_{g g'}^{v}.$ > **Injectivity.** Suppose $X \in \text{Lie}(G)$ satisfies $\varepsilon(X)=0$ (i.e. $X_{e}=0$). Then $X \equiv 0$: indeed, if $g \in G$ is arbitrary, then by left-invariance $X_{g}=(dL_{g})_{e}X_{e},$ which is zero because $X_{e}=0$. ^justification ---- #### [^1]: In justification. [^2]: Note: while $\text{dim }\mathscr{V}(G)=\infty$ in general, this establishes that $\text{dim }\text{Lie}(G)<\infty$; specifically, $\text{dim }\text{Lie }G=\text{dim }G$. ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```