The Tangent bundle of a lie group is isomorphic to a semidirect product.

Question

How can I proof that, given a Lie Group $G$ with Lie algebra $\mathfrak{g}$, the tangent group $TG$ is isomorphic to $\mathfrak{g} \rtimes_{Ad}G$ ?

Answer 1

Consider the Lie group be $(G,\circ)$. First, let's introduce the left translation $L_\bullet$ and the right translation $R_\bullet$: \begin{align*} L_q&\colon G \times G, & a&\mapsto L_q(a) = q \circ a \\ R_p&\colon G \times G, & a&\mapsto R_p(a) = a \circ p \end{align*} and their derivatives $$ dL_q(a)\colon T_qG \to T_{L_q(a)}G= T_{q\circ a}G,\qquad dR_p(a)\colon T_qG \to T_{R_p(a)}G= T_{a\circ p}G$$ for $a\in G$. Note that they are linear functions.

The left and right translation do commute $$ q\circ a\circ p = L_q(R_p(a)) = R_p(L_q(a)), $$ so for the derivatives it holds $$ dL_q(a\circ p)\; dR_p(a) = dR_p(q\circ a)\; dL_q(a).$$

We will need the following identity obtained by deriving $R_{a\circ b}(c) = R_b(R_a(c)) = c\circ a\circ b$: $$ dR_{a\circ b}(c) = dR_b(c\circ a)\; dR_a(c). $$

Now consider the tangent bundle $$ TG = \bigcup_{g\in G} \{(g,v)\colon v\in T_gG\}. $$

Now we will derive the Lie group operation: \begin{align*} \circ&\colon G\times G \to G,& (a,b)&\mapsto a\circ b = L_a(b) = R_b(a) \\ d\circ&\colon TG\times TG \to TG,& \bigl((a,v),(b,w)\bigr)&\mapsto \bigl(a\circ b, dL_a(b)\;w + dR_b(a)\;v\bigr), \end{align*} where we have basically used the product rule of differentiation. (with $da = v$, $db=w$ and $df(a,b) = \partial_1 f(a,b)\;da + \partial_2 f(a,b)\;db$) Also note that $dL_a(b)\;w\in T_{a\circ b}G$ and also $dR_b(a)\;v\in T_{a\circ b} G$.

It is well-known that for a Lie group $G$ with identity element $e\in G$, all tangent spaces are isomorphic. The mapping $dR_{a}(e)\colon T_eG \to T_aG$ is an isomorphism between the Lie algebra $T_eG=\mathfrak g$ and an arbitrary tangent space $T_aG$. This means, that we can write $T_aG = \{dR_a(e)\;V\colon V\in\mathfrak g\}$. In order to get from an element $v\in T_aG$ to its corresponding element $V\in\mathfrak g$, we can use $$ V = dR_{a^{-1}}(a)\;v,\qquad a\in G, v\in T_aG. $$

The adjoint operator is defined by $\operatorname{Ad}_a = dC_a(e)$ with the conjugation $$ C_a(b) = a\circ b\circ a^{-1} = R_{a^{-1}}\bigl(L_{a}(b)\bigr), \qquad \operatorname{Ad}_a = dC_a(e) = dR_{a^{-1}}\bigl(a)\;dL_{a}(e). $$

Now we will use this knowledge with $v=dR_a(e)\;V$, $w=dR_b(e)\;W$: \begin{align*} dL_a(b)\;w + dR_b(a)\;v &= dL_a(b)\;dR_b(e)\;W + dR_b(a)\;dR_a(e)\;V \\ &= dR_b(a)\;dL_a(e)\;W + dR_{a\circ b}(e)\;V \\ &= dR_{a\circ b}(e)\;\bigl(V + dR_{a^{-1}}(a)\;dL_a(e)\;W\bigr) \\ &= dR_{a\circ b}(e)\;\bigl(V + \operatorname{Ad}_{a}W\bigr). \end{align*} This means that we have found an Lie-group isomorphism $$\Phi\colon G\ltimes_{\operatorname{Ad}}\mathfrak g \to TG,\qquad (a,V)\mapsto (a, dR_a\;V), $$ where $TG$ has the Lie group structure induced by the Lie group structure of $G$ and the semidirect product Lie group $G\ltimes_{\operatorname{Ad}}\mathfrak g$, where the Lie group operation is defined by $$ \bigl((a,V),(b,W)\bigr) \mapsto (a\circ b, V + \operatorname{Ad}_a W),$$ because “$+$” is the Lie group operation of $\mathfrak g$.