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Let $G$ be a Lie group. I know $TG$ with the tangent maps of multiplication and inversion from $G$ has Lie group structures and is isomorphic to the semidirect product ${\frak{g}} \ltimes_{Ad} G$ as Lie groups and isomorphic to the usual product ${\frak{g}}\ltimes G$ just as tangent bundles.

Here it is proved that $Lie(G\times H) \simeq Lie(G) \oplus Lie(H).$

If we choose the Lie group operation on ${\frak{g}} \times G$ to be defined component wise, so that it is a product of Lie groups with $\frak g$ a Lie group with addition, we get $$ T({\frak{g}} \times G) \simeq ({\frak{g}} \times Lie({\frak{g}})) \times ({\frak{g}} \times G) $$ which can similarly be given a Lie group structure with operation defined component wise and $$ Lie ({\frak{g}} \times G) \simeq Lie({\frak{g}}) \oplus Lie(G) \simeq Lie({\frak{g}}) \oplus \frak{g}. $$ where the Lie bracket on $Lie(\frak{g})$ is trivial.

It seems that iterating taking the tangent bundle of ${\frak{g}} \times G$ is just a product of $\frak g$'s, $Lie(\frak g)'s$ (wchich are identical as manifolds) and one $G$. Also taking Lie algebras of those iterations is something similar to the case of ${\frak{g}} \times G.$

What if we choose the semidirect product strucutre - what is the tangent Lie group $T({\frak{g}} \ltimes_{Ad} G)$ and its Lie algebra $Lie({\frak{g}} \ltimes_{Ad} G)$?

patrini
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1 Answers1

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Since you might want to apply $T$ to $G$ more than twice, here is an answer which you might find useful. If $A$ is a commutative algebra then $\mathfrak{g}\otimes A$ is naturally a Lie algebra, with the Lie bracket given by $$[u_1\otimes a_1, u_2\otimes a_2]:=[u_1,u_1]\otimes a_1 a_2.\qquad(*)$$

To get the Lie algebra of $TG$, take $A=A_1=\mathbb R[\epsilon]/(\epsilon^2)$ (the elements of $A$ are of the form $c_1+c_2\,\epsilon$, where $c_1,c_2\in\mathbb R$, and $\epsilon^2=0$; we therefore have $\mathfrak g\otimes A= \mathfrak g \oplus \mathfrak g\,\epsilon$ and you will see easily that the Lie bracket $(*)$ gives what you expect).

To get the Lie algebra of $TTG$ we simply iterate the procedure and take $A=A_2=A_1\otimes A_1=\mathbb R[\epsilon_1, \epsilon_2]/(\epsilon_1^2, \epsilon_2^2)$. A basis of $A_2$ is $1,\epsilon_1, \epsilon_2, \epsilon_1\epsilon_2$, so $\mathfrak g\otimes A_2= \mathfrak g \oplus \mathfrak g\,\epsilon_1\oplus\mathfrak g\,\epsilon_2\oplus\mathfrak g\,\epsilon_1\epsilon_2$, with the Lie bracket given by $(*)$.

As a Lie group, $TTG$ is the semidirect product of $G$ with the nilpotent Lie group $G_2$ whose Lie algebra is $\mathfrak g\,\epsilon_1\oplus\mathfrak g\,\epsilon_2\oplus\mathfrak g\,\epsilon_1\epsilon_2$. One can describe $G_2$ explicitly as $\mathfrak g\times\mathfrak g\times\mathfrak g$ with the product $$(u_1,v_1,w_1)(u_2,v_2,w_2)=(u_1+v_1, u_2+v_2, w_1+w_2 +[u_1,v_2]/2+[u_2,v_1]/2). $$ The action of $G$ on $G_2$ (used for the semidirect product) is Ad on each $\mathfrak g$.

user8268
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