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Suppose we have a (finite dimensional) Lie group $G$ with Lie algebra $\mathfrak g$. I am interested in the properties of the tangent bundle $$TG = \bigsqcup\limits_{g \in G} T_gG \cong \mathfrak g\times G.$$

Specifically, I want to equip $TG$ with a product making it a Lie group. Of course, one could just define the product component wise, as $\mathfrak g$ is a vector space and thus an additive Lie group. But to me this seems unnatural, as it ignores the possible non-commutativity of $G$.

From The Tangent bundle of a lie group is isomorphic to a semidirect product. we obtain a more natural group structure on the tangent bundle, induced by the semi-direct outer product: $\mathfrak g \rtimes_{\operatorname{Ad}} G$.

Now for my questions:

  1. Suppose $G$ is a Matrix Lie group. Is there an easy / obvious way to represent $\mathfrak g \rtimes_{\operatorname{Ad}} G$ as a Matrix Lie group? This is easily achieved in the special euclidean group $\mathbb{SE}(3) \cong \mathbb R^ 3 \rtimes \mathbb{SO}(3)$, which seems to be similar to the abstract question.
  2. What is the Lie algebra of $TG$? An obvious choice would be $\mathfrak g \rtimes_{\operatorname{ad}} \mathfrak g$, but does this induce the correct Lie bracket? It seems to me, that depending on $G$, $\operatorname{ad}(g)(\cdot)$ may not be a automorphism of $\mathfrak g$. Then $\mathfrak g \rtimes_{\operatorname{ad}} \mathfrak g$ would not be well defined.
  3. Is there a nice representation of $\exp_{TG}$ in the given setting?
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    $\begingroup$ It might help you to know that the group structure you're talking about (that semidirect product) is simply the derivative of the group structure on $G$: if $m:G \times G \to G$ denotes the multiplication of $G$, then the tangent group is $TG$ equipped with the multiplication $Dm: T(G \times G) \simeq TG \times TG \to TG.$ $\endgroup$ – Anthony Carapetis Mar 27 '18 at 12:26
  • $\begingroup$ @AnthonyCarapetis I have not yet thought of this, thank you for that point of view! $\endgroup$ – Stefan Mar 27 '18 at 12:31
  • $\begingroup$ The tangent bundle of a Lie group is always trivial, so as you've written $TG \simeq \mathfrak{g} \times G$. He proves it by showing there can be enough independent vector fields on the Lie group to define it as a product. Haven't unpacked it though. $\endgroup$ – cactus314 Mar 27 '18 at 12:45
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    $\begingroup$ @cactus314 yes, $TG \cong \mathfrak g \times G$ as manifolds, but if one regards the $\times$ as a group operation, the induced group structure does not do $G$ justice, as the component from $\mathfrak g$ should not be treated independent of the component of $G$ (except when $G$ is abelian). $\endgroup$ – Stefan Mar 27 '18 at 12:52
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The analolgy to the group of motions in question 1 does not really work so easily. The difference is that the representation $\mathbb R^3$ of $SO(3)$ that you are forming the semidirect product with is the same representation that you have used to make $SO(3)$ into a matrix group. This will almost never be true for the adjoint representation of a matrix group $G$, since it means that you view an $N$-dimensional group as a subgroup of $GL(N,\mathbb R)$.

In general, I am not sure whether the image of $G$ under $Ad$ is automatically a closed subgroup of $GL(\mathfrak g)$. If it is and $G$ was originally a closed subgroup in $GL(n,\mathbb R)$, then you can realize $\mathfrak g\rtimes_{Ad}G$ as a closed sugroup in $GL(\mathfrak g\oplus \mathbb R^{n+1})$, by considering block matrices of the form $\begin{pmatrix} Ad(A) & 0 & X \\ 0 & A & 0 \\ 0 & 0 & 1\end{pmatrix}$ with $A\in G$ and $X\in\mathfrak g$. (So this is one of the cases in which life is much easier with Lie groups than with matrix groupes.)

Concerning the Lie algebra, you are right, it is $\mathfrak g\times\mathfrak g$ with the bracket $[(X_1,X_2),(Y_1,Y_2)]=([X_2,Y_1]-[Y_2,X_1],[X_2,Y_2])$. You don't need $ad(X)$ to be an automorphism for this to work but a derivation, and this is always true by the Jacobi identity.

I believe that the exponential map can be expressed nicely in this picture but I am not sure what the result looks like.

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  • $\begingroup$ Couldn't one use the representation you have given (if $\mathbb g \rtimes_{Ad} G$ is a closed subrgoup of $\mathbb{GL}(n, \mathbb R)$) to compute the exponential map, because then it should coincide with the usual matrix exponential? $\endgroup$ – Stefan Apr 1 '18 at 8:40
  • $\begingroup$ Sure you can do that, and in special cases (say if $A=0$ or $X=0$) this will become very simple. But it will only reproduce the result that on a subgroup exp coincides with the exponential map of the subgroup. For general elements, it could get pretty complicated, but I am not sure. $\endgroup$ – Andreas Cap Apr 2 '18 at 9:36

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