# Lie group structure of the tangent bundle

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?
• 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 $Dm: T(G \times G) \simeq TG \times TG \to TG.$ Commented Mar 27, 2018 at 12:26
• @AnthonyCarapetis I have not yet thought of this, thank you for that point of view! Commented Mar 27, 2018 at 12:31
• 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. Commented Mar 27, 2018 at 12:45
• @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). Commented Mar 27, 2018 at 12:52

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.

• 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? Commented Apr 1, 2018 at 8:40
• 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. Commented Apr 2, 2018 at 9:36

I think it's easiest to see the structure geometrically using a bit of synthetic differential geometry. Here, the tangent bundle is a function space

$$TG:=\operatorname{hom}(D,G)$$

where $$D := \{ \delta \in \mathbf{R} \mid \ \delta^2=0 \}$$. The $$\mathbf{R}$$ here is the ring of real numbers but augmented with nilpotent infinitesimals. So tangent vectors to $$X \in T_g G$$ are literally (infinitesimal) curves $$X: D \to G$$ with $$X(0)=g$$.

Now the group multiplication $$TG \times TG \to TG$$ is just pointwise multiplication along the curves, i.e.

$$(X \cdot Y)(\delta) = X(\delta) \cdot Y(\delta)$$

If your Lie group $$G$$ is a matrix Lie group then because it is embedded in the space of $$n\times n$$ matrices $$\operatorname{Mat}(n,\mathbf{R})$$ we can apply the Kock-Lawvere identity to write

$$X(\delta) = X(0) + \delta X'(0)$$

where $$X'(0) \in \mathfrak{g}$$. This is just the Taylor series, which has only terms up to first order because $$\delta^2 =0$$.

So the multiplication in your tangent group can be written $$(X\cdot Y)(\delta) = (X(0) + \delta X'(0)) \cdot (Y(0) + \delta Y'(0))= X(0)Y(0) + \delta \left( X(0)Y'(0) + X'(0)Y(0) \right).$$

You can do the same thing in the usual setting without resorting to SDG, but it all looks more messy with limits all over the place and tangent vectors only being equivalence classes of curves is annoying.

For a more classical approach, see Natural Operations in Differential Geometry by Kolar, Michor, & Slovak: