Let $G$ be a semisimple Lie group. If $G$ has a KAK decomposition then does the tangent bundle $TG$ admit an analogue where $K$ is replaced with $TK$ and $A$ is replaced with $TA$?
The tangent bundle admits a natural group structure. Write an element of it as $(1 + \varepsilon t) g$ where $t \in \mathfrak g$ and $g \in G$, and $\varepsilon$ formally squares to $0$. We have that:
$$(1 + \varepsilon t)g \cdot (1+\varepsilon t')g' = (1 + \varepsilon (t + gt'g^{-1}))gg' = (1 + \varepsilon(t + \operatorname{ad}_gt'))gg'.$$
So in other words, it has the group structure $G \rtimes_{\phi} \mathfrak g$, where $\mathfrak g$ has the additive group structure and $\phi = \operatorname{Ad}$.
It looks like the answer should be yes, but I don't know what methods could be used to prove this. The KAK decomposition is not in general differentiable, continuous or unique.