I know that $TM$ is always orientable bundle. But what is $TTM$?
How try to prove that $TTM$ is a orientable bundle on $TM$.
I know that $TM$ is always orientable bundle. But what is $TTM$?
How try to prove that $TTM$ is a orientable bundle on $TM$.
On
As often it is clearer to put the problem in a more general context.
a) Given a smooth vector bundle $\pi:E\to M$ on the differential manifold $M$, there is an exact sequence of vector bundles on $E$: $$0\to T_v(E)\to T(E)\stackrel {d\pi}{\to }\pi^*(T(M))\to 0 $$ Here $T_v(E)$ is the bundle formed by those tangent vectors which are "vertical", i.e. tangent to the fibers of $\pi$.
Recalling that for a finite-dimensional vector space $V$ (seen as a manifold) we have for each $a\in V$ a canonical isomorphism $T_a(V)=V$ we obtain the canonical isomorphism $$T_v(E)=\pi^*(E)$$ and the above exact sequence becomes the fundamental exact sequence of vector bundles on $E$ : $$0\to \pi^*(E)\to T(E)\stackrel {d\pi}{\to }\pi^*(T(M))\to 0 \quad(\bigstar) $$ Like all exact sequences of smooth vector bundles on a manifold the above exact sequence splits, alas non canonically, and we obtain an isomorphism of vector bundles on $E$: $$T(E)\cong \pi^*(E)\oplus \pi^*(T(M))$$
b) Specializing to the case $E=T(M)$ we get an isomorphism of vector bundles on $T(M)$: $$T(T(M))\cong \pi^*(T(M))\oplus \pi^*(T(M))$$
c) To conclude it suffices to recall that given any vector bundle $F\to X$ on a manifold $X$, the vector bundle $F\oplus F$ is orientable, independently of whether the bundle $F$ is or is not orientable.
This shows that $T(T(M))$ is an orientable vector bundle on the manifold $T(M)$, which exactly means that $T(M)$ is an orientable manifold, independently of whether $M$ is or is not an orientable manifold.
Remark
The exact sequence $(\bigstar)$ is beautiful, powerful and easy to derive.
Although it is certainly well known to specialists, it is regrettable that it is not mentioned in any book I know.
I would be very grateful to a user who could indicate a reference for it.
No, $TM$ is an orientable bundle if and only if $M$ is an orientable manifold.
You mean $T(TM)$. And that is always an orientable bundle: Even when $M$ is non-orientable, $T_{(x,v)}(TM) \cong T_xM \oplus T_xM$ has a canonical orientation.