I am considering the double tangent bundle $T(TM)$ of manifolds $M$. Locally, if $M=R^d$ then $T(TM)=R^{4d}=\oplus^3 TM$. My attempt is to see whether $T(TM)\cong \oplus^3 TM$ naturally for any $M$. I have two contradictory arguments, of which one says $T(TM)\not\cong \oplus^3 TM$ but the other seems say $T(TM)\cong \oplus^3 TM$.

- If we introduce local coordinates $(x^i)$ on $M$, then $(x^i, dx^i)$ can be local coordinates on $TM$, and $(x^i, dx^i, \delta x^i, d\delta x^i)$ can be local coordinates on $T(TM)$. But the rules for change of coordinates is a bit complicated for $T(TM)$, which involve the second derivatives of transition functions of $(x^i)$. So we can not hope $TTM\cong \oplus^3 TM$.
- From argument above we have three different maps $TM\to TTM$ given by $(x^i, dx^i)\mapsto (x^i, dx^i,0,0)$ or $(x^i, 0,\delta x^i,0)$ or $(x^i,0,0, d\delta x^i)$, thus we have a map $\oplus^3 TM\to TTM$ of vector bundle over $M$. This map is locally an isomorphism, as we saw $T(TR^d)=R^{4d}=\oplus^3 T R^d$, so this is also a globally isomorphism.

To put it in another way. Let $M$ be a manifold and $D=\{x\in R| x^2=0\} $ be the infinitesimal line in synthetic geometry, then $Map(D, M)$ is the tangent bundle, where $Map$ denotes the internal hom. Applying twice we get $Map(D^2, M)$, which is the twiced tangent bundle. The second argument translate to

- $Map(D^2, R)=Map(D, R)\oplus Map(D, R)\oplus Map(D, R)$, so we may replace $R$ by any microlinear superspace, hence $Map(D^2, M)=Map(D, M)\oplus Map(D, M)\oplus Map(D, M)$.

Could you point where did I make mistake(s)? I think the second is wrong, but I can not tell why. Thanks in advance.