# Double tangent bundle of manifolds, two contradictory arguments

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 $T(TM)\cong \oplus^3 TM$.

1. 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 $TTM\cong \oplus^3 TM$.
2. 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.

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Your answer 1. is local and coordinate dependent, thus 2. is not well defined. For a chart change (a diffeomorphism) $\phi$ the mapping $TT\phi$ looks as follows: $$(TT\phi)(x,dx,\delta x,d\delta x) = (\phi(x),\phi'(x)(dx),\phi'(x)(\delta x), \phi''(x)(d x,\delta x) + \phi'(x)(d\delta x)).$$ Note that the forth component transforms exactly as Christoffel symbols for a linear connection do.
Thanks Prof Michor, now I see the second argument goes wrong: I can define three maps $TM\to TTM$, but these maps are not morphisms of vector bundles ($TTM$ is not a vector bundle over $M$), thus I cannot obtain a welldefined map $\oplus^3 TM\to TTM$. I have tried the synthetic approach, and it also stuck at $M^{D^2}$ is not a vector bundle. – Ma Ming May 29 '13 at 20:03