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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.

For a description of the second tangent bundle of a manifold see 8.12 and 8.13 of here. For a comprehensive description of higher order order tangent bundles etc, viewed as product preserving functors on the category of smooth manifolds, see chapter VIII of here, or the original paper. Thsi description uses Weil algebras and is related to what you are trying to say at the end of your question.

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  • $\begingroup$ 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. $\endgroup$ – Ma Ming May 29 '13 at 20:03
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$T(TM)$ is not isomorphic to $\oplus^3 TM$, it is isomorphic to $\oplus^2 TM$. For example, if $M=\mathbb{R}$, then $TM$ is isomorphic to $\mathbb{R}^2$, and $T(TM)$ is isomorphic to $\mathbb{R}^4$.

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    $\begingroup$ The expression $\bigoplus^2 TM$ usually means the sum as vector bundles (and that is what is meant above), so it has dimension $3n$ if $n$ is the dimension of $M$. But in fact, even if you mean $TM \times TM$ when you write $\bigoplus^2 TM$, that doesn't work either. The bundle $T(TM)$ retracts down to $M$, while $TM \times TM$ retracts to $M \times M$, so for compact $M$ of postiive dimension, these are not homeomorphic. So I am not sure what you mean here. $\endgroup$ – Ben McKay Jun 3 '20 at 6:36

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