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Let's say we denote by $T^{(n,m)}M$ the vector-bundle of rank $(n,m)$ tensors on a manifold $M$ and by $\Lambda^pM$ the vector-bundle of $p$-forms on $M$. Is there a relationship (perhaps a diffeomorphism to some sort of direct sum expansion) between the various $\Lambda^p M$ (or the various $T^{(n,m)}M$) and iterated $TM$'s or $T^*M$'s? Intuitively, for example, the tensors of rank $2$ are like 2nd order terms in a Taylor expansion -- but so is $TTM$. Using physics notation, where $M$ is parametrized locally by coordinates $q$, $TM$ looks like $(q,\dot{q})$ and $TTM$ look like $(q,\dot{q},\dot{q}',\ddot{q})$ where I use the ' to denote a 2nd copy of the $\dot{q}$. If $M$ is $r$ dimensional then $TM$ has dimension $2r$, $TTM$ has dimension $4r$, etc, and $\Lambda^p M$ has dimension $({r\atop p})+r$. Is there some sort of useful morphism between $\Lambda^pM$ and a direct sum of $T^*M$, $T^*T^*M$, etc? Ditto for the $T^{(n,m)}M$? It feels like there should be some relationship between tensor bundles (or form-bundles) and iterated tangent or cotangent bundles, but I can't seem to find one (or determine an obvious ways to construct a series that even would have the correct dimension). On a similar note, are there any such relationships amongst the iterated $TM$'s and $T^*M$'s themselves? Thanks in advance for any insights.

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Well by definition $\Lambda^p M = T^{(n,0)}(M) / \Sigma_n$ (wrt to the sign action), but probably this is not what you want? –  Martin Brandenburg Jan 8 '13 at 23:28
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The relative tangent bundle to $p: TM\rightarrow M$ is $p^* TM$, so you have an exact sequence $0\rightarrow p^* TM\rightarrow TTM\rightarrow p^*TM\rightarrow 0$. For smooth manifolds it is split, hence $TTM\cong p^*TM\oplus p^*TM$. This is different from, say, (the pullback of) $T^{(2,0)}M$, which is $TM\otimes TM$. –  Pavel Safronov Jan 9 '13 at 6:10

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up vote 7 down vote accepted

The answer is 'basically, no'. The tensor bundles that you list, such as $T^{(n,m)}M$ and its quotients (such as $\Lambda^p(TM)$, etc.), are first order prolongations of $\mathrm{Diff}(M)$, whereas $TTM$ is a second order prolongation of $\mathrm{Diff}(M)$. All of these are examples of functors from the category $\mathsf{Diff}_n$ whose objects are smooth $n$-manifolds and whose morphisms are diffeomorphisms into some $\mathsf{Diff}_m$. For example, the tangent bundle functor $\mathsf{T}:\mathsf{Diff}_n\to \mathsf{Diff}_{2n}$ and the cotangent functor $\mathsf{T^\ast}:\mathsf{Diff}_n\to \mathsf{Diff}_{2n}$ are first order while $\mathsf{T\circ T}:\mathsf{Diff}_n\to \mathsf{Diff}_{4n}$ is second order. For this reason, this composition cannot be written as a first order functor, i.e., it has no natural transformation that makes it equivalent to a first order functor, such as a sum of tensor bundles (which would be a first order functor).

You can see this very concretely in the case $n=1$, if you just write out what is happening in local coordinates.

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Thanks, Robert. I'm a bit weak with Category Theory, so it took me a little digging to understand the prolongation explanation but I think I get the gist and it makes sense to me. In retrospect this may have been a somewhat naive question anyway, because there would be one "copy" of $M$ on the left side of my proposed equation and multiple copies of $M$ on the other side (one from each term) -- so the two wouldn't even be two homotopic in general. It just would have been nice to have some sort of "algebra" of these bundles. –  Kensmosis Jan 10 '13 at 16:31
    
Well, actually, there don't need to be multiple copies of the base manifold; you can always take the fiber product. For example, the fiber product $TM\oplus \Lambda^2(TM)$ is not the product of $TM$ with $\Lambda^2(TM)$. It's the set of pairs $(v,V)$ such that there exists an $x\in M$ with $v\in T_xM$ and $V\in \Lambda^2(T_xM)$. This is a vector bundle over a single copy of $M$. –  Robert Bryant Jan 11 '13 at 13:04

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