# In how many ways can an iterated tangent bundle (T^k)M be viewed as a fibre bundle over (T^(k-1))M?

Let M be a smooth manifold. The double tangent bundle, TTM,can be viewed as a fibre bundle over TM in two ways, with the projection maps given by T_πM (i.e. the derivative of the projection from TM to M) and π_TM (i.e. the standard projection onto TM). There is also a canonical involution K:TTM->TTM, which basically flips the inner two coordinates. It turns out to be a diffeomorphism and a natural transformation from T^2 to itself. In fact, Tπ_M and π_TM are related through composition with K.

My question is, what happens if you keep taking higher and higher tangent bundles? Evidently, you should have more ways to write them as fibre bundles over the lesser tangent bundle. Intuitively, to me at least, it feels that there are k ways to view (T^k)M as a fibre bundle over (T^(k-1))M: inductively, there is the derivative of all the previous projection maps, and the canonical way. Are there any others?

Is there always a diffeomorphism (or whatever the suitable notion is here) that will take one projection map to the other as in the case of the canonical involution in k=2? If not, what goes wrong, and does it have any significance?

• Incidentally, perhaps you want \&pi; not \&pi_? – Theo Johnson-Freyd Oct 23 '09 at 4:53

If we use the notation $(TM, p_M, M)$ for the tangent bundle of any manifold $M$, then you are right to think that $T^{\ k}M$ has $k$ natural vector bundle structures over $T^{\ k-1}M$ and so on down to $M$, making a diagram which is a $k$-dimensional cube. Such a structure is a $k$-fold vector bundle (See articles by Kirill Mackenzie) and $T^{\ k} M$ is a particularly symmetrical one. An easy way of writing the $k$ bundle maps would be

$$T^{\ a}(p_{T^{\ b} M})$$

for $a+b+1 = k$, where this means that we are taking the $a$-th derivative of the tangent bundle projection $T\ (T^{\ b}M)\to T^{\ b}M$, yielding a map now from $T^{\ k}$ to $T^{\ k-1}$.

To see the symmetries of $T^{\ k} M$ it is more convenient to describe the functor in a direct way rather than as a $k$-fold composition -- just as you might think of tangent vectors as infinitesimal curves, you can think of points in $T^{\ k} M$ as infinitesimal maps of a unit $k$-cube into $M$. The restrictions to the $k$ faces of the cube (going through the origin) gives your $k$ maps.

From that point of view, it is clear that you can permute the $k$ coordinate axes and get another map of a cube into $M$, so that the functor $T^{\ k} M$ has a $S_k$ group of natural-automorphisms.

Incidentally, to make the above into a definition of $T^{\ k}$, you could do the following:

Consider the fat point $fp$, which you should think of as a space whose smooth functions form the ring $\mathbb{R}[x]/(x^2)$. Then the tangent bundle $TM$ can be thought of as the space of maps from the fat point to $M$, i.e. $TM=C^\infty(fp,M)$. Such maps, by the way, are just algebra homomorphisms from the algebra $C^\infty(M,R)$ to $\mathbb{R}[x]/(x^2)$. You can check that such a map has two components $f_0 + f_1 x$, and that $f_0$ is a homomorphism to $\mathbb{R}$ defining a maximal ideal (i.e. a point $p$ in $M$) and $f_1$ defines a derivation (i.e. a vector at $p$).

In precisely the same way you can consider a cubical fat $k$-point $kfp$ with functions

$$\mathbb{R}[x_1,...,x_k]/(x_1^2,...,x_k^2)$$

and then define $T^{\ k} M = C^\infty(kfp,M)$. Then you can see the symmetries as automorphisms of the above algebra, and the $k$ maps you ask about as homomorphisms $C^\infty(kfp)\to C^\infty\big((k-1)fp\big).$

There are probably more subtle things to be said about these higher iterated bundles but I hope the above is at least correct.

• Thanks, Marco, I just have two questions: 1. Based on what you said, is it then true that any two fibre bundles in the way we've discussed above are naturally equivalent in the appropriate way (fibre bundle isomorphism? Diffeomorphism?)? And a diffeomorphism(?) of T^{k}M is induced from the corresponding element in the natural automorphism group of T^k? 2. I follow your explanation using the fat point, but I don't understand why the tangent bundle is the space of maps from it to M. Is the space just contrived to be have that property or is there a deeper interpretation? – Kirill Levin Oct 25 '09 at 18:13
• 1. the action of the symmetric group S_k preserves the structure of T^k M as a k-fold vector bundle. So, given any edge E_1 of the k-cube (this edge is a vector bundle structure) it is taken to another edge E_2, and the automorphism defines a vector bundle morphism E_1 --> E_2. 2. The fat point is just an infinitesimally small interval in the real line. So, the maps of this into a space are just the "infinitesimal" paths, which are another way of thinking of tangent vectors. Essentially the map may be viewed as an equivalence class of paths where path_1 ~ path_2 iff they agree to 1st order – Marco Gualtieri Oct 25 '09 at 21:31

AMENDED POST: Marco Gualtieri points out a fundamental mistake I made. I retract my assertion that there infinitely many such maps. It looks to me there is a surjective map $T^n M \to T^{k}M$ for each monotone injective map $\lbrace 1,\dots,k\rbrace \to \lbrace 1,\dots, n\rbrace$.

ORIGINAL POST: I believe that "infinitely many" is the answer.

We can think of $TM$ as the effect of applying an equivalence relation (two paths are equivalent if they have equal one jets at the origin) on the space of smooth paths $\Bbb R\to M$, which I'll denote as $M^{\Bbb R}$. Similarly, the $n$-fold iterated tangent bundle $T^nM$ is given by a similar equivalence relation on $M^{\Bbb R^n}$. For any linear embedding $\Bbb R^k \to \Bbb R^n$ there will be a restriction map $T^nM \to T^kM$.
This gives infinitely many maps.

Similarly, for any linear surjection $\Bbb R^n \to \Bbb R^k$ one gets an embedding $T^kM \to T^nM$.

If we use only coordinate inclusions and coordinate projections, then the structure you get is a simplicial object $X.$ with $X_k = T^{k+1}M$ augmented over $M$. This is because $M \mapsto TM$ is a triple (comonad). I've always felt that one could make sense of the de Rham complex from this point of view (somehow).

• I may be missing something, but I think the definition you are using for $T^2M$ is maps from $\mathbb{R}[x,y]/(x^2,xy,y^2)$ to $M$, which is why you see the linear symmetries. But this is not the iterated $T$ functor, which only gives $(\mathbb{R}[x]/(x^2))[y]/(y^2) = \mathbb{R}[x,y]/(x^2,y^2)$. The linear symmetries are lost in the latter case. As for comonads, I wish you would explain what you mean, but the literature in supergeometry (see Severa) says that the de Rham complex is just the functions on the graded manifold T[1]M where the differential comes from a $\mathbb{R}[1]$ action. – Marco Gualtieri Feb 11 '11 at 4:27
• I think you are right. I will modify my post accordingly (and thanks). As to comonad or triple, let me simply refer you to en.wikipedia.org/wiki/… – John Klein Feb 11 '11 at 17:43

More background to Marco Gualtieri's answer can be found in chapter VIII of here, which classifies all product preserving functors on the category of smooth manifolds and smooth mappings, and all natural transformations between them. This result is partially extended to the category of infinite dimensional smooth manifolds in section 31 of here