# Universal property of the tangent bundle

If $X$ is a scheme (over some base scheme, but which I will ignore) its tangent bundle $T(X)$ is defined as the relative spectrum of the symmetric algebra of its sheaf of differentials. Combining the universal properties of these three constructions, we get a universal property of $T(X)$, namely: Defining $U[\epsilon] := U \times \text{Spec}(\mathbb{Z}[\epsilon]/\epsilon^2)$, we have a canonical bijection

$Hom(U,T(X)) \cong Hom(U[\epsilon],X)$

Thus $T(-)$ is right adjoint to $(-)[\epsilon]$.

Question Is there a similar universal property of the tangent bundle of a manifold?

Here a manifold is assumed to be smooth and paracompact. I doubt that we can literally translate it, because $U[\epsilon]$ is $U$ as a topological space, but the structure sheaf now has the nilpotent $\epsilon$. My motivation comes from the observation that in every construction of the tangent bundle of a manifold I know, some nasty calculations with charts have to be made. I want to avoid this with the help of a universal property, proving that $T(\mathbb{R}^n)$ exists and then formally deducing the existence of $T(M)$ for every manifold. I know this can be done without using a universal property and that this would possibly not be the best construction or characterization of the tangent bundle, but it hopefully avoids irrelevant choices of charts. A similar question was asked here and here, but this seems to go in another direction.

• I think some of these issues are considered in "synthetic differential geometry"... – Qfwfq Feb 13 '11 at 0:11
• @Mariano: Sure, but often universal properties imply concrete descriptions. – Martin Brandenburg Feb 13 '11 at 0:54
• @Martin: I don't understand why you want to use a universal property to define the tangent bundle when the following suffices: derivations on $C^{\infty}(X)$ are the same thing as ring homomomorphisms $C^{\infty}(X) \to \Bbb R[\epsilon]/\epsilon^2$ and these correspond to vector fields. Likewise, if $p \in X$ is a point, then derivations on germs of smooth functions at $p$ are the same thing as ring homomorphisms from these germs to $\Bbb R[\epsilon]/\epsilon^2$ and these correspond to the space of tangent vectors at $p$. No coordinates are used here. – John Klein Feb 13 '11 at 2:59
• @Harry: I was addressing Martin's motivation. He was looking for a coordinate free description of the tangent bundle of a smooth manifold: "My motivation comes from the observation that in every construction of the tangent bundle of a manifold I know, some nasty calculations with charts have to be made." – John Klein Feb 13 '11 at 4:26
• Oh, that's not true. The tangent bundle is the total space of the tangent sheaf, which is just the sheaf of $\mathbf{R}$-derivations on the structure sheaf. – Harry Gindi Feb 13 '11 at 6:09

The definition that Martin mentions comes close to the definition of a tangent vector which I learnt as an undergraduate.

"Definition: A geometric tangent vector is an equivalence class of germs of smooth maps $\mathbb{R} \to M$ at $0$, where two germs are equivalent if their first order jets at $0$ agree."

More generally, we could say that for any test manifold $U$, a smooth map $U \to TM$ is an equivalence class of germs of maps $U \times \mathbb{R} \to M$ around $U \times \{0\}$, where two germs are equivalent if their first order jets in $\mathbb{R}$-direction are equal.

To make sense out of the equivalence relation, we need, horribile dictu, a small computation in local coordinates with the chain rule. It takes some work to show that the functor $U \mapsto \{ \text{germs of maps} U \times \mathbb{R} \to M\}$ is representable by a vector bundle $TM$. Likewise, one needs some local computations and arguments with charts for that.

The problem of course is that there does not exist a manifold $\ast[\epsilon]$ such that maps $\ast[\epsilon] \to M$ correspond to tangent vectors of $M$. So the functor $M \mapsto TM$ does not have an adjoint. It might be a good idea to use supermanifolds instead. Remeber that there is a $(0,1)$-dimensional supermanifold with function algebra $\mathbb{R} [\epsilon]/ \epsilon^2$. However, I am not expert enough to tell you something more specific about that.

I am deeply convinced that it is indeed necessary to invoke coordinates in some form and that the ''nasty'' computations with the chain rule and the fundamental theorem of calculus are absolutely crucial. If we wish to define a structure in which we can talk about derivatives of maps between manifolds, the most basic properties of the derivative of maps between euclidean spaces ought to play a role. One can of course use mathematical high-tech weaponry of all sorts to hide the local charts carefully. I will not try to do so. Anyway, if you wish to create the tangent bundle not as an object in a functor category, but as a manifold, you need to check that it is a manifold, i.e. you need charts at some point.

The construction of the tangent bundle immediately generalizes to a construction of the frame bundle $Fr(M)$ instead. Because it is so funny, we immediately define the higher frame bundles $Fr^k(M)$.

Definition: a $k$-frame is an equivalence class of germs of local diffeomorphisms $\mathbb{R}^n \to M$, where two germs are equivalent if their $k$-th order jets coincide. More generally, for a manifold $U$, let $Fr^k M (U)$ be the set of germs of smooth maps $U \times \mathbb{R}^n \to M$, which are diffeomorphic in $\mathbb{R}^n$-direction. Equivalence is defined by equality of $k$-jets."

Again, the chain rule is needed to justify the definition of the equivalence relation. Now we define the jet groups $J^k (n)$. Consider the power series ring $R:=\mathbb{R}[[x_1,\ldots ,x_n]]$ and let $G$ be the group of ring automorphisms. By the action on Taylor expansions, we get a homomorphism $Diff(\mathbb{R};0) \to G$. Let $I \subset R$ be the unique maximal ideal. Clearly, the group $G$ preserves the filtration $R \supset I \supset I^2 \supset \ldots$. Hence it acts on the finite-dimensional vector spaces $R / I^{k+1}$. The $k$th jet group $J^k (n)$ is the quotient of $G$ by the kernel of its action on $R / I^{k+1}$. This is easily seen to be a linear algebraic group. There are obvious maps $J^{k+1} (n)\to J^k (n)$ and an equally obvious isomorphism $J^1 (n) \cong GL_n (\mathbb{R})$. The extensions have nilpotent kernel and they are split (take derivatives of polynomial diffeomorphisms), but the splitting is not natural (it is natural with respect to linear maps $\mathbb{R}^n \to \mathbb{R}^n$, but not more generally).

It is easy to see that $U \mapsto Fr^k (M)(U)$ is a torsor over the group $map (U; J^k (n))$ (compose with germs of diffeomorphisms).

The functor $U \mapsto Fr^k (M)(U)$ is representable by a $J^k (n)$-principal bundle. This is done in charts and then by gluing. Again, the functor $Fr^k (M)$ certainly does not have an adjoint.

The second link that Martin gave alludes to an axiomatic characterization of the tangent bundle (and the higher frame bundles as well). It does not quite fit to the functor-of-points ideology, but I think still very useful, because it axiomatizes the gluing and hence I say some words about it as well.

Let $C_n$ be the category of smooth manifolds of dimension $n$ as objects and local diffeomorphisms as morphisms.

A natural fibre bundle on $C_n$ is the following set of data: For each $M \in \operatorname{Ob} (C_n)$, there is a smooth fibre bundle $F_M \to M$ and for each map $f:M \to N$ in $C_n$, there is a bundle map $F_M \to F_N$; plus some obvious functoriality properties. There is also a canonical notion of a morphism of natural fibre bundles. Let $F(0)$ be the fibre of $F_{\mathbb{R}^n} \to \mathbb{R}^n$ at $0$. There is an action of $Diff(\mathbb{R}^n;0)$ (diffeomorphisms fixing the origin) on $F(0)$. Here is an axiomatic characterization of the tangent bundle:

''THEOREM: The tangent bundle is the unique natural fibre bundle on $C_n$, such that $F(0)=\mathbb{R}^n$, with the action of $Diff(\mathbb{R}^n,0)$ given by the first derivative $Diff(\mathbb{R}^n,0) \to GL_n (\mathbb{R})$.''

It is clear that the tangent bundle satisfies these properties, and here is a sketch of uniqueness, in other words, that the tangent bundle is determined by these properties. Let $F_M \to M$ be a natural bundle. I show that it is determined, up to natural isomorphism, by the action on $F(0)$.

First restrict to $M = V = \mathbb{R}^n$. We know that there is an action of the diffeomorphism group $Diff(V)$ on $F_V$, covering the action on $V$. Using the translations $T_x (v):= v +x$, we get a trivialization $F_V \times V \times F(0)$. Let $x \in V$, $f \in Diff(V)$. The action $f:F(x) \to F(f(x))$ is given in this trivialization by the action of $T_{-f(x)} \circ f \circ T_x \in Diff(V;0)$ on $F(0)$, which is known by assumption. This argument shows that $F(0)$ plus the action determines $F_V$ completely.

By restriction to open subsets and naturality, the restriction of $F$ to the full subcategory $O_n$ of manifolds diffeomorphic to some open subset of $V$ is uniquely determined.

Let $M$ be an arbitrary manifold and let $U(i)_{i \in I}$ be the maximal atlas. It can be written as a diagram $U : J \to O_n$, for some indexing category $J$ (take the intersection of the charts into account). The diagram has a colimit in $C_n$, namely $M$ (of course not all colimits in $C_n$ exist). Likewise, we get a diagram $j \mapsto F_{U(j)}$, whose colimit is $F$. Thus a natural bundle is completely determined, up to natural isomorphism, by the action of $Diff (V,=)$ on the fibre $F(0)$.

The existence question for general natural fibre bundles is a bit subtler, for several reasons. There is a theorem by Palais and Terng asserting that the action on $F(0)$ automatically has finite order (it factor through $j^k (n)$ for some $k$). Also, you have to take into account the topology of the diffeomorphism group. But for the tangent bundle, these subtleties do not arise: it is known what the tangent bundle of an open subset of $\mathbb{R}^n$ has to be and how diffeomorphism act on that. The colimit procedure then produces the tangent bundle. The higher frame bundles can be constructed in a similar way.

• Great! This answers my question in the first paragraphs (universal property of $TM$), but also gives more very interesting insights. – Martin Brandenburg Feb 13 '11 at 16:44
• Some ''philosophical'' afterthoughts: there are two fundamental structures in manifold calculus, and a proper understanding requires both. The first, close in spirit to algebraic geometry, is the Taylor expansion (reflected by the derivation interpretation of tangent vector fields). The second is the homogeneity of manifolds (reflected by the geometric definition of a tangent vector). It is that homogeneity that allows us to talk about Diff-invariant structures and to use groups. – Johannes Ebert Feb 18 '11 at 0:21
• Why don't you consider also differential maps between smooth manifolds as morphisms? – C-Star-W-Star Mar 12 '14 at 20:30
• How does a smooth map give rise to a bundle map (functorial construction)? – C-Star-W-Star Mar 12 '14 at 20:34

The algebro-geometric definition does translate, almost literally. John Klein's comment to Martin's question provides one way of doing the translation.

Let $X$ be a manifold. Define a contravariant functor $F$ on manifolds by taking $F(Y)$ to be the set of pairs $(f, \delta)$ where $f : Y \rightarrow X$ is a morphism of manifolds and $\delta : C^\infty(X) \rightarrow C^\infty(Y)$ is a derivation with respect to the module structure on $C^\infty(Y)$ induced from $f$. Then $F$ is representable by $TX$.

Alternately, you may think of an element of $F(Y)$ as an extension of $f : Y \rightarrow X$ to a map of ringed spaces $Y[\epsilon] \rightarrow X$ where $Y[\epsilon]$ is the space $Y$, ringed by $C^\infty(Y)[\epsilon] / (\epsilon^2)$.

• $\delta$ is defined globally, this doesn't fit to the scheme picture (of course I know that on manifolds we can often encode local as global data, using paracompactness). – Martin Brandenburg Mar 13 '14 at 0:23
• You can define $\delta$ as a derivation of sheaves of rings, if you prefer. This is implicit in the second definition above. – Jonathan Wise Mar 13 '14 at 1:42