# Smooth manifolds as idempotent splitting completion

The nlab has a particularly interesting thing to say about the category of smooth manifolds: it is the idempotent-splitting completion of the category of open sets of Euclidean spaces and smooth maps.

After proving this, the following excerpt from a paper of Lawvere (cited below) is given.

“This powerful theorem justifies bypassing the complicated considerations of charts, coordinate transformations, and atlases commonly offered as a ”basic“ definition of the concept of manifold. For example the $2$-sphere, a manifold but not an open set of any Euclidean space, may be fully specified with its smooth structure by considering any open set $A$ in $3$-space E which contains it but not its center (taken to be $0$) and the smooth idempotent endomap of $A$ given by $e(x)=x/|x|$. All general constructions (i.e., functors into categories which are Cauchy complete) on manifolds now follow easily (without any need to check whether they are compatible with coverings, etc.) provided they are known on the opens of Euclidean spaces: for example, the tangent bundle on the sphere is obtained by splitting the idempotent $e'$ on the tangent bundle $A\times V$ of $A$ ($V$ being the vector space of translations of $E$) which is obtained by differentiating $e$. The same for cohomology groups, etc.” (Lawvere 1989, p.267)

Unfortunately the excerpt is not enough for me to understand neither the significance nor the idea behind the theorem, so I am looking for detailed, hand-holding explanations of as many parts of it as possible.

• How does this theorem justify bypassing the considerations of charts, atlases, etc?
• What are the details of the sphere example? How is the smooth structure specified by an open set containing it along with $x/|x|$?
• Why are all general constructions in fact functors into cauchy complete categories?
• What are some examples of general constructions and how do they follow easily? How does this approach circumvent messing with covers etc?
• What is meant by "the same for cohomology groups"?

Reference: F. William Lawvere, Qualitative distinctions between some toposes of generalized graphs, Contemporary Mathematics 92 (1989), 261-299.

• For starters, splitting idempotents is an absolute colimit: it is preserved by any functor whatsoever. So a functor out of a category $C$ into an idempotent complete category $D$ is the same as a functor out of the idempotent completion of $C$ into $D$. Hence if you want to know how any functor whatsoever behaves on smooth manifolds (e.g. cohomology) it suffices to know how it behaves on open subsets of $\mathbb{R}^n$. – Qiaochu Yuan Nov 5 '15 at 8:18
• (For the example of cohomology this isn't particularly compelling because it's already clear that you can reduce to the case of open subsets of $\mathbb{R}^n$ by embedding a smooth manifold into a large enough $\mathbb{R}^n$ and taking a small enough open neighborhood of it, by the tubular neighborhood theorem.) – Qiaochu Yuan Nov 5 '15 at 8:26
• Here's the article. I access it from my university, I don't know if it is accessible from anywhere. You can always find a way to contact me if needed... – Pierre Cagne Nov 5 '15 at 10:32
• It sounds like this is a categorical encoding of the uniqueness of tubular neighbourhoods theorem in a few different contexts. I think that answers your first two bullet-point questions. I suppose if you are concerned with the category of smooth manifolds this could be a useful result. – Ryan Budney Nov 26 '15 at 5:25
• I don't really know anything about this, but it sounds like this the theorem states that the category of smooth manifolds is the Karoubi envelope of the category of smooth Cartesian spaces. Maybe looking into the Karoubi envelope will be helpful? – ಠ_ಠ Mar 12 '17 at 23:45

The theorem is the following: (from 1.15 of here)

• Theorem: Let $M$ be a connected manifold and suppose that $f:M\to M$ is smooth with $f\circ f= f$. Then the image $f(M)$ of $f$ is a submanifold of $M$.

Proof: We claim that there is an open neighborhood $U$ of $f(M)$ in $M$ such that the rank of $T_yf$ is constant for $y\in U$. Then by the constant rank theorem 1.13 of loc.cit. the result follows. For $x\in f(M)$ we have $T_xf\circ T_xf = T_xf$; thus $\text{image} T_xf = \ker (Id-T_xf)$ and $\text{rank} T_xf + \text{rank} (Id-T_xf) = \dim M$. Since $\text{rank} T_xf$ and $\text{rank} (Id-T_xf)$ cannot fall locally, $\text{rank} T_xf$ is locally constant for $x\in f(M)$, and since $f(M)$ is connected, $\text{rank} T_xf = r$ for all $x\in f(M)$. But then for each $x\in f(M)$ there is an open neighborhood $U_x$ in $M$ with $\text{rank} T_yf\geq r$ for all $y\in U_x$. On the other hand $$\text{rank} T_yf = \text{rank} T_y(f\circ f) = \text{rank} T_{f(y)}f\circ T_yf\leq \text{rank} T_{f(y)}f =r$$ since $f(y)\in f(M)$. So the neighborhood we need is given by $U = \bigcup_{x\in f(M)}U_x$.

This result can also be expressed as: `smooth retracts' of manifolds are manifolds. If we do not suppose that $M$ is connected, then $f(M)$ will not be a pure manifold in general; it will have different dimensions in different connected components.

Consequences: 1. The (separable) connected smooth manifolds are exactly the smooth retracts of connected open subsets of $\mathbb R^n$'s. 2. A smooth mapping $f:M\to N$ is an embedding of a submanifold if and only if there is an open neighborhood $U$ of $f(M)$ in $N$ and a smooth mapping $r:U\to M$ with $r\circ f=Id_M$.

Proof: Any manifold $M$ may be embedded into some $\mathbb R^n$; see \nmb!{1.19} below. Then there exists a tubular neighborhood of $M$ in $R^n$ , and $M$ is clearly a retract of such a tubular neighborhood. For the second assertion we repeat the argument for $N$ instead of $\mathbb R^n$.

# Edited and extended:

Now to your questions, as I understand them: You can reconstruct the category of smooth manifolds and smooth mappings (connect, or not connected but then not pure) as follows: Objects are $(U,f)$ with $U$ open in some $\mathbb R^n$ and $f:U\to U$ smooth with $f\circ f= f$. Morphisms $h:(U,f)\to (V,g)$ are smooth maps $h:U\to V$ with $h\circ f = g\circ h$. A point of the manifold $(U,f)$ is any $x\in U$ with $x=f(x)$.

Then you have to describe diffeomorphisms and get rid of the redundancy in the description of a manifold. Note that even in the classical sense, if you write a manifold $M$, you think of it either with a special atlas, or with the atlas comprised of all possible smooth charts, or $\dots$ Somehow, we mentally identify diffeomorphic manifolds. Let me try:

The identity morphism is any $\ell:(U,f)\to (U,f)$ with $f\circ \ell = f$.

A diffeomorphism $h:(U,f)\to (V,g)$ is one which admits $\ell:(V,g)\to (U,f)$ with $f\circ \ell\circ h =f$ and $g\circ h\circ \ell = g$.

The tangent bundle of a manifold $(U,f)$ is then just $T(U,f) = (TU, Tf)$ as a manifold. For the vector bundle structure note that $TU = U \times \mathbb R^n$, and for a point $x$ in $(U,f)$, i.e., $x\in U$ with $f(x)=x$, we have $T_xf\circ T_xf = T_xf$ a projection in $\mathbb R^n$ whose image is the fiber of the tangent bundle.

I hope that I did not overlook anything.

• I didn't think the OP was seeking a proof of the theorem (which is nice for you to provide anyway; I added a remark in the nLab linking to your monograph), but something more pedagogical along the lines you began to discuss in the last two paragraphs. – Todd Trimble Nov 26 '15 at 15:21
• @ToddTrimble I am indeed looking for something more pedagogical. Prof Michor, could you continue the last two paragraphs with more details? – Arrow Nov 26 '15 at 22:18