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In order to capture the various flavors of cobordism into one concept, the notion of a "cobordism category" is introduced. This is an essentially small category $C$, together with finite coproducts, an initial object, and an additive functor $\partial$ satisfying some properties. Of course, the idea is that one should think of the objects of $C$ as manifolds of one sort or another and $\partial$ as taking the boundary. Indeed, in Tom Weston's notes on the subject he immediately restricts his attention to cobordism categories of "$(B,f)$" manifolds, which are a special type of manifold.

The question is: What are some examples of cobordism categories $(C, \partial, i)$, where the objects of $C$ are not manifolds of some kind?

The only example I know of is motivic cobordism. I'm hoping there are others!

(For the reader's convenience, here is the definition:

A cobordism category is an essentially small category $C$ with finite coproducts (including an initial object $0$), equipped with a coproduct-preserving functor $\partial: C \to C$ and a natural transformation $i: \partial \to 1_C$, such that $\partial\partial c \cong 0$ for every object $c$.)

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Back in the 70's and 80's there was quite a lot of work on various Steenrod realization problems where you replace singular bordism by bordism of various more degenerate objects -- stratified spaces with various restrictions on the singular strata. $\mathbb Z_k$-manifolds were one popular example. You could turn these kinds objects into cobordism categories. See also Buoncristiano, Rourke and Sanderson's work, etc. – Ryan Budney Mar 26 '11 at 20:00
Vaguely related:… – Qiaochu Yuan Mar 26 '11 at 20:08
Dylan, could you please add the precise definition of a cobordism category? – Martin Brandenburg Mar 26 '11 at 21:51
Thanks for adding the definition and correcting the notation, Todd! – Dylan Wilson Mar 29 '11 at 3:24
Also, Ryan, that book looks really really cool!! – Dylan Wilson Mar 29 '11 at 3:24

3 Answers 3

There is actually a reasonable amount of literature on the subject.

Check out Khovanov's $sl_3$-homology and the subsequent work by Marco Mackaay and Pedro Vaz * * *, Hao Wu, and Scott Morrison and Ari Nieh. The objects are webs, the morphisms are foams. The webs have trivalent vertices and foams are modeled on the dual complex of a tetrahedron. I also liked a paper by Serge Natanzon on Network TQFT. You an find all these papers on the Arxiv.

Early on, Frank Quinn made some forays into TQFT for $2$-complexes with the idea of detecting counterexamples to the Andrews Curtis conjecture.

There are older papers of John Baez, John Barret, and then lots of papers in the physical literature. The Landau-Ginzberg approach to quantum gravity involves webs and foams.

All of this work can be understood as being about $3+1$-dimensional TQFT.

In the $1+1$ case, TQFT's turned out to be Frobenius algebras in disguise. In the $2+1$ case, TQFT's have to do with spherical categories. The corresponding correspondence for $3+1$ has yet to be understood. I would conjecture that on the way to understanding $3+1$-dimensional TQFT, a suitable and uniformly accepted theory of TQFT based on webs and foams will be established.

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Naive question: in what sense is the correspondence for the $3 + 1$ case not settled by Lurie's proof of the cobordism hypothesis? – Qiaochu Yuan Mar 27 '11 at 16:45
@Qiaochu: this is "non-extended" TQFT, which is far more complicated. – Oscar Randal-Williams Mar 27 '11 at 18:16
As far as I know there's no reason not to expect the TQFT for Khovanov Homology to be extended. Nonetheless, working out the fully dualizable objects in some specific 4-category is hard work. (Not to mention figuring out which 4-category you want and proving that it's really a 4-category!) – Noah Snyder Mar 29 '11 at 19:41
Are you aware of any cobordism categories that use (combinatorial) simplicial complexes or simplicial sets? – Turion Oct 20 '14 at 11:00
I am not. I would guess though that you would want your simplicial sets to be at the very least pseudomanifolds, as otherwise the notion of boundary would too loose to be interesting. There must be some notion of cobordism for algebraic sets. – Charlie Frohman Oct 23 '14 at 14:30

Here is one type of example, basically inspired by the manifold-type examples but so general that they are not actually categories of manifolds. Let $B$ be any small category with finite coproducts, and let $C$ be the category of diagrams of shape

$$X \to Z \leftarrow Y$$

in $B$. Coproducts in $C$ are given pointwise by coproducts in $B$. Define

$$\partial(X \to Z \leftarrow Y) = (0 \to X + Y \leftarrow 0)$$

with the obvious extension to morphisms. It is quite clear that $\partial$ preserves coproducts and that $\partial^2 \cong 0$. Also there is a canonical natural transformation $i: \partial \to 1_C$, whose component at the object $X \to Z \leftarrow Y$ is the unique one where the arrow in the middle is the map $X + Y \to Z$ whose restrictions to $X$ and $Y$ are the given arrows $X \to Z$, $Y \to Z$ of the object.

Edit: As indicated in a comment below, it is simpler to consider instead the arrow category $B^{\mathbf{2}}$ as a cobordism category where $\partial(f: X \to Y) = (0 \to X)$. But a much more compelling reason to consider this construction is that, if I'm not mistaken, it satisfies a universal property as follows.

Let $\text{Coprod}$ be the 2-category of categories with finite coproducts and coproduct-preserving functors (and transformations between them); let $\text{Cobord}$ be the 2-category of cobordism categories and cobordism-preserving functors. There is an evident forgetful 2-functor

$$U: \text{Cobord} \to \text{Coprod}$$

In the other direction, the arrow-cobordism category construction defines a 2-functor

$$\text{Arr}: \text{Coprod} \to \text{Cobord}$$

and this is in fact a right 2-adjoint of the forgetful functor. (Notation: $U \dashv \text{Arr}$.) Thus $\text{Arr}(B) = B^{\mathbf{2}}$ defines the cofree cobordism category generated by a category with coproducts $B$.

In more detail, the unit $\eta: 1_{\text{Cobord}} \to \text{Arr} \circ U$ is defined componentwise as a cobordism-preserving functor $\eta C: C \to C^{\mathbf{2}}$ which at the object level takes an object $c$ to the object $i c: \partial c \to c$ in $C^{\mathbf{2}}$. (It is instructive to check the details of this.) The counit $\varepsilon: U \circ \text{Arr} \to 1_{\text{Coprod}}$ is defined componentwise as a coproduct-preserving functor $\varepsilon D: D^{\mathbf{2}} \to D$ which at the object level takes an object $g: d_1 \to d_2$ to the object $d_2$. It is reasonably straightforward to check that there are coherent isomorphisms

$$(U \stackrel{U \eta}{\to} U \circ \text{Arr} \circ U \stackrel{\varepsilon U}{\to} U) \cong 1_U$$

$$(\text{Arr} \stackrel{\eta \text{Arr}}{\to} \text{Arr} \circ U \circ \text{Arr} \stackrel{\text{Arr} \varepsilon}{\to} \text{Arr}) \cong 1_{\text{Arr}}$$

that make $\text{Arr}$ the right 2-adjoint of the forgetful functor $U$.

I think there's something deeper going on here than I understand at the present time.

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Well, heck, why don't I just take the arrow category instead and define $\partial (f: X \to Y) = (0 \to X)$? That works too. I guess the reason is that I was thinking also of composition of cobordisms, and how that generalizes to taking pushouts of cospans. – Todd Trimble Mar 27 '11 at 5:38
Isn't that a subcategory of the cospan category anyway? (Those are just the cospans with $Y = 0$, right?) – Qiaochu Yuan Mar 27 '11 at 5:47
@Qiaochu: yes. Either is a simple example, but I thought the arrow category was simpler. My guess is that it's well-known. – Todd Trimble Mar 27 '11 at 6:23
Thanks! Does the notion of bordism equivalence that this gives have any use for common problems? Like, in some (specific) non-geometric category, what properties of a map are bordism invariant in this case that we care about? – Dylan Wilson Mar 29 '11 at 3:32
I'll need some time to think about your question, Dylan. In the meantime, I've added some possibly more meaningful observations to my answer. – Todd Trimble Mar 29 '11 at 15:20

One famous (in my field!) example is Witt space bordism. Witt spaces are not manifolds but rather pseudomanifolds (which aren't so far off from manifolds, but they can have singularities). A pseudomanifold is a Witt space if certain local rational intersection homology groups vanish. The bordism group of Witt spaces is important because it turns out to be a geometric model for ko-homology after inverting 2. The original reference is Paul Siegel's thesis: There's a slightly fancier version due to Pardon using "IP spaces" which satisfy integral Poincare duality:

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so this is a geometric model like the "model" of $MU_*X$ where you take manifolds with maps to $X$ and look at the free abelian group modulo cobordisms? (Maybe I mean $MU^*X$) – Sean Tilson Mar 28 '11 at 22:31
Well, I don't think you need to consider free abelian groups directly. The elements of the $n$th Witt bordism group of $X$ are equivalence classes of maps $f:W\to X$ of $n$-dimensional Witt spaces to $X$. Two, say $f_1, f_2: W_1, W_2\to X$ are equivalent if there is an $n+1$ dimensional Witt space with boundary $U$ and a map $F:U\to X$ that restricts to $f_1$ and $f_2$ (with appropriate signs) on the boundary. Then the addition is just disjoint union of spaces and maps, i.e. $f_1+f_2$ is $f_1\amalg f_2: W_1\amalg W_2\to X$. So you don't need to form a free abelian group on generators for this. – Greg Friedman Mar 30 '11 at 23:37

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