I am reading about $2$-categories of bordisms, denoted $Cob_2(n)$. This paper by Lurie (see page 10) states that the objects of such a category are closed $(n-2)$-manifolds and and for $M, N \in Cob_2(n)$, the category of maps $Map_{Cob_2(n)}(M, N)$ has objects given by bordisms $B: M\to N$ and morphisms given by oriented diffeomorphism classes of oriented bordisms $X: B\to B'$. However, I don't understand how to define a bordism $X: B\to B'$ when $B$ and $B'$ are already manifolds with boundary.

For instance, we would like the cylinder $B\times [0, 1]$ to be a bordism from $B$ to itself, but its (unoriented) boundary consists of $B\times\{0\} \cup \partial B \times [0, 1] \cup B\times \{1\}$, not just $B\times \{0\} \cup B\times \{1\}$ as one might prefer. Using this example as inspiration, I have come up with the following attempt at a definition:

A bordism $X: B\to B'$ between bordisms $B, B': M \to N$ is an oriented $n$-manifold with corners together with an orientation-preserving diffeomorphism from $\partial X$ to $$\overline{B} \coprod_{\overline{\partial B}}(\partial B \times [0, 1]) \coprod_{\partial B} B'$$ where we have used the given orientation-preserving diffeomorphism $\partial B\simeq \partial B'$.

Is my definition equivalent to the standard one? Or is there something I'm missing here?

  • $\begingroup$ "The paper": what paper? There are multiple possible definitions for extended bordism categories. $\endgroup$ – Najib Idrissi Sep 6 '15 at 18:16
  • $\begingroup$ @NajibIdrissi Edited to include the source. $\endgroup$ – Alex G. Sep 6 '15 at 18:18
  • 1
    $\begingroup$ Oh, you're starting with the heavy artillery already. This survey by Schommer-Pries is perhaps more readable for a beginner. Also this paper by Freed. $\endgroup$ – Najib Idrissi Sep 6 '15 at 18:20

Yes, you got that right as far as I can tell. More generally given two integers $0 \le k < n$, one can define a $(n-k)$-category $\mathsf{Cob}_k(n)$ of "$n$-dimensional cobordisms extended down to dimension $k$" (in your case, $k=2$). The definition is as follows:

  • Objects are $(n-k)$-dimensional closed oriented manifolds.
  • 1-morphisms $X \to Y$ are cobordisms $X \to Y$, that is $(n-k+1)$-dimensional compact oriented manifolds $M$ together with a diffeomorphism $\phi :\bar{X} \sqcup Y \to \partial M$ (the bar denoting reversing of orientation).
  • Given two cobordisms $(M, \phi)$ and $(N, \psi)$ from $X$ to $Y$, a 2-morphism between the two is a $(n-k+2)$-dimensional compact oriented manifold with corners $Z$, together with a diffeomorphism $$\theta : \bar{M} \cup_{X \sqcup \bar{Y}} \bigl( (\bar{X} \sqcup Y) \times [0,1] \bigr) \cup_{\bar{X} \cup Y} N \to \partial Z.$$
  • 3-morphisms are defined similarly, with manifolds with higher corners (every point has a neighborhood diffeomorphic to an open subset of $\mathbb{R}_+^3 \times \mathbb{R}^{n-k}$), etc.
  • $k$-morphisms are a special case: they are cobordisms as above, thus an $n$-dimensional compact oriented manifold with higher corners together with an identification of the boundary, but you actually need to consider diffeomorphism classes of such manifolds (the diffeomorphisms must be compatible with the boundaries).

The image you need to think about is this one, taken from Freed's paper The cobordism hypothesis (Bull. Amer. Math. Soc. (N.S.), 2013, vol. 50, pp. 57--92, DOI): elementary cobordism This represents a cobordism $I \sqcup I \to I \sqcup I$, where $I = [0,1]$ is the interval. The boundary of $I \sqcup I$ is four points, and you see that the cobordism does contain something of the form $\{\text{four points}\} \times [0,1]$. The source is the "bottom" with the two parallel intervals, and the target is the "top".

  • $\begingroup$ Is the "formula" for the boundary of an $l$-morphism similar to that for $2$-morphisms? I.e. given an $l$-morphism $Z: X \to Y$ between $(l-1)$-morphisms, do we implicitly have a diffeo. $\partial Z \simeq \overline{X} \coprod_{\overline{\partial X}} (\partial X \times [0, 1]) \coprod_{\partial X} Y$? $\endgroup$ – Alex G. Sep 6 '15 at 18:43
  • $\begingroup$ Yes that's it @AlexG. $\endgroup$ – Najib Idrissi Sep 6 '15 at 18:51

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