0
$\begingroup$

I read here the following:

Given a manifold $M$, the double of $M$ is the boundary of $M \times [0,1]$. This gives doubles a special role in cobordism.

  1. What is this special role?
  2. Moreover, is it true that the double of a manifold $M$ with boundary (which should be $\partial M \times [0,1] \cup M \times \{0,1\}$) is the boundary of $M \times [0,1]$?
$\endgroup$
2
  • 2
    $\begingroup$ 1. $M \times [0,1]$ is the identity bordism from $M$ to $M$. Equivalently, $M \times [0,1]$ is a nullbordism of the double of $M$. $\endgroup$ Apr 3 at 1:11
  • $\begingroup$ Thank you very much $\endgroup$ Apr 3 at 1:21

1 Answer 1

1
$\begingroup$
  1. Yes, $\partial(M \times [0,1]) = \partial M \times [0,1] \cup M \times \{0, 1\}$. Picture a standard cylinder, $D^2 \times [0,1]$, whose boundary is a cylindrical tube $S^1 \times [0,1]$ together with the disks on either end, $D^2 \times \{0, 1\}$
$\endgroup$
2
  • $\begingroup$ But how can I see it is the double of $M$? Don't I need to identify the two boundaries of $M$ in the definition of double of a manifold? $\endgroup$ Apr 3 at 1:30
  • 1
    $\begingroup$ @CrashBandicoot Correct, to be precise this is homotopy equivalent to the double. The cylinder part $\partial M \times [0,1]$ is like a collar neighborhood of the gluing of the two copies along the boundary. If you contracted that down ($\partial M \times [0,1] \sim \partial M \times \{0\}$) then you get the double by your definition. $\endgroup$
    – nkm
    Apr 5 at 16:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .