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Proof of Cobordism Hypothesis

3 Inductive formulation

Recall that there is an map \(\Bord _{n-1}^{X_0,\xi _0} \to \Bord _n^{X,\xi }\), where \(X_0\) is the unit sphere bundle of \(\xi \). Let \(C\) be as in the cobordism hypothesis, and consider a field theory on \(\Bord _{n-1}^{X_0,\xi _0}\) we would like to extend along \(\Bord _n^{X,\xi }\). For each \(x \in X\), the unit disk of \(\xi \) at \(x\), \(D^n_x\) is a bordism from \(\phi \) to \(S^{n-1}_x\), the fibre of \(X_0 \to X\) at \(x\). The sphere \(S^{n-1}_x\) can be broken into two hemispheres, making it the composite \(D_+\circ D_-\). We say that \(D^n_x: \phi \to D_+\circ D_-\) exhibits \(D_+\) as right adjoint to \(D_-\). This is a condition that doesn’t depend on the way we broke up \(S^{n-1}_x\) into hemispheres.
  • Theorem 3.1. Let \(Z_0: Bord_{n-1}^{X_0,\xi _0} \to C\) be symmetric monoidal, and \(C\) an \((\infty ,n)\)-category with duals. Then the following data are equivalent:

    • • An extension \(Z\) of \(Z_0\) to \(Bord_n^{X,\xi }\).

    • • Families of nondegenerate morphisms \(1 \to Z_0(S_x^{n-1})\) parameterized by \(X\).

  • Proposition 3.2. Theorem 3.1 in dimensions \(\leq n\) and Theorem 1.1 in dimensions \(<n\) imply Theorem 1.1 in dimension \(n\).

  • Proof. It suffices to assume that \(X\) is a point, because Theorem 3.1 implies that the left hand side of the cobordism hypothesis sends colimits of spaces over \(BO(n)\) to limits. But spaces over \(BO(n)\) is generated under colimits by a point.

    When \(X\) is a point, \(X_0,\xi \) is \(S^{n-1}\) with its tangent bundle. Thus we need to show that giving the data of \(Z_0\) as well as a nondegenerate morphism \(\eta : 1 \to Z_0(S^{n-1})\) is equivalent to giving an object of \(C\).

    By Theorem 1.1 in dimension \(n-1\), giving the data of \(Z_0\) is equivalent to giving an \(O(n-1)\)-equivariant map from \(O(n)\) to \(C^{\cong }\). The action comes from the fibration \(O(n-1) \to O(n) \to S^{n-1}\). We can break up \(O(n)\) into the fibres at the north and south poles, and everything in between. \(O(n)\) also is acted on by \(O(n-1)\) from the right. While this action doesn’t preserve the fibres, it preserves the north and south pole, and relates fibres along each \(S^{n-2}\) slice of the rest of the sphere. From this description, we get that an \(O(n-1)\)-equivariant map from \(O(n)\) is the same as two \(O(n-1)\)-equivariant maps from \(O(n-1)\) (corresponding to the north pole with the inverse right action and south pole with usual action), and an \(O(n-2)\)-equivariant homotopy from the first to the second conjugated by the map in \(O(n-1)\) reflecting along a direction.

    By Theorem 1.1, this is equivalent to two functors \(Z_- \to C, Z_+ \to C^{op}\) out of \(\Bord _{n-1}^{*}\), where \(C^{op}\) is the opposite on the level of \(n-1\)-morphisms, along with an isomorphism of their restrictions to \(\Bord _{n-2}^{S^{n-2}}\) (the underlying \((\infty ,n-2)\)-categories of \(C\) and \(C^{op}\) agree).

    By Theorem 3.1 in dimension \(n-1\), this is equivalent to giving one functor \(Z':\Bord _{n-2}^{S^{n-2}} \to C\) and a nondegenerate morphism \(f:1 \to Z'(S^{n-2})\) and \(g:1 \to Z'(S^{n-2})\) in \(C\) and \(C^{op}\) respectively. Thus we can think of \(g\) as a morphism \(Z'(S^{n-2}) \to 1\). But the composite \(g \circ f\) is exactly the \(Z_0(S^{n-1})\), and so the data of \(\eta \) and \(g\) are redundant since they exhibit \(f\) as a right adjoint of \(g\), and right adjoints are essentially unique. Thus we are left with just the data of \(f\) and \(Z'\).

    Via Theorem 3.1 again, this is equivalent to the data of functor \(Z_-: \Bord _{n-1}^* \to C\), which by Theorem 1.1 in dimension \(n-1\) is equivalent to just an object of \(C\).