# Relating bordism generators in d and d+2 dimensions --- an explicit example

This is an attempt to make my relation between bordism invariants in $$d$$ and $$d+2$$ dimensions, following a previous attempt more explicit. This counts as a different question, since some more specific and tailor-made statement is given here.

Let us consider the mapping the bordism invariants of eq.1 and eq.2 $$\Omega_{O}^{d}(B(PSU(2^n)\rtimes\mathbb{Z}_2)), \tag{eq.1}$$ where the semi-direct product ($$\rtimes$$) is defined here, and $$\Omega_{O}^{d+2}(K(\mathbb{Z}/{2^n},2)). \tag{eq.2}$$ Here $$K(G,2)$$ is the Eilenberg–MacLane space. We can take $$d=3$$ here. We can also take $$n=1$$ here for $$\Omega_{O}^{d}(B(SO(3)\times\mathbb{Z}_2))$$.

• In the case $$d=3$$, we are considering ALL the possible nontrivial (nonzero) maps from the $$d+2$$-manifold (a $$5$$-manifold) $$M^5$$ which is a manifold generator of $$\Omega_{O}^{d}(B(SO(3)\times\mathbb{Z}_2))=\Omega_{O}^{d}(B(O(3)))$$; mapping to a dimensional reduced $$d$$-manifold (a $$3$$-manifold) $$\Sigma^3$$, which is a manifold generator of $$\Omega_{O}^{5}(K(\mathbb{Z}/{2},2))$$.

My questions are:

1. What are precise relations between the manifold generators of $$M^5$$ and $$\Sigma^3$$ that we can say?

2. What are precise relations between the cobordism generators $$S_5$$ and $$S_3$$ (topological terms like invertible TQFTs) we can say that associated to $$M^5$$ and $$\Sigma^3$$ that we can say? Let us say we have the corresponding cobordism generators $$S_5$$ and $$S_3$$. Then $$\langle S_5, M^5\rangle$$ and $$\langle S_3, \Sigma^3\rangle$$ are the pairing between the topological terms/TQFT with the fundamental classes of manifolds.

p.s. Here are some random thoughts (may not be rigorous, need experts' inputs). My tentative answer is that if the 5d (5-dimensional) manifold can be viewed as $$M^5 =\Sigma^3 \times V^2$$ where $$V^2$$ is another 2-manifold. We may view that $$\Sigma^3$$ as a Poincare dual to $$V^2$$.

The tangent bundle is 3d respect to $$\Sigma^3$$, and the normal bundle is 2d respect to $$\Sigma^3$$. How do we construct say a $$PSU(N)$$ bundle (here $$N=2$$ or $$N=2^n$$), which has a dimensions of $$(N^2-1)$$?

1. Is it true that we need a trivial $$( N^2-3)$$ bundle with the normal bundle 2d in the direct sum, so that we can have the dimensions of bundle match? (Or are other methods to construct the bundles?)

$$(N^2 - 1) = 2 + (N^2 - 3)?$$ $$\text{dim(PSU(N) bundle)}= \text{dim(normal bundle)}+ \text{dim(virtual bundle)}?$$