8
$\begingroup$

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?)

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.