5
$\begingroup$

Consider the extension
$$1\to SU(2)\to X\to O\to1,$$

there are 4 possibilities for $X$: $X=O\times SU(2)$ or $E\times_{\mathbb{Z}_2}SU(2)$ or $Pin^+\times_{\mathbb{Z}_2}SU(2)$ or $Pin^-\times_{\mathbb{Z}_2}SU(2)$ where $E$ is defined in Freed-Hopkins's work1 as the colimit of the group $E(d)$, the group $E(d)$ is defined to be the subgroup of $O(d)\times\mathbb{Z}_4$ consisting of the pairs $(A,j)$ such that $\det A=j^2$, where $\mathbb{Z}_4=\{\pm1,\pm\sqrt{-1}\}$ is the multiplicative group of order 4.

Here the notation $G_1\times_{\mathbb{Z}_2} G_2 :=\frac{G_1\times G_2}{\mathbb{Z}_2} $ is defined as mod out the common $\mathbb{Z}_2$ of $G_1\times G_2$.

The question is about computing $MT(E(d)\times_{\mathbb Z_2} SU(2))$ and the bordism group $\Omega_d^{E \times_{\mathbb Z_2}SU(2)}$.

(1) There is a short exact sequence of groups: $1\to SO(d)\to E(d)\to\mathbb{Z}_4\to 1$. So naively, people may suspect that $$MT(E(d)\times_{\mathbb Z_2} SU(2))=MT E(d)\wedge\Sigma^{-3}M SO(3)=MSO(d)\wedge\Sigma^{-2}M\mathbb Z_4\wedge\Sigma^{-3}M SO(3).$$ However, this is likely to be incorrect.

(2) The space $B(E \times_{\mathbb Z_2}SU(2))$ sits in a homotopy pullback square: a map $M \to B(E \times_{\mathbb Z_2}SU(2))$ is determined by two maps $M \to BO$ and $M\to BSO(3)$ which correspond to bundles $TM$ and $V_{SO(3)}$ such that $w_1(TM)^2=w_2(V_{SO(3)})$.

To compute the bordism group $\Omega_d^{E \times_{\mathbb Z_2}SU(2)}$, we need to know the Madsen-Tillmann spectrum $MT(E \times_{\mathbb Z_2}SU(2))$ and decompose it as the wedge sum or smash product of some familiar spectra.

The figure attachment here is my own attempt,

enter image description here

but the map $f$ is not a homotopy equivalence. I actually obtain an identification $$ \text{Thom$(B(E \times_{\mathbb Z_2}SU(2)),-2V)=MT(Pin^+ \times_{\mathbb Z_2}SU(2))$} $$ which is already known in 1604.06527 paper, but we need to know $$ \text{Thom$(B(E \times_{\mathbb Z_2}SU(2)),-V)=MT(E \times_{\mathbb Z_2}SU(2))$,}$$

where $V$ is the induced virtual bundle of dimension 0 by $B(E \times_{\mathbb Z_2}SU(2)) \to BO$.

  • Is Thom$(B(E \times_{\mathbb Z_2}SU(2)),-2V)$=smash product of Thom$(B(E \times_{\mathbb Z_2}SU(2)),-V)$ and Thom$(B(E \times_{\mathbb Z_2}SU(2)),-V)$? If so, how to obtain Thom$(B(E \times_{\mathbb Z_2}SU(2)),-V)$ as the "square root" of Thom$(B(E \times_{\mathbb Z_2}SU(2)),-2V)$?

1 Reflection positivity and invertible topological phases Daniel S. Freed, Michael J. Hopkins, arXiv:1604.06527

$\endgroup$
2

1 Answer 1

5
$\begingroup$

Let $Y$ be a space, $V$ be a virtual bundle of dimension $0$ over $Y$ (this $V$ is your $-V$). Then $Thom(Y,2V)$ is almost never (except when $Y$ is contractible, or something like that) a smash product of the form $Thom(Y,V)$ with itself. You can see this by looking at the homology: the homology of the former is more or less isomorphic to that of $Y$, whereas the latter is more or less isomorphic to its square.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.