Timeline for U(1) vs. BZ and representations of 2-groups
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 11, 2013 at 20:09 | vote | accept | Ryan Thorngren | ||
| Nov 24, 2013 at 15:06 | comment | added | Urs Schreiber | You need to do the homotopy pullback in higher stacks. (This is what the boldface of the "B" is meant to remind us of.) Of course the statement does remain true under geometric realization as a statement about homotopy pullbacks in homotopy types of topological spaces, too. | |
| Nov 24, 2013 at 4:38 | comment | added | Ryan Thorngren | Hi Urs. Is it true that the ordinary loopspace $\Omega \mathbf{B}^{d+1} U(1) = \mathbf{B}^d U(1)$, or in general does one need to use a smooth homotopy pullback? | |
| Aug 7, 2013 at 10:38 | comment | added | Urs Schreiber | ...for instance the smooth 2-group $\mathbf{B}U(1)$ is equivalent to that given by the crossed module $\mathbf{B}[\mathbb{Z} \to \mathbb{R}] = [\mathbb{Z} \to \mathbb{R} \to 1]$ (all under Dold-Kan). | |
| Aug 7, 2013 at 10:38 | comment | added | Urs Schreiber | Okay, so what you have in mind now seems to be the presentation of the group $U(1)$ as equivalently the 2-group that comes from the crossed module $[\mathbb{Z} \hookrightarrow \mathbb{R}]$. A cocycle in this latter 2-group is given by elements in $\mathbb{R}$ that on triple intersections differ by an element in $\mathbb{Z}$. This equivalent 2-group is also one way to see that the geometric realization of $U(1)$ is $B \mathbb{Z} = [\mathbb{Z} \to 1]$, because $\mathbb{R}$ is topologically contractible. Morever, all this remains true under arbitrary further delooping. So for instance... | |
| Aug 7, 2013 at 8:03 | comment | added | Ryan Thorngren | Thanks, Urs. Let me see if I understand you. $BA$ is typically a group up to homotopy, and for Lie groups, every element is homotopic to the identity. However, choosing such a homotopy for every element is going to introduce some strange relations. For example, if elements $a$ of $U(1)$ are cancelled out using the homotopy $h_a$ that just rotates clockwise without any extra winding, then $h_a h_b$ differs from $h_{ab}$ by some element in $\pi_1(U(1)) = \mathbb{Z}$, and this is the relationship between $\pi_1$ and the H-space structure. These homotopies are the cohesion $\Pi$ cares about. | |
| Aug 7, 2013 at 7:51 | history | edited | Urs Schreiber | CC BY-SA 3.0 |
added 199 characters in body
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| Aug 7, 2013 at 7:42 | history | answered | Urs Schreiber | CC BY-SA 3.0 |