Timeline for What is the interpretation of the Gerstenhaber bracket?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Aug 15, 2018 at 7:57 | comment | added | Lukas Woike | May I ask about more details: My $E_2$-algebra $A$ is a chain complex (not cochain), and I have a homotopy between multiplication and opposite multiplication; it consists of maps $h_{p,q} : A_p \otimes A_q \to A_{p+q+1}$. Now what would your pre-Lie relation be and how can I build the braces or $[-,-]$ from $h$? | |
| Aug 13, 2018 at 11:04 | vote | accept | Lukas Woike | ||
| Aug 13, 2018 at 8:15 | comment | added | Pavel Safronov | For the first question: yes, if the homotopy satisfies the pre-Lie relation (possibly up to homotopy). For the second question: I guess that's indeed how you can think about it. If you imagine multiplications in $E_2$ parametrized by $S^1$, then the homotopy $a\{b\}$ is the class of a semicircle from the basepoint to its antipodal point. The homotopy $b\{a\}$ is the class of the other semicircle. Adding them together you get the fundamental class of $S^1$. Here I am using that $Lie\{1\}\rightarrow H_\bullet(E_2)$ in arity 2 sends the bracket to the fundamental class. | |
| Aug 13, 2018 at 7:46 | comment | added | Lukas Woike | Is there also a possibility to relate the Gerstenhaber bracket to "swapping factors twice" (to some kind of double braiding, using the terminology of $E_2$-categories)? | |
| Aug 13, 2018 at 7:46 | comment | added | Lukas Woike | Thank you for your answer. So if I have the homotopy which makes the multiplication commutative, can I construct the Gerstenhaber bracket? | |
| Aug 12, 2018 at 19:22 | history | answered | Pavel Safronov | CC BY-SA 4.0 |