Timeline for Rank vanishing in tensor categories
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 30, 2014 at 7:19 | vote | accept | Martin Brandenburg | ||
| Mar 30, 2014 at 1:35 | comment | added | Qiaochu Yuan | Right, the conclusion is already false for chain complexes of rational vector spaces; here the dualizable objects are the bounded chain complexes of finite-dimensional vector spaces, with dimension the Euler characteristic. | |
| Mar 30, 2014 at 0:25 | comment | added | Martin Brandenburg | I am interested in char. $0$. If $\mathcal{L}$ is an invertible object, then we don't necessarily have $\mathrm{rk}(\mathcal{L})=1$? In fact, the rank equals the signature of $\mathcal{L}$, which is an involution of $1$, and $-1$ is possible e.g. when considering $\mathbb{Z}$-graded objects of $\mathcal{C}$ (twisted symmetry). If $X$ is $1_\mathcal{C}$ concentrated in degree $1$, then every dualizable graded object has the form $M=\sum_n M_n \otimes X^{\otimes n}$ with $M_n$ dualizable (almost all $0$), $\mathrm{rk}(X)=-1$, hence $\mathrm{rk}(M) = \sum_n (-1)^n \mathrm{rk}(M_n)$. Correct? | |
| Mar 29, 2014 at 22:41 | history | answered | Dustin Clausen | CC BY-SA 3.0 |