Timeline for $\lambda$-ring structure defined for a graded ring in Fulton–Lang's book
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 8, 2012 at 1:21 | comment | added | JBorger | 'Universal' just emphasizes that the polynomial is independent of the ring. For instance, the Leibniz rule $d(xy)=xdy+ydx$ for derivations can be expressed as $d(xy)=P(x,y,dx,dy)$, where $P(a,b,c,d)=ad+bc$, and the polynomial $P$ is independent of the ring on which you have a derivation. You can imagine other kind of operators where $d(xy)$ is a polynomial in $x,y,dx,dy$ but that polynomial can depend on the ring. | |
| Oct 7, 2012 at 23:14 | comment | added | Mahdi Majidi-Zolbanin | What is the definition of universal as in universal polynomials? I mean I know how they are defined, but why are they called universal? Does it simply mean that the polynomials have integral coefficients or is it more to it? | |
| Sep 30, 2012 at 14:32 | vote | accept | Mahdi Majidi-Zolbanin | ||
| Sep 30, 2012 at 6:09 | comment | added | JBorger | Yes, I was saying that and also that the universal polynomials are the ones that describe the Chern classes of tensor products. Note that the polynomials are not completely universal -- they do depend on the ranks of the factors. The usual way around this is to assume both factors have rank zero (which is why you get non-unital rings). | |
| Sep 30, 2012 at 5:26 | comment | added | Mahdi Majidi-Zolbanin | Dear James: Are you saying that one can correct the issue by using a different set of universal polynomials to define multiplication in the graded ring that they consider? | |
| Sep 30, 2012 at 0:05 | history | answered | JBorger | CC BY-SA 3.0 |