bio | website | math.cornell.edu/m/People/… |
---|---|---|
location | Cornell University | |
age | 24 | |
visits | member for | 2 years, 1 month |
seen | Apr 30 at 14:54 | |
stats | profile views | 350 |
Grad student at Cornell.
Apr 2 |
awarded | Yearling |
Mar 5 |
asked | Errata for Getzler-Kapranov “Cyclic operads and Cyclic homology” |
Aug 19 |
comment |
Can we “complete” model categories to compute derived functors in the usual way?
More precisely, the axiomatics itself makes the cofibrant resolutions to exist always, in any model category. But it is still not clear to me, why can we compute the derived functor by applying it to a cofibrant replacement, unless it is not a part of a Quillen pair. So I was interested, maybe we can extend the categories and the functor so that it will become Quillen, and so we know we can compute the derived functor by applying it to a resolution. |
Aug 19 |
comment |
Can we “complete” model categories to compute derived functors in the usual way?
@PhilippeGaucher Yes, you are right. In fact, if we cansider model category of complexes over an abelian category $\mathcal{A}$ (with enough projectives) then taking cofibrant replacements of objects in $\mathcal{A}\subset Com(\mathcal{A})$ is exactly taking projective resolutions. What I was asking is, in abelian categories we can't sometimes compute a derived functor by simply applying the functor to the resolution. Then, we extend our categories so that we can compute the derived functor in the usual way (by applying it to a resolution). I was asking if this is a case in model categories. |
Aug 12 |
comment |
Can we “complete” model categories to compute derived functors in the usual way?
@ZhenLin I am not asking for minimal such extensions. Maybe one can embed the model categories $\mathcal{C}$ and $\mathcal{D}$ into something much bigger... |
Aug 12 |
revised |
Can we “complete” model categories to compute derived functors in the usual way?
edited tags |
Aug 12 |
comment |
Can we “complete” model categories to compute derived functors in the usual way?
@DavidWhite No, I don't think so... At least for the first question. For the second question I guess we need to assume something about $F$. For example, it should commute with colimits or limits (if we want it to be left or right Quillen). Maybe we can assume it already has an adjoint if it is necessary. |
Aug 12 |
revised |
Can we “complete” model categories to compute derived functors in the usual way?
edited tags; edited title |
Aug 12 |
asked | Can we “complete” model categories to compute derived functors in the usual way? |
Aug 8 |
accepted | Noncommutative HKR theorem |
Jul 25 |
revised |
Noncommutative HKR theorem
added 593 characters in body |
Jul 23 |
comment |
Noncommutative HKR theorem
@MarianoSuárez-Alvarez Is it obvious why the two definitions are the same? I am still confused about that... |
Jul 22 |
comment |
Noncommutative HKR theorem
@VahidShirbisheh Yes, I know this book. But this section deals only with the case when $A$ is commutative and is smooth in the commutative world, which is different than being smooth in the noncommutative world. |
Jul 22 |
comment |
Noncommutative HKR theorem
@DanielPomerleano I don't know.. I have seen only the definition where $\Omega^1_{nc}(A)=ker(m\colon A\otimes A\to A)$ is the kernel of the multiplication map, and then $\Omega^\bullet(A)=T_A^\bullet(\Omega^1_{nc}(A))$ is the tensor algebra of $\Omega^1_{nc}(A)$. |
Jul 21 |
asked | Noncommutative HKR theorem |
Jul 2 |
awarded | Curious |
Apr 2 |
awarded | Yearling |
Mar 8 |
accepted | Why is “naive” definition of non-commutative spectrum bad? |
Mar 8 |
awarded | Nice Question |
Mar 5 |
asked | Why is “naive” definition of non-commutative spectrum bad? |