26
votes
Accepted
Existence and uniqueness of Haar measure on compacta; a cohomological approach
Fix a compact group $G$ and consider its category of Banach representations:
the objects are (complex) Banach spaces $X$ endowed with a $G$-action by automorphims (not necessarily isometries) such ...
- 10k
18
votes
Accepted
Abelian category with enough injectives but not functorially
Since the dual of an abelian category is also an abelian category, the question is equivalent to the same question for projective resolutions.
I will show that the category $\mathbf{Ab}^{\...
- 20.4k
16
votes
Accepted
Unifying "cohomology groups classify extensions" theorems
$\newcommand{\cA}{\mathcal{A}}\newcommand{\Ext}{\mathrm{Ext}}\newcommand{\Hom}{\mathrm{Hom}}$Let $\cA$ be an abelian category; then, $\Ext_\cA^i(A,B)$ is literally $\Hom_{D(\cA)}(A, B[i])$, where $B[i]...
- 5,133
15
votes
Accepted
Example of an additive functor admitting no right derived functor
Let ${\cal C}$ be the category of finite dimensional ${\bf Z}/2$-vector spaces equipped with a ${\bf Z}/2$ action, let ${\cal C'}$ be the category of finite dimensional ${\bf Z}/2$-vector spaces and ...
- 8,856
15
votes
Accepted
When the restriction of a derived functor to a subcategory is the derived functor of the restriction
In the example where $\mathcal{D}$ is the category of abelian groups and $\mathcal{C}$ is the category of finite abelian groups, take $F(X)=X\otimes_\mathbb{Z}\mathbb{Q}/\mathbb{Z}$. Then the ...
- 30k
14
votes
Accepted
Derived functors - homotopical vs homological approach
EDIT Corrected a couple of inaccuracies and mistakes, added some references.
For the sake of clarity, let me work with non-negatively graded cochain complexes, and analyze the case of a left exact ...
- 481
12
votes
Grothendieck spectral sequence when one of the functors is contravariant
I think this case is actually not so obvious. The issue is that to derive $R\mathscr Hom$ in the first variable you would need to use a locally free resolution while $Rf_*$ being a covariant right ...
- 40.9k
9
votes
Accepted
Fourier-Mukai functors being identity on objects
If $X$ is smooth and projective, then any such FM functor is in fact naturally isomorphic to the identity functor. This follows immediately from Corollary 5.23 of Huybrechts' book on Fourier-Mukai ...
- 4,856
9
votes
Accepted
Different definitions of derived functors
The total right derived functor ${\bf R}F(-)$ contains a bit more information than just its individual cohomologies ${\bf R}^iF(-) = H^i({\bf R}F(-))$. This information can indeed be described as a ...
- 8,856
8
votes
Accepted
The Mittag-Leffler condition as necessary and sufficient
This is due to Emmanouil as far as I know. See this.
- 96
8
votes
Accepted
Grothendieck-Verdier duality without the noetherian condition
Does one have the duality in this setting?
Yes, we do have duality in a very general setting. Your question is equivalent to asking for the existence of a right adjoint to the derived pushforward ...
- 2,736
8
votes
Accepted
Can we define derived functors in model categories without functorial factorisations?
This depends on whether one insists on derived functors landing in the original model category (as is the case with modern approaches of Hinich and Dwyer–Hirschhorn–Kan–Smith), or in its homotopy ...
- 30.9k
7
votes
Accepted
Question on condition for a sheaf to be locally free in Orlov 2004
The question is local, so it is enough to show that if $A$ is a Noetherian local ring with maximal ideal $\mathfrak{m}$ and $M$ is a finitely generated module such that $Ext^i(M,A/\mathfrak{m}) = 0$ ...
- 31.9k
7
votes
Accepted
Two basic questions on derived categories
First of all, the functor $Rf_\ast$ is a red herring. Take an injective resolution $I^{\bullet,\bullet}$ of $F^\bullet$ and apply $f_\ast$; then you just have two questions regarding double complexes. ...
- 37.5k
7
votes
What is integration along the fibers in D-module theory?
I think, for a locally constant $D$-module it basically cashes out to
Defining a formal solution to the differential equation.
Defining a formal integral along the fibers.
Writing down all the ...
- 117k
7
votes
Accepted
Derived Nakayama for complete modules
Let $A$ be a commutative ring and $I\subset A$ be a finitely generated ideal. The basic facts are:
For any complex of derived $I$-complete $A$-modules $C^\bullet$, the cohomology modules $H^*(C^\...
- 14.5k
6
votes
Accepted
Derived pullback of the coarse moduli morphism
If both $\mathcal{X}$ and $X$ are locally Noetherian and regular, then $f$ is flat. Then $Lf^*$ is the usual pullback $f^*$. If $\mathcal{X}$ is tame, then the natural transformation $$\theta:\text{...
6
votes
Motivation/intuition behind the definition of delta-functors and related concepts
To be honest, the question is a bit borderline for MO, but I'll make a few comments anyway. I would argue that there is nothing inherently geometric about the definition of (universal) delta functors, ...
- 31.9k
6
votes
Accepted
Questions about $\text{Perf}(A)$ of dg algebra $A$
As explained a little bit further in Elagin and Lunts' paper, the category $Perf(A)$ consists of the compact objects of $D(A)$, this is exactly what happens in the usual situation in algebraic ...
- 599
6
votes
Accepted
How to complete $i^*i_*\mathcal{F}\to \mathcal{F}$ into an exact triangle for a smooth divisor $i: X\hookrightarrow Y$?
The cone of $i^*i_*\mathcal{F} \to \mathcal{F}$ is isomorphic to $\mathcal{F} \otimes \mathcal{O}_X(-X)[2]$.
EDIT. Let me write an argument for a sheaf $F$. Consider the distinguished triangle
$$
i^*...
- 31.9k
6
votes
Accepted
Is the composite of absolute derived functors a derived functor?
Here is a somewhat degenerate example that illustrates what can go wrong.
Let $\textbf{Ab}$ be the category of abelian groups, considered as a homotopical category where the weak equivalences are the ...
- 13.3k
5
votes
Accepted
The naive approach to deriving profunctors - What's wrong with it?
The problem with this definition is that the formula $\mathrm{colim}_{X \to Z \in \mathcal{W}} F(Z)$ does not, in general, depend functorially on $X$. For it to depend functorially on $X$ you need a ...
- 8,856
5
votes
Accepted
$Lf^*$ is fully faithful
Smoothness is not necessary. What is important is that $f$ has finite $Tor$-dimension (otherwise $Lf^*$ does not preserve boundedness); a sufficient (but not necessary) condition for this is ...
- 31.9k
5
votes
Accepted
multiplicative structure of Ext
So, maybe I should format this as an answer.
Consider any pair of objects $A, B$ in an abelian category with, say, enough projective objects. Then, the extension group $Ext^i(A,B)$ is defined as $H^i(...
- 1,812
5
votes
Accepted
Subspace inclusion with non-vanishing higher direct images
You expect the higher direct images of $f: X \to Y$ to be nonzero if for arbitrarily small neighborhoods $y_0 \in U$ the space $f^{-1}(U)$ has non-vanishing higher cohomology. I.e. $R^i f_* \mathbb Z$...
- 2,847
5
votes
Accepted
Action on group $\operatorname{Ext}^i(\mathcal{L}, \mathcal{M})$ by scalar multiplication
Remark. Exact sequence $0 \to L \to E \to M \to 0$ corresponds to $Ext^1(M,L)$, not to $Ext^1(L,M)$.
Q1. $a \in k^\times$ acts on $Ext^1(L,M)$ via pullback along $a:L \to L$ or via pushout along $a: ...
- 31.9k
5
votes
Conflicting definitions of RHom
The bifunctor
$$R\operatorname{Hom}^\bullet
\colon D(\mathcal{A}) ^{op} \times D(\mathcal{A}) \to D(\operatorname{Ab})
$$
is understood —whenever $\mathcal{A}$ has enough injective objects (e.g. $\...
- 8,261
5
votes
Accepted
Derived functor of functor tensor product
The answer is yes if you assume enough things. In particular, the notion of a left flat object of $\mathcal A$ comes up :
Definition: An object $L\in\mathcal A$ is left flat if $-\otimes L$ is exact.
...
- 8,860
4
votes
Grothendieck spectral sequence when one of the functors is contravariant
If $I^*$ and $J^*$ are global sections of appropriate injective resolutions of $f_*{\cal O}_X$ and ${\cal O}_S$, then you have a double complex that looks in part like this:
$$\matrix{
Hom(I^q,J^p)&...
- 20.9k
4
votes
Accepted
Balanced dualizing complexes according to A. Yekutieli
I happened to be looking at this very issue a few days ago! Here is my explanation to myself...
Indeed, in the notation above, there is a homomorphism of complexes of graded modules (not derived)
$...
- 540
Only top scored, non community-wiki answers of a minimum length are eligible