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The Lumenary · Research pathways

Derived categories

Name: ______________________

Date: ______________________

You will learn to

Construct the homotopy category $K(\mathcal{A})$ and the derived category $D(\mathcal{A})$ , and explain why localising at quasi-isomorphisms needs care
Prove that homotopic chain maps agree on cohomology, and show that a quasi-isomorphism need not have an inverse chain map
Prove $\operatorname{Ext}^n_{\mathcal{A}}(M, N) \cong \operatorname{Hom}_{D(\mathcal{A})}(M, N[n])$ when $\mathcal{A}$ has enough projectives, and compute $R\operatorname{Hom}$ and $\otimes^{L}$ in examples
Show that objects of $D(\mathcal{A})$ are not determined by their cohomology in general, and when they are
Describe derived equivalences (tilting, Fourier–Mukai) and separate established results from open problems

1.
Which of these chain maps are quasi-isomorphisms? (Complexes of abelian groups; a single group means a complex concentrated in degree $0$ . Choose all that apply.)
- A. $p : (\Z \xrightarrow{2} \Z) \to \Z/2$ , reduction mod $2$ in degree $0$ (the left $\Z$ in degree $-1$ )
- B. $0 \to (\Z \xrightarrow{1} \Z)$ , the zero complex mapping to an acyclic complex
- C.multiplication by $2$ , $\Z \to \Z$ , in degree $0$
- D.the inclusion $\Z/2 \hookrightarrow \Z/4$ in degree $0$
2.
Compute the order of $\operatorname{Ext}^1_{\Z}(\Z/12, \Z/18)$ .
3.
The derived tensor product $\Z/8 \otimes^{L}_{\Z} \Z/12$ is an object of $D(\mathrm{Ab})$ . What is the order of its cohomology group in degree $-1$ ?
4.
Which of these Hom-groups in $D(\mathrm{Ab})$ are non-zero? (Choose all that apply.)
- A. $\operatorname{Hom}(\Z/2, \Z[1])$
- B. $\operatorname{Hom}(\Z/2, \Z)$
- C. $\operatorname{Hom}(\Z, \Z/2[1])$
- D. $\operatorname{Hom}(\Z/3, \Z/3[2])$
- E. $\operatorname{Hom}(\Z/2[1], \Z/2)$
5.
Let $R = \Z/4$ . What is the order of $\operatorname{Hom}_{D(R)}(\Z/2, \Z/2[5]) = \operatorname{Ext}^5_R(\Z/2, \Z/2)$ ?
6.
In the proof of Lemma 2 ( $\operatorname{Hom}_K(P, Z) = 0$ for $P$ a bounded-above complex of projectives and $Z$ acyclic), we put $g = f^n - h^{n+1} d_P$ . Where is the projectivity of $P^n$ used?
- A.To lift $g : P^n \to \operatorname{im} d_Z^{n-1}$ along the epimorphism $Z^{n-1} \to \operatorname{im} d_Z^{n-1}$
- B.To prove $d_Z g = 0$
- C.To show that $g$ lands in $\operatorname{im} d_Z^{n-1}$
- D.To start the downward induction
7.
Let $p : P = (\Z \xrightarrow{2} \Z) \to \Z/2$ be the resolution map. Which statement is correct?
- A. $p$ is a quasi-isomorphism with no homotopy inverse, and $Q(p)$ is an isomorphism in $D(\mathrm{Ab})$
- B. $p$ is a homotopy equivalence
- C. $Q(p)$ is not an isomorphism, because $p$ has no inverse chain map
- D. $p$ is an isomorphism of complexes
8.
“Every object $X$ of $D^b(\mathcal{A})$ is isomorphic to $\bigoplus_n H^n(X)[-n]$ .” Which hypothesis on $\mathcal{A}$ makes this true, together with an example showing that some hypothesis is needed?
- A. $\mathcal{A}$ hereditary ( $\operatorname{Ext}^2 = 0$ ); fails for $X = (\Z/4 \xrightarrow{2} \Z/4)$ over $R = \Z/4$
- B. $\mathcal{A}$ has enough projectives; fails over $\Z$
- C.No hypothesis is needed; it holds in every abelian category
- D. $\mathcal{A}$ must be semisimple; fails over $\Z$
9.
Which Lean statement formalises “homotopic chain maps become equal in the derived category”? (Here f g : K ⟶ L are maps of cochain complexes and Q is DerivedCategory.Q.)
- A.theorem t (ho : Homotopy f g) : Q.map f = Q.map g
- B.theorem t (ho : Homotopy f g) : f = g
- C.theorem t [QuasiIso f] : Q.map f = Q.map g
- D.theorem t (h : Q.map f = Q.map g) : Nonempty (Homotopy f g)
10.
Write ext_orders(a, b) returning the list [|Ext^0|, |Ext^1|, |Ext^2|] for $\operatorname{Ext}^n_{\Z}(\Z/a, \Z/b)$ , computed from the resolution $0 \to \Z \xrightarrow{a} \Z \to \Z/a \to 0$ : apply $\operatorname{Hom}(-, \Z/b)$ to get $\Z/b \xrightarrow{a} \Z/b$ and count the kernel and the cokernel of $x \mapsto ax \bmod b$ (do not use gcd).
Computational — Practice

Answer key — Derived categories

1.
(a) and (b). In (a), exactness of $0 \to \Z \xrightarrow{2} \Z \to \Z/2 \to 0$ says exactly that $p$ is a quasi-isomorphism. In (b), both complexes are acyclic. In (c) and (d) the complexes live in degree $0$ only, where a quasi-isomorphism is the same as an isomorphism, and neither map is one.
2.
$R\operatorname{Hom}(\Z/12, \Z/18)$ is $\Z/18 \xrightarrow{12} \Z/18$ . The image of multiplication by $12$ is the subgroup generated by $\gcd(12, 18) = 6$ , of index $6$ . So $\operatorname{Ext}^1(\Z/12, \Z/18) \cong \Z/6$ , of order $6$ — and $\operatorname{Hom}$ (the kernel) also has order $6$ .
3.
$\Z/8 \otimes^{L} \Z/12$ is $\Z/12 \xrightarrow{8} \Z/12$ in degrees $-1, 0$ . Its $H^{-1}$ is $\ker(8) = \{x : 8x \equiv 0 \bmod 12\} = \{0, 3, 6, 9\} \cong \Z/4 = \operatorname{Tor}_1(\Z/8, \Z/12)$ , of order $\gcd(8, 12) = 4$ .
4.
Only (a). Translating each into Ext: (a) $\operatorname{Ext}^1(\Z/2, \Z) = \Z/2$ ; (b) $\operatorname{Ext}^0(\Z/2, \Z) = 0$ ; (c) $\operatorname{Ext}^1(\Z, \Z/2) = 0$ ; (d) $\operatorname{Ext}^2(\Z/3, \Z/3) = 0$ ; (e) $\operatorname{Ext}^{-1}(\Z/2, \Z/2) = 0$ .
5.
The resolution $\cdots \xrightarrow{2} R \xrightarrow{2} R \to \Z/2$ is periodic. $\operatorname{Hom}_R(R, \Z/2) = \Z/2$ , and the induced maps are multiplication by $2$ , i.e. zero. So every cohomology group is $\Z/2$ : $\operatorname{Ext}^n_R(\Z/2, \Z/2) \cong \Z/2$ for all $n \ge 0$ , and the order is $2$ . In particular $\Z/4$ has infinite global dimension.
6.
(a). Boundedness starts the induction, the identity $d_Z g = 0$ is pure algebra, acyclicity of $Z$ turns cycles into boundaries, and projectivity of $P^n$ lifts $g$ through $Z^{n-1} \twoheadrightarrow \operatorname{im} d^{n-1}_Z$ , producing $h^n$ .
7.
(a). $p$ induces isomorphisms on cohomology, so $Q(p)$ is invertible by definition of $D(\mathrm{Ab})$ . A homotopy inverse would be a chain map $q : \Z/2 \to P$ with $pq \simeq \mathrm{id}$ ; but $q^0 : \Z/2 \to \Z$ is zero, so $pq = 0$ , and $0 \simeq \mathrm{id}_{\Z/2}$ fails since there are no non-zero homotopies between complexes concentrated in degree $0$ . (This is the content of the Lean theorem hom_zmod_two_int_eq_zero.)
8.
(a). If $\mathcal{A}$ is hereditary, an induction on the length of $X$ using truncation triangles $\tau_{\le n-1}X \to \tau_{\le n} X \to H^n(X)[-n] \to$ shows that the connecting maps lie in $\operatorname{Ext}^{\ge 2} = 0$ , so the triangles split. Over $\Z/4$ , $\operatorname{Ext}^2(\Z/2, \Z/2) \ne 0$ and $X = (\Z/4 \xrightarrow{2} \Z/4)$ does not split: $X \otimes^{L} \Z/2$ has cohomology in two degrees, whereas $(\Z/2[1] \oplus \Z/2) \otimes^{L} \Z/2$ has cohomology in infinitely many.
9.
(a). Option (d) is the converse, which fails: for example, the periodic complex $\cdots \xrightarrow{2} \Z/4 \xrightarrow{2} \Z/4 \xrightarrow{2} \cdots$ is acyclic, so it is isomorphic to $0$ in $D$ and $Q(\mathrm{id}) = Q(0)$ , but it is not contractible ( $\mathrm{id}$ is not null-homotopic, since that would split the inclusions $2\Z/4 \hookrightarrow \Z/4$ ).
10.
```
def ext_orders(a, b):
    kernel = [x for x in range(b) if (a * x) % b == 0]
    image = {(a * x) % b for x in range(b)}
    return [len(kernel), b // len(image), 1]
```
The answers agree with $\gcd(a, b)$ for both $\operatorname{Ext}^0 = \operatorname{Hom}$ and $\operatorname{Ext}^1$ — the kernel and cokernel of an endomorphism of a finite group always have the same size. $\operatorname{Ext}^2 = 0$ because the resolution has length $1$ : every object of $D^b(\mathrm{Ab})$ is a sum of shifted groups. The counts are evidence for these cases; Theorem 3 explains why this complex computes $\operatorname{Hom}_{D}(\Z/a, \Z/b[n])$ .