Homological Algebra
Homological algebra¶
Example: \(\operatorname{Hom}\), \(\mathscr{H}\!om\), \(\operatorname{Ext}^{i}\), \(\mathscr{E}\!xt^{i}\), and \(\operatorname{Tor}_{i}\) for lines in \(\mathbb{P}^3\)¶
We compute the sheaf-Ext groups \(\mathscr{E}\!xt^i(\mathcal{O}_{L_1}, \mathcal{O}_{L_2})\) for two lines \(L_1, L_2 \subset \mathbb{P}^3_k\) over an algebraically closed field \(k\), illustrating the distinction between global and local homological functors.
Global vs. local functors. For coherent sheaves \(\mathscr{F}, \mathscr{G}\) on a scheme \(X\), the global \(\operatorname{Hom}\) and its derived functors \(\operatorname{Ext}^i\) are abelian groups (or \(k\)-vector spaces), while the sheaf \(\mathscr{H}\!om\) and its derived functors \(\mathscr{E}\!xt^i\) are sheaves on \(X\). They are related by the local-to-global spectral sequence
At the level of stalks, \(\mathscr{E}\!xt^i(\mathscr{F}, \mathscr{G})_x \cong \operatorname{Ext}^i_{\mathcal{O}_{X,x}}(\mathscr{F}_x, \mathscr{G}_x)\), so \(\mathscr{E}\!xt^i\) detects local extension phenomena.
Setup. Let \(S = k[x_0, x_1, x_2, x_3]\) be the homogeneous coordinate ring of \(\mathbb{P}^3\). Take two lines:
- \(L_1 = V(x_0, x_1)\), with ideal \(I_1 = (x_0, x_1)\) and structure sheaf \(\mathcal{O}_{L_1} = \widetilde{S/I_1}\).
- \(L_2 = V(x_2, x_3)\), with ideal \(I_2 = (x_2, x_3)\) and structure sheaf \(\mathcal{O}_{L_2} = \widetilde{S/I_2}\).
These two lines are disjoint: \(L_1 \cap L_2 = V(x_0, x_1, x_2, x_3) = \emptyset\) in \(\mathbb{P}^3\).
Resolution of \(\mathcal{O}_{L_1}\). The ideal \(I_1 = (x_0, x_1)\) is generated by a regular sequence of length 2, so \(\mathcal{O}_{L_1}\) admits the Koszul resolution
Computing \(\mathscr{E}\!xt^i(\mathcal{O}_{L_1}, \mathcal{O}_{L_2})\). Apply \(\mathscr{H}\!om(-, \mathcal{O}_{L_2})\) to the Koszul resolution. Since \(\mathscr{H}\!om(\mathcal{O}_{\mathbb{P}^3}(a), \mathcal{O}_{L_2}) \cong \mathcal{O}_{L_2}(-a)\), the resulting complex is
Now \(L_2 = V(x_2, x_3)\), so \(x_0\) and \(x_1\) restrict to the homogeneous coordinates on \(L_2 \cong \mathbb{P}^1\), and in particular are not zero on \(L_2\). The first map \(\mathcal{O}_{L_2} \to \mathcal{O}_{L_2}(1)^{\oplus 2}\) given by \(f \mapsto (x_0 f, x_1 f)\) is injective (since \(x_0, x_1\) are not both zero at any point of \(L_2\)). Taking cohomology of the complex:
-
\(\mathscr{E}\!xt^0(\mathcal{O}_{L_1}, \mathcal{O}_{L_2}) = \ker\left(\mathcal{O}_{L_2} \xrightarrow{(x_0, x_1)^T} \mathcal{O}_{L_2}(1)^2\right)\). Since the lines are disjoint, there are no nonzero morphisms \(\mathcal{O}_{L_1} \to \mathcal{O}_{L_2}\): any such morphism factors through \(\mathcal{O}_{L_1 \cap L_2} = 0\). Alternatively, the map is injective as noted, so \(\mathscr{E}\!xt^0 = 0\).
-
\(\mathscr{E}\!xt^1(\mathcal{O}_{L_1}, \mathcal{O}_{L_2})\) is the cohomology at the middle term. On \(L_2 \cong \mathbb{P}^1\) with coordinates \([x_0, x_1]\), the sequence \(0 \to \mathcal{O}_{L_2} \to \mathcal{O}_{L_2}(1)^2 \to \mathcal{O}_{L_2}(2) \to 0\) is precisely the Koszul complex of the regular sequence \((x_0, x_1)\) on \(\mathbb{P}^1\), twisted from the Euler sequence. This complex is exact, so \(\mathscr{E}\!xt^1(\mathcal{O}_{L_1}, \mathcal{O}_{L_2}) = 0\).
-
\(\mathscr{E}\!xt^2(\mathcal{O}_{L_1}, \mathcal{O}_{L_2}) = \operatorname{coker}\left(\mathcal{O}_{L_2}(1)^2 \xrightarrow{(-x_1, x_0)} \mathcal{O}_{L_2}(2)\right)\). By the exactness of the Koszul complex, the map is surjective, so \(\mathscr{E}\!xt^2 = 0\) as well.
In summary, when \(L_1 \cap L_2 = \emptyset\):
This is a general phenomenon: \(\mathscr{E}\!xt^i(\mathscr{F}, \mathscr{G})\) is supported on \(\operatorname{Supp}(\mathscr{F}) \cap \operatorname{Supp}(\mathscr{G})\), so disjoint supports force all sheaf-Ext's to vanish.
When the lines meet. Now suppose \(L_1\) and \(L_2\) intersect at a single point \(p\). For instance, take \(L_1 = V(x_0, x_1)\) and \(L_2 = V(x_0, x_2)\), which meet at \(p = [0:0:0:1]\). The support of \(\mathscr{E}\!xt^i(\mathcal{O}_{L_1}, \mathcal{O}_{L_2})\) is contained in \(\{p\}\), and a local computation at \(p\) (or a Macaulay2 computation over the homogeneous coordinate ring) gives
where \(k_p\) denotes the skyscraper sheaf at \(p\) with stalk \(k\). The local computation uses the Koszul resolution of \(\mathcal{O}_{L_1}\) as before, but now \(x_0\) restricts to zero on \(L_2\) (since \(x_0 \in I_2\)), so the maps in the dual complex have nontrivial kernel and cokernel at \(p\).
The Tor side. For completeness, the derived tensor product \(\operatorname{Tor}_i^{\mathcal{O}_{\mathbb{P}^3}}(\mathcal{O}_{L_1}, \mathcal{O}_{L_2})\) computes the "intersection" of \(L_1\) and \(L_2\) in the derived sense. Tensoring the Koszul resolution of \(\mathcal{O}_{L_1}\) with \(\mathcal{O}_{L_2}\):
For disjoint lines, this complex is exact (by the same argument as above: \(x_0, x_1\) form a regular sequence on \(L_2\)), so \(\operatorname{Tor}_i(\mathcal{O}_{L_1}, \mathcal{O}_{L_2}) = 0\) for all \(i\), consistent with \(L_1 \cap L_2 = \emptyset\). For intersecting lines meeting at \(p\), one obtains \(\operatorname{Tor}_0 \cong k_p\) (the structure sheaf of the intersection point) and \(\operatorname{Tor}_1 \cong k_p\) (detecting the excess intersection).
Example: \(\operatorname{Ext}^{1}\) and extensions of line bundles on \(\mathbb{P}^1\)¶
We classify extensions of line bundles on \(\mathbb{P}^1_k\) via \(\operatorname{Ext}^1\), using the cohomology of line bundles and Grothendieck's splitting theorem to determine all possible middle terms.
The Ext computation. For line bundles \(\mathcal{O}(a)\) and \(\mathcal{O}(b)\) on \(\mathbb{P}^1\),
The cohomology of line bundles on \(\mathbb{P}^1\) is:
(The second formula follows from Serre duality: \(H^1(\mathbb{P}^1, \mathcal{O}(n)) \cong H^0(\mathbb{P}^1, \mathcal{O}(-n-2))^{\vee}\), using \(\omega_{\mathbb{P}^1} = \mathcal{O}(-2)\).) Therefore
This is nonzero precisely when \(a \geq b + 2\), i.e., when the "source" line bundle has degree at least 2 more than the "target."
| \(a - b\) | \(\dim \operatorname{Ext}^1(\mathcal{O}(a), \mathcal{O}(b))\) | Extensions |
|---|---|---|
| \(\leq 1\) | \(0\) | All split |
| \(2\) | \(1\) | Unique nonsplit (up to scalar) |
| \(3\) | \(2\) | \(\mathbb{P}^1\)-family of nonsplit |
| \(d \geq 2\) | \(d - 1\) | \((d-1)\)-dimensional |
A trivial case: \(\operatorname{Ext}^1(\mathcal{O}(-2), \mathcal{O}(2))\). Here \(a - b = -4 < 2\), so \(\operatorname{Ext}^1(\mathcal{O}(-2), \mathcal{O}(2)) = 0\). Every short exact sequence
splits, giving \(\mathscr{E} \cong \mathcal{O}(2) \oplus \mathcal{O}(-2)\).
A nontrivial case: \(\operatorname{Ext}^1(\mathcal{O}(2), \mathcal{O}(-2))\). Now \(a - b = 4\), so \(\operatorname{Ext}^1(\mathcal{O}(2), \mathcal{O}(-2)) \cong H^1(\mathbb{P}^1, \mathcal{O}(-4))\). By Serre duality,
since \(H^0(\mathbb{P}^1, \mathcal{O}(2))\) has the basis \(\{x_0^2, x_0 x_1, x_1^2\}\). So there is a 3-dimensional space of extensions
Classifying the middle terms. By Grothendieck's theorem, every vector bundle on \(\mathbb{P}^1\) splits as a direct sum of line bundles. The middle term of any extension is \(\mathscr{E} \cong \mathcal{O}(a') \oplus \mathcal{O}(b')\) for some integers \(a' \geq b'\). Two constraints determine the possibilities:
-
Determinant constraint: \(\det(\mathscr{E}) = \mathcal{O}(a' + b') \cong \mathcal{O}(-2 + 2) = \mathcal{O}\), so \(a' + b' = 0\), i.e., \(\mathscr{E} \cong \mathcal{O}(m) \oplus \mathcal{O}(-m)\) for some \(m \geq 0\).
-
Quotient constraint: There must be a surjection \(\mathscr{E} \twoheadrightarrow \mathcal{O}(2)\). For \(\mathscr{E} = \mathcal{O}(m) \oplus \mathcal{O}(-m)\), a surjection onto \(\mathcal{O}(2)\) requires \(\operatorname{Hom}(\mathcal{O}(m) \oplus \mathcal{O}(-m), \mathcal{O}(2)) = H^0(\mathcal{O}(2-m)) \oplus H^0(\mathcal{O}(2+m))\) to contain an element whose image generates \(\mathcal{O}(2)\) at every point. The component \(H^0(\mathcal{O}(2-m))\) is nonzero only when \(m \leq 2\), and the surjectivity condition requires \(m \leq 2\).
The possible middle terms are therefore \(\mathcal{O}(m) \oplus \mathcal{O}(-m)\) for \(m = 0, 1, 2\):
| \(m\) | \(\mathscr{E}\) | Extension type |
|---|---|---|
| \(2\) | \(\mathcal{O}(2) \oplus \mathcal{O}(-2)\) | Split (the zero class in \(\operatorname{Ext}^1\)) |
| \(1\) | \(\mathcal{O}(1) \oplus \mathcal{O}(-1)\) | Nonsplit |
| \(0\) | \(\mathcal{O} \oplus \mathcal{O}\) | Nonsplit |
To see that \(m = 1\) and \(m = 0\) both occur, note that extension classes in \(\operatorname{Ext}^1(\mathcal{O}(2), \mathcal{O}(-2)) \cong k^3\) are acted on by \(\operatorname{Aut}(\mathcal{O}(2)) \times \operatorname{Aut}(\mathcal{O}(-2)) \cong k^* \times k^*\), which rescales but preserves the splitting type of the middle term. A general (generic) nonzero class gives \(m = 0\), while special classes on a certain discriminant locus give \(m = 1\). Explicitly, the extension class corresponding to a nonzero element \(\xi \in H^1(\mathbb{P}^1, \mathcal{O}(-4))\) gives \(\mathscr{E} \cong \mathcal{O}(1) \oplus \mathcal{O}(-1)\) when the Čech cocycle representing \(\xi\) has a zero (i.e., the corresponding map \(\mathcal{O}(2) \to \mathcal{O}(-2)\) on the overlap \(U_{01}\) vanishes at some point), and \(\mathscr{E} \cong \mathcal{O} \oplus \mathcal{O}\) for a generic cocycle.
Verification. For the \(m = 0\) case, the Euler exact sequence on \(\mathbb{P}^1\) provides a concrete example. The tangent bundle \(T_{\mathbb{P}^1} \cong \mathcal{O}(2)\) sits in the Euler sequence
Twisting by \(\mathcal{O}(-2)\) gives \(0 \to \mathcal{O}(-2) \to \mathcal{O}(-1)^{\oplus 2} \to \mathcal{O} \to 0\), and dualizing yields \(0 \to \mathcal{O} \to \mathcal{O}(1)^{\oplus 2} \to \mathcal{O}(2) \to 0\) again. Taking the sequence \(0 \to \mathcal{O}(-2) \to \mathcal{O}(-1)^2 \to \mathcal{O} \to 0\), this is an extension of \(\mathcal{O}\) by \(\mathcal{O}(-2)\) with middle term \(\mathcal{O}(-1)^2\), which is an instance of \(m = 1\) in the \(\operatorname{Ext}^1(\mathcal{O}(0), \mathcal{O}(-2))\) case (\(a - b = 2\), one-dimensional Ext).
Remark: \(\operatorname{Ext}^1_{\mathbb{Z}}\) for abelian groups¶
We compute \(\operatorname{Ext}^1_{\mathbb{Z}}(A, B)\) for various finitely generated and classical abelian groups, illustrating the interplay of projectivity, injectivity (divisibility), and free resolutions.
Vanishing from projectivity. A free abelian group \(F\) is projective, so \(\operatorname{Ext}^i_{\mathbb{Z}}(F, B) = 0\) for all \(i \geq 1\) and all \(B\). Since \(\mathbb{Z}\) is a PID, projective \(\Leftrightarrow\) free \(\Leftrightarrow\) torsion-free for finitely generated modules, so this gives
Vanishing from injectivity. Over \(\mathbb{Z}\), a module is injective if and only if it is divisible (Baer's criterion). The groups \(\mathbb{Q}\), \(\mathbb{Q}/\mathbb{Z}\), and \(\mathbb{R}\) are divisible, hence injective. Therefore
Computation via the free resolution of \(\mathbb{Z}/n\mathbb{Z}\). The cyclic group \(\mathbb{Z}/n\mathbb{Z}\) has the free resolution
Applying \(\operatorname{Hom}_{\mathbb{Z}}(-, B)\) to the deleted resolution \(0 \to \mathbb{Z} \xrightarrow{\cdot\, n} \mathbb{Z} \to 0\) gives the complex
since \(\operatorname{Hom}(\mathbb{Z}, B) \cong B\) canonically and the induced map is multiplication by \(n\). Therefore
Since \(\mathbb{Z}\) has global dimension 1, \(\operatorname{Ext}^i(\mathbb{Z}/n\mathbb{Z}, B) = 0\) for \(i \geq 2\).
Specializing to particular targets:
| \(A\) | \(B\) | \(\operatorname{Ext}^1(A, B)\) | Reason |
|---|---|---|---|
| \(\mathbb{Z}/n\mathbb{Z}\) | \(\mathbb{Z}\) | \(\mathbb{Z}/n\mathbb{Z}\) | \(\mathbb{Z}/n\mathbb{Z} = \mathbb{Z}/n\mathbb{Z}\) |
| \(\mathbb{Z}/n\mathbb{Z}\) | \(\mathbb{Z}/m\mathbb{Z}\) | \(\mathbb{Z}/\gcd(m,n)\mathbb{Z}\) | \((\mathbb{Z}/m\mathbb{Z})/n(\mathbb{Z}/m\mathbb{Z}) \cong \mathbb{Z}/\gcd(m,n)\mathbb{Z}\) |
| \(\mathbb{Z}/n\mathbb{Z}\) | \(\mathbb{Q}\) | \(0\) | \(\mathbb{Q}/n\mathbb{Q} = 0\) (\(\mathbb{Q}\) is divisible) |
| \(\mathbb{Z}\) | any \(B\) | \(0\) | \(\mathbb{Z}\) is free |
For the entry \(\operatorname{Ext}^1(\mathbb{Z}/n\mathbb{Z}, \mathbb{Z}/m\mathbb{Z})\): the image of \(n\) in \(\mathbb{Z}/m\mathbb{Z}\) is \(\bar{n} = n \bmod m\), and \((\mathbb{Z}/m\mathbb{Z}) / \langle \bar{n} \rangle \cong \mathbb{Z}/\gcd(m,n)\mathbb{Z}\). This is because the image of the multiplication-by-\(n\) map on \(\mathbb{Z}/m\mathbb{Z}\) is \(\gcd(m,n) \cdot (\mathbb{Z}/m\mathbb{Z}) \cong \mathbb{Z}/(m/\gcd(m,n))\mathbb{Z}\), so the cokernel has order \(\gcd(m,n)\).
The case \(\operatorname{Ext}^1(\mathbb{Q}, \mathbb{Z}/m\mathbb{Z}) = 0\). This requires a different argument since \(\mathbb{Q}\) is not finitely generated. For any nonzero integer \(n\), the multiplication map \(n : \mathbb{Q} \to \mathbb{Q}\) is an isomorphism. By functoriality of \(\operatorname{Ext}\), the induced map \(n : \operatorname{Ext}^1(\mathbb{Q}, \mathbb{Z}/m\mathbb{Z}) \to \operatorname{Ext}^1(\mathbb{Q}, \mathbb{Z}/m\mathbb{Z})\) is also an isomorphism. This gives \(\operatorname{Ext}^1(\mathbb{Q}, \mathbb{Z}/m\mathbb{Z})\) the structure of a \(\mathbb{Q}\)-vector space (since every nonzero integer acts invertibly).
On the other hand, the target \(\mathbb{Z}/m\mathbb{Z}\) is \(m\)-torsion, and \(\operatorname{Ext}^1(\mathbb{Q}, \mathbb{Z}/m\mathbb{Z})\) is a quotient of a module built from \(\operatorname{Hom}\) groups into \(\mathbb{Z}/m\mathbb{Z}\). More precisely, from any free resolution \(0 \to F_1 \to F_0 \to \mathbb{Q} \to 0\) (e.g., the one where \(F_0 = \bigoplus_{n \geq 1} \mathbb{Z}\) maps to \(\mathbb{Q}\) by sending the \(n\)-th generator to \(1/n!\)), the group \(\operatorname{Ext}^1\) is a subquotient of \(\operatorname{Hom}(F_1, \mathbb{Z}/m\mathbb{Z})\), which is \(m\)-torsion. An abelian group that is simultaneously a \(\mathbb{Q}\)-vector space and \(m\)-torsion must be zero:
The subtle case \(\operatorname{Ext}^1(\mathbb{Q}, \mathbb{Z})\). By contrast, \(\operatorname{Ext}^1_{\mathbb{Z}}(\mathbb{Q}, \mathbb{Z})\) is nonzero and quite large. Consider the short exact sequence \(0 \to \mathbb{Z} \to \mathbb{Q} \to \mathbb{Q}/\mathbb{Z} \to 0\) and apply \(\operatorname{Hom}(-, \mathbb{Z})\). Since \(\operatorname{Hom}(\mathbb{Q}, \mathbb{Z}) = 0\) (no nonzero homomorphism from a divisible group to \(\mathbb{Z}\)) and \(\operatorname{Hom}(\mathbb{Z}, \mathbb{Z}) = \mathbb{Z}\), the long exact sequence gives
The group \(\operatorname{Ext}^1(\mathbb{Q}/\mathbb{Z}, \mathbb{Z})\) can be computed from \(\mathbb{Q}/\mathbb{Z} \cong \bigoplus_p \mathbb{Z}[p^{-1}]/\mathbb{Z} = \bigoplus_p \mathbb{Q}_p/\mathbb{Z}_p\):
the profinite completion of \(\mathbb{Z}\). (Here \(\operatorname{Ext}^1(\mathbb{Q}_p/\mathbb{Z}_p, \mathbb{Z}) \cong \mathbb{Z}_p\) follows from taking the inverse limit of \(\operatorname{Ext}^1(\mathbb{Z}/p^n\mathbb{Z}, \mathbb{Z}) \cong \mathbb{Z}/p^n\mathbb{Z}\).) Therefore
This is an uncountable, torsion-free, divisible abelian group. Its divisibility can be checked directly: given \(x \in \widehat{\mathbb{Z}}/\mathbb{Z}\) and a nonzero integer \(n\), the multiplication-by-\(n\) map on \(\widehat{\mathbb{Z}}\) is surjective (since it is surjective on each \(\mathbb{Z}_p\)), so \(x\) has a preimage. The torsion-freeness follows because \(\widehat{\mathbb{Z}}\) is torsion-free and \(\mathbb{Z} \hookrightarrow \widehat{\mathbb{Z}}\) is a pure subgroup. As a \(\mathbb{Q}\)-vector space, \(\widehat{\mathbb{Z}}/\mathbb{Z}\) has dimension \(2^{\aleph_0}\) (the cardinality of the continuum).