Characteristic Classes
Characteristic classes and numbers¶
Example: Horrocks–Mumford bundle¶
The Horrocks–Mumford bundle \(\mathscr{F}\) is a rank-2 vector bundle on \(\mathbb{P}^4\) with Chern classes \(c_1(\mathscr{F}) = 5h\) and \(c_2(\mathscr{F}) = 10h^2\), where \(h\) is the hyperplane class. It is the unique (up to twist and automorphisms) indecomposable rank-2 bundle on \(\mathbb{P}^n\) for \(n \geq 3\).
We compute \(\chi(\mathscr{F}(n-5))\) via the Hirzebruch–Riemann–Roch theorem:
Chern character of \(\mathscr{F}(t)\) with \(t = n-5\). The twisted bundle has \(c_1(\mathscr{F}(t)) = (2t + 5)h\) and \(c_2(\mathscr{F}(t)) = (t^2 + 5t + 10)h^2\). For a rank-2 bundle, the Chern character components are determined by Newton's identities:
- \(\operatorname{ch}_0 = 2\)
- \(\operatorname{ch}_1 = c_1 = (2t+5)h\)
- \(\operatorname{ch}_2 = \frac{c_1^2 - 2c_2}{2} = \frac{(2t+5)^2 - 2(t^2+5t+10)}{2} = \frac{2t^2+10t+5}{2}\; h^2\)
- \(\operatorname{ch}_3 = \frac{c_1^3 - 3c_1 c_2}{6}\). We have \((2t+5)^3 = 8t^3+60t^2+150t+125\) and \(3(2t+5)(t^2+5t+10) = 6t^3+45t^2+135t+150\), so \(\operatorname{ch}_3 = \frac{2t^3+15t^2+15t-25}{6}\; h^3\).
- \(\operatorname{ch}_4 = \frac{c_1^4 - 4c_1^2 c_2 + 2c_2^2}{24}\). Computing the three terms:
- \((2t+5)^4 = 16t^4+160t^3+600t^2+1000t+625\)
- \(4(2t+5)^2(t^2+5t+10) = 16t^4+160t^3+660t^2+1300t+1000\)
- \(2(t^2+5t+10)^2 = 2t^4+20t^3+90t^2+200t+200\)
So \(\operatorname{ch}_4 = \frac{2t^4+20t^3+30t^2-100t-175}{24}\; h^4\).
Todd class of \(\mathbb{P}^4\). The tangent bundle \(T_{\mathbb{P}^4}\) has \(c_i = \binom{5}{i} h^i\) from the Euler sequence \(0 \to \mathcal{O} \to \mathcal{O}(1)^{\oplus 5} \to T_{\mathbb{P}^4} \to 0\). The Todd class components are:
- \(\operatorname{td}_0 = 1\)
- \(\operatorname{td}_1 = \frac{c_1}{2} = \frac{5}{2}\)
- \(\operatorname{td}_2 = \frac{c_1^2 + c_2}{12} = \frac{25+10}{12} = \frac{35}{12}\)
- \(\operatorname{td}_3 = \frac{c_1 c_2}{24} = \frac{50}{24} = \frac{25}{12}\)
- \(\operatorname{td}_4 = \frac{-c_1^4 + 4c_1^2 c_2 + 3c_2^2 + c_1 c_3 - c_4}{720} = \frac{720}{720} = 1\)
The Euler characteristic. We extract the degree-4 component \(\chi = \sum_{i+j=4} \operatorname{ch}_i \cdot \operatorname{td}_j\). Writing each term over the common denominator 24:
| Term | Expression | Over 24 |
|---|---|---|
| \(\operatorname{ch}_0 \cdot \operatorname{td}_4\) | \(2 \cdot 1\) | \(48\) |
| \(\operatorname{ch}_1 \cdot \operatorname{td}_3\) | \((2t+5) \cdot \frac{25}{12}\) | \(100t + 250\) |
| \(\operatorname{ch}_2 \cdot \operatorname{td}_2\) | \(\frac{2t^2+10t+5}{2} \cdot \frac{35}{12}\) | \(70t^2 + 350t + 175\) |
| \(\operatorname{ch}_3 \cdot \operatorname{td}_1\) | \(\frac{2t^3+15t^2+15t-25}{6} \cdot \frac{5}{2}\) | \(20t^3 + 150t^2 + 150t - 250\) |
| \(\operatorname{ch}_4 \cdot \operatorname{td}_0\) | \(\frac{2t^4+20t^3+30t^2-100t-175}{24}\) | \(2t^4 + 20t^3 + 30t^2 - 100t - 175\) |
Collecting by degree: \(2t^4 + 40t^3 + 250t^2 + 500t + 48\). So
Substituting \(t = n - 5\). We expand and collect:
| Power | \((n-5)^4\) | \(20(n-5)^3\) | \(125(n-5)^2\) | \(250(n-5)\) | \(+24\) | Total |
|---|---|---|---|---|---|---|
| \(n^4\) | \(1\) | \(1\) | ||||
| \(n^3\) | \(-20\) | \(20\) | \(0\) | |||
| \(n^2\) | \(150\) | \(-300\) | \(125\) | \(-25\) | ||
| \(n^1\) | \(-500\) | \(1500\) | \(-1250\) | \(250\) | \(0\) | |
| \(n^0\) | \(625\) | \(-2500\) | \(3125\) | \(-1250\) | \(24\) | \(24\) |
The vanishing of the \(n^3\) and \(n^1\) coefficients reflects Serre duality: since \(\omega_{\mathbb{P}^4} = \mathcal{O}(-5)\) and \(c_1(\mathscr{F}) = 5\), we have \(\mathscr{F}^{\vee} \cong \mathscr{F}(-5)\), so \(\chi(\mathscr{F}(t)) = \chi(\mathscr{F}(-t-5))\) by Serre duality, which forces the polynomial in \(n = t+5\) to be even. Hence
Verification.
- \(n = 5\) (i.e., \(\mathscr{F}\) itself): \(\chi = \frac{24 \cdot 1}{12} = 2\). This matches the known cohomology \(h^0(\mathscr{F}) = 4\), \(h^1(\mathscr{F}) = 2\), \(h^i(\mathscr{F}) = 0\) for \(i \geq 2\), giving \(\chi = 4 - 2 = 2\). ✓
- \(n = 0\) (i.e., \(\mathscr{F}(-5) \cong \mathscr{F}^{\vee}\)): \(\chi = \frac{24}{12} = 2 = \chi(\mathscr{F})\) by Serre duality. ✓
- \(n = \pm 1\): \(\chi = 0\). Indeed, \(\mathscr{F}(-4)\) and \(\mathscr{F}(-6)\) have all cohomology groups vanishing, which can be verified from the monad description \(0 \to \mathcal{O}(-1)^5 \to \Omega^2(2)^2 \to \mathcal{O}(1)^5 \to 0\).