If $\Omega$ is the curvature 2-form on a $n-$manifold, then I would think that the Chern classes (forms), $c_k$ are defined as,

$det(I + \frac{it\Omega}{2\pi}) = \sum c_k t^k$

  • I would like to know of a local coordinate expression for the above so that I can clearly see as to how each $c_k$ on the RHS turns out to be a rank $2k$ form.

(..naively it seems that one has to think of the determinant on the LHS to be that of a "matrix" each of whose entries are 2-forms themselves and the probably the "multiplication" that is being done among the elements of the matrix to evaluate the determinant is taking a wedge product -- but I would like to know of something explicit..)

  • What does it mean when one talks of a Chern class being positive or negative? Is there an invariant meaning to may be the integral over the manifold of that form?

  • One particular case which I am interested in is this,

Consider the zero-set in $\mathbb{CP}^n$ of a homogeneous degree $k$ polynomial in $n$ variables. Firstly when is it guaranteed that this is going to be a manifold? Now how does one understand that the "sign" of the Chern class (whetever that means) depends on whether $k> n$ or $k<n$?

(..I often see the claim that for $k=n$ the zero-set is a Calabi-Yau manifold since then the first Chern class vanishes..)


1 Answer 1


Let $M$ be an $n$-dimensional manifold and $E$ a rank $m$ complex vector bundle over $M$. The Chern-Weil description of the Chern classes of $E$ is, as you have said, given by the generating function $$\det\left( \mathbb{I}_E + \frac{it}{2\pi}\Omega_A \right) = \sum c_k(E)t^k,$$ where $\Omega_A$ is the curvature of an arbitrarily chosen connection $A$ on $E$; the cohomology class of each $c_k(E)$ is independent of the choice of $A$. Now $\Omega_A$ can be viewed as an $\mathrm{End}(E)$-valued $2$-form, or, as you guessed, an $m \times m$ matrix of complex-valued $2$-forms. When taking products of the entries of $\Omega_A$, we use the wedge product. Note that with this interpretation, $i\Omega_A/2\pi$ is a skew-symmetric matrix and is hence diagonalizable. Let $x_1, \dots, x_m$ be the eigenvalues of $\Omega_A$ , and note that they are complex-valued $2$-forms. Then we have $$\det\left(\mathbb{I}_E + \frac{it}{2\pi}\Omega_A\right) = \det(\mathrm{diag}(1 + tx_1, \dots, 1 + tx_m)) = \sum_{k = 0}^m S_k(x_1, \dots, x_m)t^k,$$ where $S_k(x_1, \dots, x_m)$ is the $k^\text{th}$ elementary symmetric polynomial in $x_1, \dots, x_m$, i.e. $$S_0(x_1, \dots, x_m) = 1,$$ $$S_1(x_1, \dots, x_m) = \sum_{i = 1}^m x_i,$$ $$S_2(x_1, \dots, x_m) = \sum_{i < j} x_i x_j,$$ and so on until $$S_m(x_1, \dots, x_m) = x_1 \cdots x_m.$$ Using this, it is clear that $c_k(E)$ is a form of degree $2k$. Furthermore, this permits calculation of the Chern classes: $$c_1(E) = \frac{i}{2\pi} \mathrm{Tr}(\Omega_A),$$ $$c_2(E) = \frac{1}{2} \left( \frac{i}{2\pi} \mathrm{Tr}(\Omega_A) \right)^2 - \frac{1}{2} \mathrm{Tr}\left( \frac{i\Omega_A}{2\pi}\right)^2 = \frac{1}{2}c_1(E)^2 + \frac{1}{8\pi^2} \mathrm{Tr}(\Omega_A \wedge \Omega_A),$$ and so on until $$c_m(E) = \left( \frac{i}{2\pi} \right)^m \det(\Omega_A).$$

Let $M$ be a complex manifold with complex structure $J$ on its tangent bundle. Then the bundle $S^{1,1}M$ of real symmetric $J$-invariant forms on $TM$ and the bundle $\wedge^{1,1} T^\ast M$ of $2$-forms of type $(1,1)$ on $M$ can be put in one-to-one correspondence by associating to $b \in S^{1,1}M$ the $2$-form $\beta \in \wedge^{1,1} T^\ast M$ defined by $$\beta(X,Y) = b(JX,Y)$$ for all $X, Y \in T_x M$ and $x \in M$; if $b$ is positive definite (resp. negative definite), then $\beta$ is called positive (resp. negative). Then a cohomology class $\alpha \in H^2(M;\mathbb{R})$ is positive (resp. negative) if it can be represented by a real positive (resp. negative) $2$-form of type $(1,1)$. This notion of positivity/negativity depends only on the complex structure $J$.

The zero set in $\mathbb{C}P^n$ of homogeneous degree $k$ polynomial $p$ is a smooth manifold when $p$ satisfies $(\nabla p)(Z) \neq 0$ for any $Z \neq 0$ such that $p(Z) = 0$. We call the zero set of such a polynomial the non-singular complex hypersurface of degree $k$ in $\mathbb{C}P^n$ and denote it by $V^n(k)$; using Ehresmann's fibration theorem one can show that the diffeomorphism class of $V^n(k)$ depends only on $n$ and $k$ (and not on the polynomial $p$). Using the adjunction formula, one finds that $$c_1(V^n(k)) = (n + 1 - k)\omega,$$ where $\omega$ is the pullback of the Fubini-Study form on $\mathbb{C}P^n$ to $V^n(k)$. The Fubini-Study form is of course positive. Hence one sees that $c_1(V^n(k))$ is positive for $k \leq n$, zero for $k = n + 1$, and negative for $k \geq n + 2$. In particular, $V^n(n+1)$ is a Calabi-Yau manifold (not $V^n(n)$ as you said).


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