### Invariants of projective space II: Cycles and Bundles

Thursday, December 06th, 2012 | Author:

I want to explain a particularly easy example of a motivic cellular decomposition: That of $n$-dimensional projective space. The discussion started with cohomology (part 1) and in this part 2, we discuss intersection-theoretic and bundle-theoretic invariants. In part 3 we will see the motivic stuff.

The invariants discussed in this article are closely related and I wrote an article about the divisorial jungle before.

#### Divisor class group, Picard group

Since projective space is an integral scheme of finite type, smooth over the base, we have four different isomorphic characterizations of the Picard group. It is the group of invertible sheaves (lines bundles) with tensor product, up to isomorphism $Pic(\mathbb{P}^n)$. It is isomorphic to the Cartier class group $CaCl(\mathbb{P}^n)$ of Cartier divisors modulo principal divisors, which is isomorphic to the sheaf cohomology group $H^1(\mathbb{P}^n,\mathcal{O}_{\mathbb{P}^n}^\times)$. It is the Class group $Cl(\mathbb{P}^n)$ of Weil divisors modulo principal divisors.

Any hypersurface $D$ of degree $d$ in projective $n$-space is linearly equivalent to the Weil divisor $d H$, where $H = \{[x_0:\cdots:x_n] \in \mathbb{P}^n : x_0=0\}$ is a (generic) hyperplane. In particular, all hyperplanes are linearly equivalent to each other. The degree of $H$ is $1$, so the degree homomorphism $Cl(\mathbb{P}^n) \to \mathbb{Z}$ is an isomorphism. In particular, $Cl(\mathbb{P}^1)$ is non-trivial. One can show $D \sim d H$ by first proving $deg(f)=0$ for $(f)$ the divisor of a non-zero rational function $f \in \mathcal{K}_{\mathbb{P}^n}^\times$ and then writing $D = (g)-(h)$ for $g,h$ some rational functions, so $D-dH = (g/x_0^d h)$.

The line bundles on $\mathbb{P}^n_k$, for $k$ a field, are generated (up to isomorphism) by the tautological bundle $\mathcal{O}(-1)$, its dual $\mathcal{O}(1)$ and their tensor powers $\mathcal{O}(n)$, which are all called twisting sheaves. The usual isomorphism $Cl(X) \to Pic(\mathbb{P}^n)$ identifies $H$ with $\mathcal{O}(1)$.

#### Chow ring

The Chow ring $A^\bullet(\mathbb{P}^n)$ of $\mathbb{P}^n$ is the graded ring whose $k$-th graded component is the Chow group $A^k(\mathbb{P}^n)$ of codimension $k$ cycles up to rational equivalence, with ring structure coming from the intersection product. Here, we consider Chow groups with integer coefficients.

Let us first look at $A^\bullet(\mathbb{P}^1)$. Can we come up with any cycles, hopefully non-trivial ones? In dimension $0$, that is codimension $1$, we have the points $p \in \mathbb{P}^1$ which are all rationally equivalent, since any two points $p,q$ in $\mathbb{P}^1$ can be joined by a $\mathbb{P}^1$ via the map $\phi : [x_0:x_1] \mapsto [x_0p_0+x_1q_0,x_0p_1+x_1q_1]$ (check $\phi([1:0]) = p, \phi([0:1]) = q$). In dimension $1$, that is codimension $0$, there is $\mathbb{P}^1$ and that's the only closed subvariety, since $\mathbb{P}^1$ is irreducible. In higher or lower codimensions, there is nothing.

The trick that shows that two points on $\mathbb{P}^1$ are rationally equivalent actually works for hyperplanes in $\mathbb{P}^n$, and we have seen (in the section above) that one can show (using linear equivalence of divisors) that these are all classes of cycles. Since rational equivalence of codimension 1 cycles coincides with linear equivalence of divisors, this shows that $A^\bullet(\mathbb{P}^1) = \mathbb{Z}[H]/(H^2)$, for $H$ the class of a hyperplane $\{[x_0:x_1] : x_0=0\}$ (in fact, a point).

The structure of $A^k(\mathbb{P}^n)$ looks less clear at first, but one can show that it is generated by any codimension $k$ linear subspace via the following "excision lemma":

Lemma: Let $i : Y \to X$ be a closed immersion, $j : U := X \setminus Y \to X$ the open complement, then the sequence $A_k(Y) \to A_k(X) \to A_k(U) \to 0$ is exact for all $k$ (where the arrows are pushforward by $i$ and $j$, so we need to take the grading by dimension).

We apply this with $k=n-1$, $X = \mathbb{P}^n$, $Y = H$ a hyperplane, and then $U = \mathbb{A}^n$. Then we use $A_k(\mathbb{A}^n)=0$ for $k < n$ to get that $i_\ast : A_{n-1}(H) \to A_{n-1}(\mathbb{P}^n)$ is a surjection. In fact, also $A_k(H) \to A_k(\mathbb{P}^n)$ is a surjection for $k < n$, and because of $H \simeq \mathbb{P}^{n-1}$ we can use induction. A small remark: the excision lemma can be improved to a long exact localization/Gysin sequence for higher Chow groups, and the next term on the left would be $A_k(U,1)$, which gives a much smoother proof. The localization sequence, on the other hand, is very very hard to prove (at least that's what Voevodsky writes, so I believe it). To see that there are no relations in $A^k(\mathbb{P}^n)$ (for $0 \leq k \leq n$), i.e. $A^k(\mathbb{P}^n) = \mathbb{Z} [H^k]$ with $H^k$ the class of a codimension $k$ linear subspace, we look at the cases $k > 1$ (since the case $k=0$ is clear and we have seen the case $k=1$ above, that was the divisor-setting). Any relation would be of the form $d H^k = \sum n_i [div(r_i)]$ for $r_i \in \mathcal{K}^\times(V_i)$ some rational functions. Let $Z := \bigcup V_i$ and $f : Z \to \mathbb{P}^{n-k+1}$ the projection to a linear $k-2$-dimensional subspace disjoint from $Z$. Then $f$ is proper, so there is proper pushforward $f_\ast : A_\bullet(Z) \to A_\bullet(\mathbb{P}^{n-k+1})$ which can not annull $d H^k$ because of compatibility with the degree morphism $deg : A_\bullet(Z) \to \mathbb{Z}$ (and $d H^k$ has degree $d$ over $\mathbb{P}^{n-k+1}$), but has to annul $\sum n_i [div(r_i)]$.

We see that two hyperplanes in $\mathbb{P}^n$ in general position intersect to a linear subspace of codimension $2$, and in general $H^i \cdot H^j = H^{i+j} \in A^\bullet(\mathbb{P}^n)$, so we have
$A^\bullet(\mathbb{P}^n) = \mathbb{Z}[H]/(H^{n+1})$.

#### Algebraic K-Theory

From the general theory, we can use smoothness of $\mathbb{P}^n_k$ over a perfect field $k$ to deduce that the Chern character with rational coefficients $K^\bullet(\mathbb{P}^n_k)_{\mathbb{Q}} \to A^\bullet(\mathbb{P}^n_k)_{\mathbb{Q}}$ is an isomorphism. I think this also works for $\mathbb{P}^n$ over $\mathbb{Z}$, since only regularity is used. However, the rational coefficients are quite unsatisfying. At least from $K^\bullet(\mathbb{P}^n_k)_{\mathbb{Q}} \simeq \mathbb{Q}[H]/(H^{n+1})$ we can guess how $K^\bullet(\mathbb{P}^n)$ might look like. Or can we? Actually it is way more complicated than the rationalized picture, and I think the general answer is still unknown!

Serre proved a theorem, that every coherent sheaf on $\mathbb{P}^n$ admits a surjective map of some $\mathcal{O}(k)^m$ for positive $k,m$. This shows that $K_0(\mathbb{P}^n)$ is generated by the set of all $\mathcal{O}(k)$. Furthermore, one can show that there is a relation in $K_0$ between any set of $n+2$ coherent sheaves on $\mathbb{P}^n$, by analyzing the Koszul complex on $\mathbb{A}^{n+1}$.

For $\mathcal{E} \to X$ a rank $k+1$ vector bundle on a quasiprojective variety $X$, one can look at the projective bundle $\mathbb{P}(\mathcal{E}) \to X$ which is fiber-wise a projective space of the fiber. On the total space $\mathbb{P}(\mathcal{E})$ one has canonically the tautological bundle and its tensor powers $\mathcal{O}(-i)$, which generate (for $i=0,...,k$) $K_0(\mathbb{P}(\mathcal{E})$. One can take a vector bundle $V \to X$, pull it back to $\mathbb{P}(\mathcal{E})$ and twist it (i.e. tensor it with some $\mathcal{O}(-i)$). This gives functors $u_i : VB(X) \to VB(\mathbb{P}(\mathcal{E}))$.

The projective bundle theorem for algebraic K-Theory says that these functors $u_i$ induce an equivalence $K(X)^{r+1} \simeq K(\mathbb{P}(\mathcal{E}))$ and $K_\bullet(X) \otimes_{K_0(X)} K_0(\mathbb{P}(\mathcal{E})) \to K_\bullet(\mathbb{P}(\mathcal{E}))$ is a ring isomorphism.

If we apply this to a trivial rank $k+1$ vector bundle $\mathcal{E}$, where $\mathbb{P}(\mathcal{E}) = \mathbb{P}^k_X$, we see that $K_\bullet(\mathbb{P}^k_X) \simeq K_\bullet(X)[z]/[z^{k+1}]$.

In particular, $K_\bullet(\mathbb{P}^k) \simeq K_\bullet(\mathbb{Z})[z]/[z^{k+1}]$. But the higher algebraic K-groups of the integers are not known! One might naively guess that it becomes better by taking a field as base, but fields (even algebraically closed ones) have a very rich higher K-theory, too.

Still, the projective bundle theorem shows us that the K-Theory of projective space isn't really complicated, since all the complexity lies in the K-Theory of the base.

#### Some afterthoughts

By the way, I realise during writing this that I would love to hear something about other isomorphism invariants I'm missing, like the Kodaira dimension, existence of a spin structure on the associated complex manifold, volume of the Fubini-Study metric, geodesic completeness, Lusternik-Schnirelman category, whatever... I fell in love with the idea of computing everything of $\mathbb{P}^n$. Please tell me your favourite isomorphism invariant of $\mathbb{P}^n$ or $\mathbb{CP}^n$ or $\mathbb{RP}^n$ in the comments!

Category: English, Mathematics