Preparation

Adaquate Equivalence Relations

On a smooth curve,

Zero Cycles on Surfaces

Investigating \(T(X)\) in different settings.

\(A_0(X)\) and \(T(X)\) are divisible.

First Example: Hyperelliptic

Let \(E\) and \(F\) be elliptic curves, with \(\eta\in F\) a point of order \(2\). Define an involution \(\sigma\) on \(F\times E\) by \(\sigma(f,g)=(-f,g+\eta)\), and let \(X=(F\times E)/\sigma\). Let \(E'=E/\{1,\eta\}\) we then conclude a diagram

therefore all the fibres of \(\rho\) should be isomorphic to \(F\). To see this, track the inverse image of the origin \(o\in E'\). \(o\) goes upward to \(\{o,\eta\}\in E\) then back to two copies of \(F\) in \(F\times E\), which are identified under quotient of the involution \(\sigma\) (reflection).

Note that the Albanese variety to \(X\) is defined as \(\mathrm{H}^0(X,\Omega_{X}^1)^*/\mathrm{H}_1(X,\mathbb{Z})\), which is the initial object in the category of abelian varieties under \(X\). That is, \(\mathrm{Alb}(X)\) must be the largest abelian variety admitting a morphism from \(X\), in the sense of dimension. Here \[ \begin{aligned} \mathrm{H}^0(X,\Omega_{X}^1)&=\mathrm{H}^0(F\times E,\Omega_{F\times E}^1)/\sigma\ (to\ be\ explained)\\ &=\left(\mathrm{H}^0(F,\Omega_{F}^1)\oplus\mathrm{H}^0(E,\Omega_{E}^1)\right)/\sigma\\ &=\mathrm{H}^0(E,\Omega_{E}^1)=\mathbb{C}. \end{aligned} \] Hence \(\mathrm{Alb}(X)\) has dimension \(1\), and the induced morphism \(\mathrm{Alb}(X)\rightarrow E'\) is quasi-finite. Since \(\rho\) has connected fibres, \(\mathrm{Alb}(X)\rightarrow E'\) is an isomorphism.

Claim \(A_0(X)\cong E'=\mathrm{Alb}(X)\), or \(T(x)=0\). If \(z\in T(X)\), we have \[ \pi^*z=\sum_i r_i\left((q_i,p_i)+(-q_i,p_i+\eta)\right) \] where \(\sum r_i=0\) (algebraically equivalent to \(0\)) and \(\sum 2r_ip_i=0\) (maps to \(0\in E'\)). According to the arithmetic of elliptic curves, \[ (p)+^{pic}(q)\sim (p+^{ellip}q)+(\infty) \] so \[ \begin{aligned} 2(q,p)&\sim 2(q,p+\eta)\\ (q,p)+(-q,p)&\sim 2(0,p) \end{aligned} \] on \(F\times E\). Therefore \[ \begin{aligned} 2\pi^*z&\sim\sum 2r_i\left((q_i,p_i)+(-q_i,p_i+\eta)\right)\\ &\sim\sum 4r_i(0,p_i)\\ &\sim [(0)\times E]\cdot[F\times\sum 4r_i(p_i)]\\ &\sim 0. \end{aligned} \] It follows that \(0\sim 2\pi_*\pi^*z=4z\) (is a cover with two branches), hence \(4T(x)=0\), which implies that \(T(X)=0\) since \(T(X)\) is divisible.

Second Example: Surface of degree \(5\)

Let \(Y\) be the subvariety of \(\mathbb{P}^3\) cut out by \(T_0^5+T_1^5+T_2^5+T_3^5=0\). Define an action of \(\mathbb{Z}/5\mathbb{Z}\) on \(Y\) by \[ (t_0,t_1,t_2,t_3)\mapsto (t_0,\omega t_1,\omega^2 t_2,\omega^3 t_3) \] where \(\omega\) is a primitive \(5\)-th root of unity. This action is fixed point free, so the quotient \(X=Y/(\mathbb{Z}/5\mathbb{Z})\) is smooth.

Claim \(A_0(X)=\{0\}\).

Let \(\pi:Y\rightarrow X\) be the projection. Since \(A_0(X)\) is divisible, it suffices to show \(5A_0(X)=0\), or \(\pi_*\pi^*A_0(X)=0\). Since \(\pi_*\) is surjective, it suffices to show \(\pi^*\pi_*:A_0(Y)\rightarrow A_0(Y)\) is the zero map.

Define an action of \((\mathbb{Z}/5\mathbb{Z})^{\oplus 3}\) on \(Y\) by \[ e_i^*T_j=\omega^{\delta_{ij}}T_j. \]

This gives a representation \(\mathbb{Z}[\mathbb{Z}/5\mathbb{Z}^{\oplus 3}]\rightarrow\mathrm{End}(A_0(Y))\) sending \(x\) to \(\bar{x}\), which image is denoted by \(R\subset\mathrm{End}(A_0(Y))\). It follows immediately That \[ \pi^*\pi_*=1+\bar{e}_1\bar{e}_2^2\bar{e}_3^3+(\bar{e}_1\bar{e}_2^2\bar{e}_3^3)^2+\cdots+(\bar{e}_1\bar{e}_2^2\bar{e}_3 ^3)^4. \]

Observation The quotient of the surface \(Y\) under any of the following groups

• \(<1,e_i,\cdots,e_i^4>\)
• \(<1,e_ie_i,\cdots,e_i^4e_i^4>\)
• \(<1,e_1e_2e_3,\cdots,e_1^4e_2^4e_3^4>\)

is rational (birational to \(\mathbb{P}^2\)). To prove it…

As a result, \(A_0(Y/G)\) must be trivial when \(Y/G\) is rational, which could be seen easily using moving lemma (just to push the zero cycle away from the exceptional loci). As a result, every zero cycle of zero degree is rational. Denote by \(\pi_G\) the projection from \(Y\) to \(Y/G\), for each zero cycle \(z\in Y/G\) we have \[ \int_{g\in G} g^*z=\pi_{G*}\pi^*z \] so \[ \sum_{n=1}^4\bar{e}_i^n=\sum_{n=1}^4\bar{e}_i^n\bar{e}_j^n=\sum_{n=1}^4\bar{e}_1^n\bar{e}_2^n\bar{e}_3^n=0 \] in \(R\subset\mathrm{End}(A_0(Y))\).

Now go back to the proof of the claim.

Family of Lines on a Hypersurface

Case 1

\(X\subset\mathbb{P}^4\) a smooth hypersurface of degree \(3\) (cubic threefold). The family of lines on \(X\) forms a smooth connected surface \(S\subset\mathrm{Gr}(2,5)\) called the Fano surface (not Del Pezzo surface). (A. N. Tyurin, Five lectures on three-dimensional varieties (in Russian), Uspehi Mat. Nauk, 27 (1972), no. 5, (167) 3–50. [Translation: Russian Math. Surveys, 27 (1972), no. 5, 1–53.])

The Fano surface \(S\) of a smooth cubic threefold \(F\) into \(\mathbb{P}^4\) carries many remarkable geometric properties. The surface \(S\) is naturally embedded into the grassmannian of lines \(\mathrm{Gr}(2,5)\) of \(\mathbb{P}^4\). Let \(U\) be the restriction to \(S\) of the universal rank \(2\) bundle on \(G\). We have the:

Tangent bundle Theorem (Fano, Clemens-Griffiths, Tyurin) The tangent bundle of \(S\) is isomorphic to \(U\).

Torelli-type Theorem Let \(g'\) be the natural morphism from \(S\hookrightarrow \mathrm{Gr}(2,5)\) defined by the cotangent sheaf of \(S\) generated by its 5-dimensional space of global sections (to add some recall on morphism to grassmannians). Let \(F'\) be the union of the lines corresponding to \(g'(S)\). The threefold \(F'\) is isomorphic to \(F\).

Case 2

\(X\subset\mathbb{P}^4\) a smooth hypersurface of degree \(4\). The family of lines on \(X\) is known to form a connected (in general, smooth) curve \(F\subset\mathrm{Gr}(2,5)\). (S. Bloch and J. P. Murre, On the Chow groups of certain types of Fano threefolds, Compositio Math., 39 (1979), 47–105.)

Geometry of the Fano Surface

(Example 1.7) On the Fano Surface, Incidence Divisor

On a Fano surface \(S\), intersection of divisors gives bilinear maps \[ \mathrm{Pic}(S)\otimes_{\mathbb{Z}}\mathrm{Pic}^0(S)\rightarrow A_0(S).\quad\tag*{(A)} \] Restrict the first variable to divisors algebraically equivalent to \(0\), we can see the morphism \[ \mathrm{Pic}^0(S)\otimes_{\mathbb{Z}}\mathrm{Pic}^0(S)\xrightarrow{\varphi} A_0(S)\rightarrow\mathrm{Alb}(S) \] is a continuous bilinear map between compact tori, hence \(\mathrm{Mor}(\mathrm{Pic}^0(S),\mathrm{Alb}(S))\) is descrete, which implies that the image of \(\varphi\) lies in \(T(S)\).

Claim Both the morphism (A) and \[ \mathrm{Pic}^0(S)\otimes_{\mathbb{Z}}\mathrm{Pic}^0(S)\rightarrow T(S)\quad\tag*{(B)} \] are surjective.

proof

Intermediate Jacobian

Main example: smooth hypersurface \(X\subset\mathbb{P}^4\) of degree \(4\), which witnesses which \(A^2\) concides with \(J^2\). Here \(J^2\) is the intermediate jacobian.

TBS: the \(r\)-th Intermediate jacobian on \(X\) is defined as \[ J^r(X)=(F^{n-r+1}\mathrm{H}^{2n-2r+1}(X,\mathbb{C}))^*/\mathrm{H}_{2n-2r+1}(X,\mathbb{Z}) \] where \(F^\bullet\) is the Hodge filtration. According to the Hodge decomposition \[ \mathrm{H}^k(X,\mathbb{C})=\mathrm{H}^{k,0}\oplus\cdots\oplus\mathrm{H}^{0,k} \] the \(r\)-th filtration \(F^r\mathrm{H}^k(X,\mathbb{C})\) is \[ F^r\mathrm{H}^k(X,\mathbb{C})=\mathrm{H}^{k-r,r}\oplus\cdots\oplus\mathrm{H}^{k,0}. \]

In the case of \(r=n\) the intermediate jacobian \(J^n(X)\) is just \[ (F^1\mathrm{H}^1(X,\mathbb{C}))^*/\mathrm{H}_{1}(X,\mathbb{Z})=\mathrm{H}^0(X,\Omega_{X}^1)^*/\mathrm{H}_{1}(X,\mathbb{Z})=\mathrm{Alb}(X) \] which coincides with the ordinary Jacobian.

Why is Intermediate Jacobian a Complex Torus

An alternative definition of the intermediate jacobian is \[ J^r(X)=\mathrm{H}^{2r-1}(X,\mathbb{C})/(F^r) \]

To see such a construction is indeed a complex torus, one first translate using Poincaré duality, where