Covariant Motives
(Just the inverse category of the category of pure motives, but let’s rephrase it)
Structure of the Pure Motive of a Curve
Over a field \(k\). Say there is a (geometrically integral) smooth proper curve \(C/k\). Hence the field of constants of \(C\) is just \(k\). Suppose there is a rational point
\[ k\xrightarrow{\alpha} C\xrightarrow{struc} k \]
We see immediately that \(h(C)\) admits a direct summand of \(h(k)\). Moreover, take the transportation of the morphisms above, we have
\[ k\xrightarrow{struc^t} C\xrightarrow{\alpha^t} k \]
Let \(d=\mathrm{dim}(C)\)(this must be \(1\) but let’s just use this signature for wider generality), by passing to the category of covariant motives we have
\[ h(struc^t)\in\mathrm{Corr}_{d}(k,C)\\ h(\alpha^t)\in\mathrm{Corr}_{-d}(C,k) \]
so we have another composition calculates to identity, namely
\[ h(k)\xrightarrow{h(struc^t)} h(C)(-d)\xrightarrow{h(\alpha^t)} h(k) \]
So we find another summand \(h(k)(d)\subset h(C)\).
These two summands are denoted by \(h^0(C)\) and \(h^2(C)\) repspectively. From the definition it is easy to see that they are cut out by projectors \(p_0=[C\times\{x\}]\) (the graph collapsed to a single point, indeed) and \(p_2=[\{x\}\times C]\) respectively. In fact (I still feel like explaining further),
\[h(C)\xrightarrow{h(struc)}h(k)\xrightarrow{h(\alpha)}h(C)\]
defines the projectors by
\[ p_0=h(\alpha)\circ h(struc)\in\mathrm{Corr}_0(C,C) \\ p_2=h(struc^t)\circ h(\alpha^t)\in\mathrm{Corr}_0(C(-d),C(-d))=\mathrm{Corr}_0(C,C). \]
(Omit the calculation here)
For some curves, all parts they have is \(h^0\) and \(h^2\). For example, \(h(\mathbb{P}^1)=h^0(\mathbb{P}^1)\oplus h^2(\mathbb{P}^1)=\mathbb{1}\oplus\mathbb{L}\). This is because
\[ [\mathbb{P}^1\times \{x\}] + [\{x\}\times \mathbb{P}^1] \sim_{rat} \Delta_{\mathbb{P}^1\times\mathbb{P}^1} \]
in fact, the divisor \(\Delta-[\mathbb{P}^1\times \{x\}] - [\{x\}\times \mathbb{P}^1]\) is the principle divisor defined by \((XZ-YW)/ZW\) if the coordinations are denoted by \([X:Y]\) and \([X:W]\) respectively.
In general, we could define \(h^1(C)=h(C)-h^0(C)-h^2(C)\) (or strictly, the part cut off by \(\mathrm{id}_C-p_0-p_2\)).
Preliminaries on Line Bindles
Degree problem, degree \(0\) criterion
CHATGPT START
Great question! On an abelian variety \(A/k\) there are several equivalent ways to say a line bundle is “degree 0.” The condition you wrote
\[ m^{*}L\;\simeq\;p^{*}L\otimes q^{*}L\qquad\text{on }A\times A \]
is one of the standard characterisations of algebraically trivial / translation–invariant line bundles, i.e. of \(\Pic^0(A)\). Here’s the clean picture and why this really matches the “degree 0” intuition.
1) Why \(m^{*}L\simeq p^{*}L\otimes q^{*}L\) singles out \(\Pic^0(A)\)
For any line bundle \(L\) on \(A\), define the translation-difference map
\[ \phi_L:A\longrightarrow \Pic^0(A)=A^\vee,\qquad a\longmapsto T_a^{*}L\otimes L^{-1}. \]
By Weil’s theorem of the square/cube, \(\phi_L\) is a group homomorphism and the line bundle
\[ \Lambda(L):=m^{*}L\otimes p^{*}L^{-1}\otimes q^{*}L^{-1} \]
is exactly the pullback \((1\times\phi_L)^{*}\mathcal P\) of the Poincaré bundle \(\mathcal P\) on \(A\times A^\vee\). Hence
\[ \Lambda(L)\text{ is trivial }\Longleftrightarrow \phi_L=0 \Longleftrightarrow m^{*}L\simeq p^{*}L\otimes q^{*}L. \]
But \(\phi_L=0\) holds precisely when \(L\) is algebraically equivalent to \(\mathcal O_A\), i.e. \(L\in\Pic^0(A)\). So your condition is equivalent to “\(L\) lies in \(\Pic^0(A)\).” (A slick proof uses that \(\Lambda(L)=(1\times\phi_L)^*\mathcal P\); if \(\phi_L=0\) then \(\Lambda(L)\) is trivial, giving the displayed identity, and conversely.) ([Virtual Math][1], [Southwest Center for Arithmetic Geometry][2])
A quick sanity check: restrict the identity along \(i_a:A\to A\times A,\ x\mapsto(a,x)\). Then
\[ i_a^*(m^*L)\simeq T_a^*L,\quad i_a^*(p^*L)\simeq\mathcal O_A,\quad i_a^*(q^*L)\simeq L, \]
so you get \(T_a^*L\simeq L\) for all \(a\), i.e. translation invariance, which is another common description of \(\Pic^0(A)\). ([Yifeng Huang’s Homepage][3], [Climbing Mount Bourbaki][4])
2) So… why call these “degree 0”?
On a curve \(C\), \(\Pic^0(C)\) literally means “line bundles of degree \(0\).” In higher dimension there isn’t a single integer degree, but the right replacement is numerical triviality (all intersection numbers with curves vanish), or equivalently \(c_1(L)=0\) in the Néron–Severi group:
\[ \Pic^0(A)=\ker\big(\Pic(A)\twoheadrightarrow \mathrm{NS}(A)\big),\qquad \mathrm{NS}(A)=\Pic(A)/\Pic^0(A). \]
For abelian varieties, algebraic and numerical equivalence coincide for divisors, and \(\mathrm{NS}(A)\) is torsion-free. Thus \(L\in\Pic^0(A)\) iff \(c_1(L)\) is numerically \(0\); with any polarization \(H\) one has \(\deg_H(L):=c_1(L)\cdot H^{g-1}=0\). That’s why these bundles are universally called “degree 0.” ([Wikipedia][5], [arXiv][6], [jmilne.org][7])
There’s a third, very useful “degree” viewpoint: the homomorphism \(\phi_L:A\to A^\vee\) above.
- If \(L\in\Pic^0(A)\), then \(\phi_L=0\) (so in this sense its “degree” is \(0\)).
- If \(L\) is ample, then \(\phi_L\) is an isogeny; its (finite) degree is the usual “degree” of a polarization, and equals a top intersection number of \(c_1(L)\). ([Yifeng Huang’s Homepage][3])
3) Summary of equivalent characterizations
For a line bundle \(L\) on an abelian variety \(A\), the following are equivalent:
- \(L\in\Pic^0(A)\) (connected component of the identity in the Picard scheme / algebraically trivial). ([MIT Mathematics][8])
- \(m^{*}L\simeq p^{*}L\otimes q^{*}L\) on \(A\times A\) (biextension condition). ([Virtual Math][1])
- \(T_a^{*}L\simeq L\) for all \(a\in A\) (translation invariance). ([Yifeng Huang’s Homepage][3])
- \(\phi_L=0:A\to A^\vee\) (the translation–difference map is zero). ([Virtual Math][1])
- \(c_1(L)=0\) in \(\mathrm{NS}(A)\), i.e. \(L\) is numerically trivial (hence “degree 0”
Jacobian of a general curve (not necessarily having a complex structure): having points corresponding to
CHATGPT END
Poinc'are bundle on \(X\times\mathrm{Pic}_X\):
Abel-Jacobi map \(X\rightarrow J_X\) for a curve:
Weil’s Formula on Jacobians and Picard Groups
A classical theorem (Weil) says that divisor classes on a product of curves (we may assume the curves both admit a rational point, but maybe this condition could be relaxed) split as
\[ \mathrm{Pic}(X\times X')\;\cong\;\mathrm{Pic}(X)\ \oplus\ \mathrm{Pic}(X')\ \oplus\ \mathrm{Hom}(J,J'). \]
Intuitively:
- the first two summands are the pullbacks from the two factors, and
- the cross-term corresponds to line bundles on \(X\times X'\) that restrict trivially to each fibre; by the seesaw theorem these are exactly the line bundles coming from homomorphisms \(J\to P_{X'}\) (and for curves \(P_{X'}\cong J'\)).
An observation: each fibre \(X\times x'\) where \(x'\in X'\)
[Statement]
(Better state it clearly here)
Proposition a
If \(X\), \(X'\) are curves with Jacobian varieties \(J\), \(J'\), then
\[\mathrm{Hom}(h^1(X),h^1(X'))=\mathrm{Hom}(J,J')\otimes\mathbb{Q}.\]
To see it, note that \(\mathrm{Hom}(h^1(X),h^1(X'))\) is just
\[ (\Delta_{X'}-p_{0,X'}-p_{2,X'})\circ\mathrm{Corr}_0(X,X')\circ(\Delta_X-p_{0,X}-p_{2,X}) \]
The central term \(\mathrm{Corr}_0(X,X')=\mathrm{A}_1(X\times X')\) calculates to three partitions
- \(\mathrm{Pic}(X)\) pulling back to \(X\times X'\). This is a divisor concentrating on the first projection, or a cylinder \(c\) of type \((cycle\times X')/\sim\). In this case \(\Delta_{X'}\circ c=c=p_{0,X'}\circ c\) and \(p_{2,X'}\circ c=0\), so such term vanishes in composition.
- Symmetrically another, also vanishes.
- \(\mathrm{Hom}(J,J')\otimes\mathbb{Q}\). For a class \(c\) in such part, there must be \(p_{0,X'}\circ c=p_{2,X'}\circ c=c\circ p_{0, X}=c\circ p_{2,X}\). In fact, (fact what?)
Proposition a’
Furthermore,
Higher Case of the Decomposition
Remaining Problem: Identifying \(\mathrm{Hom}(J,J')\) for a Curve. What is this group viewed as a subgroup of cycles?
Associated to a variety \(X/k\) there are two abelian varieties: the Albanese variety \(\mathrm{J}_{X/k}\) and the Picard variety \(\mathrm{Pic}_{X/k}\), when the former one serves as the initial object of the category of the morphisms from \(X\) to abelian varieties over \(k\), given a distinguished point \(x_0\in X\), and the later functor being the \(fppf\)-sheafication of
\[ S/k\mapsto\{\mathcal{L}\in\mathrm{Pic}(X\times_k S)/p^*\mathrm{Pic}(S)\} \]
Moreover, when \(X(S)\) is nonempty, \(\mathrm{Pic}_{X/k}(S)\) remains still after sheafication. Such functor is representable as an abelian variety (condition?).
What we care most is the degree \(0\) part of the picard variety, namely \(P_{X/k}\).
Theorem
Let \(X\) (connnected) and \(Y\) be of dimensions \(d\) and \(e\) respectively.
- \[\mathrm{Hom}(J_X,P_Y)\otimes\mathbb{Q}=\frac{A^1(X\times Y)}{p_1^*A^1(X)+p_2^*A^1(Y)};\]
- Let \(\zeta\in A_0(X)\), \(\eta\in A_0(Y)\) be zero cycles of positive degree. Then there is an isomophism \[ \Omega:\mathrm{Hom}(J_X,P_Y)\otimes\mathbb{Q}\rightarrow \{c\in A^1(X\times Y)\;|\;c\circ\zeta_*=0\;and\;\eta^*\circ c=0\}.\]
What is (ii) talking about? In fact, \(c\) could be viewed as \[c\in A^1(X\times Y)=\mathrm{Corr}_{e-1}(X,Y)\] (in the case of curve this is correspondence of degree \(0\)), while \(\xi\in A_0(X)=\mathrm{Corr}_0(1,h(X))\) and \(\eta\in A_0(Y)=A^e(Y)=\mathrm{Corr}_0(h(Y),\mathbb{L}^e)\).
Here an maps to arrow with degree \(e\) means it sends a degree \(0\) motive to a degree \(0+e\) motive.
((ii) is derived from (i), we pretend to know it right now.)
For (i), suppose \(X(k)\) and \(Y(k)\) are both nonempty. Pick an \(x_0\in X\) and
\[ \mathrm{Hom}(J_X,P_Y)=\{\phi:X\rightarrow P_Y\;|\;\phi(x_0)=0\}\\ =\frac{\{\mathcal{L}\in\mathrm{Pic}(X\times Y)\;|\;\mathcal{L}_{x_0\times Y}\cong\mathcal{O}_Y\}}{p^*\mathrm{Pic}(X)}\\ =(seesaw\ theorem)\frac{\mathrm{Pic}(X\times Y)}{p_1^*\mathrm{Pic}(X)+p_2^*\mathrm{Pic}(Y)} \]
Then take the tensor product with \(\mathbb{Q}\).
The point: since we require a morphism from \(J_X\) to \(P_Y\) instead of \(X\) to \(P_Y\), the induced morphism \(X\rightarrow P_Y\) must cover the origin of the picard variety, so the image should lie in the degree-\(0\) part naturally. Typically, the Albanese variety is connected (citation needed?).
PY
Another Topic
Chow Group Filtration
For \(X/k\) equidimensional, write \(\mathbf{t}^m\mathrm{CH}^p(X_L)\) for the image of the pullbacks
\[ \mathrm{CH}^p(X_K)\rightarrow\mathrm{CH}^p(X_L) \]
where \(K\hookrightarrow L\) over \(k\) ranges over all intermediate fields \(k\subset K\subset L\) where \(\mathrm{tr.deg.}(K/k)\leq m\). Such trancation gives a filtration
\[ \varnothing=\mathbf{t}^{-1}\mathrm{CH}^p(X_L)\subset\mathbf{t}^0\mathrm{CH}^p(X_L)\subset\cdots \]
Then we have
[Proposition] The increasing filtration tower stablizes at \(\mathbf{t}^p\mathrm{CH}^p(X_L)\).
In fact, \(\mathbf{t}^p\mathrm{CH}^p(X_L)=\mathrm{CH}^p(X_L)\). First we do spread-out: For a cycle \(\mathfrak{p}\in X_L\), take \(V_{\mathfrak{p}}\) to be it’s closure in \(X_L\) and $$
We then have graded components \(\mathrm{Gr}_t^m\mathrm{CH}^p(X_L)=\mathbf{t}^m\mathrm{CH}^p(X_L)/\mathbf{t}^{m-1}\mathrm{CH}^p(X_L)\).
[Property] The action of correspondences preserves the transcendental filtration on Chow groups and induces the action on the corresponding graded pieces.
To be precise this is \(\mathrm{Corr}^i(X_L, Y_L)\) pulling back from truncated \(\mathrm{CH}^*(X_L)\) to \(\mathrm{CH}^*(Y_L)\) given a fixed \(L/k\).