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 \(L/k\). (I have to be pretty sure?)
Transcendental Filtration on Cycles
Assume we have a universal field \(\Omega\) chosen.
\[ \varnothing=\mathbf{t}^{-1}\mathrm{CH}^*(X_L)\subset \mathbf{t}^0\mathrm{CH}^*(X_L)\subset\cdots \]
Let \(C\) be a smooth dim \(1\) cycle of \(X\) \[ P_X\rightarrow P_C \]
pulling back: Sending a divisor/~ (or a line bundle) to a point/~ of \(C\)
\[ P_C\rightarrow J_C \]
Identifying Picard & Abanese varietties of a curve, there is nothing to mention
\[ J_C\rightarrow J_X \]
Pushing out a point/~ to a point/~.
In other words this is just a duality between \(CH^1\) and \(CH_1\). However, the question is, why is this morphsim an isogeny (morphism between \(CH^1\otimes\mathbb{Q}\) and \(CH_1\otimes\mathbb{Q}\))? (Andre Weil) Also, do we have other dualities?
We could say that such a morphism sends a hyperplane class to its intersection with the given (smooth) curve. Q2: Why is this morphism independent with the choice of the curve?
- The section \(C\) is chosen to be a linear section! But in order to talk about what is linear, we need an ample bundle on \(X\), or an immersion from \(X\) to the projective space. The isogeny should depend on the choice of the ample bundle.
Degenerate (Balanced) Correspondences
Consider the ring \(\mathrm{CH}^d(X\times X)\) for \(\mathrm{dim}X=d\). We want to see if the degenerate (balanced) correspondences (resp. \(\mathbb{Q}\)-correspondences) form a ideal.
For \(c\in \mathrm{CH}^d(X\times X)\) balanced on the left (supported on some \(\mathrm{CH}^d(V\times X)\)), the composition holds
\[ \mathrm{CH}^d(X\times X)\circ \mathrm{CH}^d(V\times X)\in \mathrm{CH}^d(V\times X) \]
since the intersection product is supported on \(V\times X\times X\). Composition on the other side supports on
$ XVX $
(skip)
Hey let’s come back to Dr.Guletskii’s “factorization” criterion!
Lemma 2
Let \(X\), \(Y\) and \(Z\) be three smooth projective varieties over \(k\), where \(X\) is equidimensional, and \(f:Y\rightarrow Z\) be a morphism. let \(\alpha\in\mathrm{CH}^m(Y,X)\) be a corresponence, and \(\gamma=(f\times id_X)_*\alpha\)