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Hodge Computer

A collective effort to advance our understanding of the generalized Hodge conjecture. Let's work on explicit constructions for an important subcase: sextic fourfolds with an involution. This case was isolated by Voisin; a reduction carried out by Diamond makes its prediction computationally effective. See more at hodge.computer.

If you manage to complete a mission, please mirror your work back to hodge.computer. Let's get to work!

16 missions

Missions

1–16 of 16
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Algebraic Geometry·Captain: benediamond

hodge.computer: Eight planes · contact arrangementOpen Problem

The hodge.computer project studies a case of the generalized Hodge conjecture for sextic fourfolds with an involution, posed by Voisin. Diamond's reduction turns this Hodge-theoretic prediction into the explicit rational-vector-field problem below; it is the basis of the certificate formalism used in these missions.

Write z=(X0,X1,X2,Y0,Y1,Y2)z=(X_0,X_1,X_2,Y_0,Y_1,Y_2)z=(X0​,X1​,X2​,Y0​,Y1​,Y2​). For a fixed homogeneous sextic FFF, pole parameter q∈{1,2}q\in\{1,2\}q∈{1,2}, and homogeneous polynomial AAA of degree 6q6q6q, prove that there exists a rational vector field v=(v0,…,v5)v=(v_0,\ldots,v_5)v=(v0​,…,v5​) over C\mathbb CC, with components homogeneous of degree 6q−56q-56q−5 and a common denominator relatively prime to FFF, satisfying the key equation

Fq+1∣A−q⋅dF(v)+F⋅div⁡(v).F^{q+1}\mid A-q\cdot dF(v)+F\cdot\operatorname{div}(v).Fq+1∣A−q⋅dF(v)+F⋅div(v).

Here dF(v)=∑i=05vi∂iFdF(v)=\sum_{i=0}^5v_i\partial_iFdF(v)=∑i=05​vi​∂i​F and div⁡(v)=∑i=05∂ivi\operatorname{div}(v)=\sum_{i=0}^5\partial_i v_idiv(v)=∑i=05​∂i​vi​. Divisibility is taken in the ring of rational functions whose denominators are relatively prime to FFF. There is no bound on the denominator degree.

This instance: "Eight planes · contact arrangement"

F=F0+∑i=025pi⋅Qi⋅Hi.F=F_0+\sum_{i=0}^2\frac5{p_i}\cdot Q_i\cdot H_i.F=F0​+i=0∑2​pi​5​⋅Qi​⋅Hi​. q=1,A=X03⋅Y03.q=1,\qquad A=X_{0}^{3} \cdot Y_{0}^{3}.q=1,A=X03​⋅Y03​.

where

F0=∑i=02(Xi6+Yi6)Qi=Xi2+Yi2,i=0,1,2Hi=ℓi3⋅ni+mi3⋅oi+ℓi2⋅oi2ℓi=17⋅∑j=02(j+i+1)⋅Xjmi=111⋅∑j=02(2⋅j+i+1)⋅Yjni=113⋅∑j=02(j+1)i+1⋅Xjoi=117⋅∑j=02(j+2)i+1⋅Yj(p0,p1,p2)=(7,11,13)\begin{aligned} F_0&=\sum_{i=0}^{2}(X_i^6+Y_i^6) \\ Q_i&=X_i^2+Y_i^2,\quad i=0,1,2 \\ H_i&=\ell_i^3\cdot n_i+m_i^3\cdot o_i+\ell_i^2\cdot o_i^2 \\ \ell_i&=\frac17\cdot\sum_{j=0}^2(j+i+1)\cdot X_j \\ m_i&=\frac1{11}\cdot\sum_{j=0}^2(2\cdot j+i+1)\cdot Y_j \\ n_i&=\frac1{13}\cdot\sum_{j=0}^2(j+1)^{i+1}\cdot X_j \\ o_i&=\frac1{17}\cdot\sum_{j=0}^2(j+2)^{i+1}\cdot Y_j \\ (p_0,p_1,p_2)&=(7,11,13) \end{aligned}F0​Qi​Hi​ℓi​mi​ni​oi​(p0​,p1​,p2​)​=i=0∑2​(Xi6​+Yi6​)=Xi2​+Yi2​,i=0,1,2=ℓi3​⋅ni​+mi3​⋅oi​+ℓi2​⋅oi2​=71​⋅j=0∑2​(j+i+1)⋅Xj​=111​⋅j=0∑2​(2⋅j+i+1)⋅Yj​=131​⋅j=0∑2​(j+1)i+1⋅Xj​=171​⋅j=0∑2​(j+2)i+1⋅Yj​=(7,11,13)​

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Algebraic Geometry·Captain: benediamond

hodge.computer: Quadratic pullback · residual densityOpen Problem

The hodge.computer project studies a case of the generalized Hodge conjecture for sextic fourfolds with an involution, posed by Voisin. Diamond's reduction turns this Hodge-theoretic prediction into the explicit rational-vector-field problem below; it is the basis of the certificate formalism used in these missions.

Write z=(X0,X1,X2,Y0,Y1,Y2)z=(X_0,X_1,X_2,Y_0,Y_1,Y_2)z=(X0​,X1​,X2​,Y0​,Y1​,Y2​). For a fixed homogeneous sextic FFF, pole parameter q∈{1,2}q\in\{1,2\}q∈{1,2}, and homogeneous polynomial AAA of degree 6q6q6q, prove that there exists a rational vector field v=(v0,…,v5)v=(v_0,\ldots,v_5)v=(v0​,…,v5​) over C\mathbb CC, with components homogeneous of degree 6q−56q-56q−5 and a common denominator relatively prime to FFF, satisfying the key equation

Fq+1∣A−q⋅dF(v)+F⋅div⁡(v).F^{q+1}\mid A-q\cdot dF(v)+F\cdot\operatorname{div}(v).Fq+1∣A−q⋅dF(v)+F⋅div(v).

Here dF(v)=∑i=05vi∂iFdF(v)=\sum_{i=0}^5v_i\partial_iFdF(v)=∑i=05​vi​∂i​F and div⁡(v)=∑i=05∂ivi\operatorname{div}(v)=\sum_{i=0}^5\partial_i v_idiv(v)=∑i=05​∂i​vi​. Divisibility is taken in the ring of rational functions whose denominators are relatively prime to FFF. There is no bound on the denominator degree.

This instance: "Quadratic pullback · residual density"

F=∑i=05Qi3.F=\sum_{i=0}^5 Q_i^3.F=i=0∑5​Qi3​. q=1,A=X03⋅Y03.q=1,\qquad A=X_{0}^{3} \cdot Y_{0}^{3}.q=1,A=X03​⋅Y03​.

where

Q0=X02+5⋅X1⋅X27+5⋅Y0⋅Y17Q1=5⋅X0⋅X211+X12+5⋅Y1⋅Y211Q2=5⋅X0⋅X113+X22+5⋅Y0⋅Y213Q3=5⋅X0⋅X117+Y02+5⋅Y1⋅Y217Q4=5⋅X1⋅X219+5⋅Y0⋅Y219+Y12Q5=5⋅X0⋅X223+5⋅Y0⋅Y123+Y22\begin{aligned} Q_0&=X_{0}^{2} + \frac{5 \cdot X_{1} \cdot X_{2}}{7} + \frac{5 \cdot Y_{0} \cdot Y_{1}}{7} \\ Q_1&=\frac{5 \cdot X_{0} \cdot X_{2}}{11} + X_{1}^{2} + \frac{5 \cdot Y_{1} \cdot Y_{2}}{11} \\ Q_2&=\frac{5 \cdot X_{0} \cdot X_{1}}{13} + X_{2}^{2} + \frac{5 \cdot Y_{0} \cdot Y_{2}}{13} \\ Q_3&=\frac{5 \cdot X_{0} \cdot X_{1}}{17} + Y_{0}^{2} + \frac{5 \cdot Y_{1} \cdot Y_{2}}{17} \\ Q_4&=\frac{5 \cdot X_{1} \cdot X_{2}}{19} + \frac{5 \cdot Y_{0} \cdot Y_{2}}{19} + Y_{1}^{2} \\ Q_5&=\frac{5 \cdot X_{0} \cdot X_{2}}{23} + \frac{5 \cdot Y_{0} \cdot Y_{1}}{23} + Y_{2}^{2} \end{aligned}Q0​Q1​Q2​Q3​Q4​Q5​​=X02​+75⋅X1​⋅X2​​+75⋅Y0​⋅Y1​​=115⋅X0​⋅X2​​+X12​+115⋅Y1​⋅Y2​​=135⋅X0​⋅X1​​+X22​+135⋅Y0​⋅Y2​​=175⋅X0​⋅X1​​+Y02​+175⋅Y1​⋅Y2​​=195⋅X1​⋅X2​​+195⋅Y0​⋅Y2​​+Y12​=235⋅X0​⋅X2​​+235⋅Y0​⋅Y1​​+Y22​​

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5 thms2 active usersReviewed
Algebraic Geometry·Captain: benediamond

hodge.computer: Eight powers · linear-section remainderOpen Problem

The hodge.computer project studies a case of the generalized Hodge conjecture for sextic fourfolds with an involution, posed by Voisin. Diamond's reduction turns this Hodge-theoretic prediction into the explicit rational-vector-field problem below; it is the basis of the certificate formalism used in these missions.

Write z=(X0,X1,X2,Y0,Y1,Y2)z=(X_0,X_1,X_2,Y_0,Y_1,Y_2)z=(X0​,X1​,X2​,Y0​,Y1​,Y2​). For a fixed homogeneous sextic FFF, pole parameter q∈{1,2}q\in\{1,2\}q∈{1,2}, and homogeneous polynomial AAA of degree 6q6q6q, prove that there exists a rational vector field v=(v0,…,v5)v=(v_0,\ldots,v_5)v=(v0​,…,v5​) over C\mathbb CC, with components homogeneous of degree 6q−56q-56q−5 and a common denominator relatively prime to FFF, satisfying the key equation

Fq+1∣A−q⋅dF(v)+F⋅div⁡(v).F^{q+1}\mid A-q\cdot dF(v)+F\cdot\operatorname{div}(v).Fq+1∣A−q⋅dF(v)+F⋅div(v).

Here dF(v)=∑i=05vi∂iFdF(v)=\sum_{i=0}^5v_i\partial_iFdF(v)=∑i=05​vi​∂i​F and div⁡(v)=∑i=05∂ivi\operatorname{div}(v)=\sum_{i=0}^5\partial_i v_idiv(v)=∑i=05​∂i​vi​. Divisibility is taken in the ring of rational functions whose denominators are relatively prime to FFF. There is no bound on the denominator degree.

This instance: "Eight powers · linear-section remainder"

F=F0+577⋅((L+M)6+(L−M)6).F=F_0+\frac5{7^7}\cdot\bigl((L+M)^6+(L-M)^6\bigr).F=F0​+775​⋅((L+M)6+(L−M)6). q=1,A=X03⋅Y03.q=1,\qquad A=X_{0}^{3} \cdot Y_{0}^{3}.q=1,A=X03​⋅Y03​.

where

F0=∑i=02(Xi6+Yi6)L=X0+2⋅X1+3⋅X2M=Y0+3⋅Y1+5⋅Y2\begin{aligned} F_0&=\sum_{i=0}^{2}(X_i^6+Y_i^6) \\ L&=X_0+2\cdot X_1+3\cdot X_2 \\ M&=Y_0+3\cdot Y_1+5\cdot Y_2 \end{aligned}F0​LM​=i=0∑2​(Xi6​+Yi6​)=X0​+2⋅X1​+3⋅X2​=Y0​+3⋅Y1​+5⋅Y2​​

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Algebraic Geometry·Captain: benediamond

hodge.computer: Reflection · complementary characterOpen Problem

The hodge.computer project studies a case of the generalized Hodge conjecture for sextic fourfolds with an involution, posed by Voisin. Diamond's reduction turns this Hodge-theoretic prediction into the explicit rational-vector-field problem below; it is the basis of the certificate formalism used in these missions.

Write z=(X0,X1,X2,Y0,Y1,Y2)z=(X_0,X_1,X_2,Y_0,Y_1,Y_2)z=(X0​,X1​,X2​,Y0​,Y1​,Y2​). For a fixed homogeneous sextic FFF, pole parameter q∈{1,2}q\in\{1,2\}q∈{1,2}, and homogeneous polynomial AAA of degree 6q6q6q, prove that there exists a rational vector field v=(v0,…,v5)v=(v_0,\ldots,v_5)v=(v0​,…,v5​) over C\mathbb CC, with components homogeneous of degree 6q−56q-56q−5 and a common denominator relatively prime to FFF, satisfying the key equation

Fq+1∣A−q⋅dF(v)+F⋅div⁡(v).F^{q+1}\mid A-q\cdot dF(v)+F\cdot\operatorname{div}(v).Fq+1∣A−q⋅dF(v)+F⋅div(v).

Here dF(v)=∑i=05vi∂iFdF(v)=\sum_{i=0}^5v_i\partial_iFdF(v)=∑i=05​vi​∂i​F and div⁡(v)=∑i=05∂ivi\operatorname{div}(v)=\sum_{i=0}^5\partial_i v_idiv(v)=∑i=05​∂i​vi​. Divisibility is taken in the ring of rational functions whose denominators are relatively prime to FFF. There is no bound on the denominator degree.

This instance: "Reflection · complementary character"

F=F0+t⋅P.F=F_0+t\cdot P.F=F0​+t⋅P. q=1,A=X03⋅Y03.q=1,\qquad A=X_0^3\cdot Y_0^3.q=1,A=X03​⋅Y03​.

where

F0=X06+X16+X26+Y06+Y16+Y26t=57P=X03⋅X1⋅Y0⋅Y1+X0⋅X1⋅Y03⋅Y1+X0⋅X1⋅Y02⋅Y22+2⋅X0⋅X1⋅Y12⋅Y22+3⋅X0⋅X23⋅Y0⋅Y1+2⋅X13⋅X2⋅Y0⋅Y1+X1⋅X2⋅Y0⋅Y13\begin{aligned} F_0&=X_0^6+X_1^6+X_2^6+Y_0^6+Y_1^6+Y_2^6 \\ t&=\frac{5}{7} \\ P&=X_{0}^{3} \cdot X_{1} \cdot Y_{0} \cdot Y_{1} + X_{0} \cdot X_{1} \cdot Y_{0}^{3} \cdot Y_{1} + X_{0} \cdot X_{1} \cdot Y_{0}^{2} \cdot Y_{2}^{2} + 2 \cdot X_{0} \cdot X_{1} \cdot Y_{1}^{2} \cdot Y_{2}^{2} + 3 \cdot X_{0} \cdot X_{2}^{3} \cdot Y_{0} \cdot Y_{1} + 2 \cdot X_{1}^{3} \cdot X_{2} \cdot Y_{0} \cdot Y_{1} + X_{1} \cdot X_{2} \cdot Y_{0} \cdot Y_{1}^{3} \end{aligned}F0​tP​=X06​+X16​+X26​+Y06​+Y16​+Y26​=75​=X03​⋅X1​⋅Y0​⋅Y1​+X0​⋅X1​⋅Y03​⋅Y1​+X0​⋅X1​⋅Y02​⋅Y22​+2⋅X0​⋅X1​⋅Y12​⋅Y22​+3⋅X0​⋅X23​⋅Y0​⋅Y1​+2⋅X13​⋅X2​⋅Y0​⋅Y1​+X1​⋅X2​⋅Y0​⋅Y13​​

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Algebraic Geometry·Captain: benediamond

hodge.computer: One reflection · skew couplingOpen Problem

The hodge.computer project studies a case of the generalized Hodge conjecture for sextic fourfolds with an involution, posed by Voisin. Diamond's reduction turns this Hodge-theoretic prediction into the explicit rational-vector-field problem below; it is the basis of the certificate formalism used in these missions.

Write z=(X0,X1,X2,Y0,Y1,Y2)z=(X_0,X_1,X_2,Y_0,Y_1,Y_2)z=(X0​,X1​,X2​,Y0​,Y1​,Y2​). For a fixed homogeneous sextic FFF, pole parameter q∈{1,2}q\in\{1,2\}q∈{1,2}, and homogeneous polynomial AAA of degree 6q6q6q, prove that there exists a rational vector field v=(v0,…,v5)v=(v_0,\ldots,v_5)v=(v0​,…,v5​) over C\mathbb CC, with components homogeneous of degree 6q−56q-56q−5 and a common denominator relatively prime to FFF, satisfying the key equation

Fq+1∣A−q⋅dF(v)+F⋅div⁡(v).F^{q+1}\mid A-q\cdot dF(v)+F\cdot\operatorname{div}(v).Fq+1∣A−q⋅dF(v)+F⋅div(v).

Here dF(v)=∑i=05vi∂iFdF(v)=\sum_{i=0}^5v_i\partial_iFdF(v)=∑i=05​vi​∂i​F and div⁡(v)=∑i=05∂ivi\operatorname{div}(v)=\sum_{i=0}^5\partial_i v_idiv(v)=∑i=05​∂i​vi​. Divisibility is taken in the ring of rational functions whose denominators are relatively prime to FFF. There is no bound on the denominator degree.

This instance: "One reflection · skew coupling"

F=F0+X03X1Y0Y1+X13X2Y0Y1+17X0X1Y0Y1(Y02−Y12).F=F_0 + X_0^3 X_1 Y_0 Y_1 + X_1^3 X_2 Y_0 Y_1 + \frac17 X_0 X_1 Y_0 Y_1 (Y_0^2-Y_1^2).F=F0​+X03​X1​Y0​Y1​+X13​X2​Y0​Y1​+71​X0​X1​Y0​Y1​(Y02​−Y12​). q=1,A=X03Y03.q=1,\qquad A=X_{0}^{3} Y_{0}^{3}.q=1,A=X03​Y03​.

where

F0=X06+X16+X26+Y06+Y16+Y26\begin{aligned} F_0&=X_0^6+X_1^6+X_2^6+Y_0^6+Y_1^6+Y_2^6 \end{aligned}F0​​=X06​+X16​+X26​+Y06​+Y16​+Y26​​

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5 thms2 active usersReviewed
Algebraic Geometry·Captain: benediamond

hodge.computer: Quadric contact · residual sectorOpen Problem

The hodge.computer project studies a case of the generalized Hodge conjecture for sextic fourfolds with an involution, posed by Voisin. Diamond's reduction turns this Hodge-theoretic prediction into the explicit rational-vector-field problem below; it is the basis of the certificate formalism used in these missions.

Write z=(X0,X1,X2,Y0,Y1,Y2)z=(X_0,X_1,X_2,Y_0,Y_1,Y_2)z=(X0​,X1​,X2​,Y0​,Y1​,Y2​). For a fixed homogeneous sextic FFF, pole parameter q∈{1,2}q\in\{1,2\}q∈{1,2}, and homogeneous polynomial AAA of degree 6q6q6q, prove that there exists a rational vector field v=(v0,…,v5)v=(v_0,\ldots,v_5)v=(v0​,…,v5​) over C\mathbb CC, with components homogeneous of degree 6q−56q-56q−5 and a common denominator relatively prime to FFF, satisfying the key equation

Fq+1∣A−q⋅dF(v)+F⋅div⁡(v).F^{q+1}\mid A-q\cdot dF(v)+F\cdot\operatorname{div}(v).Fq+1∣A−q⋅dF(v)+F⋅div(v).

Here dF(v)=∑i=05vi∂iFdF(v)=\sum_{i=0}^5v_i\partial_iFdF(v)=∑i=05​vi​∂i​F and div⁡(v)=∑i=05∂ivi\operatorname{div}(v)=\sum_{i=0}^5\partial_i v_idiv(v)=∑i=05​∂i​vi​. Divisibility is taken in the ring of rational functions whose denominators are relatively prime to FFF. There is no bound on the denominator degree.

This instance: "Quadric contact · residual sector"

F=F∗+t⋅X0⋅X1⋅Y0⋅Y1⋅(X22+2⋅Y22).F=F_*+t\cdot X_0\cdot X_1\cdot Y_0\cdot Y_1\cdot(X_2^2+2\cdot Y_2^2).F=F∗​+t⋅X0​⋅X1​⋅Y0​⋅Y1​⋅(X22​+2⋅Y22​). q=1,A=X03Y03.q=1,\qquad A=X_{0}^{3} Y_{0}^{3}.q=1,A=X03​Y03​.

where

F0=X06+X16+X26+Y06+Y16+Y26t=57F∗=F0+5⋅(X04⋅Y02+X02⋅Y04+(X02+Y02)⋅Q)Q=X12⋅Y12+X1⋅X2⋅Y1⋅Y2+X22⋅Y22\begin{aligned} F_0&=X_0^6+X_1^6+X_2^6+Y_0^6+Y_1^6+Y_2^6 \\ t&=\frac{5}{7} \\ F_*&=F_0+5\cdot\bigl(X_0^4\cdot Y_0^2+X_0^2\cdot Y_0^4+(X_0^2+Y_0^2)\cdot Q\bigr) \\ Q&=X_1^2\cdot Y_1^2+X_1\cdot X_2\cdot Y_1\cdot Y_2+X_2^2\cdot Y_2^2 \end{aligned}F0​tF∗​Q​=X06​+X16​+X26​+Y06​+Y16​+Y26​=75​=F0​+5⋅(X04​⋅Y02​+X02​⋅Y04​+(X02​+Y02​)⋅Q)=X12​⋅Y12​+X1​⋅X2​⋅Y1​⋅Y2​+X22​⋅Y22​​

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Algebraic Geometry·Captain: benediamond

hodge.computer: Cyclic triple cover · complementary characterOpen Problem

The hodge.computer project studies a case of the generalized Hodge conjecture for sextic fourfolds with an involution, posed by Voisin. Diamond's reduction turns this Hodge-theoretic prediction into the explicit rational-vector-field problem below; it is the basis of the certificate formalism used in these missions.

Write z=(X0,X1,X2,Y0,Y1,Y2)z=(X_0,X_1,X_2,Y_0,Y_1,Y_2)z=(X0​,X1​,X2​,Y0​,Y1​,Y2​). For a fixed homogeneous sextic FFF, pole parameter q∈{1,2}q\in\{1,2\}q∈{1,2}, and homogeneous polynomial AAA of degree 6q6q6q, prove that there exists a rational vector field v=(v0,…,v5)v=(v_0,\ldots,v_5)v=(v0​,…,v5​) over C\mathbb CC, with components homogeneous of degree 6q−56q-56q−5 and a common denominator relatively prime to FFF, satisfying the key equation

Fq+1∣A−q⋅dF(v)+F⋅div⁡(v).F^{q+1}\mid A-q\cdot dF(v)+F\cdot\operatorname{div}(v).Fq+1∣A−q⋅dF(v)+F⋅div(v).

Here dF(v)=∑i=05vi∂iFdF(v)=\sum_{i=0}^5v_i\partial_iFdF(v)=∑i=05​vi​∂i​F and div⁡(v)=∑i=05∂ivi\operatorname{div}(v)=\sum_{i=0}^5\partial_i v_idiv(v)=∑i=05​∂i​vi​. Divisibility is taken in the ring of rational functions whose denominators are relatively prime to FFF. There is no bound on the denominator degree.

This instance: "Cyclic triple cover · complementary character"

F=X26+t⋅X23⋅C+H.F=X_2^6+t\cdot X_2^3\cdot C+H.F=X26​+t⋅X23​⋅C+H. q=1,A=X03⋅Y03.q=1,\qquad A=X_0^3\cdot Y_0^3.q=1,A=X03​⋅Y03​.

where

t=57C=X03+X0⋅Y1⋅Y2+X1⋅Y0⋅Y1H=X06+57⋅X03⋅X1⋅Y0⋅Y1+57⋅X02⋅Y0⋅Y1⋅Y22+57⋅X0⋅X1⋅Y13⋅Y2+X16+Y06+Y16+Y26\begin{aligned} t&=\frac57 \\ C&=X_{0}^{3} + X_{0} \cdot Y_{1} \cdot Y_{2} + X_{1} \cdot Y_{0} \cdot Y_{1} \\ H&=X_{0}^{6} + \frac{5}{7}\cdot X_{0}^{3}\cdot X_{1}\cdot Y_{0}\cdot Y_{1} + \frac{5}{7}\cdot X_{0}^{2}\cdot Y_{0}\cdot Y_{1}\cdot Y_{2}^{2} + \frac{5}{7}\cdot X_{0}\cdot X_{1}\cdot Y_{1}^{3}\cdot Y_{2} + X_{1}^{6} + Y_{0}^{6} + Y_{1}^{6} + Y_{2}^{6} \end{aligned}tCH​=75​=X03​+X0​⋅Y1​⋅Y2​+X1​⋅Y0​⋅Y1​=X06​+75​⋅X03​⋅X1​⋅Y0​⋅Y1​+75​⋅X02​⋅Y0​⋅Y1​⋅Y22​+75​⋅X0​⋅X1​⋅Y13​⋅Y2​+X16​+Y06​+Y16​+Y26​​

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Algebraic Geometry·Captain: benediamond

hodge.computer: Simultaneous permutations · standard sectorOpen Problem

The hodge.computer project studies a case of the generalized Hodge conjecture for sextic fourfolds with an involution, posed by Voisin. Diamond's reduction turns this Hodge-theoretic prediction into the explicit rational-vector-field problem below; it is the basis of the certificate formalism used in these missions.

Write z=(X0,X1,X2,Y0,Y1,Y2)z=(X_0,X_1,X_2,Y_0,Y_1,Y_2)z=(X0​,X1​,X2​,Y0​,Y1​,Y2​). For a fixed homogeneous sextic FFF, pole parameter q∈{1,2}q\in\{1,2\}q∈{1,2}, and homogeneous polynomial AAA of degree 6q6q6q, prove that there exists a rational vector field v=(v0,…,v5)v=(v_0,\ldots,v_5)v=(v0​,…,v5​) over C\mathbb CC, with components homogeneous of degree 6q−56q-56q−5 and a common denominator relatively prime to FFF, satisfying the key equation

Fq+1∣A−q⋅dF(v)+F⋅div⁡(v).F^{q+1}\mid A-q\cdot dF(v)+F\cdot\operatorname{div}(v).Fq+1∣A−q⋅dF(v)+F⋅div(v).

Here dF(v)=∑i=05vi∂iFdF(v)=\sum_{i=0}^5v_i\partial_iFdF(v)=∑i=05​vi​∂i​F and div⁡(v)=∑i=05∂ivi\operatorname{div}(v)=\sum_{i=0}^5\partial_i v_idiv(v)=∑i=05​∂i​vi​. Divisibility is taken in the ring of rational functions whose denominators are relatively prime to FFF. There is no bound on the denominator degree.

This instance: "Simultaneous permutations · standard sector"

F=F0+57⋅O1+511⋅O2+513⋅O3.F=F_0+\frac57\cdot O_1+\frac5{11}\cdot O_2+\frac5{13}\cdot O_3.F=F0​+75​⋅O1​+115​⋅O2​+135​⋅O3​. q=1,A=X03⋅Y03−X13⋅Y13.q=1,\qquad A=X_{0}^{3} \cdot Y_{0}^{3} - X_{1}^{3} \cdot Y_{1}^{3}.q=1,A=X03​⋅Y03​−X13​⋅Y13​.

where

F0=∑i=02(Xi6+Yi6)P={(i,j,k):{i,j,k}={0,1,2}}O1=∑(i,j,k)∈PXi3⋅Xj⋅Yi⋅YjO2=∑(i,j,k)∈PXi3⋅Xj⋅Yk2O3=∑(i,j,k)∈PYi3⋅Yj⋅Xk2\begin{aligned} F_0&=\sum_{i=0}^{2}(X_i^6+Y_i^6) \\ \mathcal P&=\{(i,j,k):\{i,j,k\}=\{0,1,2\}\} \\ O_1&=\sum_{(i,j,k)\in\mathcal P}X_i^3\cdot X_j\cdot Y_i\cdot Y_j \\ O_2&=\sum_{(i,j,k)\in\mathcal P}X_i^3\cdot X_j\cdot Y_k^2 \\ O_3&=\sum_{(i,j,k)\in\mathcal P}Y_i^3\cdot Y_j\cdot X_k^2 \end{aligned}F0​PO1​O2​O3​​=i=0∑2​(Xi6​+Yi6​)={(i,j,k):{i,j,k}={0,1,2}}=(i,j,k)∈P∑​Xi3​⋅Xj​⋅Yi​⋅Yj​=(i,j,k)∈P∑​Xi3​⋅Xj​⋅Yk2​=(i,j,k)∈P∑​Yi3​⋅Yj​⋅Xk2​​

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Algebraic Geometry·Captain: benediamond

hodge.computer: Four plus two · suspension remainderOpen Problem

The hodge.computer project studies a case of the generalized Hodge conjecture for sextic fourfolds with an involution, posed by Voisin. Diamond's reduction turns this Hodge-theoretic prediction into the explicit rational-vector-field problem below; it is the basis of the certificate formalism used in these missions.

Write z=(X0,X1,X2,Y0,Y1,Y2)z=(X_0,X_1,X_2,Y_0,Y_1,Y_2)z=(X0​,X1​,X2​,Y0​,Y1​,Y2​). For a fixed homogeneous sextic FFF, pole parameter q∈{1,2}q\in\{1,2\}q∈{1,2}, and homogeneous polynomial AAA of degree 6q6q6q, prove that there exists a rational vector field v=(v0,…,v5)v=(v_0,\ldots,v_5)v=(v0​,…,v5​) over C\mathbb CC, with components homogeneous of degree 6q−56q-56q−5 and a common denominator relatively prime to FFF, satisfying the key equation

Fq+1∣A−q⋅dF(v)+F⋅div⁡(v).F^{q+1}\mid A-q\cdot dF(v)+F\cdot\operatorname{div}(v).Fq+1∣A−q⋅dF(v)+F⋅div(v).

Here dF(v)=∑i=05vi∂iFdF(v)=\sum_{i=0}^5v_i\partial_iFdF(v)=∑i=05​vi​∂i​F and div⁡(v)=∑i=05∂ivi\operatorname{div}(v)=\sum_{i=0}^5\partial_i v_idiv(v)=∑i=05​∂i​vi​. Divisibility is taken in the ring of rational functions whose denominators are relatively prime to FFF. There is no bound on the denominator degree.

This instance: "Four plus two · suspension remainder"

F=H+B.F=H+B.F=H+B. q=1,A=X03⋅Y03.q=1,\qquad A=X_{0}^{3} \cdot Y_{0}^{3}.q=1,A=X03​⋅Y03​.

where

H=X06+57⋅X05⋅X1+57⋅X03⋅X1⋅Y0⋅Y1+107⋅X0⋅X15+107⋅X0⋅X1⋅Y03⋅Y1+X16+Y06+57⋅Y05⋅Y1+157⋅Y0⋅Y15+Y16B=X26+157⋅X24⋅Y22+107⋅X22⋅Y24+Y26\begin{aligned} H&=X_{0}^{6} + \frac{5}{7}\cdot X_{0}^{5}\cdot X_{1} + \frac{5}{7}\cdot X_{0}^{3}\cdot X_{1}\cdot Y_{0}\cdot Y_{1} + \frac{10}{7}\cdot X_{0}\cdot X_{1}^{5} + \frac{10}{7}\cdot X_{0}\cdot X_{1}\cdot Y_{0}^{3}\cdot Y_{1} + X_{1}^{6} + Y_{0}^{6} + \frac{5}{7}\cdot Y_{0}^{5}\cdot Y_{1} + \frac{15}{7}\cdot Y_{0}\cdot Y_{1}^{5} + Y_{1}^{6} \\ B&=X_{2}^{6} + \frac{15}{7}\cdot X_{2}^{4}\cdot Y_{2}^{2} + \frac{10}{7}\cdot X_{2}^{2}\cdot Y_{2}^{4} + Y_{2}^{6} \end{aligned}HB​=X06​+75​⋅X05​⋅X1​+75​⋅X03​⋅X1​⋅Y0​⋅Y1​+710​⋅X0​⋅X15​+710​⋅X0​⋅X1​⋅Y03​⋅Y1​+X16​+Y06​+75​⋅Y05​⋅Y1​+715​⋅Y0​⋅Y15​+Y16​=X26​+715​⋅X24​⋅Y22​+710​⋅X22​⋅Y24​+Y26​​

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Algebraic Geometry·Captain: benediamond

hodge.computer: Broken paired signsOpen Problem

The hodge.computer project studies a case of the generalized Hodge conjecture for sextic fourfolds with an involution, posed by Voisin. Diamond's reduction turns this Hodge-theoretic prediction into the explicit rational-vector-field problem below; it is the basis of the certificate formalism used in these missions.

Write z=(X0,X1,X2,Y0,Y1,Y2)z=(X_0,X_1,X_2,Y_0,Y_1,Y_2)z=(X0​,X1​,X2​,Y0​,Y1​,Y2​). For a fixed homogeneous sextic FFF, pole parameter q∈{1,2}q\in\{1,2\}q∈{1,2}, and homogeneous polynomial AAA of degree 6q6q6q, prove that there exists a rational vector field v=(v0,…,v5)v=(v_0,\ldots,v_5)v=(v0​,…,v5​) over C\mathbb CC, with components homogeneous of degree 6q−56q-56q−5 and a common denominator relatively prime to FFF, satisfying the key equation

Fq+1∣A−q⋅dF(v)+F⋅div⁡(v).F^{q+1}\mid A-q\cdot dF(v)+F\cdot\operatorname{div}(v).Fq+1∣A−q⋅dF(v)+F⋅div(v).

Here dF(v)=∑i=05vi∂iFdF(v)=\sum_{i=0}^5v_i\partial_iFdF(v)=∑i=05​vi​∂i​F and div⁡(v)=∑i=05∂ivi\operatorname{div}(v)=\sum_{i=0}^5\partial_i v_idiv(v)=∑i=05​∂i​vi​. Divisibility is taken in the ring of rational functions whose denominators are relatively prime to FFF. There is no bound on the denominator degree.

This instance: "Broken paired signs"

F=F0+X03X1Y0Y12+X02X1X2Y0Y23+X13X2Y1Y22.F=F_0 + \frac{X_{0}^{3} X_{1} Y_{0} Y_{1}}{2} + \frac{X_{0}^{2} X_{1} X_{2} Y_{0} Y_{2}}{3} + \frac{X_{1}^{3} X_{2} Y_{1} Y_{2}}{2}.F=F0​+2X03​X1​Y0​Y1​​+3X02​X1​X2​Y0​Y2​​+2X13​X2​Y1​Y2​​. q=1,A=X03Y03.q=1,\qquad A=X_{0}^{3} Y_{0}^{3}.q=1,A=X03​Y03​.

where

F0=X06+X16+X26+Y06+Y16+Y26\begin{aligned} F_0&=X_0^6+X_1^6+X_2^6+Y_0^6+Y_1^6+Y_2^6 \end{aligned}F0​​=X06​+X16​+X26​+Y06​+Y16​+Y26​​

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Algebraic Geometry·Captain: benediamond

hodge.computer: Dense rational · full coefficient supportOpen Problem

The hodge.computer project studies a case of the generalized Hodge conjecture for sextic fourfolds with an involution, posed by Voisin. Diamond's reduction turns this Hodge-theoretic prediction into the explicit rational-vector-field problem below; it is the basis of the certificate formalism used in these missions.

Write z=(X0,X1,X2,Y0,Y1,Y2)z=(X_0,X_1,X_2,Y_0,Y_1,Y_2)z=(X0​,X1​,X2​,Y0​,Y1​,Y2​). For a fixed homogeneous sextic FFF, pole parameter q∈{1,2}q\in\{1,2\}q∈{1,2}, and homogeneous polynomial AAA of degree 6q6q6q, prove that there exists a rational vector field v=(v0,…,v5)v=(v_0,\ldots,v_5)v=(v0​,…,v5​) over C\mathbb CC, with components homogeneous of degree 6q−56q-56q−5 and a common denominator relatively prime to FFF, satisfying the key equation

Fq+1∣A−q⋅dF(v)+F⋅div⁡(v).F^{q+1}\mid A-q\cdot dF(v)+F\cdot\operatorname{div}(v).Fq+1∣A−q⋅dF(v)+F⋅div(v).

Here dF(v)=∑i=05vi∂iFdF(v)=\sum_{i=0}^5v_i\partial_iFdF(v)=∑i=05​vi​∂i​F and div⁡(v)=∑i=05∂ivi\operatorname{div}(v)=\sum_{i=0}^5\partial_i v_idiv(v)=∑i=05​∂i​vi​. Divisibility is taken in the ring of rational functions whose denominators are relatively prime to FFF. There is no bound on the denominator degree.

This instance: "Dense rational · full coefficient support"

F=F0+19973∑α∈EcαZα.F=F_0+\frac{1}{9973}\sum_{\alpha\in\mathcal E}c_\alpha Z^\alpha.F=F0​+99731​α∈E∑​cα​Zα. q=1,A=X03Y03.q=1,\qquad A=X_{0}^{3} Y_{0}^{3}.q=1,A=X03​Y03​.

where

E={α∈N6:∣α∣=6, α3+α4+α5≡0(mod2), max⁡iαi<6}cα=1+(∑i=05(i+3)(αi+1)2 mod 17)Z=(X0,X1,X2,Y0,Y1,Y2)F0=∑i=02(Xi6+Yi6)\begin{aligned} \mathcal E&=\{\alpha\in\mathbb N^6:|\alpha|=6,\ \alpha_3+\alpha_4+\alpha_5\equiv0\pmod2,\ \max_i\alpha_i<6\} \\ c_\alpha&=1+\Bigl(\sum_{i=0}^{5}(i+3)(\alpha_i+1)^2\bmod17\Bigr) \\ Z&=(X_0,X_1,X_2,Y_0,Y_1,Y_2) \\ F_0&=\sum_{i=0}^{2}(X_i^6+Y_i^6) \end{aligned}Ecα​ZF0​​={α∈N6:∣α∣=6, α3​+α4​+α5​≡0(mod2), imax​αi​<6}=1+(i=0∑5​(i+3)(αi​+1)2mod17)=(X0​,X1​,X2​,Y0​,Y1​,Y2​)=i=0∑2​(Xi6​+Yi6​)​

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Algebraic Geometry·Captain: benediamond

hodge.computer: Polarized split · K3 remainderOpen Problem

The hodge.computer project studies a case of the generalized Hodge conjecture for sextic fourfolds with an involution, posed by Voisin. Diamond's reduction turns this Hodge-theoretic prediction into the explicit rational-vector-field problem below; it is the basis of the certificate formalism used in these missions.

Write z=(X0,X1,X2,Y0,Y1,Y2)z=(X_0,X_1,X_2,Y_0,Y_1,Y_2)z=(X0​,X1​,X2​,Y0​,Y1​,Y2​). For a fixed homogeneous sextic FFF, pole parameter q∈{1,2}q\in\{1,2\}q∈{1,2}, and homogeneous polynomial AAA of degree 6q6q6q, prove that there exists a rational vector field v=(v0,…,v5)v=(v_0,\ldots,v_5)v=(v0​,…,v5​) over C\mathbb CC, with components homogeneous of degree 6q−56q-56q−5 and a common denominator relatively prime to FFF, satisfying the key equation

Fq+1∣A−q⋅dF(v)+F⋅div⁡(v).F^{q+1}\mid A-q\cdot dF(v)+F\cdot\operatorname{div}(v).Fq+1∣A−q⋅dF(v)+F⋅div(v).

Here dF(v)=∑i=05vi∂iFdF(v)=\sum_{i=0}^5v_i\partial_iFdF(v)=∑i=05​vi​∂i​F and div⁡(v)=∑i=05∂ivi\operatorname{div}(v)=\sum_{i=0}^5\partial_i v_idiv(v)=∑i=05​∂i​vi​. Divisibility is taken in the ring of rational functions whose denominators are relatively prime to FFF. There is no bound on the denominator degree.

This instance: "Polarized split · K3 remainder"

F=12⋅(f(X+Y)+f(X−Y)).F=\frac12\cdot\bigl(f(X+Y)+f(X-Y)\bigr).F=21​⋅(f(X+Y)+f(X−Y)). q=1,A=X0⋅X1⋅X2⋅Y0⋅Y1⋅Y2.q=1,\qquad A=X_{0} \cdot X_{1} \cdot X_{2} \cdot Y_{0} \cdot Y_{1} \cdot Y_{2}.q=1,A=X0​⋅X1​⋅X2​⋅Y0​⋅Y1​⋅Y2​.

where

X=(X0,X1,X2),Y=(Y0,Y1,Y2)t=57f(a,b,c)=a6+b6+c6+t⋅(a5⋅b+2⋅b5⋅c+3⋅c5⋅a+a2⋅b3⋅c)\begin{aligned} X&=(X_0,X_1,X_2),\quad Y=(Y_0,Y_1,Y_2) \\ t&=\frac57 \\ f(a,b,c)&=a^6+b^6+c^6+t\cdot(a^5\cdot b+2\cdot b^5\cdot c+3\cdot c^5\cdot a+a^2\cdot b^3\cdot c) \end{aligned}Xtf(a,b,c)​=(X0​,X1​,X2​),Y=(Y0​,Y1​,Y2​)=75​=a6+b6+c6+t⋅(a5⋅b+2⋅b5⋅c+3⋅c5⋅a+a2⋅b3⋅c)​

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Algebraic Geometry·Captain: benediamond

hodge.computer: Coupled Klein · nontrivial characterOpen Problem

The hodge.computer project studies a case of the generalized Hodge conjecture for sextic fourfolds with an involution, posed by Voisin. Diamond's reduction turns this Hodge-theoretic prediction into the explicit rational-vector-field problem below; it is the basis of the certificate formalism used in these missions.

Write z=(X0,X1,X2,Y0,Y1,Y2)z=(X_0,X_1,X_2,Y_0,Y_1,Y_2)z=(X0​,X1​,X2​,Y0​,Y1​,Y2​). For a fixed homogeneous sextic FFF, pole parameter q∈{1,2}q\in\{1,2\}q∈{1,2}, and homogeneous polynomial AAA of degree 6q6q6q, prove that there exists a rational vector field v=(v0,…,v5)v=(v_0,\ldots,v_5)v=(v0​,…,v5​) over C\mathbb CC, with components homogeneous of degree 6q−56q-56q−5 and a common denominator relatively prime to FFF, satisfying the key equation

Fq+1∣A−q⋅dF(v)+F⋅div⁡(v).F^{q+1}\mid A-q\cdot dF(v)+F\cdot\operatorname{div}(v).Fq+1∣A−q⋅dF(v)+F⋅div(v).

Here dF(v)=∑i=05vi∂iFdF(v)=\sum_{i=0}^5v_i\partial_iFdF(v)=∑i=05​vi​∂i​F and div⁡(v)=∑i=05∂ivi\operatorname{div}(v)=\sum_{i=0}^5\partial_i v_idiv(v)=∑i=05​∂i​vi​. Divisibility is taken in the ring of rational functions whose denominators are relatively prime to FFF. There is no bound on the denominator degree.

This instance: "Coupled Klein · nontrivial character"

F=K+Y06+Y16+Y26+t⋅P.F=K+Y_0^6+Y_1^6+Y_2^6+t\cdot P.F=K+Y06​+Y16​+Y26​+t⋅P. q=1,A=X03⋅Y03.q=1,\qquad A=X_{0}^{3} \cdot Y_{0}^{3}.q=1,A=X03​⋅Y03​.

where

t=57K=X05⋅X1+X15⋅X2+X25⋅X0P=2⋅X03⋅X2⋅Y1⋅Y2+X0⋅X13⋅Y0⋅Y1+3⋅X1⋅X23⋅Y0⋅Y2\begin{aligned} t&=\frac57 \\ K&=X_0^5\cdot X_1+X_1^5\cdot X_2+X_2^5\cdot X_0 \\ P&=2 \cdot X_{0}^{3} \cdot X_{2} \cdot Y_{1} \cdot Y_{2} + X_{0} \cdot X_{1}^{3} \cdot Y_{0} \cdot Y_{1} + 3 \cdot X_{1} \cdot X_{2}^{3} \cdot Y_{0} \cdot Y_{2} \end{aligned}tKP​=75​=X05​⋅X1​+X15​⋅X2​+X25​⋅X0​=2⋅X03​⋅X2​⋅Y1​⋅Y2​+X0​⋅X13​⋅Y0​⋅Y1​+3⋅X1​⋅X23​⋅Y0​⋅Y2​​

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Algebraic Geometry·Captain: benediamond

hodge.computer: Permutation symmetry · residual sectorOpen Problem

The hodge.computer project studies a case of the generalized Hodge conjecture for sextic fourfolds with an involution, posed by Voisin. Diamond's reduction turns this Hodge-theoretic prediction into the explicit rational-vector-field problem below; it is the basis of the certificate formalism used in these missions.

Write z=(X0,X1,X2,Y0,Y1,Y2)z=(X_0,X_1,X_2,Y_0,Y_1,Y_2)z=(X0​,X1​,X2​,Y0​,Y1​,Y2​). For a fixed homogeneous sextic FFF, pole parameter q∈{1,2}q\in\{1,2\}q∈{1,2}, and homogeneous polynomial AAA of degree 6q6q6q, prove that there exists a rational vector field v=(v0,…,v5)v=(v_0,\ldots,v_5)v=(v0​,…,v5​) over C\mathbb CC, with components homogeneous of degree 6q−56q-56q−5 and a common denominator relatively prime to FFF, satisfying the key equation

Fq+1∣A−q⋅dF(v)+F⋅div⁡(v).F^{q+1}\mid A-q\cdot dF(v)+F\cdot\operatorname{div}(v).Fq+1∣A−q⋅dF(v)+F⋅div(v).

Here dF(v)=∑i=05vi∂iFdF(v)=\sum_{i=0}^5v_i\partial_iFdF(v)=∑i=05​vi​∂i​F and div⁡(v)=∑i=05∂ivi\operatorname{div}(v)=\sum_{i=0}^5\partial_i v_idiv(v)=∑i=05​∂i​vi​. Divisibility is taken in the ring of rational functions whose denominators are relatively prime to FFF. There is no bound on the denominator degree.

This instance: "Permutation symmetry · residual sector"

F=F0+t⋅(QX2⋅QY+2⋅QX⋅QY2).F=F_0+t\cdot(Q_X^2\cdot Q_Y+2\cdot Q_X\cdot Q_Y^2).F=F0​+t⋅(QX2​⋅QY​+2⋅QX​⋅QY2​). q=1,A=X0⋅X1⋅X2⋅Y0⋅Y1⋅Y2.q=1,\qquad A=X_{0} \cdot X_{1} \cdot X_{2} \cdot Y_{0} \cdot Y_{1} \cdot Y_{2}.q=1,A=X0​⋅X1​⋅X2​⋅Y0​⋅Y1​⋅Y2​.

where

F0=X06+X16+X26+Y06+Y16+Y26t=57QX=X0⋅X1+X1⋅X2+X2⋅X0QY=Y0⋅Y1+Y1⋅Y2+Y2⋅Y0\begin{aligned} F_0&=X_0^6+X_1^6+X_2^6+Y_0^6+Y_1^6+Y_2^6 \\ t&=\frac{5}{7} \\ Q_X&=X_0\cdot X_1+X_1\cdot X_2+X_2\cdot X_0 \\ Q_Y&=Y_0\cdot Y_1+Y_1\cdot Y_2+Y_2\cdot Y_0 \end{aligned}F0​tQX​QY​​=X06​+X16​+X26​+Y06​+Y16​+Y26​=75​=X0​⋅X1​+X1​⋅X2​+X2​⋅X0​=Y0​⋅Y1​+Y1​⋅Y2​+Y2​⋅Y0​​

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Algebraic Geometry·Captain: benediamond

hodge.computer: Rank-one coupling · biquartic pencilOpen Problem

The hodge.computer project studies a case of the generalized Hodge conjecture for sextic fourfolds with an involution, posed by Voisin. Diamond's reduction turns this Hodge-theoretic prediction into the explicit rational-vector-field problem below; it is the basis of the certificate formalism used in these missions.

Write z=(X0,X1,X2,Y0,Y1,Y2)z=(X_0,X_1,X_2,Y_0,Y_1,Y_2)z=(X0​,X1​,X2​,Y0​,Y1​,Y2​). For a fixed homogeneous sextic FFF, pole parameter q∈{1,2}q\in\{1,2\}q∈{1,2}, and homogeneous polynomial AAA of degree 6q6q6q, prove that there exists a rational vector field v=(v0,…,v5)v=(v_0,\ldots,v_5)v=(v0​,…,v5​) over C\mathbb CC, with components homogeneous of degree 6q−56q-56q−5 and a common denominator relatively prime to FFF, satisfying the key equation

Fq+1∣A−q⋅dF(v)+F⋅div⁡(v).F^{q+1}\mid A-q\cdot dF(v)+F\cdot\operatorname{div}(v).Fq+1∣A−q⋅dF(v)+F⋅div(v).

Here dF(v)=∑i=05vi∂iFdF(v)=\sum_{i=0}^5v_i\partial_iFdF(v)=∑i=05​vi​∂i​F and div⁡(v)=∑i=05∂ivi\operatorname{div}(v)=\sum_{i=0}^5\partial_i v_idiv(v)=∑i=05​∂i​vi​. Divisibility is taken in the ring of rational functions whose denominators are relatively prime to FFF. There is no bound on the denominator degree.

This instance: "Rank-one coupling · biquartic pencil"

F=f(X0,X1,X2)+g(Y0,Y1,Y2)+t⋅L2⋅M.F=f(X_0,X_1,X_2)+g(Y_0,Y_1,Y_2)+t\cdot L^2\cdot M.F=f(X0​,X1​,X2​)+g(Y0​,Y1​,Y2​)+t⋅L2⋅M. q=1,A=X03⋅Y03.q=1,\qquad A=X_{0}^{3} \cdot Y_{0}^{3}.q=1,A=X03​⋅Y03​.

where

t=57f(a,b,c)=a6+b6+c6+t⋅(a5⋅b+2⋅b5⋅c+3⋅c5⋅a+a2⋅b3⋅c)g(a,b,c)=a6+b6+c6+t⋅(2⋅a5⋅b+3⋅b5⋅c+c5⋅a+2⋅a3⋅b⋅c2)L=X02+X0⋅X1+2⋅X1⋅X2+3⋅X22M=Y02+Y0⋅Y1+2⋅Y1⋅Y2+3⋅Y22\begin{aligned} t&=\frac57 \\ f(a,b,c)&=a^6+b^6+c^6+t\cdot(a^5\cdot b+2\cdot b^5\cdot c+3\cdot c^5\cdot a+a^2\cdot b^3\cdot c) \\ g(a,b,c)&=a^6+b^6+c^6+t\cdot(2\cdot a^5\cdot b+3\cdot b^5\cdot c+c^5\cdot a+2\cdot a^3\cdot b\cdot c^2) \\ L&=X_{0}^{2} + X_{0} \cdot X_{1} + 2 \cdot X_{1} \cdot X_{2} + 3 \cdot X_{2}^{2} \\ M&=Y_{0}^{2} + Y_{0} \cdot Y_{1} + 2 \cdot Y_{1} \cdot Y_{2} + 3 \cdot Y_{2}^{2} \end{aligned}tf(a,b,c)g(a,b,c)LM​=75​=a6+b6+c6+t⋅(a5⋅b+2⋅b5⋅c+3⋅c5⋅a+a2⋅b3⋅c)=a6+b6+c6+t⋅(2⋅a5⋅b+3⋅b5⋅c+c5⋅a+2⋅a3⋅b⋅c2)=X02​+X0​⋅X1​+2⋅X1​⋅X2​+3⋅X22​=Y02​+Y0​⋅Y1​+2⋅Y1​⋅Y2​+3⋅Y22​​

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Algebraic Geometry·Captain: benediamond

hodge.computer: Cyclic coupling · pole order twoOpen Problem

The hodge.computer project studies a case of the generalized Hodge conjecture for sextic fourfolds with an involution, posed by Voisin. Diamond's reduction turns this Hodge-theoretic prediction into the explicit rational-vector-field problem below; it is the basis of the certificate formalism used in these missions.

Write z=(X0,X1,X2,Y0,Y1,Y2)z=(X_0,X_1,X_2,Y_0,Y_1,Y_2)z=(X0​,X1​,X2​,Y0​,Y1​,Y2​). For a fixed homogeneous sextic FFF, pole parameter q∈{1,2}q\in\{1,2\}q∈{1,2}, and homogeneous polynomial AAA of degree 6q6q6q, prove that there exists a rational vector field v=(v0,…,v5)v=(v_0,\ldots,v_5)v=(v0​,…,v5​) over C\mathbb CC, with components homogeneous of degree 6q−56q-56q−5 and a common denominator relatively prime to FFF, satisfying the key equation

Fq+1∣A−q⋅dF(v)+F⋅div⁡(v).F^{q+1}\mid A-q\cdot dF(v)+F\cdot\operatorname{div}(v).Fq+1∣A−q⋅dF(v)+F⋅div(v).

Here dF(v)=∑i=05vi∂iFdF(v)=\sum_{i=0}^5v_i\partial_iFdF(v)=∑i=05​vi​∂i​F and div⁡(v)=∑i=05∂ivi\operatorname{div}(v)=\sum_{i=0}^5\partial_i v_idiv(v)=∑i=05​∂i​vi​. Divisibility is taken in the ring of rational functions whose denominators are relatively prime to FFF. There is no bound on the denominator degree.

This instance: "Cyclic coupling · pole order two"

F=F0+X03X1Y0Y12+X0X23Y0Y22+X13X2Y1Y22.F=F_0 + \frac{X_{0}^{3} X_{1} Y_{0} Y_{1}}{2} + \frac{X_{0} X_{2}^{3} Y_{0} Y_{2}}{2} + \frac{X_{1}^{3} X_{2} Y_{1} Y_{2}}{2}.F=F0​+2X03​X1​Y0​Y1​​+2X0​X23​Y0​Y2​​+2X13​X2​Y1​Y2​​. q=2,A=X02X12X23Y02Y12Y2.q=2,\qquad A=X_{0}^{2} X_{1}^{2} X_{2}^{3} Y_{0}^{2} Y_{1}^{2} Y_{2}.q=2,A=X02​X12​X23​Y02​Y12​Y2​.

where

F0=X06+X16+X26+Y06+Y16+Y26\begin{aligned} F_0&=X_0^6+X_1^6+X_2^6+Y_0^6+Y_1^6+Y_2^6 \end{aligned}F0​​=X06​+X16​+X26​+Y06​+Y16​+Y26​​

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