Discriminant of a K3 surface

$$\tau:\operatorname{Gal}(k^{\mathrm{al}}/k)\longrightarrow O(V:=T(1)\otimes\Bbb{Q}_\ell)$$

$\det\tau=1\Longleftrightarrow\operatorname{im}\tau\subset SO(V)$
$\det\tau\neq1$ $\Rightarrow$ nontrivial quadratic character

$\rho(X^{\mathrm{al}})$ even; $\varphi:=\operatorname{Frob}_p|V$

$$\det\varphi=-1\Rightarrow\rho(X_p^{\mathrm{al}})\geq\rho(X^{\mathrm{al}})+2$$

$D_X:=\Delta_{H^2}(X)\in\Bbb{Q}^{\times}/(\Bbb{Q}^{\times})^2$: determinant-character square class

$D_X\in\Z\setminus\{0\}$ a representative; $p$ good, $p\nmid2D_X$

Theorem (Deligne; Suh)

The functional equation of Frobenius on $H^2(X)$ has the plus sign iff $D_X$ is square mod $p$.

$$\varepsilon_p=\det(-\operatorname{Frob}_p\mid H^2_{\mathrm{et}}(X^{\mathrm{al}},\Bbb{Q}_\ell(1)))=\left(\frac{D_X}{p}\right)$$

$D_X$ represents the quadratic extension cut out by the determinant on $H^2(1)$. It is neither the Picard intersection discriminant nor an unspecified equation discriminant.

Dimension twenty-two gives $\det(-\operatorname{Frob})=\det(\operatorname{Frob})$. C-Elsenhans-Jahnel, Proposition 2.1, attributes the projective sign statement to Deligne; Suh treats the proper nonprojective extension.

References: Costa-Elsenhans-Jahnel, "On the distribution of the Picard ranks of the reductions of a K3 surface" (2020), arXiv:1610.07823v3, Proposition 2.1, including the attribution to Deligne.

The Tate twist makes the pairing orthogonal. The determinant detects a quotient of order two, not all components of monodromy.

Forced excess can survive determinant one: quadratic RM with $\dim_E T=3$ still forces two Tate classes.

For Kummer surfaces the universal SO assertion is false. Serre, Lectures on N_X(p), Section 8.5.6.4, removes one polarization from $H^2(A)(1)$. Removing all divisor classes gives determinant equal to the algebraic determinant. For $A=(y^2=x^3-x)^2$, complex conjugation on the CM field gives a nontrivial character.

For $X=\operatorname{Kum}(A)$, the determinant on $V$ equals the determinant on $(\operatorname{Pic}(A^{\mathrm{al}})/\operatorname{Pic}^0(A^{\mathrm{al}}))_{\Bbb{Q}_\ell}$. It can be nontrivial.

Remove the reciprocal pairs other than $\pm1$. Determinant minus one makes the multiplicity of minus one odd. Even dimension then makes the multiplicity of plus one odd.

The two eigenvalues become one over a finite residue extension. Tate identifies the new geometric classes. In odd dimension minus one is forced but plus one need not be.

NEEDS APPROVAL: [s20-m01] Recommendation: Use "Theorem (Deligne; C-Elsenhans-Jahnel 2020)"; keep Suh in the notes.
GPT 6 astra: Credit Deligne visibly for the projective sign theorem; Suh belongs in broader notes.
GPT 5.6 sol: Keep the split credit Deligne-Suh for the sign and C-Elsenhans-Jahnel for the character.
Source: CEJ 2020, Prop. 2.1.

NEEDS APPROVAL: [s20-m02] Recommendation: Define D_X as the cohomological determinant square class before its residue symbol; choose a nonzero integer representative.
Source: CEJ 2020, Def. 2.4, Thm. 2.15.

AUTHOR'S CALL: [s20-m04] Recommendation: Keep the cohomological D_X and the explicit Galois-fixed Picard premise on slide 23.
GPT 6 astra: Credit Deligne visibly for the projective sign theorem; Suh belongs in broader notes.
GPT 5.6 sol: Keep the split credit Deligne-Suh for the sign and C-Elsenhans-Jahnel for the character.
Source: CEJ 2020, Def. 2.4 and Thm. 2.15.

NEEDS APPROVAL: [s20-m05] Recommendation: Keep the quoted candidate wording, formulas and arrangement on this slide.
Exact candidate text: "Discriminant of a K3 surface" | "$X/\Bbb{Q}$ quartic K3" | "$D_X:=\Delta_{H^2}(X)\in\Bbb{Q}^{\times}/(\Bbb{Q}^{\times})^2$: determinant-character square class" | "$D_X\in\Z\setminus\{0\}$ a representative; $p$ good, $p\nmid2D_X$" | "Theorem (Deligne; C-Elsenhans-Jahnel)" | "The functional equation of the Frobenius action on $H^2(X)$ has the plus sign if and only if $D_X$ is square mod $p$." | "$$\varepsilon_p=\det(-\operatorname{Frob}_p\mid H^2_{\mathrm{et}}(X^{\mathrm{al}},\Bbb{Q}_\ell(1)))=\left(\frac{D_X}{p}\right)$$" | "$\operatorname{Gal}(\Bbb{Q}^{\mathrm{al}}/\Bbb{Q})$ fixes $\operatorname{Pic}(X^{\mathrm{al}})$ $\Rightarrow$ $\Delta_{\operatorname{Pic}}(X)=1$" | "Theorem (C-Elsenhans-Jahnel)" | "$$\rho(X^{\mathrm{al}})=2r,\quad\left(\frac{D_X}{p}\right)=-1\quad\Rightarrow\quad\rho(X_p^{\mathrm{al}})\geq2r+2$$"
Speaker notes proposed: "$D_X$ represents the quadratic extension cut out by the determinant on $H^2(1)$. It is neither the Picard intersection discriminant nor an unspecified equation discriminant." | "Dimension twenty-two gives $\det(-\operatorname{Frob})=\det(\operatorname{Frob})$. C-Elsenhans-Jahnel, Proposition 2.1, attributes the projective sign statement to Deligne; Suh treats the proper nonprojective extension." | "$\Delta_{\operatorname{Pic}}$ is the square class of the Picard representation determinant. Galois fixing every geometric class makes it one; an integral descent equality is unnecessary."
SETTLED mathematical source: I15:L626-646; C-Elsenhans-Jahnel 2020, Prop. 2.1, Def. 2.4 and Thm. 2.15. Exact teaching arrangement still needs approval.

AUTHOR'S CALL: [s20-m90] Keep the prose plus-sign criterion, the determinant/Legendre-symbol display, or both? The display also identifies the determinant; both formulations are preserved pending your choice.

NEEDS APPROVAL: [s18-m02] Recommendation: Keep "Forced excess can survive determinant one" spoken on slide 21.
GPT 6 astra: Keep the forced-excess qualification in notes.
GPT 5.6 sol: Keep the forced-excess qualification in notes.
Source: Charles 2014, Prop. 15.

APPROVED (2026-09-14): [s18-m03] replace the quoted closing sentence with "An easy way to explain some jumps: O vs SO." The author calls it slide 15; the exact sentence is on slide 21 (historical slide 18). Apply at the text location; preserve the author's quoted numbers. Author, verbatim: "I am also unsure what is the purpose of Slide 17, in particular given Slide 16, some of teh questions are already answered in the previous slide. On slide 15, not sure we should write "The jump criterion on the slides that immediately follow is a consequence of that one fact.". Intead, we should point, there is an easy way to explain some jumps, O vs SO . And maybe there one should point out that for kummer varieties we always land in SO (check this for me please)"

AUTHOR'S CALL: [s18-m04] Recommendation: Show the representation first, then the determinant-one and nontrivial-character alternatives, then the approved closing line.
Source: CEJ 2020, Prop. 2.13.

NEEDS APPROVAL: [s18-m05] Recommendation: Keep the quoted candidate wording, formulas and arrangement on this slide.
Exact candidate text: "O or SO?" | "$V:=T_\ell(1)$; cup-product pairing" | "$$\tau:\operatorname{Gal}(k^{\mathrm{al}}/k)\longrightarrow O(V)$$" | "$\det\tau=1\Longleftrightarrow\operatorname{im}\tau\subset SO(V)$" | "$\det\tau\neq1$ $\Rightarrow$ nontrivial quadratic character" | "An easy way to explain some jumps: $O$ vs $SO$."
Speaker notes proposed: "The Tate twist makes the pairing orthogonal. The determinant detects a quotient of order two, not all components of monodromy." | "Forced excess can survive determinant one: quadratic RM with $\dim_E T=3$ still forces two Tate classes." | "For Kummer surfaces the universal SO assertion is false. Serre, Lectures on N_X(p), Section 8.5.6.4, removes one polarization from $H^2(A)(1)$. Removing all divisor classes gives determinant equal to the algebraic determinant. For $A=(y^2=x^3-x)^2$, complex conjugation on the CM field gives a nontrivial character."
SETTLED mathematical source: C-Elsenhans-Jahnel 2020, Prop. 2.13; Serre, Sec. 8.5.6.4. Exact teaching arrangement still needs approval.
SETTLED author instruction, s18-m03: "An easy way to explain some jumps: O vs SO." Approval is requested for the additional content and arrangement.

NEEDS APPROVAL: [s15-m04] Recommendation: Keep "An easy way to explain some jumps: O vs SO." on slide 21 only.
GPT 6 astra: Keep the approved line at its actual location on the O/SO frame.
GPT 5.6 sol: Keep the O/SO frame and its approved transition; the earlier report proposed a second pointer.
Source: author decision on s18-m03.

NEEDS APPROVAL: [s15-m06] Recommendation: Keep the qualified Kummer determinant explanation in the notes.
GPT 6 astra: The universal Kummer-to-SO claim is false; a swapping lift need not have a fixed spectrum.
GPT 5.6 sol: The final jury agrees the universal claim is false; its earlier fixed-spectrum argument is superseded.
Source: Serre, Lectures on N_X(p), Sec. 8.5.6.4; CEJ 2020, Ex. 2.36.

AUTHOR'S CALL: [s19-m01] Recommendation: Keep slide 22 as the four-step proof of the two new Tate classes.
GPT 6 astra: KEEP the elementary determinant proof.
GPT 5.6 sol: KEEP the elementary determinant proof.
Source: CEJ 2020, Prop. 2.13.

NEEDS APPROVAL: [s19-m02] Recommendation: Keep the even-rank hypothesis in the setup and the bound rho(X_p^al) >= rho(X^al)+2 in the final step.
GPT 6 astra: Put the even-rank hypothesis in the setup.
GPT 5.6 sol: Say "In this even-rank case" in the final step.
Source: CEJ 2020, Prop. 2.13.

NEEDS APPROVAL: [s19-m03] Recommendation: Keep "Orthogonality $\Rightarrow$ lambda and lambda^{-1}, with equal multiplicities."
GPT 6 astra: State reciprocal pairing explicitly; absolute value one alone is insufficient.
GPT 5.6 sol: Rely on the preceding orthogonal representation; omit the repeated word.
Source: CEJ 2020, Prop. 2.13.

AUTHOR'S CALL: [s19-m04] Recommendation: Show reciprocal pairing first; reveal minus one, plus one and Tate at 0/1/2.
Source: CEJ 2020, Prop. 2.13.

AUTHOR'S CALL: [s19-m05] See s19-m01 for this identical recommendation and its evidence.

NEEDS APPROVAL: [s19-m06] Recommendation: Keep the quoted candidate wording, formulas and arrangement on this slide.
Exact candidate text: "What $\det=-1$ costs you" | "$\rho(X^{\mathrm{al}})$ even; $\varphi:=\operatorname{Frob}_p|T_\ell(1)$; $\det\varphi=-1$" | "Orthogonality $\Rightarrow$ $\lambda$ and $\lambda^{-1}$, with equal multiplicities." | "Other pairs: determinant $+1$; multiplicity of $-1$ odd." | "$\dim T_\ell(1)$ even $\Rightarrow$ multiplicity of $+1$ odd." | "Tate: $+1,-1$ give two new geometric divisor classes. $$\rho(X_p^{\mathrm{al}})\geq\rho(X^{\mathrm{al}})+2$$"
Speaker notes proposed: "Remove the reciprocal pairs other than $\pm1$. Determinant minus one makes the multiplicity of minus one odd. Even dimension then makes the multiplicity of plus one odd." | "The two eigenvalues become one over a finite residue extension. Tate identifies the new geometric classes. In odd dimension minus one is forced but plus one need not be."
SETTLED mathematical source: C-Elsenhans-Jahnel 2020, Prop. 2.13. Exact teaching arrangement still needs approval.

AUTHOR'S CALL: [s18-m01] Recommendation: Keep the separate four-step determinant proof on slide 22.
GPT 6 astra: KEEP the proof; it explains the increase by two.
GPT 5.6 sol: KEEP the proof; the earlier drop recommendation is superseded.
Source: CEJ 2020, Prop. 2.13.

AUTHOR'S CALL: [s19-m90] Which proof-placement recommendation should remain: s18-m01 or s19-m01? Their wording differs; both are preserved pending your choice.