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Link to @TorstenEkedahl's comment while this is on the front page
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LSpice
LSpice
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My answer will consist largely of an elaboration of Torsten's comment aboveabove. Give $\Lambda^\bullet V = \bigoplus V^{\otimes \bullet} / \langle v_1\otimes v_2 - v_2 \otimes v_1 \rangle$ the $\mathbb Z$-grading corresponding to polynomial degree. Then the wedge multiplication is commutative in the signed sense. The algebra can moreover be given a Hopf structure (in the category of graded vector spaces, with the signed braiding) with comultiplication extending

$$ v \mapsto v\otimes 1 + 1\otimes v $$

for $v\in \Lambda^1 V$.

The dual Hopf algebra is therefore also commutative and cocommutative. It contains $V^\ast$ (in $\mathbb Z$-grading $-1$) as primitive elements, and so receives a map from $\Lambda^\bullet V^\ast$.

The problem is, assuming I have not made a calculation error, that the Hopf algebra map $\Lambda^\bullet V^\ast \to (\Lambda^\bullet V)^\ast$ is not an iso if $(\dim V)!$ is not invertible in the ground ring. The ring $\Lambda^\bullet V^\ast$ is the "polynomial ring" in generators $V$, whereas $(\Lambda^\bullet V)^\ast$ is the "divided power ring".

So this will obstruct finding a "functorial" reason for the Brian Conrad's pairing: it is not a Hopf pairing for the canonical Hopf structures on the two sides.

My answer will consist largely of an elaboration of Torsten's comment above. Give $\Lambda^\bullet V = \bigoplus V^{\otimes \bullet} / \langle v_1\otimes v_2 - v_2 \otimes v_1 \rangle$ the $\mathbb Z$-grading corresponding to polynomial degree. Then the wedge multiplication is commutative in the signed sense. The algebra can moreover be given a Hopf structure (in the category of graded vector spaces, with the signed braiding) with comultiplication extending

$$ v \mapsto v\otimes 1 + 1\otimes v $$

for $v\in \Lambda^1 V$.

The dual Hopf algebra is therefore also commutative and cocommutative. It contains $V^\ast$ (in $\mathbb Z$-grading $-1$) as primitive elements, and so receives a map from $\Lambda^\bullet V^\ast$.

The problem is, assuming I have not made a calculation error, that the Hopf algebra map $\Lambda^\bullet V^\ast \to (\Lambda^\bullet V)^\ast$ is not an iso if $(\dim V)!$ is not invertible in the ground ring. The ring $\Lambda^\bullet V^\ast$ is the "polynomial ring" in generators $V$, whereas $(\Lambda^\bullet V)^\ast$ is the "divided power ring".

So this will obstruct finding a "functorial" reason for the Brian Conrad's pairing: it is not a Hopf pairing for the canonical Hopf structures on the two sides.

My answer will consist largely of an elaboration of Torsten's comment above. Give $\Lambda^\bullet V = \bigoplus V^{\otimes \bullet} / \langle v_1\otimes v_2 - v_2 \otimes v_1 \rangle$ the $\mathbb Z$-grading corresponding to polynomial degree. Then the wedge multiplication is commutative in the signed sense. The algebra can moreover be given a Hopf structure (in the category of graded vector spaces, with the signed braiding) with comultiplication extending

$$ v \mapsto v\otimes 1 + 1\otimes v $$

for $v\in \Lambda^1 V$.

The dual Hopf algebra is therefore also commutative and cocommutative. It contains $V^\ast$ (in $\mathbb Z$-grading $-1$) as primitive elements, and so receives a map from $\Lambda^\bullet V^\ast$.

The problem is, assuming I have not made a calculation error, that the Hopf algebra map $\Lambda^\bullet V^\ast \to (\Lambda^\bullet V)^\ast$ is not an iso if $(\dim V)!$ is not invertible in the ground ring. The ring $\Lambda^\bullet V^\ast$ is the "polynomial ring" in generators $V$, whereas $(\Lambda^\bullet V)^\ast$ is the "divided power ring".

So this will obstruct finding a "functorial" reason for the Brian Conrad's pairing: it is not a Hopf pairing for the canonical Hopf structures on the two sides.

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Theo Johnson-Freyd
Theo Johnson-Freyd
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My answer will consist largely of an elaboration of Torsten's comment above. Give $\Lambda^\bullet V = \bigoplus V^{\otimes \bullet} / \langle v_1\otimes v_2 - v_2 \otimes v_1 \rangle$ the $\mathbb Z$-grading corresponding to polynomial degree. Then the wedge multiplication is commutative in the signed sense. The algebra can moreover be given a Hopf structure (in the category of graded vector spaces, with the signed braiding) with comultiplication extending

$$ v \mapsto v\otimes 1 + 1\otimes v $$

for $v\in \Lambda^1 V$.

The dual Hopf algebra is therefore also commutative and cocommutative. It contains $V^\ast$ (in $\mathbb Z$-grading $-1$) as primitive elements, and so receives a map from $\Lambda^\bullet V^\ast$.

The problem is, assuming I have not made a calculation error, that the Hopf algebra map $\Lambda^\bullet V^\ast \to (\Lambda^\bullet V)^\ast$ is not an iso if $(\dim V)!$ is not invertible in the ground ring. The ring $\Lambda^\bullet V^\ast$ is the "polynomial ring" in generators $V$, whereas $(\Lambda^\bullet V)^\ast$ is the "divided power ring".

So this will obstruct finding a "functorial" reason for the Brian Conrad's pairing: it is not a Hopf pairing for the canonical Hopf structures on the two sides.