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Angelo
Angelo
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It seems to me that the involution of $Q \otimes Q$ that exchanges $a \otimes b$ and $b\otimes a$ is inner, which means that the homomorphism you describe should always be trivial.

Here is the argument, I hope it is correct. Suppose that $Q$ is trivial, that is, there is a projective $k$-module $V$ such that $Q \simeq \operatorname{End}V$. Then $Q \otimes Q \simeq \operatorname{End}(V\otimes V)$. Consider the operator $\tau\colon V\otimes V \simeq V\otimes V$ that exchanges $v \otimes w$ and $w \otimes v$. We can think of $\tau$ as an element of $Q \otimes Q$; conjugation by $\tau$ exchanges $a \otimes b$ and $b\otimes a$.

Now, choose another projective $k$-module $W$ and an isomorphism $Q \simeq \operatorname{End}W$. Consider the induced isomorphism $\phi\colon\operatorname{End}V \simeq \operatorname{End}W$. Then there exists an invertible $k$-module $L$ and an isomorphism $\psi\colon W \simeq L \otimes V$ such that $\phi$$\psi$ that induces $\phi$. This implies that $\tau$ is independent of the isomorphism $Q \simeq \operatorname{End}V$, and it is a canonical element of $Q\otimes Q$.

If $Q$ is not trivial, choose a faithfully flat extension $k'/k$ such that $Q_{k'}$ is trivial. Because it is canonical, by descent theory the element $\tau' \in Q_{k'}\otimes_{k'}Q_{k'}$ constructed above descends to an element $\tau \in Q \otimes_k Q$ which will induce the involution $Q \otimes Q$ that exchanges $a \otimes b$ and $b\otimes a$.

It seems to me that the involution of $Q \otimes Q$ that exchanges $a \otimes b$ and $b\otimes a$ is inner, which means that the homomorphism you describe should always be trivial.

Here is the argument, I hope it is correct. Suppose that $Q$ is trivial, that is, there is a projective $k$-module $V$ such that $Q \simeq \operatorname{End}V$. Then $Q \otimes Q \simeq \operatorname{End}(V\otimes V)$. Consider the operator $\tau\colon V\otimes V \simeq V\otimes V$ that exchanges $v \otimes w$ and $w \otimes v$. We can think of $\tau$ as an element of $Q \otimes Q$; conjugation by $\tau$ exchanges $a \otimes b$ and $b\otimes a$.

Now, choose another projective $k$-module $W$ and an isomorphism $Q \simeq \operatorname{End}W$. Consider the induced isomorphism $\phi\colon\operatorname{End}V \simeq \operatorname{End}W$. Then there exists an invertible $k$-module $L$ and an isomorphism $\psi\colon W \simeq L \otimes V$ such that $\phi$ that induces $\phi$. This implies that $\tau$ is independent of the isomorphism $Q \simeq \operatorname{End}V$, and it is a canonical element of $Q\otimes Q$.

If $Q$ is not trivial, choose a faithfully flat extension $k'/k$ such that $Q_{k'}$ is trivial. Because it is canonical, by descent theory the element $\tau' \in Q_{k'}\otimes_{k'}Q_{k'}$ constructed above descends to an element $\tau \in Q \otimes_k Q$ which will induce the involution $Q \otimes Q$ that exchanges $a \otimes b$ and $b\otimes a$.

It seems to me that the involution of $Q \otimes Q$ that exchanges $a \otimes b$ and $b\otimes a$ is inner, which means that the homomorphism you describe should always be trivial.

Here is the argument, I hope it is correct. Suppose that $Q$ is trivial, that is, there is a projective $k$-module $V$ such that $Q \simeq \operatorname{End}V$. Then $Q \otimes Q \simeq \operatorname{End}(V\otimes V)$. Consider the operator $\tau\colon V\otimes V \simeq V\otimes V$ that exchanges $v \otimes w$ and $w \otimes v$. We can think of $\tau$ as an element of $Q \otimes Q$; conjugation by $\tau$ exchanges $a \otimes b$ and $b\otimes a$.

Now, choose another projective $k$-module $W$ and an isomorphism $Q \simeq \operatorname{End}W$. Consider the induced isomorphism $\phi\colon\operatorname{End}V \simeq \operatorname{End}W$. Then there exists an invertible $k$-module $L$ and an isomorphism $\psi\colon W \simeq L \otimes V$ such that $\psi$ that induces $\phi$. This implies that $\tau$ is independent of the isomorphism $Q \simeq \operatorname{End}V$, and it is a canonical element of $Q\otimes Q$.

If $Q$ is not trivial, choose a faithfully flat extension $k'/k$ such that $Q_{k'}$ is trivial. Because it is canonical, by descent theory the element $\tau' \in Q_{k'}\otimes_{k'}Q_{k'}$ constructed above descends to an element $\tau \in Q \otimes_k Q$ which will induce the involution $Q \otimes Q$ that exchanges $a \otimes b$ and $b\otimes a$.

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Angelo
Angelo
  • 27k
  • 6
  • 92
  • 112

It seems to me that the involution of $Q \otimes Q$ that exchanges $a \otimes b$ and $b\otimes a$ is inner, which means that the homomorphism you describe should always be trivial.

Here is the argument, I hope it is correct. Suppose that $Q$ is trivial, that is, there is a projective $k$-module $V$ such that $Q \simeq \operatorname{End}V$. Then $Q \otimes Q \simeq \operatorname{End}(V\otimes V)$. Consider the operator $\tau\colon V\otimes V \simeq V\otimes V$ that exchanges $v \otimes w$ and $w \otimes v$. We can think of $\tau$ as an element of $Q \otimes Q$; conjugation by $\tau$ exchanges $a \otimes b$ and $b\otimes a$.

Now, choose another projective $k$-module $W$ and an isomorphism $Q \simeq \operatorname{End}W$. Consider the induced isomorphism $\phi\colon\operatorname{End}V \simeq \operatorname{End}W$. Then there exists an invertible $k$-module $L$ and an isomorphism $\psi\colon W \simeq L \otimes V$ such that $\phi$ that induces $\phi$. This implies that $\tau$ is independent of the isomorphism $Q \simeq \operatorname{End}V$, and it is a canonical element of $Q\otimes Q$.

If $Q$ is not trivial, choose a faithfully flat extension $k'/k$ such that $Q_{k'}$ is trivial. Because it is canonical, by descent theory the element $\tau' \in Q_{k'}\otimes_{k'}Q_{k'}$ constructed above descends to an element $\tau \in Q \otimes_k Q$ which will induce the involution $Q \otimes Q$ that exchanges $a \otimes b$ and $b\otimes a$.