# On a morphism from the Brauer group to the Picard group

Suppose that $$k$$ is a commutative ring and that $$A$$ is an Azumaya $$k$$-algebra. Then there is a well-known morphism from $$Aut_{Alg_k}(A)$$, the group of algebra automorphisms, to the Picard group $$Pic(k)$$. One way to phrase it is as follows. Every automorphism $$\phi$$ determines a $$k$$-linear autoequivalence $$\phi^*: Mod_A \to Mod_A$$ However, since $$A$$ is Azumaya, every $$k$$-linear autoequivalence of the category of (right) $$A$$-modules is of the form $$M \mapsto M \otimes_k J$$ for some $$J$$ in the Picard group of $$k$$, unique up to isomorphism. Moreover, this assignment takes composition to tensor product. This assignment is part of the Rosenberg-Zelinsky exact sequence $$0 \to k^\times \to A^\times \to Aut_{Alg_k}(A) \to Pic(k).$$ It is possible to express $$J$$ as $$HH^0(A,A^\tau)$$ where $$A^\tau$$ is $$A$$, but with half of its bimodule structure pulled back along $$\phi$$.

In particular, given an Azumaya $$k$$-algebra $$Q$$, let $$A = Q \otimes_k Q$$. Then A is also an Azumaya $$k$$-algebra with automorphism $$\phi(a \otimes b) = b \otimes a$$, and thus $$Q$$ determines a (2-torsion) element in $$Pic(k)$$. This assignment turns out to be Morita invariant and it respects the tensor product; this makes it a (2-torsion) homomorphism from the Brauer group $$Br(k)$$ to the Picard group $$Pic(k)$$.

Are there examples of rings $$k$$ for which this homomorphism is nontrivial?

• Is the homomorphism you wrote down the same as the map $\mathrm{Br}(k) = \pi_0 \mathbf{Br}(k) \to \pi_1 \mathbf{Br}(k) = \mathrm{Pic}(k)$ determined by $\eta\in \pi_1(S^0)$ where $\mathbf{Br}(k)$ is the Brauer space of $k$? – skd Jun 17 at 18:53
• @skd Yes, it is. (In fact, that's my original motivation for asking.) – Tyler Lawson Jun 17 at 20:30

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 $$b\otimes a$$.

• Wonderful, thank you! What I find curious about this is that it uses that the twist $L \otimes L \to L \otimes L$ is the identity for $L$ invertible; I had verified this separately and it was what led me to wonder if the same were true of the twist in the Brauer group. – Tyler Lawson Jun 19 at 6:26
• Very nice! The element $\tau$ is known as the Goldman element of $Q$, see Propostion 5.1 in Saltman's "Lectures on Division Algebras", which also proves it is canonical. – Uriya First Jun 19 at 6:46

Here is another way to see that the given map should be zero. It corresponds to a map of sheaves of grouplike $$\mathbb{E}_\infty$$-spaces $$K(\mathbb{G}_m,2)\rightarrow K(\mathbb{G}_m,1)$$. All such maps are nullhomotopic (for example by delooping to sheaves of spectra and using a $$t$$-structure argument).

I would also guess that the map is given by $$A\mapsto\mathrm{HH}(A/k)$$. Since Hochschild homology is symmetric monoidal, $$\mathrm{HH}(A/k)$$ is a line bundle. The argument above shows that it is trivial.

On the other hand, I suspect there are derived Azumaya algebras where this is non-trivial. If we look at the derived Brauer sheaf (on the étale site of an ordinary commutative ring) it is an extension $$K(\mathbb{G}_m,2)\rightarrow\mathbf{dBr}\rightarrow K(\mathbb{Z},1).$$ Taking Hochschild homology gives a map of sheaves of grouplike $$\mathbb{E}_\infty$$-spaces $$\mathbf{dBr}\rightarrow\mathbf{dPic}$$, where $$\mathbf{dPic}$$ fits into a fiber sequence $$K(\mathbb{G}_m,1)\rightarrow\mathbf{dPic}\rightarrow K(\mathbb{Z},0).$$ The map $$\mathbf{dBr}\rightarrow\mathbf{dPic}$$ thus canonically factors through a map $$K(\mathbb{Z},1)\rightarrow K(\mathbb{G}_m,1)$$ of grouplike $$\mathbb{E}_\infty$$-spaces. I believe this is the map that sends $$1$$ to $$-1\in\mathbb{G}_m(\mathbb{Z})$$. Indeed, the copy of $$\mathbb{Z}$$ is coming from suspension in the derived category.

To make this concrete, one needs a ring $$k$$ where the induced map $$H^1(k,\mathbb{Z})\rightarrow\mathrm{Pic}(k)$$ is non-zero. I don't have an example, but I'd be surprised if this didn't exist.

• I believe that the ring $\mathbb C[x,y](y^2 - x^2(x-1))$ (the ring of the standard affine nodal cubic) should have this property. – Angelo Jun 19 at 16:05