# Does a nonabelian Picard group exist?

Over a noncommutative algebra $A$ we have no problem in defining invertible bimodules (as in the book by Bass on algebraic $K$-theory) - corresponding to line bundles over topological spaces $X$ if $A=C(X)$. Further we can make these into a group (the Picard group) by taking the group operation $\otimes_A$. However after quite a bit of searching I cannot find examples where the resulting Picard group is itself nonabelian.

I would be very pleased if people could point out examples where Picard group is nonabelian - I assume that they do exist, since there is no obvious reason why it should be abelian (I think... I may be wrong...). Examples where $A$ is a star algebra would be especially welcome.

This question came from considering the properties of `line bundles' over non commutative spaces, and noncommutative algebraic geometry. Apologies if it is well known in some circles!

• This construction does not recover line bundles when $A$ is commutative. In general, if $A$ is a $k$-algebra where $k$ is commutative, you can consider $(A, A)$-bimodules over $k$, and you only recover line bundles if $A$ is commutative and $k = A$. – Qiaochu Yuan Jun 18 '16 at 22:11
• In any case, I think it's misleading to think of this construction as a noncommutative analogue of line bundles. Line bundles ought to pull back, and these guys don't. (I think there is in fact no noncommutative analogue of line bundles. There is an $E_2$ analogue of line bundles, though.) – Qiaochu Yuan Jun 18 '16 at 22:14
• A better analogy is that invertible bimodules are analogous to Fourier-Mukai transforms. – Qiaochu Yuan Jun 18 '16 at 22:25

By the Eilenberg-Watts theorem, the 2-group of invertible $(A, A)$-bimodules is naturally equivalent to the 2-group of automorphisms of $\text{Mod}(A)$. There is a natural map

$$\text{Aut}(A) \to \pi_0 \text{Aut}(\text{Mod}(A))$$

whose kernel is the subgroup of inner automorphisms. So it suffices to find $A$ whose outer automorphism group is nonabelian, and these are plentiful. (This conclusion does not require the Eilenberg-Watts theorem to reach, but it's convenient to phrase things this way.)

For a commutative example, we can let $A$ be a Galois extension of some field with nonabelian Galois group. For a noncommutative example we can let $A$ be a matrix algebra over such a thing.

• A similar thing is true in the context of $C^*$-algebras, when one looks at Morita-Rieffel equivalences. The Picard group of $A$ is isomorphic to $Aut(A \otimes \mathbb{K})/Inn(A \otimes \mathbb{K})$, where the inner automorphisms use unitaries from the multiplier algebra of $A \otimes \mathbb{K}$. – Ulrich Pennig Jun 18 '16 at 22:13
• Thanks - a useful answer and a useful comment. Looks like I still have some work to do. – Edwin Beggs Jun 19 '16 at 20:45

Let me just try a few more comments on this and on the answer of @Qiaochu Yuan:

One reasons for confusion might be the actual definition of the Picard group. If I understand you correct then you take iso classes of invertible bimodules without the assumption that the underlying algebra $A$ is commutative. In this case, one has a canonical short exact sequence \begin{equation} 1 \longrightarrow \mathrm{InnAut}(A) \longrightarrow \mathrm{Aut}(A) \longrightarrow \mathrm{Pic}(A) \end{equation} and hence the outer automorphisms always contribute to the Picard group. IN particular, if this group is non-abelian, the Picard group is non-abelian as well. This is completely classical and can be found in all kind of books on Morita theory.

Now the catch comes when $A$ is actually commutative. In this case we can consider invertible bimodules which are in addition central: for $a \in A$ and $x \in E$ we have $ax = xa$, i.e. the left and right module structures coincide. If you think of a geometric situation then this is what you usually have in mind when taking about sections of line bundles: functions are multiplied to section from left or right without much thinking. But in principle you can twist the left module structure by means of an automorphism while keeping the right module structure.

Now taking only central invertible bimodules gives a subgroup of the Picard group, sometimes called the central/commutative/static Picard group $\mathrm{SPic}(A)$. This is the one which really corresponds to line bundles. It is always commutative! Moreover, in this case ($A$ commutative) one has a semi-direct product structure \begin{equation} \mathrm{Pic}(A) = \mathrm{Aut}(A) \ltimes \mathrm{SPic}(A) \end{equation} where the autormorphisms (are always outer!) act on the bimodules in the obvious way. So the full Picard group has this decomposition and hence its non-abelianness comes only from the automorphism group.

To make the confusion complete: in geometry people call $\mathrm{SPic}(A)$ sometimes the Picard group (since it corresponds to line bundles), which therefore is always abelian. This is perhaps the reasons that some people say that the Picard group is abelian.

A last remark: there are notions of Morita equivalence for $^*$-algebras as well, even beyond the $C^*$-algebraic case. Here you have similar statements on Picard groups (which now take into account $^*$-involutions and notions of positivity). You can find such things in a lecture note by myself on my homepage as well as in several papers also there.