MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\mathcal C$ be a symmetric monoidal category — for example, $\mathcal C$ might be the category of $R$-modules for a commutative ring $R$. At a minimum, I request further that:

  1. $\mathcal C$ is closed under coequalizers. (I generally assume moreover that all of my categories are closed under all colimits.)
  2. the tensor structure distributes over coequalizers. (I generally assume moreover that my monoidal categories are closed monoidal, meaning that $X\otimes$ has a left adjoint for all objects $X$; this implies that tensoring distributes over all colimits.)

Suppose that $X \in \mathcal C$ is a dualizable object, with dual $X^*$. Then the endomorphism algebra of $X$ is $A = X^* \otimes X \in \mathcal C$, with multiplication $$ A \otimes A = X^* \otimes X \otimes X^* \otimes X \overset{ X^* \otimes \mathrm{ev} \otimes X}\longrightarrow X^* \otimes \mathbf 1 \otimes X = A $$ and unit $\mathrm{coev}: \mathbf 1 \to X^* \otimes X = A$. Note in particular that $A$ is a unital associative algebra object in $\mathcal C$.

For any associative algebra $B \in \mathcal C$, recall that a left module for $B$ is an object $M \in \mathcal C$ along with an action $B \otimes M \to M$ which is compatible with the unit and multiplication on $B$ in the natural way. Similarly, one can define bimodules. Suppose that $M$ is a $B$–$B'$ bimodule, and $M'$ is a $B'$–$B''$ bimodule. The bimodule tensor product is $$ M \underset{B'}\otimes M' = \mathrm{coequalizer}\bigl( M \otimes B' \otimes M' \rightrightarrows M \otimes M' \bigr) $$ where the two arrows are the actions of $B'$ either on $M$ or $M'$. The result is a $B$–$B''$ bimodule. My requests above assure that this coequalizer exists, and moreover that bimodule tensor product is associative.

This allows one to define the Morita bicategory of $\mathcal C$, whose objects are associative algebras in $\mathcal C$, 1-morphisms are bimodules, and 2-morphisms are intertwiners. In particular, it provides the notion of Morita equivalence: two algebras $B,B'$ are Morita equivalent if there exist bimodules $_B M _{B'}$ and $_{B'} M' _B$ such that $M \otimes_{B'} M' \cong B$ as a $B$–$B$ bimodule and $M' \otimes_B M \cong B'$ as a $B'$–$B'$ bimodule. The unit object $\mathrm 1$ is, of course, an associative algebra. I will say that $B$ is Morita trivial if it is Morita equivalent to $\mathrm 1$.

Let me return now to the algebra $A = X^*\otimes X$. Is it Morita trivial? Note that $X$ is naturally a distinguished $\mathrm 1$–$A$ bimodule (i.e. right $A$-module), and $X^*$ is a distinguished $A$–$\mathrm 1$ bimodule. Moreover, it is almost trivial to check that $X^* \otimes_{\mathrm 1} X \cong A$ as an $A$–$A$ bimodule. The evaluation map moreover induces a homomorphism $X \otimes_A X^* \to \mathrm 1$ as a $\mathrm 1$–$\mathrm 1$ bimodule. If this induced homomorphism is an isomorphism, then yes, $A$ is Morita trivial.

Question: What are some examples of dualizable $X$ in categories $\mathcal C$ such that $A = X^* \otimes X$ is not Morita trivial?

Lest you think that there are no such examples, here's a trivial example. Let $\mathcal C$ be the category of $R$-modules for $R$ a commutative ring, and let $X = \mathrm 0$ be the zero $R$-module. Then $\mathrm 0$ is certainly dualizable, with dual $\mathrm 0^* = \mathrm 0$, and its endomorphism algebra is $\mathrm 0$, which is not Morita trivial unless $R$ is the zero ring.

On the other hand, here are some cases where $A$ is Morita trivial. If $X$ admits an endomorphism $f: X \to X$ with invertible trace then $A$ is Morita trivial — this is a fun exercise. This condition is sufficient but not necessary. For example, consider the category of representations over $\mathbb F_2$ of the cyclic group of order $3$. Let $X$ denote the unique $2$-dimensional irrep. Its dual is $X^* \cong X$. Since $X$ is an irrep, it has a unique non-zero endomorphism, which has trace $\dim X = 2 = 0$. Nevertheless, direct calculation shows that $A = X^* \otimes X$ is Morita-trivial (indeed, $X,X^*$ witness the Morita equivalence).

In general, I think that Morita-nontrivial endomorphism algebras determine "proper direct summands of $\mathcal C$ as a $\mathcal C$-module". I expect this means that if $R = S\oplus T$ is a direct sum of commutative rings, then $S$ and $T$ each determine Morita-nontrivial endomorphism algebras in the category of $R$-modules. Are there other examples?

share|cite|improve this question
There must be lots of examples in the stable homotopy category - many nontrivial ring spectra arise as $X\wedge DX$ – მამუკა ჯიბლაძე Mar 3 '14 at 8:05
$ (X \otimes A X^*) \otimes_1 (X \otimes_A X^*) = X \otimes_A ( X^* \otimes_1 X) \otimes_A X^*= X \otimes_A A \otimes_A X^*= X \otimes X_*$. So this object is an idempotent, and I think it is self-dual, which I think means one can without loss of generality take $A$. Can one argue from there to split the category into a product category? I am thinking taking the map $e \to 1$ and the dual $1 \to e$ and using them to define the complement in each object $Y$ of $e \otimes Y$. I haven't checked that all the diagrams and such commute, though. – Will Sawin Mar 4 '14 at 2:21
@WillSawin Yes, I believe that will work. So the reason I tend to believe (mistakenly) that all endomorphism algebras are Morita trivial is because Vect does not split as a product of categories (so the only non-Morita-trivial endomorphism algebra is 0). – Theo Johnson-Freyd Mar 4 '14 at 17:36
@მამუკაჯიბლაძე I don't know much about ring spectra, but I'd love to learn. Maybe you can describe some such examples in an answer? – Theo Johnson-Freyd Mar 4 '14 at 17:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.