Re-seating a monad - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T09:02:29Z http://mathoverflow.net/feeds/question/38863 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/38863/re-seating-a-monad Re-seating a monad Aleks Kissinger 2010-09-15T19:52:56Z 2011-12-02T18:20:55Z <p>Let $\mathcal C$ and $\mathcal D$ be categories with suitable limits and colimits for the following discussion. Is it possible to re-interpret, or "re-seat" a monad $T : \mathcal C \to \mathcal C$ as a monad over $\mathcal D$? When $T$ is finitary, I know at least one way to do this. Compute the Kleisli category $\mathcal C_T$ and consider $(\mathcal C_T)^{op}$ as a Lawvere theory. Then models of $(\mathcal C_T)^{op}$ on $\mathcal C$ are the same as $T$-algebras. If we then consider models of $(\mathcal C_T)^{op}$ on $\mathcal D$, we have a canonical, monadic forgetful functor $[(\mathcal C_T)^{op},\mathcal D] \to \mathcal D$ from which we can build a new finitary monad $T' : \mathcal D \to \mathcal D$. My question is as follows.</p> <blockquote> <p>Is there an direct way to obtain $T'$ from $T$ (i.e. that doesn't go via the Lawvere theory)? If so, can it be extended to work for arbitrary monads?</p> </blockquote> http://mathoverflow.net/questions/38863/re-seating-a-monad/82497#82497 Answer by Tom Leinster for Re-seating a monad Tom Leinster 2011-12-02T18:20:55Z 2011-12-02T18:20:55Z <p>I like the question, Aleks - and I like the term "re-seating".</p> <p>Yes, there is a direct way, at least in suitable circumstances. First suppose that we're beginning with a finitary monad $T$ on $\text{Set}$. For each set $X$, we have $$T(X) = \int^n T(n) \times X^n$$ where the coend is over the category of finite sets. (This is essentially the definition of finitariness, in disguise. If you haven't seen this before, think of $T(n)$ as the set of $n$-ary operations in the theory $T$.) The symbol $\times$ is of course product, but it could also be interpreted as copower: $T(n) \times X^n$ is a copower of the object $X^n$ of $\text{Set}$. </p> <p>Now take a category $\mathcal{D}$ with suitable colimits. For $Y \in \mathcal{D}$, put $$T'(Y) = \int^n T(n) \times Y^n$$ where again the coend is over the category of finite sets, and $\times$ is copower. (We can no longer interpret $\times$ as product, because $T(n)$ and $Y^n$ belong to different categories.) You can give $T'$ the structure of a monad, at least under mild assumptions on the behaviour of colimits in $\mathcal{D}$. That's the same monad $T'$ you'd get if you followed the kind of procedure you describe.</p> <p>Incidentally, I don't think your description of the procedure is quite right. If, for example, $\mathcal{C} = \text{Set}$, then $(\mathcal{C}_T)^\text{op}$ is <em>not</em> the Lawvere theory of $T$. The Lawvere theory is the full subcategory of $(\mathcal{C}_T)^\text{op}$ consisting of just the (free algebras on) <em>finite</em> sets. </p> <p>I think the coend formula for re-seating should work when $\mathcal{C}$ is an arbitrary locally finitely presentable category, not necessarily $\text{Set}$. You'd have to replace the category of finite sets by the category <code>$\mathcal{C}_{\text{fp}}$</code> of finitely presentable objects of $\mathcal{C}$. And you'd need <code>$\mathcal{C}_{\text{fp}}$</code> to act on $\mathcal{D}$, so that the copower defining $T'$ made sense.</p> <p>Generalizing in a different direction, there's probably nothing special about finitariness: you could do it with any old rank. But I suspect you do need the monad to have rank in order to re-seat a monad in the kind of way I'm suggesting. The category of finite (or suitably small) sets provides a bridge between $\mathcal{C}$ and $\mathcal{D}$, and that's what enables the monad to be re-seated. </p>