Re: Tensor of monads

["Prof. Peter Johnstone" <P.T.Johnstone@dpmms.cam.ac.uk>]

Dear Richard,

Your (*) is not an additional condition. Being a sheaf for both J and J'
is equivalent to being a sheaf for their join (which I presume is what you
mean by J n J'). For a proof, see A4.5.16 in the Elephant.

Peter
---------------------------
On Sun, 1 Aug 2010, Richard Garner wrote:

> Further to my earlier question:
>
>   -- Given idempotent monads S, T on a category C for which we can speak of
>> the tensor of S and T, is it always the case that S * T is isomorphic to S +
>> T?
>>
>
> I think I'm now happy that the answer is "no". Consider, as in my previous
> message, a presheaf category [D^op, Set]. Let S and T be the idempotent
> monads corresponding to two
> Grothendieck topologies J and J' on [D^op, Set]. Then S + T is the monad
> whose algebras are presheaves which are simultaneously J-sheaves and
> J'-sheaves. On the other hand,
> S * T has as algebras those presheaves X which are both J-sheaves and
> J'-sheaves, but which satisfy an additional axiom (*). This axiom may be
> expressed most expediently when D has finite products; so let us assume that
> now. The condition says:
>
> (*) Let f_i : U_i --> U be J-covering, and let g_k : V_k --> V be
> J'-covering. Let ( x_ik \in X(U_i x V_k) ) be a compatible family for ( f_i
> x g_k : U_i x V_j --> U x V ). Then the two natural ways of patching to an
> element of X(U x V) agree.
>
> These two ways of patching are as follows. For the first, note that since (
> f_i x V_k | i \in I ) is J-covering for each k in K, we may patch to obtain
> elements ( y_k \in X(U x V_k) | k \in K ). Then since ( U x g_k | k \in K )
> is J'-covering, we may patch these to obtain an element z \in X(U x V). For
> the second way of patching, we proceed entirely analogously, but this time
> going via a family ( y'_i \in X(U_i x V) | i \in I).
>
> Now (*) is a genuine extra condition which as far as I can see is not a
> consequence of being both a J-sheaf and a J'-sheaf, so that S + T algebras
> are not the same as S * T algebras. Note, however, that (*) _is_ a
> consequence of being a (J n J')-sheaf, since ( f_i x g_j ) is covering in J
> n J'. On the other hand, I'm not sure if (*) implies being a (J n J')-sheaf,
> as I conjectured in my previous message; I don't have an Elephant to hand to
> check.
>
> Richard


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