Hi,

Simon King wrote:
>>> I think quite often one is in a situation that one has two different
>>> sets (rings, groups, ...) S1, S2, such that there is only one action
>>> of S1 on S2 from the left, and thus if one has s1 from S1 and s2
>>> from S2, then s1*s2 is not ambiguous.
>>
>> I guess it depends what exactly you mean by "only one action of S1 on
>> S2":
> 
> I wrote: "quite often one is in a situation that...".

Sorry, apparently I misinterpreted your "thus...".

>>think of S1 = K[x][d/dx] and S2 = K[x]. Differential operators act
>> on polynomials (e.g., d/dx(x) = 1), but there is also an embedding of
>> K[x] into K[x][d/dx] under which d/dx*x = x*d/dx+1.
> 
> Do you see what you just did? You use "call notation" for one action,
> i.e., d/dx(x), but use multiplicative notation for the other action,
> i.w., d/dx*x.

Yes, except that the second one is not an action of S1 on S2 from the 
left, but an action of S2 on S1 from the right, isn't it? That's all I 
wanted to point out: when trying to decide whether there is a single 
action that makes sense, one has to be careful to take into account 
actions from both sides and/or actions on larger sets to which the 
elements coerce. Apologies if I'm stating the obvious :-)

>> The internal
>> multiplication of K[x][d/dx] applied in this context is not a left
>> action of K[x][d/dx] _on K[x]_, but I would definitely expect
>> (d/dx)*x to coerce x to K[x][d/dx] and multiply there.
> 
> No, I'd rather expect to coerce x into a *non-commutative* version of
> the polynomial ring K[x][d/dx].

Yes, what I denoted K[x][d/dx] was the ring of differential operators 
with polynomial coefficients. Or do you have something else in mind?

Best wishes,

-- 
Marc

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