Hi Marc,

On 2013-08-14, Marc Mezzarobba <m...@mezzarobba.net> wrote:
> 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...". Of course,
sometimes there are two or more different actions to be considered at
the same time. But I would expect that far more often one has just one
action.

>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.

You can do exactly the same with Sage/Python. I guess if you call a
differential operator, then it means to apply the operator. And if you
define a (multiplicative) action, then "*" will use this action.

> 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]. This ring, by the way, is implemented in Sage.
And actually I would expect that the parent of d/dx *is* this ring. So,
instead of using register_action, it would be enough to tell this
non-commutative ring that there is a coercion from K[x].

Best regards,
Simon


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