infinite products of commutators

Stream: maths

[Joachim Breitner]

I was very dissatisfied that I only managed to prove

⁅subgroup.pi I H, subgroup.pi I K⁆ = subgroup.pi I (λ i, ⁅H i, K i⁆)

for finite families of groups, as it seemed to obviously hold also for infinite families. But I came across https://mathoverflow.net/q/35397 and if I draw the right conclusion and that post is right then the equality above does not actually hold. That's a relief (I can stop trying to find a proof), but I also find it highly curious that something that looks so “generic” like the direct product should behave differently depending on the groups involved.

[Damiano Testa]

This reminds me of a fact that I learned recently. Let $K=k \langle x,y \rangle$ be the polynomial ring over a field $k$ in two non-commuting variables. Finally, let $K[[t]]$ be the ring of Laurent polynomials with coefficients in $K$. The quotient of $K[[t]]$ by the ideal $(xy-yx)$ is not the ring of Laurent series in two commuting variables. For instance the elements $\sum x^ny^n t^n$ and $\sum (xy)^n t^n$ are not equal in the quotient.

If you take an appropriate completion, everything works out, though!

[Kevin Buzzard]

This is the kind of stuff you find out when formalising!

[Reid Barton]

Is that because you took the quotient by a non-closed ideal?

[Reid Barton]

This kind of thing usually happens when you borrow left adjoint/colimit-type constructions from the "wrong" algebraic category. e.g., you probably knew already that $K[[x]] \otimes K[[y]]$ is not $K[[x,y]]$ unless you complete it.

[Damiano Testa]

Indeed! The completion bit comes "unnoticed" when talking about such things, but when you actually want to write something literally correct, you realize your mistakes!

As usual, you know how to fix it, once you see it is broken. The difficult bit is seeing that there is an error.

[Damiano Testa]

Now that I think of it, this might also be a similar issue to what happens with the Krull topology on Galois groups and the discussion that took place yesterday about it.

[Riccardo Brasca]

It remembers me that $K((x, y)) \subsetneq K((x))((y))$, but finding an explicit in the latter that is not in the former is quite subtle. (Here $R((\cdot))$ is by definition the field of fraction of $R\llbracket\cdot\rrbracket$, the ring of power series.)