Re: A Differential Equation

What are your reasons for believing it's true? Or is that a secret? ;)

Re: A Differential Equation

Quote:

Originally Posted by

**Bruno J.** What are your reasons for believing it's true? Or is that a secret? ;)

no, it's not a secret! (Happy) it should be true because it is consistent with something which is expected to be true about the structure of centralizers in the second complex Weyl algebras. i'm not giving more details because it's a very long story!

Re: A Differential Equation

I am confused as to what "$\displaystyle A\setminus \mathbb{C}$" means! You had already defines A to include only continuous functions so what is left if you remove continuous functions?

Re: A Differential Equation

I think by

"$\displaystyle A=\mathbb{C}[x,y,z]$ with the relations $\displaystyle yx=xy+1, \ yz=zy, \ xz=zx$"

NonCommAlg means the free $\displaystyle \mathbb{C}$-ring on 3 indeterminates, modded out by the relations provided.

It's the ring of complex polynomials over $\displaystyle \mathbb{C}[x,y,z]$, except without commutivity except as those relations dictate.

The partial derivative, I assume, is just a formalization of the analytic notion of a partial derivative. It's the same as the partial derivative for an ordinary polynomial over $\displaystyle \mathbb{C}$ in x, y, and z, except that the non-commutivity now has to be respected.

Thus $\displaystyle A - \mathbb{C}$ means an element of A, a funky non-commuting polynomial in x, y, z, that isn't just a constant $\displaystyle \mathbb{C}$ value. It must include at least some power of x, y, or z - at least, after those relations are taken into account (meaning yx-xy is not in $\displaystyle A - \mathbb{C}$, because under these relations, it's a constant polynomial).

The $\displaystyle A - \mathbb{C}$ is required because otherwise b=1 (actually, any b in $\displaystyle \mathbb{C}$) trivially works.

That's what I take the description to mean.