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Just as we know, $w_\infty$:=span {${z^\alpha }\partial _z^\beta|\alpha,\beta\in\mathbb{Z}, \beta\geq0$ }.

But, what's the name of the following algebra, span {$\{{z^{\alpha_1}}{y^{\alpha_2}}\partial _z^{\beta_1}\partial _y^{\beta_1}|\alpha_i,\beta_i\in\mathbb{Z}, \beta\geq0\}$ }?

Is it isomorphic to $w_\infty \times w_\infty$?

Can one provide some references about this algebra? Thanks very much!

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If I'm not mistaken, this is the associative algebra of algebraic differential operators on the torus $\mathbb{G}_{m,\mathbb{C}}^2$. It is the tensor product of two copies of $w_\infty$, not the direct product. That is, you should replace $\times$ with $\otimes$.

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    $\begingroup$ I'm very confused by this question and this answer. In which sense is $w_\infty$ an algebra in this context? I have always understood it as a Lie algebra. It may simply come down to what "span" means in this context. Is it meant as a generating set? or as a linear basis? From your answer it seems you take it to mean the former. Is this standard? $\endgroup$
    – José Figueroa-O'Farrill
    Commented Dec 19, 2010 at 15:34
  • $\begingroup$ Usually, I think $w_\infty$ is an Lie algebra, though it seems that $w_\infty$ can be viewed as an associative algebra. The algebra spanned by {$z^{\alpha_1}z^{\alpha_2}\partial_{z}^{\beta_1}\partial_{y}^{\beta_2}$} should be Lie algebra. "span" means as a linear basis. $\endgroup$
    – Jack Cheng
    Commented Dec 20, 2010 at 1:26
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    $\begingroup$ Thanks for this. I think that my initial objection to your question was misguided and I've deleted my comment. Sorry about the noise! I now think that indeed what you wrote down is a Lie algebra. It's the Lie algebra obtained from the associative algebra in Scott's answer via the commutator. $\endgroup$
    – José Figueroa-O'Farrill
    Commented Dec 20, 2010 at 3:49

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