11 votes

A simple proof of the Weyl algebra's rigidity.

Since the Weyl algebra is a deformation of the polynomial ring in two variables, there is a short transparent proof using deformation theory, cf. M. Gerstenhaber and A. Giaquinto, On the cohomology of ...
Murray Gerstenhaber's user avatar
9 votes
Accepted

von Neumann algebra of canonical commutation relations

The C$^*$-algebra generated by the exponentials $e^{isQ}$ and $e^{itP}$, $s,t \in \mathbb{R}$, is the CCR algebra. The von Neumann algebra they generate in this representation is all of $B(L^2(\mathbb{...
Nik Weaver's user avatar
  • 42.2k
8 votes
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A Leibniz-like formula for $(f(x) \frac{d}{dx})^n f(x)$?

Revamped Feb. 12, 2022: I posted an answer to this (perennial) question in detail in the old MO-Q "Formula for n-th iteration of dx/dt=B(x)" and pointed out a common conflation of related ...
Tom Copeland's user avatar
  • 9,931
8 votes
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Infinite dimensional irreducible representations of a tensor product

Nate's suggestion on math.SE works. We'll show that if $A = k[x, \partial_x]$ and $B = k[y, \partial_y]$ are both taken to be the Weyl algebra, then the module over $A_2 = A \otimes B \cong k[x, \...
Qiaochu Yuan's user avatar
7 votes
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A non-commutative analog of: two polynomials are algebraically dependent iff their Jacobian is zero

The reverse implication is true in a considerably more general setting (Burchnall-Chaundy theory). Namely, for any pair $(U,V)$ of commuting meromorphic coefficient differential operators in one ...
Victor Protsak's user avatar
7 votes

A Leibniz-like formula for $(f(x) \frac{d}{dx})^n f(x)$?

Suppose that we don't reorder the parts within terms and that we always post-multiply the $f$. So e.g. $$\begin{eqnarray*}f_1 &=& f \\ f_2 &=& (Df)f \\ f_3 &=& (D^2f)ff + (Df)(...
Peter Taylor's user avatar
  • 6,571
5 votes

A Leibniz-like formula for $(f(x) \frac{d}{dx})^n f(x)$?

In the paper 'Nested Derivatives:' A simple method for computing series expansions of inverse function' by D. Dominici, with arXiv version https://arxiv.org/pdf/math/0501052.pdf has something very ...
skbmoore's user avatar
  • 884
4 votes

Gelfand-Kirillov dimension of the first Weyl algebra

Let $R=\mathbb{F}[h]$. That $GK(R)=1$ comes straight from the definition of Gelfand-Kirillov dimension. Let $R_n$ be the $\mathbb{F}$-subspace of $R$ spanned by monomials of degree at most $n$. Set $f(...
J. Gaddis's user avatar
3 votes

Gelfand-Kirillov dimension of generalized Weyl algebras

This is addressed in the section "GK Dimension" of Ebrahim - The prime spectrum and representation theory of the 2×2 reflection equation algebra. If the automorphism $\sigma$ is locally algebraic, ...
ebrahim's user avatar
  • 161
3 votes
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Weyl algebra as an Azumaya algebra over its centre

But my question remains the same: is there a direct way to prove the claim of the paper in the case when 𝑋 is the affine 𝑛-space (or even the affine line) over 𝑘. Yes, read the proof of ...
lieven lebruyn's user avatar
3 votes

A Leibniz-like formula for $(f(x) \frac{d}{dx})^n f(x)$?

This, too, is not an answer, just a preview of a paper I probably won't be writing for a while (one of only 9 papers on my immediate to-do list). It so happens I derived a related formula a few weeks ...
darij grinberg's user avatar
2 votes

A Leibniz-like formula for $(f(x) \frac{d}{dx})^n f(x)$?

It also related to the so-called elementary differentials which appear in the algebraic setting for Runge-Kutta methods. See for example the related chapter in the book Hairer, Wanner and Lubich.
AntoineL's user avatar
  • 131
2 votes

A Leibniz-like formula for $(f(x) \frac{d}{dx})^n f(x)$?

While not pretending to answer the OP, the following is too long to fit in a comment while it might contain elements of interest to the poster. If $f$ is a convergent object (smooth or analytic), then ...
Loïc Teyssier's user avatar
1 vote

Koszul duality between Weyl and Clifford algebras?

If Koszul dual is used for computing resolutions (for $k$ as $A$-module or for $A$ as $A$-bimodule), you can use this idea to find a small resolution for the Weyl algebra. In this way we computed ...
Marco Farinati's user avatar
1 vote

A non-commutative analog of: two polynomials are algebraically dependent iff their Jacobian is zero

The first Weyl algebra over $C$ is isomorphic to the algebra of polynomials $C[x_1,x_2]$ equipped with a new multiplication, as follows: define a linear operator $L$ on $C[x_1,x_2,y_1,y_2]$ to be the ...
Gavin Wraith's user avatar
1 vote
Accepted

What are the fixed points of $\alpha^n-\mu_j$ for a fixed $j$?

Let $L$ be the line generated by $\underline{\mu} = (\mu_1,\dots,\mu_s)$ in $V = \mathbb{C}^s$, and consider a projection $p : V \rightarrow V$ onto $L$, with kernel $H$. Any polynomial function of ...
js21's user avatar
  • 7,199
1 vote

Gelfand-Kirillov dimension of generalized Weyl algebras

You can find your answer in Zhao, Mo, and Zhang - Gelfand–Kirillov dimension of generalized Weyl algebras, Lemma 3.2.
Farrokh's user avatar
  • 111

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