Настроение:	bored
Музыка:	Brahms, Piano quartet no. 1 in g, op. 25

Делать мне нечего
Тут некоторые спрашивали, и, по слухам, Леня Макар-Лиманов уже ответил, но все равно пусть будет.



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\title{The question of xaxam}
\author{Sasha Anan$'$in}
\date{24/12/2013}

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Consider the Weyl algebra $\Bbb C[x,d]$, i.e., the associative $\Bbb C$-algebra generated by $x$ and $d$ with the defining relation $[d,x]:=dx-xd=1$. It is well known that it lives inside some skew field.

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{\bf Problem.} Describe $C:=\big\{L\in\Bbb C[x,d]\mid d^{-n}Ld^n\in\Bbb C[x,d]\big\}$.

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Let $i\ge0$, denote $y_i=\frac{(-1)^i}{i!}x^i$. Since $[d,x]=1$, we have $[d,y_i]=-y_{i-1}$ for $i>0$. From $d\sum_{i=0}^ky_id^i=\sum_{i=0}^ky_id^{i+1}+\sum_{i=0}^k[d,y_i]d^i=\sum_{i=1}^ky_id^{i+1}-\sum_{i=1}^ky_{i-1}d^i=y_kd^{k+1}$, we conclude $d^{-1}y_kd=\sum_{i=0}^ky_{k-i}d^{-i}$ for $k\ge0$. By induction on $n$, we obtain
$$d^{-n}y_kd^n=\sum_{i=0}^k{i+n-1\choose n-1}y_{k-i}d^{-i}.$$
Indeed,
$$d^{-1}\Bigg(\sum_{i=0}^k{i+n-1\choose n-1}y_{k-i}d^{-i}\Bigg)d=\sum_{i=0}^k{i+n-1\choose n-1}\bigg(\sum_{j=0}^{k-i}y_{k-i-j}d^{-j}\bigg)d^{-i}=$$
$$=\sum_{l=0}^k\Bigg(\sum_{i=0}^l{i+n-1\choose n-1}\Bigg)y_{k-l}d^{-l}=\sum_{l=0}^k{l+n\choose n}y_{k-l}d^{-l}.$$
As $y_id^j\in C$ for $i\le j$, the conjugation by $d^n$ preserves the degree $\deg(y_id^j):=i-j$, and $Cd\subset C$, it suffices to find a basis of $C_k$ modulo $C_{k+1}d$ for all $k>0$, where $C_k$ stands for the homogeneous component of $C$ of degree $k$.

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{\bf Lemma.} {\sl For every\/ $k>0$, there are\/ $f_k=y_{2k}d^k+\sum_{i=0}^{k-1}a_{k,i}y_{k+i}d^i$ and\/ $g_k=y_{2k+1}d^k+\sum_{i=0}^{k-1}b_{k,i}y_{k+i+1}d^i$ with\/ $a_{k,i},b_{k,i}\in\Bbb C$ such that\/ $f_k$ and\/ $g_kd$ form a basis of\/ $C_k$ modulo\/ $C_{k+1}d$.}

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In fact, the conditions $f_k,g_kd\in C_k$ determine uniquely the coefficients $a_{k,i},b_{k,i}\in\Bbb C$. Indeed,
$$d^{-n}y_{2k}d^kd^n=\sum_{j=0}^{2k}{j+n-1\choose n-1}y_{2k-j}d^{k-j}\equiv\sum_{j=1}^k{j+k+n-1\choose n-1}y_{k-j}d^{-j}\mod\Bbb C[x,d],$$
$$d^{-n}y_{k+i}d^id^n=\sum_{j=0}^{k+i}{j+n-1\choose n-1}y_{k+i-j}d^{i-j}\equiv\sum_{j=1}^k{j+i+n-1\choose n-1}y_{k-j}d^{-j}\mod\Bbb C[x,d],$$
$$d^{-n}y_{2k+1}d^{k+1}d^n=\sum_{j=0}^{2k+1}{j+n-1\choose n-1}y_{2k+1-j}d^{k+1-j}\equiv\sum_{j=1}^k{j+k+n\choose n-1}y_{k-j}d^{-j}\mod\Bbb C[x,d],$$
$$d^{-n}y_{k+i+1}d^{i+1}d^n=\sum_{j=0}^{k+i+1}{j+n-1\choose n-1}y_{k+i+1-j}d^{i+1-j}\equiv\sum_{j=1}^k{j+i+n\choose n-1}y_{k-j}d^{-j}\mod\Bbb C[x,d].$$
So, the claim follows from the fact that the matrix with the components ${i+j+n-2\choose n-1}$ and the matrix with the components ${i+j+n-1\choose n-1}$, where $1\le i,j\le k$, are both nondegenerate.

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{\bf Proof.} Let $f=\sum_{i=0}^lc_iy_{k+i}d^i\in C_k$. Suppose that $l\ge k$. If $k+l=2s$, then $s\ge k$, and we can diminish $l$ by taking $f-c_lf_sd^{s-k}$ in place of $f$. If $k+l=2s+1$, then $s\ge k$, and we can diminish $l$ by taking $f-c_lg_sdd^{s-k}$ in place of $f$. Hence, we can assume that $l\le k-1$. Considering $d^{-n}fd^n\mod\Bbb C[x,d]$ and using the above linear independence, we obtain $c_i=0$ for all $i$.

Let $fd\in C_{k+1}d$ be a nontrivial linear combination of $f_k$ and $g_kd$. Then $f=\sum_{i=0}^kc_iy_{k+1+i}d^i\in C_{k+1}$. Again, considering $d^{-n}fd^n\mod\Bbb C[x,d]$ and the above linear independence for $k+1$ in place of $k$, we arrive at a contradiction
$_\blacksquare$

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