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Weyl algebra
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=== Derivation === {{See|Derivation (differential algebra)}}{{Math theorem| | math_statement = The derivations of <math display="inline">A_n</math> are in bijection with the elements of <math display="inline">A_n</math> up to an additive scalar.{{sfn | Dirac | 1926 | pp=415β417}}}} That is, any derivation <math display="inline">D</math> is equal to <math display="inline">[\cdot, f]</math> for some <math display="inline">f \in A_n</math>; any <math display="inline">f\in A_n</math> yields a derivation <math display="inline">[\cdot, f]</math>; if <math display="inline">f, f' \in A_n</math> satisfies <math display="inline">[\cdot, f] = [\cdot, f']</math>, then <math display="inline">f - f' \in F</math>. The proof is similar to computing the potential function for a conservative polynomial vector field on the plane.{{sfn | Coutinho | 1997 | p=597}} {{Collapse top|title=Proof}} Since the commutator is a derivation in both of its entries, <math display="inline">[\cdot, f]</math> is a derivation for any <math display="inline">f\in A_n</math>. Uniqueness up to additive scalar is because the center of <math display="inline">A_n</math> is the ring of scalars. It remains to prove that any derivation is an inner derivation by induction on <math display="inline">n</math>. Base case: Let <math display="inline">D: A_1 \to A_1</math> be a linear map that is a derivation. We construct an element <math display="inline">r</math> such that <math display="inline">[p, r] = D(p), [q,r] = D(q)</math>. Since both <math display="inline">D</math> and <math display="inline">[\cdot, r]</math> are derivations, these two relations generate <math display="inline">[g, r] = D(g)</math> for all <math display="inline">g\in A_1</math>. Since <math display="inline">[p, q^mp^n] = mq^{m-1}p^n</math>, there exists an element <math display="inline">f = \sum_{m,n} c_{m,n} q^m p^n</math> such that <math display="block"> [p, f] = \sum_{m,n} m c_{m,n} q^m p^n = D(p) </math> <math display="block"> \begin{aligned} 0 &\stackrel{[p, q] = 1}{=} D([p, q]) \\ &\stackrel{D \text{ is a derivation}}{=} [p, D(q)] + [D(p), q] \\ &\stackrel{[p,f] = D(p)}{=} [p, D(q)] + [[p,f], q] \\ &\stackrel{\text{Jacobi identity}}{=} [p, D(q) - [q, f]] \end{aligned} </math> Thus, <math display="inline">D(q) = g(p) + [q, f]</math> for some polynomial <math display="inline">g</math>. Now, since <math display="inline">[q, q^m p^n] = -nq^mp^{n-1}</math>, there exists some polynomial <math display="inline">h(p)</math> such that <math display="inline">[q, h(p)] = g(p)</math>. Since <math display="inline">[p, h(p)] = 0</math>, <math display="inline">r = f + h(p)</math> is the desired element. For the induction step, similarly to the above calculation, there exists some element <math display="inline">r \in A_n</math> such that <math display="inline">[q_1, r] = D(q_1), [p_1, r] = D(p_1)</math>. Similar to the above calculation, <math display="block"> [x, D(y) - [y, r]] = 0 </math> for all <math display="inline">x \in \{p_1, q_1\}, y \in \{p_2, \dots, p_n, q_2, \dots, q_n\}</math>. Since <math display="inline">[x, D(y) - [y, r]]</math> is a derivation in both <math display="inline">x</math> and <math display="inline">y</math>, <math display="inline">[x, D(y) - [y, r]] = 0</math> for all <math display="inline">x\in \langle p_1, q_1\rangle</math> and all <math display="inline">y \in \langle p_2, \dots, p_n, q_2, \dots, q_n\rangle</math>. Here, <math display="inline">\langle \rangle</math> means the subalgebra generated by the elements. Thus, <math display="inline">\forall y \in \langle p_2, \dots, p_n, q_2, \dots, q_n\rangle</math>, <math display="block"> D(y) - [y, r] \in \langle p_2, \dots, p_n, q_2, \dots, q_n\rangle </math> Since <math display="inline">D - [\cdot, r]</math> is also a derivation, by induction, there exists <math display="inline">r' \in \langle p_2, \dots, p_n, q_2, \dots, q_n\rangle</math> such that <math display="inline">D(y) - [y, r] = [y, r']</math> for all <math display="inline">y \in \langle p_2, \dots, p_n, q_2, \dots, q_n\rangle</math>. Since <math display="inline">p_1, q_1</math> commutes with <math display="inline">\langle p_2, \dots, p_n, q_2, \dots, q_n\rangle</math>, we have <math display="inline">D(y) = [y, r + r']</math> for all <math>y \in \{p_1, \dots, p_n, q_1, \dots, q_n\}</math>, and so for all of <math>A_n</math>. {{Collapse bottom}}
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