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I am reading an article at the moment and there is one step that I am having trouble understanding.

The article proves that if $P$ and $Q$ are commuting differential operators there is a non-trivial polynomial $f(s,t)$ such that $f(P,Q)=0$.

In order to to this it considers the eigenvalue problem $Py=Ey$ for any number $E$. This problem has $n$ ($n$ being the order of $P$) linearly independent solutions by basic existence and uniqueness theorem. These span a space, $V_E$. The article takes as a basis of this space the solutions $y_i$ with $\frac{d^j y_i}{dx^j}(0)=\delta_{ij}$.

If $Q$ commutes with $P$ then $Q$ maps $V_E$ into itself. Now the article claims, and this is the step I do not understand, that the matrix elements will be polynomials in $E$.

Anyone who can explain this will get my gratitude.

Edit: The article I am reading is "Methods of algebraic geometry in the theory of non-linear equations" by Krichever in Russian Math Surveys 32, 1977 . The operators the article studies are ordinary differential operators acting on $C^\infty(\mathbb{R},\mathbb{C})$. Operators are assumed to have constant leading coefficient. (So $P=\sum_{i=0}^n a_i x^i$ where $a_n$ is a non-zero constant.)

The theorem is Theorem 2.1 on page 9 of the pdf. It is specifically the second sentence of the proof that I have problems with.

I know that there actually is such a polynomial as claimed in the article as I know algebraic proofs of this fact. I am however trying to understand the analytic proof given by Krichever.

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Am I missing something, but [P,Q] is already a sort of a polynomial. [P,Q] = PQ - QP = 0. – valdo Sep 8 '11 at 10:10
@valdo: They mean a polynomial in which $P$ and $Q$ are already considered as commuting, so that would be the trivial polynomial. – joriki Sep 8 '11 at 10:24
The article is "Methods of algebraic geometry in the theory of non-linear equations" by Krichever in Russian Math Surveys 32, 1977. – Johan Sep 8 '11 at 11:17
Here's a freely accessible version of the article: – joriki Sep 8 '11 at 11:38
Perhaps Thm 7.4.5 of can apply here. – John M Sep 8 '11 at 12:55
up vote 3 down vote accepted

Consider the linear operator $D$ on $V_E$ that maps every solution $y$ to the solution $Dy$ having the same $0$-th through $(n-1)$-th derivatives at $0$ (or $x_0$ in the article) as $y'$. Let's determine the matrix of this operator in the given basis. Since $(Dy)^{(k)}(0)=y^{(k+1)}(0)$, the matrix elements in the first $n-1$ rows are simply $D_{ij}=\delta_{i+1,j}$. That leaves only the last row, corresponding to $(Dy)^{(n-1)}$, to be determined. Since $Py=Ey$ on $V_E$, we have


So the matrix elements are all constants independent of $E$, except for $D_{n-1,0}=(E-a_0(0))/a_n$. But the action of $Q=\sum_{i=0}^{m}b_i\frac{\mathrm d^i}{\mathrm dx^i}$ on $V_E$ coincides with that of $\sum_{i=0}^{m}b_i(0)D^i$. Since this is a polynomial in $D$ and the matrix elements of $D$ are polynomials in $E$, so are the matrix elements of $Q$.

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Thank you! $ \ $ – Johan Sep 9 '11 at 9:14

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