Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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: mathit.unco.edu/vi2011/pdfs/1975-1979/1977-MOAGITTONE0.pdf –  joriki Sep 8 '11 at 11:38
    
Perhaps Thm 7.4.5 of scribd.com/doc/58838869/66/Two-Commuting-Linear-Operators can apply here. –  John M Sep 8 '11 at 12:55
add comment

1 Answer

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

$$(Dy)^{(n-1)}(0)=y^{(n)}(0)=\frac1{a_n}\left(Py-\sum_{i=0}^{n-1}a_iy^{(i)}\right)(0)=\frac1{a_n}\left(Ey(0)-\sum_{i=0}^{n-1}a_i(0)y^{(i)}(0)\right)\;.$$

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 $Q$.

share|improve this answer
    
Thank you! $ \ $ –  Johan Sep 9 '11 at 9:14
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.