I have an operator acting on the polynomial algebra $\mathbb{C}[x,y,z]$ that I would like to find the eigenvalues/eigenvectors of. More specifically, let $P(x_1, \ldots, x_6)$ be a homogeneous polynomial, my operator has the form $P(x,y,z, \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z})$. Are there any general strategies that could help me? For instance, say my operator were: $$z^2y\frac{\partial^3}{\partial x^2 \partial y} + y^3z\frac{\partial^4}{\partial y^2 \partial z^2} + xz^2\frac{\partial^3}{\partial x \partial z^2}.$$ My actual operator is degree 6 and more complicated, but other than that the same type of object. Any thoughts or references on how to attack this type of problem will be very welcome. Thanks a lot!

EDIT: My first example-operator was very poorly chosen, since every monomial would automatically be an eigenvector. I have now altered it a little to avoid this. Keep in mind this was only an example to show what type of object I am considering, and I am looking for general strategies. None the less, thanks for the quick response.

EDIT 2: I had misread the degree of my actual operator - it is 6.

  • $\begingroup$ In your example, $P$ is such that the differential operator is homogeneous (if you apply it to an homogeneous function, the result is also homogeneous of the same degree). Is this true in the general case you are interested in? $\endgroup$ – Mariano Suárez-Álvarez Jan 29 '10 at 17:44
  • $\begingroup$ My operator preserves the total degree of every monomial, but not the x-degree, y-degree and z-degree respectively. $\endgroup$ – user3628 Jan 29 '10 at 17:51
  • $\begingroup$ removed the "operator algebras" tag as that moniker has (to my slight chagrin) become dedicated to a domain which has little direct bearing on the present question. $\endgroup$ – Yemon Choi Jan 29 '10 at 21:56

If the polynomial $P(x_1, \ldots, x_6)$ is homogeneous of degree zero (where $x_1, x_2, x_3$ are considered of degree 1 and $x_4, x_5, x_6$ of degree -1), the computation of the eigenvalues/eigenvectors reduces to linear algebra. Indeed, since the operator $P(x,y,z, \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z})$ preserves total degree, it is enough to consider each graded piece $\mathbb{C}[x,y,z]_d$ separately, and the latter are finite-dimensional vector spaces on which the operator acts linearly. Your explicit example is of this kind.

If $P$ is of homogeneous of negative degree, then it can only have zero as an eigenvalue, since the associated operator is nilpotent. For $P$ of positive degree it also seems clear that the only possible is zero.

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    $\begingroup$ Notice that in a comment AK said the polynomial is of zero degree in your grading. $\endgroup$ – Mariano Suárez-Álvarez Jan 29 '10 at 18:32
  • $\begingroup$ In the example, yes. But then he says he's actually interested in a degree 5 operator that is more complicated. I assume he means that it raises total degree by 5. $\endgroup$ – Alberto García-Raboso Jan 29 '10 at 18:37
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    $\begingroup$ Yeah, but there are infinitely many degrees. Can we, in some sense, get all the degrees at once by a finite computation? $\endgroup$ – David E Speyer Jan 29 '10 at 18:38
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    $\begingroup$ I'm sorry, AK, but I don't quite understand: how do you get an operator to preserve the total degree from a polynomial of degree 5? There's no way to get zero as the sum of five $\pm 1$. $\endgroup$ – Alberto García-Raboso Jan 29 '10 at 20:58
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    $\begingroup$ @darji: there are two different degrees here. First, the degree of P as a homogeneous polynomial in which you consider the variables x_1,...,x_6 to have degree one. Then there is the degree of P(x,y,z,\partial_x,\partial_y,\partial_z) as an operator on the graded algebra C[x,y,z]. E.g., the homogeneous polynomial of degree two x_1 x_4 yields the operator x \partial_x, which takes C[x,y,z]_d to C[x,y,z]_d and is hence of degree zero as an operator. $\endgroup$ – Alberto García-Raboso Jan 29 '10 at 23:56

NB: This answers an old version of the question...

The differential operator in your example is very, very special: each monomial in $x$, $y$ and $z$ is an eigenvector, so it is in fact diagonalizes is the basis of monomials of $k[x,y,z]$!


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