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on page 177 in book Orthogonal Polynomials of Several Variables 2nd.ed. is written , [where from prior definition(s) R is a vector in d dimensional space and member of Coxeter group W generated by the root system R is subgroup of O(d) meaning real orthogonal group, generated by {$\sigma_u:u\in R$} and $\sigma_u$ is reflection in the hyperplane perpendicular to vector u.]

For any reduced root system R,discriminant, or alternating polynomial is

$a_R(x)=\prod_{u\in R_{+}}\!<\!x,u\!>$

So i say that is error. First of all discriminant has to do with the b^2-4ac term of which the square root is taken and then + or - it in solution of 2 roots of quadratic which is totally unrelated to this. Ok let's assume there are 2 definitions of discriminant then still since <x,u> is scalar product then the rhs is not in the form of polynomial or if it were then it is of single variable which would make no sense wrt what is written in book afterwards such as taking the Laplacian of rhs and other operations. The x in lhs , $a_R(x)$ , is i assume from what is written prior in the book or more to the point what is not written about it , an arbitrary d=dimension polynomial(s) of any degree. So now i am trying to figure what he should have written instead to define $a_R(x)$ .

Next in book is written. For $v\in R$ and $w\in W(R)$ [where from prior the W(R) means the subgroup of O(d),orthogonal group, generated by the reflections in $R_+$]

$a_R(x\sigma_v)=-a_R(x)$ and $a_R(xw)=det\,w\;a_R(x)$

where I would assume means the determinant of only w by itself , det(w) , and not det of the entire expression $w a_R(x)$

So now I am trying to figure what he should written or means on the rhs of $a_R(x)$. Could he mean to have written the product of to apply the reflection op's of every member of R to every $x_i$ [0<i<d+1] in the polynomial. Or product of replace every $x_i$ in the polynomial by <x,u>u/<u,u>. If so would either of these agree with later statements in the book as follows.

Assume u is non 0 vector in $R^d$ and polynomial $p(x)$ is polynomial such that $p((x\sigma_u)=-p(x)$ for all $x\in R^d$. Then p(x) is divisible by <x,u>. I don't see right off how that is possible but anyway assume that is true. Then next is the Laplacian , also call it $\Delta$, commutes with $R_{\sigma_u}$. Ok that seems irrelevant to the issue but is used in later statements that do. Can anyone prove that ? Seems if so then it should be true for any reflection at all ? Another in the book is written $\Delta a_R=0$ for any reduced root system. How could that be ? And another for the specific $A_{d-1}$ system is written.

$a_R(x)=\prod_{u\in R_{+}}\!<\!x,u\!>=\prod_{i<j}(x_i-x_j)$

Can anyone surmise or guestimate what is meant by this absurd definition incorrect definition $a_R(x)$ ? Should there be restrictions on x that the authors did not mention ? Has anyone ever seen anything remotely similar the development here ?

user158293
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1 Answers1

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What you see in the book is a fairly standard use of the term discriminant, applying to more situations than just solving quadratic equations. There is also a discriminant of any polynomial, being the product of differences of the roots, similar to your $\prod_{i\not=j}(x_i-x_j)$. Notice that this expression is invariant under permutations of the indices... Permutation groups are the simplest non-trivial Coxeter groups.

It also has the geometric sense used in your book, which, algebraically, is this kind of product of differences.

paul garrett
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  • Yea i have seen that in relation to some type or what called Vandermonde or something like that determinant. Have seen that and other expressions but they do not come from products of scalar products alone but using those scalar product in other relations.- to form other expressions. – user158293 Dec 31 '24 at 00:43
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    @ paul garrett So you would agree that the expression written by the author for $a_R(x)$ is wrong ? But you haven't told me exactly what he meant , such as exactly what operation on the x in lhs $a_R(x)$ is supposed to be for arbitrary x to get the correct rhs whatever that is because what he wrote $aR(x)=\prod_{u\in R^+} <x,u>$ is wrong. – user158293 Dec 31 '24 at 01:54
  • And besides that he writes different expressions not of the form u wrote but for type Bd system he writes $a R(x)=\prod{i=1}^d x_i \prod_{i<j} (x_i^2-x_j^2)$ and another place for type G2 as $a_R(x)= 2^{-m} i^{-1} (z^m-{\bar z}^m)$ – user158293 Dec 31 '24 at 05:41
  • In matrix form $\sigma_u=I_d- 2u^Tu/<!u,u!>$ .Both u and x are single row and x acts on the left to give $x \sigma_u =x-2<!x,u!>u/||u||^2$ where $||u||^2=<!u,u!>=uu^T$ . The <..,.> scalar product in not soley used by itself but only with other matter for example in above and i would expect same in his $a_R(x)$ if it is to have the meanings and satisfy the relations he writes. So what is the correct definition or correct way to write what he means – user158293 Dec 31 '24 at 06:11
  • You were talking about permutation groups. But it seems the simplest of all would be just a single reflection since it's square is the identity would be 2 elements in the group is not allowed as a root system as neither is the cyclic group in the plane I don't think so the simplest of that would be the 180 degree rotation again only 2 elements . Both of these could be just permutation of 2 elements but neither is allowed to be a root system i don't think so $a_R(x)$ could not be of this simple form which could be thought of as a permutation on 2 objects. – user158293 Dec 31 '24 at 12:29
  • was mistaken in that last comment because there is root system that does correspond to a reflection and identity isomorphic to about the simplest group of permutation group of order 2. The references usually refer to it as A1 with the root in the plane generator as e2-e1 or e1-e2 where the e2 and e1 are unit vectors say along y and x axis. But i would think it could also be e2+e1 or more simply just e1 or e2 or any of the infinity of directions in the first quadrant for example. It could be that it is defined that way to be consistent with the notation A_{d-1} system for any dimension d. – user158293 Jan 04 '25 at 15:25