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I recently came across some polynomials with some remarkable properties. A polynomial $P(u,v) \in \mathbb{R}[u,v]$ in 2 variables is remarkable if the set of solutions to the system $P(u,v)=P(v,u)=0$ is a finite number of points in $\mathbb{C}^2$ such that each point is of the form $(x, \overline{x}).$

Here are some examples of such polynomials:

$v^2-u$

$-2 u v+v^3+1$

$u^2-3 u v^2+v^4+2 v$

$3 u^2 v-4 u v^3-2 u+v^5+3 v^2$

$-u^3+6 u^2 v^2-5 u v^4-6 u v+v^6+4 v^3+1$

$-4 u^3 v+10 u^2 v^3+3 u^2-6 u v^5-12 u v^2+v^7+5 v^4+3 v$

$u^4-10 u^3 v^2+15 u^2 v^4+12 u^2 v-7 u v^6-20 u v^3-3 u+v^8+6 v^5+6 v^2$

A quite complicated algorithm is behind this sequence, any help identifying a formula for these polynomials would be helpful. Also, a proof that these polynomials are remarkable would be nice, these are only checked by numeric computations in Mathematica.

Question: Can one classify the set of remarkable polynomials?

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Could you please give some more terms of your sequence? In the given sequence p(n)-vp(n-1)+up(n-2)-p(n-3) is 2u,-2uv,2u^2,-2u. Perhaps there is some regularity in this sequence? –  Johann Cigler Mar 21 '11 at 17:59
    
@Johann: That relation is related to the characters of $SU(3)$, relative to my nomenclature below, $Sym^k(V_2) = v\cdot Sym^{k-1}(V_2)\ominus u\cdot Sym^{k-2}(V_2)\oplus Sym^{k-3}(V_2)$. Did you use the answer about SU(3) as inspiration to look at this relation, or did you see this relation some other way? –  ARupinski Mar 21 '11 at 18:38
    
I am not familiar with group representations. I simply looked for some regularities. The obvious one is for u=0 which gives p(n)=vp(n-1)+p(n-3) if the rest of the sequence behaves in the same way. –  Johann Cigler Mar 21 '11 at 20:24
    
My first remark is not correct. I have overlooked a term in p(5). See my answer. –  Johann Cigler Mar 22 '11 at 7:55
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2 Answers

up vote 8 down vote accepted

Let $V_1$ be the defining 3-dimensional representation of SU(3) with character $\chi_1$. Likewise, let $V_2$ be the conjugate representation with character $\chi_2 = \overline{\chi_1}$. Then every irreducible representation of SU(3) has a character which is a $\mathbb{Z}-$polynomial in $\chi_1$ and $\chi_2$.

What does this have to do with your problem? Well, letting $u=\chi_1$ and $v=\chi_2$, your polynomials are simple linear combinations of the polynomials of the irreducible characters of SU(3). For example the polynomial $v^2-u$ corresponds to the character of the irreducible representation which is the symmetric square of $V_2$, which is the (0,2)-representation. Similarly $v^3-2uv+1$ corresponds to the character of the (0,3)-representation which is the third symmetric power of $V_2$. (Here the $(a,b)$-representation is the representation with highest weight $a\omega_1 + b\omega_2$ with $\omega_1$ and $\omega_2$ the highest weights of $V_1$ and $V_2$ respectively).

Now since the character $\chi_{[a,b]}$ of the $(a,b)$-representation is the complex conjugate of the character of the (b,a)-representation, the polynomial $P_{a,b}(\chi_1,\chi_2)$ expressing the character of the $(a,b)$-representation satisfies

$P_{a,b}(\chi_1,\chi_2) = \overline{P_{a,b}(\chi_2,\chi_1)} = P_{b,a}(\chi_2,\chi_1)$

This implies the polynomials $P_{a,b}$ satisfy your condition (if you want a short argument of this fact, give me a little while to think of something coherent).

But the short of it is, your polynomials correspond to characters of SU(3).

Edit: Inspired by Johann's and comment about the recurrence, I looked more closely at the polynomials you listed and realized all of them $are$ in fact polynomials corresponding to the (0,k)-representations of SU(3) (for k=2..8) which is also the irreducible representation $Sym^k(V_2)$. In particular I am pretty sure one can extend this to say that $any$ polynomial corresponding to an irreducible representation of SU(3) will satisfy the property of being 'remarkability'. I think therefore that the next logical step is to look for polynomials which do not correspond to irreducible representations (for example, take simple linear combinations of the above polynomials) and see if they too have the 'remarkable' property.

Edit 2: Just realized that the above statement cannot be entirely true as I have already pointed out in a comment to Johann's answer that $q = uv-1$ is the polynomial associated to the 8-dimensional irreducible adjoint representation; clearly $q(u,v) = q(v,u) = 0$ has infinitely many solutions $(u,v)$, all of the form $(x,\overline{x})$. This is the (1,1)-representation of SU(3), and similarly for every $k$ one has $P_{k,k}(u,v)$ is symmetric with respect to $u$ and $v$; aside from these polynomials I don't see any obvious obstruction to a general $P_{a,b}(u,v)$ satisfying your conditions as long as $a\neq b$.

Edit 3: Last edit I promise. One thing I overlooked in the previous edit; namely that $P_{k,k}$ divides $P_{a,b}$ whenever $a\equiv b\equiv -1\mod k+1$; in particular, aside from the $P_{k,k}$, there are a lot of other $P_{a,b}$ which do not have the property of remarkability. In such cases the structure of the cofactor $P_{a,b}/P_{k,k}$ is actually pretty well-behaved, and based on what I know about this structure it ought to be true that if $k+1 = gcd(a+1,b+1)$ then this cofactor has the property of remarkability. Some simple examples of such cofactors are $u^2-2v$ and $u^3-3uv+3$, so you might test these examples too (unfortunately I don't currently have access to Maple or Mathematica to do it myself).

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Incidentally, once you know the first few of the $P_{a,b}$ you can use Klimyk's formula to recursively derive many more, for example one has: $P_{a+1,b} = P_{1,0}\cdot P_{a,b} - P_{a-1,b+1} - P_{a,b-1}$ –  ARupinski Mar 21 '11 at 17:15
    
This is awesome! Can one see SU(3) in the original problem itself somehow? Also, is every polynomial satisfying the required properties a character of SU(3)? –  Daniel Litt Mar 21 '11 at 18:12
    
Offhand, without knowing the algorithm or motivation behind the polynomials, its hard to say whether one can easily see SU(3). As for whether every polynomial satisfying the properties is a character of SU(3), a partial answer is that the irreducible characters of SU(3) form a $\mathbb{Z}$-basis of $\mathbb{Z}[u,v]$, so on the one hand every 2-variable polynomial does correspond to the character of some SU(3) representation. –  ARupinski Mar 21 '11 at 18:28
    
On the other hand, the characters of SU(3) carry a natural $\mathbb{Z}/3\mathbb{Z}$-grading: give the variable $u$ a grade of 1, $v$ a grade of 2, and constants a grade of 0. Each of the polynomials listed is homogenous with respect to this grading (i.e. all terms have the same grade mod 3). Perhaps this grading might be what is needed to make a polynomial "remarkable" in the sense of the question. –  ARupinski Mar 21 '11 at 18:51
    
Thanks for the answer! There is a polynomial in $v$, such that all roots of $v$ admits a solution to the bivariate system (one may find these by using groebner bases). The sequence one gets is a sequence of polynomials of degree n(n+1)/2. A recurrence for this sequence would be very nice, however, I doubt there is a reasonable nice one... –  Per Alexandersson Mar 25 '11 at 9:49
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Your sequence $p_n (u,v)$ can be defined by $p_n (u,v) = vp_{n - 1} (u,v) - up_{n - 2} (u,v) + p_{n - 3} (u,v)$ with initial values $p_0 (u,v) = 1,p_1 (u,v) = v,p_2 (u,v) = v^2 - u.$ In my above remark I have overlooked a term in the fifth polynomial. This is now the same as the formula given by ARupinski.

Added later: Extend the sequence $p_n (u,v)$ to negative indices by $p_{ - 1} (u,v) = p_{ - 2} (u,v) = 0$ and $p_{ - n} (u,v) = p_{ n - 3} (v,u)$ for $n >2 .$ Define a new sequence of polynomials $r_n (u,v)$ by the same recurrence and initial values $r_{ - 1} (u,v) = 1,r_0 (u,v) = 0,r_1 (u,v) = - u.$ Extend it to negative values by $r_{ - n} (u,v) = r_{n - 2} (v,u).$

Let $A$ be the matrix with rows $(0,1,0),(0,0,1),(1, - u,v).$

Then $A^n$ is the matrix with the following rows: $\left( {p_{n - 3 + j} (u,v),r_{n - 2 + j} (u,v),p_{n - 2 + j} (u,v} \right)$ for $0 \le j \le 2.$

It seems that the sequence $r_n (u,v)$ or the sequence $r_{2n} (u,v)/(1 - uv)$ has analogous properties with respect to the zeroes. Is it also related to the group representation?

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I thought about it a bit more, and your $r_n$'s are also related to the $P_{a,b}$'s I defined by $r_k(u,v) = -P_{1,k-1}(u,v)$. In light of my third edit, this implies the $r_{2k}$ are all divisible by $(1-uv)$ as you already noted; but since $1-uv$ does not satisfy remarkability, the $r_{2k}$ cannot satisfy it either; however in light of my third edit, I am pretty sure you are right that the cofactors $r_{2k}(u,v)/(1-uv)$ do satisfy remarkability. –  ARupinski Mar 23 '11 at 21:50
    
@ARupinski. As I already wrote I am not familiar with group representations. So I did not quite understand your argument that the above sequence is indeed "remarkable". Is there perhaps also a direct way using only the recurrence and initial values to prove this? The matrix $A$ is a useful construction associated with such a recurrence. But I am astonished that all entries of $A^n$ are related to representations of $SU(3)$. What about the matrix $A$ itself or the sequence $Tr(A^n)$? Are they also in some form related to $SU(3)$ or are the above results only accidental coincidencies? –  Johann Cigler Mar 24 '11 at 15:27
    
@Johann: Based on your results on $A^n$, its almost certain that $A$ is somehow closely related to the representation theory of $SU(3)$ but offhand I'm not sure how as I have never really looked at such matrices (although based on your results it seems that I should sometime). As far as using the recurrence and initial values to prove remarkability, there is probably some way (independent of the relationship to $SU(3)$), but offhand I don't know how. Definitely something I will keep thinking about when I have free time. –  ARupinski Mar 25 '11 at 1:47
    
So I thought about it and $A$ is indeed closely related to the representations of $SU(3)$ for (almost) trivial reasons. Despite this, I think I might be able to use the general idea of such matrices to solve a few open questions I've been pondering for awhile, so thanks for the inspiration. –  ARupinski Mar 27 '11 at 19:06
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