I would rather transform the polynomials to symmetric functions and
then do the expansion there. Something like
sage: Sym = SymmetricFunctions(QQ['t'])
sage: P = Sym.hall_littlewood().P()
sage: R = PolynomialRing(QQ['t'], 'x', 4)
sage: x = R.gens()
sage: Sym.from_polynomial(x[0]+x[1]+x[2]+x[3])
m[1]
sage: P(Sym.from_polynomial(x[0]+x[1]+x[2]+x[3]))
HLP[1]
Best,
Anne
On 12/4/13 11:04 PM, Travis Scrimshaw wrote:
> Hey Dan,
> This is only a part of the way, but I'd start by doing the computations by
> having t in a polynomial ring R (over QQ say), and the basic variables x_i
> coming from a power series ring over R:
>
> sage: R.<t> = QQ[]
> sage: S.<x1,x2,x3> = PowerSeriesRing(R)
> sage: (1 - t*x1*x3) / (1 - x1*x3)
> 1 + (-t + 1)*x1*x3 + (-t + 1)*x1^2*x3^2 + (-t + 1)*x1^3*x3^3 + (-t +
> 1)*x1^4*x3^4 + (-t + 1)*x1^5*x3^5 + O(x1, x2, x3)^12
>
> Of course there's the reverse too:
>
> sage: R.<x1,x2,x3> = PowerSeriesRing(QQ)
> sage: S.<t> = R[]
> sage: f = (1 - t*x1*x3) / (1 - x1*x3); f
> (-x1*x3 - x1^2*x3^2 - x1^3*x3^3 - x1^4*x3^4 - x1^5*x3^5 - x1^6*x3^6 + O(x1,
> x2, x3)^14)*t + 1 + x1*x3 + x1^2*x3^2 + x1^3*x3^3 + x1^4*x3^4 + x1^5*x3^5 +
> O(x1, x2, x3)^12
>
> And if you don't want any grouping:
>
> sage: R.<x1,x2,x3,t> = PowerSeriesRing(QQ)
> sage: (1 - t*x1*x3) / (1 - x1*x3)
> 1 + x1*x3 - x1*x3*t + x1^2*x3^2 - x1^2*x3^2*t + x1^3*x3^3 - x1^3*x3^3*t +
> x1^4*x3^4 - x1^4*x3^4*t + x1^5*x3^5 - x1^5*x3^5*t + O(x1, x2, x3, t)^12
>
> Now to get the HL P polynomials into the correct variables:
>
> sage: Sym = SymmetricFunctions(R)
> sage: HLP = Sym.hall_littlewood(t).P()
> sage: p21 = HLP[2,1].expand(3)
> sage: p21 = p21.substitute(x0=x1, x1=x2, x2=x3)
> sage: p21
> x1^2*x2 + x1^2*x3 + x1*x2^2 + (-t^2 - t + 2)*x1*x2*x3 + x1*x3^2 + x2^2*x3 +
> x2*x3^2
> sage: p21.parent()
> Multivariate Power Series Ring in x1, x2, x3 over Univariate Polynomial Ring
> in t over Rational Field
>
> Then from this point you could do some linear algebra using a "nice" ordering
> of the basis:
>
> sage: R.<t> = QQ[]
> sage: S.<x1,x2,x3> = PowerSeriesRing(R) # If you aren't getting enough
> terms, add "default_prec=20" to increase the degrees computed
> sage: f = (1 - t*x1*x2) / (1 - x1*x2) * (1 - t*x1*x3) / (1 - x1*x3) * (1 -
> t*x2*x3) / (1 - x2*x3)
> sage: Sym = SymmetricFunctions(R)
> sage: HLP = Sym.hall_littlewood(t).P()
> sage: p11 = HLP[1,1].expand(3)
> sage: p11 = p11.substitute(x0=x1, x1=x2, x2=x3)
> sage: c = f.coefficients()
> sage: f[2] - p11 * c[x1*x2] # I only took the degree 2 terms for easier
> viewing/proof of concept
> 0
> sage: f - p11 * c[x1*x2]
> 1 + (-t + 1)*x1^2*x2^2 + (t^2 - 2*t + 1)*x1^2*x2*x3 + (-t + 1)*x1^2*x3^2 +
> (t^2 - 2*t + 1)*x1*x2^2*x3 + (t^2 - 2*t + 1)*x1*x2*x3^2 + (-t + 1)*x2^2*x3^2
> + (-t + 1)*x1^3*x2^3 + (t^2 - 2*t +
> 1)*x1^3*x2^2*x3 + ... + O(x1, x2, x3)^12
>
> So from this point it would be wrapping this up in a function that iterates
> over partitions in some "nice" way (Sage goes by lex largest to smallest) and
> returns a dictionary of index/value as
> partition/coefficient, for example [1, 1]: -t + 1 (which comes from c[x1*x2]).
>
> Here's the more automated, but less explicit, version:
>
> sage: n = 3 # This could be the input to a function
> sage: R.<t> = QQ[]
> sage: S = PowerSeriesRing(R, n, 'x', default_prec=n+1)
> sage: Sym = SymmetricFunctions(R)
> sage: HLP = Sym.hall_littlewood(t).P()
> sage: d = S.gens_dict()
> sage: make_poly = lambda la: HLP[la].expand(n).substitute(**d)
> sage: lead_term = lambda la: prod(S.gen(i)**v for i,v in enumerate(la))
>
> sage: f = prod( (1 - t*S.gen(i)*S.gen(j)) / (1 - S.gen(i)*S.gen(j)) for i in
> range(n) for j in range(i+1,n) )
> sage: f
> 1 + (-t + 1)*x0*x1 + (-t + 1)*x0*x2 + (-t + 1)*x1*x2 + O(x0, x1, x2)^4
>
> sage: ret = {}
> sage: for k in range(n+1): # For the default precision
> ....: for la in Partitions(k):
> ....: c = f.coefficients()
> ....: coeff = c.get(lead_term(la), 0)
> ....: if coeff != 0:
> ....: ret[la] = coeff
> ....: f -= coeff * make_poly(la)
> ....:
> sage: ret # This would be the output of that function
> {[]: 1, [1, 1]: -t + 1}
> sage: f
> 0 + O(x0, x1, x2)^4
>
> Same code but with n = 5:
>
> sage: ret
> {[]: 1, [1, 1]: -t + 1, [1, 1, 1, 1]: t^4 - t^3 - t + 1, [2, 2]: -t + 1}
> sage: f
> 0 + O(x0, x1, x2, x3, x4)^6
>
> Same thing but with n = 8 and the default_prec = 6:
>
> sage: ret
> {[]: 1, [1, 1]: -t + 1, [1, 1, 1, 1]: t^4 - t^3 - t + 1, [2, 2]: -t + 1}
> sage: f
> 0 + O(x0, x1, x2, x3, x4, x5, x6, x7)^6
>
> Is this what you were looking for? It's not the fastest (and IDK how it would
> compare to your implementation), but it should work.
>
> Best,
> Travis
>
>
> On Wednesday, December 4, 2013 2:59:08 PM UTC-8, Dan Betea wrote:
>
> Hi there,
>
> Is there a quick way (in Sage) to expand a product like \prod_{i<j}
> \frac{1-t x_i x_j}{1- x_i x_j} in Hall Littlewood polynomials in the
> appropriate number of variables? More precisely, given a
> partition of length <= to the number of variables, I want to compute the
> coefficient of the corresponding P or Q HL polynomial.
>
> Note this example is known precisely (one of the Littlewood identities
> for HL polynomials).
>
> The product I'm interested in is more difficult and contains an extra set
> of variables y (together with their inverses). It contains univariate factors
> like \prod_{i} (1-t_0 x_i)(1-t_1 x_i)/(1-t
> x_i^2) along with "i<j" factors of the above form. Through some magic
> using Littlewood Richardson coefficients, Kostka polynomials and a deep look
> into Macdonald, one can obtain the coefficient of
> P_{\lambda}. However, the process takes a lot of setup and is *slow* (so
> not very prone to the experimentation we want to try). I imagine Sage would
> have this functionality when dealing with
> symmetric functions.
>
> Thanks,
>
> db
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