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Why is sage not recognizing symmetric polynomials as symmetric?

asked 2020-04-18 22:36:04 -0600

masseygirl gravatar image

I am trying to decompose symmetric polynomials into polynomial combinations of elementary symmetric polynomials. It has been driving me absolutely up the wall, and I would be very grateful for any help.

It works perfectly with 2 variables.

S.<x0,x1>=QQ[]
f = (x0-x1)*(x1-x0)
Sym = SymmetricFunctions(QQ)
g = Sym.from_polynomial(f)
print g

For 3 variables it fails. Sage throws a type error and says the polynomial f is not symmetric, even though it is certainly invariant under the action of the symmetric group of order 3.

S.<x0,x1,x2>=QQ[]
f = (x0-x1)*(x1-x2)*(x2-x0)
Sym = SymmetricFunctions(QQ)
g = Sym.from_polynomial(f)
print g

The type error is as folliows:

/sage/combinat/sf/sf.pyc in from_polynomial(self, f)
   1372             ValueError: x0 + 2*x1 + x2 is not a symmetric polynomial
   1373         """
-> 1374         return self.m().from_polynomial(f)
   1375 
   1376     def register_isomorphism(self, morphism, only_conversion=False):

However, the following does not throw a type error, perhaps because it is a linear combination of the elementary basis rather than a polynomial one. I don't understand what is going on. Why is this type error being triggered for a valid symmetric polynomial? Thank you very much for your time.

S.<x0,x1,x2>=QQ[]
f = (x0+x1)*(x1+x2)*(x2+x0)
Sym = SymmetricFunctions(QQ)
g = Sym.from_polynomial(f)
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answered 2020-04-19 00:55:32 -0600

FrédéricC gravatar image

Because it is not a symmetric polynomial:

sage: f==f(x0=x1,x1=x0)
False
sage: f==-f(x0=x1,x1=x0)
True
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Comments

Ah! Thank you very much.

masseygirl gravatar imagemasseygirl ( 2020-04-19 13:54:05 -0600 )edit

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Asked: 2020-04-18 22:36:04 -0600

Seen: 212 times

Last updated: Apr 19