在 sympy 中与 kronecker delta 签约

Question

I'm trying to do some tensor calculations in sympy, but I can't seem to get it to simplify any contractions of tensors against the kronecker delta, ie with the minimal example:

from sympy import *

n = Idx('n')
i = Idx('i',(1,n))
j = Idx('j',(1,n))

x = IndexedBase('x')
print(Sum(KroneckerDelta(i,j)*x[j],(j,1,n)))

here, n is the dimension of the space, and i,j, are indices that run from 1 to n. You would expect the sum to evaluate to x[i], except sympy doesn't do any simplification whatsoever, despite hitting it with the simplify command

Answer 1

I think I found the solution to your question while I posted a similar question

What you want is the sympy.concrete.delta.deltasummation in place of Sum

print(sympy.concrete.delta.deltasummation(KroneckerDelta(i,j)*x[j],(j,1,n)))
>>> Piecewise((x[i], (n >= i) & (i >= 1)), (0, True))

This simplifies further if the KronckerDelta function specifies bounds

print(sympy.concrete.delta.deltasummation(KroneckerDelta(i,j,(1,n))*x[j],(j,1,n)))
>>> x[i]

Unfortunately I don't know how to automatically replace the Sum with the deltasummation, but at least this is a starting point for you.

There are a few other simplifications to be found in sympy.concrete and sympy.concrete.delta

edit: link to my somewhat related question Sympy simplify sum of Kronecker delta