The Wolfram Language & System Documentation Center page for DirichletCharacter indicates Dirichlet characters modulo k form a group:

G[k_] := Table[DirichletCharacter[k, j, n], {j, EulerPhi[k]}]

The group operation is defined in terms of the k and j indices as follows:

add[DirichletCharacter[k_, j1_, n_], 
  DirichletCharacter[k_, j2_, n_]] := 
 DirichletCharacter[k, Mod[j1 + j2, EulerPhi[k], 1], n]

Question: What is the mathematical operation corresponding to this definition of the group operation?

I tried the following definition which seems to produce a different result than the definition above. Whereas the definition above is based on k and j indices the definition below assumes char1 and char2 are of the form defined by char below.

addChar[char1_, char2_] := char1 char2
char[k_, j_] := Table[DirichletCharacter[k, j, n], {n, 1, k}]

The following evaluation illustrates "addChar" produces a different result than "add".

Block[{k = 5, j1 = 2, j2 = 3, addResult},
 addResult = 
  addOld[DirichletCharacter[k, j1, n], DirichletCharacter[k, j2, n]];
 {Table[addResult, {n, 1, k}], addChar[char[k, j1], char[k, j2]]}]
{{1, 1, 1, 1, 0}, {1, -I, I, -1, 0}}

Here is the list of Dirichlet characters modulo k=5:



The documentation states

DirichletCharacter[k, j, n] picks a particular ordering for possible Dirichlet characters modulo k.

Different conventions can give different orderings for the possible characters.

What this means in practice is that you should treat the index j as a "black box" and undocumented. Even if it was completely documented, the group of characters is, in general, not cyclic, thus trying to understand the operation on charcters given by simply adding their indices modulo k is not going to be group theoretically sensible. The correct group operation is your addChar[] .

  • $\begingroup$ Thanks for the response. I was beginning to wonder if the index approach was arbitrary, but I was hoping for more insight. The page I linked above indicates "Dirichlet characters are labeled in an increasing order of the number of factors", but I didn't understand this statement either or if it somehow related to the definition of the group based on the j and k indexes. $\endgroup$ – Steven Clark May 29 at 1:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.