# Revision history [back]

I am a new student in SAGE. I read the following discussion:
evaluation of character of symmetric group
and then also read the manual.

However, I am still confused about some fundamental problem:
(I cannot find these function in "Sage Reference Manual: Groups, Release 8.2". Are both new functions?).

SymmetricGroupRepresentation(partition, implementation='specht', ring=None, cache_matrices=True)


I am confused about "partition". Suppose for $S_3$, and partition $=[2,1]$. What does it mean? (It seems $[1,2]$ is not valid)

spc = SymmetricGroupRepresentation([2,1], 'orthogonal')
spc.representation_matrix(Permutation([1,2,3]))


When I use

spc.representation_matrix(Permutation([1,2]))


error pops out. However, as far as I know, $(1,2)$ is a valid permutation, which represent the matrix representation: $$\begin{bmatrix}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}$$

I cannot find "Permutation" in "Sage Reference Manual: Group". Where can I find this function?

I am a new student in SAGE. I read the following discussion:
evaluation of character of symmetric group
and then also read the manual.

However, I am still confused about some fundamental problem:
(I cannot find these function in "Sage Reference Manual: Groups, Release 8.2". Are both new functions?).

SymmetricGroupRepresentation(partition, implementation='specht', ring=None, cache_matrices=True)


I am confused about "partition". Suppose for $S_3$, and partition $=[2,1]$. What does it mean? (It seems $[1,2]$ is not valid)

spc = SymmetricGroupRepresentation([2,1], 'orthogonal')
'specht')
spc.representation_matrix(Permutation([1,2,3]))


When I use

spc.representation_matrix(Permutation([1,2]))


error pops out. However, as far as I know, $(1,2)$ is a valid permutation, which represent the matrix representation: $$\begin{bmatrix}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}$$

I cannot find "Permutation" in "Sage Reference Manual: Group". Where can I find this function?

I am a new student in SAGE. I read the following discussion:
evaluation of character of symmetric group
and then also read the manual.

However, I am still confused about some fundamental problem:
(I cannot find these function in "Sage Reference Manual: Groups, Release 8.2". Are both new functions?).

SymmetricGroupRepresentation(partition, implementation='specht', ring=None, cache_matrices=True)


I am confused about "partition". Suppose for $S_3$, and partition $=[2,1]$. What does it mean? (It seems $[1,2]$ is not valid)

spc = SymmetricGroupRepresentation([2,1], 'specht')
spc.representation_matrix(Permutation([1,2,3]))


When I use

spc.representation_matrix(Permutation([1,2]))


error pops out. However, as far as I know, $(1,2)$ is a valid permutation, which represent the matrix representation: $$\begin{bmatrix}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}$$

$$\begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}$$

But I test another permutation: $[1,2,3]$, the answer is the same. However, $[1,2]$ and $[1,2,3]$ are in the different conjugacy classes; they should not have the same character.

I cannot find "Permutation" in "Sage Reference Manual: Group". Where can I find this function?

I am a new student in SAGE. I read the following discussion:
evaluation of character of symmetric group
and then also read the manual.

However, I am still confused about some fundamental problem:
(I cannot find these function in "Sage Reference Manual: Groups, Release 8.2". Are both new functions?).

SymmetricGroupRepresentation(partition, implementation='specht', ring=None, cache_matrices=True)


I am confused about "partition". Suppose for $S_3$, and partition $=[2,1]$. What does it mean? (It seems $[1,2]$ is not valid)

spc = SymmetricGroupRepresentation([2,1], 'specht')
spc.representation_matrix(Permutation([1,2,3]))


When I use

spc.representation_matrix(Permutation([1,2]))


error pops out. However, as far as I know, $(1,2)$ is a valid permutation, which represent the matrix representation: $$\begin{bmatrix}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}$$

$$\begin{bmatrix}0 & 1$$\begin{bmatrix}1 & 0 \\ 1 & 0\end{bmatrix}$$0 & 1\end{bmatrix}$$
But I test another permutation: $[1,2,3]$, the answer is the same. However, $[1,2]$ and $[1,2,3]$ are in the different conjugacy classes; they should not have the same character.