Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Frequently in the literature on Hecke algebras for the symmetric group and their generalisations, one encounters references to Young's seminormal form and Young's orthogonal form. I have a good understanding of the seminormal form, which gives simple formulae for the actions of simple transpositions on Specht modules for which the matrices are "nearly triangular".

What I don't understand is how the orthogonal form is different. Are these just two names for the same thing?

share|improve this question
add comment

2 Answers 2

up vote 3 down vote accepted

The only difference between the two is rescaling the basis vectors, i.e. conjugating by a diagonal matrix.

For instance with the representations of the symmetric group, the usual choice for the seminormal representation would be to have matrices of the form

$$ \begin{pmatrix} -1/k & 1 \\ 1 - 1/k^2 & 1/k \end{pmatrix} $$

for switching between two tableaux by a transposition $s_i$ of two boxes with axial distance $k$ between them. (See for instance question 66602 which attempts to provide motivation for this particular convention.)

On the other hand, the orthogonal representation is what you would expect, giving matrices

$$ \begin{pmatrix} -1/k & \sqrt{1 - 1/k^2} \\ \sqrt{1 - 1/k^2} & 1/k \end{pmatrix}. $$

share|improve this answer
add comment

The main difference is that in the seminormal form, or Specht module the representing matrices have integer entries (which can be reduced mod p) and in the orthogonal form the representing matrices have entries which are rational numbers.

The other difference is that although both have an invariant symmetric inner product for the orthogonal form the Gram matrix is diagonal (so the basis vectors are orthogonal) whereas the Gram matrix for the seminormal form is far from diagonal.

Then there is the orthonormal form which is obtained from the orthogonal form by scaling the basis vectors to be orthonormal. This introduces square roots which I find inelegant but this is the form that seems to be preferred in the physics literature.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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