Young’s projection operators

Associated to a Young tableau with $n$ boxes, we have two elementsof the group ring of the permutation group on $n$ symbols. Toconstruct the operators, we first construct the antisymmetrizingoperators corresponding to the columns and the symmetrizingoperators corresponding to the rows. Then one operator correspondingto the tableau consists of the product of the symmetrizing operatorscorresponding to the rows multiplied by the product of theantisymmetrizing operators corresponding to the columns and theother consists of the product of the antisymmetrizing operatorscorresponding to the columns multiplied by the symmetrizing operatorscorresponding to the rows.

How this works may be illustrated with a simple example. Considerthe tableau

\\begin{matrix}1&2\\\\3&4\\end{matrix}.

Corresponding to the first row, we have the symmetrization operator

1+(1\\,2).

Corresponding to the second row, we have the symmetrization operator

1+(3\\,4)

Multiplying these two symmetrization operators (the order does notmatter because they involve permutations of different elements)produces

1+(1\\,2)+(3\\,4)+(1\\,2)(3\\,4)

Corresponding to the first column, we have the antisymmetrizationoperator

1-(1\\,3).

Corresponding to the second column, we have the antisymmetrizationoperator

1-(2\\,4).

Multiplying these two antisymmetrization operators (the order does notmatter because they involve permutations of different elements)produces

1-(1\\,3)-(2\\,4)+(1\\,3)(2\\,4).

To obtain one Young projector, we multiply the product ofthe symmetrization operators by the product of theantisymmetrizationoperators.

\\left(1+(1\\,2)+(3\\,4)+(1\\,2)(3\\,4)\\right)\\left(1-(1\\,3)-(2\\,4)+(1\\,3)(2\\,4)%\\right)=

	$\\displaystyle 1$	$\\displaystyle+$	$\\displaystyle(1\\,2)+(3\\,4)+(1\\,2)(3\\,4)-(1\\,3)-(1\\,2\\,3)-(1\\,3\\,4)-(1\\,2\\,3\\,4)-$
	$\\displaystyle(2\\,4)$	$\\displaystyle-$	$\\displaystyle(1\\,4\\,2)-(2\\,4\\,3)-(1\\,4\\,3\\,2)+(1\\,3)(2\\,4)+(1\\,4\\,2\\,3)+(1\\,3%\\,2\\,4)+(1\\,4)(2\\,3)$

To obtain the other projector, we multiply in the other order.

\\left(1-(1\\,3)-(2\\,4)+(1\\,3)(2\\,4)\\right)\\left(1+(1\\,2)+(3\\,4)+(1\\,2)(3\\,4)%\\right)=

	$\\displaystyle 1$	$\\displaystyle+$	$\\displaystyle(1\\,2)+(3\\,4)+(1\\,2)(3\\,4)-(1\\,3)-(1\\,3\\,2)-(1\\,4\\,3)-(1\\,4\\,3\\,2)-$
	$\\displaystyle(2\\,4)$	$\\displaystyle-$	$\\displaystyle(1\\,2\\,4)-(2\\,3\\,4)-(1\\,2\\,3\\,4)+(1\\,3)(2\\,4)+(1\\,3\\,2\\,4)+(1\\,4%\\,2\\,3)+(1\\,4)(2\\,3)$