Consider $V^{\otimes n}$ where $V$ is vector space and the representation of GL(V) acting in the usual way. Now if I consider tensor products or plethysms of irreducible spaces, this is not in general irreducible.

As an example Let $U \equiv (\bigwedge^3 V) \otimes (Sym^2 V) \subset V^{\otimes 5}$.

The Littlewood-Richardson algorithm can applied in this case to find the irreducible spaces contained in $U$. For and arbitrary element $ X \equiv ( v_1 \wedge v_2 \wedge v_3 )\otimes (w_1 \otimes_s w_2) \in (\bigwedge^3 V) \otimes (Sym^2 V)$. The algorithm generates a standard tableau, for concreteness one of them will be, $T_{\lambda} = (v_1 w_1, v_2, v_3, w_2)$ where comma separates the rows. The young symmetriser associated with this is $c_{\lambda}$ and acts in the obvious manner on $( v_1 \wedge v_2 \wedge v_3 )\otimes (w_1 \otimes_s w_2) $. Thus span of $c_\lambda X$ will generate an irreducible subspace of $U$.

Question 1) Given a decomposition of a reducible space into irreducibles. Any element can be uniquely decomposed as the sum of elements each living in a irreducible space. My question is what is the projection operator $P_{\lambda}$ associated with a standard young tableau generated by Littlewood-Richardson, for a given vector in the reducible space ?

I understand that young symmetriser is an idempotent and will give one of the elements in the irreducible space, the span of which generates the whole irreducible space. However I don't believe this projects the 'correct' component, given an arbitrary element.

In our running example, $c_{\lambda}X$ would be one of the elements in the irrep. However in general it is not true that $c_\lambda X$ is the orthogonal component of $X$ inside the irreducible rep characterised by the standard tableau $\lambda$ generated by Littlewood-richardson algorithm.

Question 2) In general is there a procedure find out the 'orthogonal components' living in irrep characterised by the standard Young tableau ?


2 Answers 2


For the number of boxes less than five, the young idempotents for standard tableaux of the same shape are mutually orthogonal projectors. Once $n\ge5$ the $Y=NP$ projectors are not in general mutually orthogonal --- i.e $Y_iY_j$ is not always equal to $\delta_{ij} Y_j$. There are various methods to obtain mutually orthogonal projectors. These are described in many places --- for example the book of Littlewood on group representations.

Young himself introduced a recursive construction of mutually orthogonal primative idempotents in his seminormal representation. The lecture notes of Garsia are a good source and available online: http://www.math.ucsd.edu/~garsia/somepapers/Youngseminormal.pdf


Let us start at the beginning. Suppose that $G$ is a finite group. Let $\widehat{G}$ be the set of irreducible finite dimensional representations of $G$. If $ \lambda \in \widehat{G} $, then write $V_{\lambda}$ for the corresponding irreducible representation. Semisimplicity implies that we have an isomorphism $$ \mathbb{C}[G] \cong \prod_{\lambda \in \widehat{G}} {\rm End}(V_{\lambda}) $$ It sends a group element $g$, to the tuple of matricies $ v \mapsto gv $. If you want to understand how a group element $g$ acts on an irreducible, then you map it across the isomorphism and pick out the corresponding matrix. In ${\rm End}(V_{\lambda})$, we have the identity matrix. This gives us a central idempotent $i_{\lambda}$ in $ \mathbb{C}[G]$. Given any representation $W$, we can talk about the $ \lambda$-isotypic component ${\rm I}_{\lambda} W$ which is the sum of all copies of $V_{\lambda} $ occouring inside $W$. We have $$V = \bigoplus_{\lambda \in \widehat{G}} {\rm I}_{\lambda} w $$ and $i_{\lambda} $ is the projection onto the isotypic component $I_{\lambda} W$.

Now assume that $P $ is a projection matrix in $ {\rm End}(V_{\lambda}) $ with rank 1. Then we get a corresponding element $ P \in \mathbb{C}[G]$. The young symmetrizers in the symmetric group algebra are of this form. If $U $ is a representation and $u \in U$, then $P u $ is either $ 0 $ or generates an irreducible subrepresentation isomorphic to $V_{\lambda}$. When choosing $P$, there is no canonical choose. Indeed, there is an entire projective space of elements which will have the same effect. Before we move onto the lie group case, let me close by saying that once you have constructed explicit models for each irreducible representation, then the Peter-Weyl isomorphism becomes explicit, and everything I said above can be carried out algorithmically. Constructing explicit irreps can be pretty difficult in general, and that is why tools like character theory or young symmetrizers are useful.

Now lets talk about semisimple connected algebraic groups. In this case, we replace the group algebra by the universal enveloping algebra $U(\mathfrak{g})$. Unfortunately, the isomorphism which was so useful in the finite group case is no longer true for the universal enveloping algebra. However, the Harish-Chandra isomorphism allows us to compute the center $Z(U(\mathfrak{g}))$. It is isomorphic to $ {\rm Sym}(\mathfrak{h})^W $. When $ \mathfrak{g} = \mathfrak{sl}_2 $ then the center is a polynomial ring generated by the Casmir element. Therefore, we can't have a projection for each irreducible. What you can do is compute the multiplicity space of any irreducible. If $ \lambda $ is an integral dominant weight, then the vector space of $ \lambda $-highest weights is exactly the multiplicity space of the irreducible corresponding to $\lambda$.

There is still one issue to resolve: How does the symmetric group interact with the general linear group? The answer is Schur-Weyl duality, but this answer is already long...


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