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This is my first post on MathSE, since I could not find a helpful answer to my question on here yet.

Let $(L,[\cdot,\cdot])$ be a finite-dimensional semisimple complex Lie-Algebra, $H\subseteq L$ a Cartan-Subalgebra, $\Phi$ the roots of $L$ and $\rho \colon L\to \mathfrak{gl}(V)$ a finite-dimensional representation of $L$ on $V$.

According to Weyl's Theorem $V$ decomposes into a direct sum of irreducible subrepresentations.

Let $$ L=H\oplus\bigoplus_{\alpha \in \Phi} L_\alpha $$ be the Cartan-Decomposition of $L$ and $\Phi=\Phi^+\cup\Phi^-$ an ordering of the roots. We then have

$\textbf{Theorem.}$ If $v$ is a highest weight vector of some representation of $L$, then the subspace $U$ generated by successive applications to $v$ of the root spaces $L_\alpha$ with $\alpha \in \Phi^-$ is an irreducible subrepresentation of $L$.

Does this in any way help me to find a decomposition into irreducible subrepresentations?

Here is an example I would like to understand: Let $L=\mathfrak{sp}_4(\mathbb C)$ be the symplectic Lie-Algebra, $W=\lbrace A\in \mathfrak{gl}_4(\mathbb C)\mid A=-A^t\rbrace$ the Lie-Subalgebra of skew-symmetric Matrices and for $x\in L, \, A\in W$ let $\rho\colon L\to \mathfrak{gl}(W)$ be defined as $\rho(x)(A)=xA+Ax^t$. Finally, let $H$ be the Cartan-Subalgebra of diagonal matrices in $L$. The roots of $L$ are (after canonical identification) $$ \Phi=\lbrace \pm 2e_1, \pm 2e_2, e_1+e_2,-e_1+e_2,-e_1-e_2, e_1-e_2\rbrace. $$ Furthermore we have that $$ \Delta=\lbrace 2e_1,-e_1+e_2\rbrace $$ is a basis of $\Phi$ and $$ \Phi^+=\lbrace 2e_1,-e_1+e_2,e_1+e_2,2e_2\rbrace. $$ Additionally I know that $\rho$ has maximal weight $\lambda=e_1+e_2$.

How would one decompose $W$ into a direct sum of irreducible subrepresentations?

Thank you very much for your time and help!

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First thing I would say is that I would call that decomposition a "root space decomposition". I think I have seen it referred to as a Cartan decomposition before but usually that means the decomposition associated to a Cartan involution.

A non-irreducible representation will have multiple "highest" weight spaces in that they are killed by the action of all the positive root spaces. Taking the weight space you mention you will get a irreducible subrepresentation of codimension 1. This leaves a trivial representation. Its highest weight is $0$ so that may be why it isn't obvious at first especially since there is a $0$ weight space contained in the bigger subrepresentation.

I think the best way to see this that the symplectic group is precisely the invertible linear transformations that preserve some symplectic form $\omega$ on our vector space $V$. Now a symplectic form is precisely an element of $\bigwedge^2V^*$ (of maximal rank so it is non-degenerate). Now if the symplectic group preserves $\omega$ that means that the symplectic Lie algebra must send it to $0$ by differentiating the action. Thus $\langle\omega\rangle$ is a trivial 1-dimensional subrepresentation of $\bigwedge^2V^*$. Note that $\omega$ gives a identification of $V$ with $V^*$ and likewise of $\bigwedge^2V$ with $\bigwedge^2V^*$ so we can view this in $\bigwedge^2V$ if we want.

Then the bigger subrepresentation is all the elements of $\bigwedge^2V$ that contract with $\omega$ to give $0$ (aka the annihilator of $\omega$). This is often denoted $\bigwedge^2_0V$. If we wanted to work in $\bigwedge^2V^*$ this corresponds to the orthocomplement of $\langle\omega\rangle$.

To get some good practise with what these kinds of decompositions look like I think it's useful to use LiE (Online version here) a program that does exactly these kinds of calculations for you. To use it you will need to know how to relate the irreducible representations to coefficients of fundamental weights as that's how it refers to them.

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  • $\begingroup$ Thank you for your insights. If $U$ is the irreducible subrepresentation of codimension $1$ you mention this leaves only a trivial subrepresentation. Trivial meaning of dimension $1$, I guess? How could I find a vector spanning this trivial subrepresentation? How do I know it's a complement for $U$? $\endgroup$ Jul 18, 2021 at 17:18
  • $\begingroup$ Trivial meaning the action of the the Lie algebra sends everything to 0. It is of dimension 1 as well but that's a separate thing. You can find such a vector by taking $\omega$ to its corresponding element in $\bigwedge^2 V$. Alternatively $\omega$ defines a bilinear form on $\bigwedge^2 V$ in a natural way and the trivial subrepresentation is the orthocomplement so you just need to find a vector orthogonal to U. This also tells us it is a complement. $\endgroup$
    – Callum
    Jul 18, 2021 at 18:58

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