Return to Answer

deleted 89 characters in body

Source Link

edited Mar 28, 2017 at 16:50

52.9k
4
120
241

As the comments suggest, there don't seem to be any easily stated necessary and sufficient conditions for the adjoint representation to occur as a summand of the tensor product of two irreducibles (say with highest weights $\lambda$ and $\mu$). Of course, one has to fix a simple system of roots to speak of "highest weight".

However, there is a simple necessary condition. Start with the classical fact that 0 is a weight of an irreducible representation precisely when the highest weight lies in the root lattice. In particular, this applies to the adjoint representation, whose highest weight is the highest root; here the 0 weight space corresponds to a Cartan subalgebra. In turn, classical results such as Brauer's early method for decomposing tensor products (later extended by Klimyk) show that the highest weights $\nu$ of all irreducible summands lie below the sum of the two given highest weights in the usual partial ordering of weights. It follows immediately that the adjoint representation occurs as a summand of the tensor product only if $\lambda + \mu$ lies in the root lattice.

In some Lie types such as $G_2$, the root lattice equals the weight lattice. But here one sees readily that not all tensor products of two irreducibles have the 14-dimensional adjoint representation as a summand. I don't know of an easily stated sufficient condition in general.

As the comments suggest, there don't seem to be any easily stated necessary and sufficient conditions for the adjoint representation to occur as a summand of the tensor product of two irreducibles (say with highest weights $\lambda$ and $\mu$). Of course, one has to fix a simple system of roots to speak of "highest weight".

However, there is a simple necessary condition. Start with the classical fact that 0 is a weight of an irreducible representation precisely when the highest weight lies in the root lattice. In particular, this applies to the adjoint representation, whose highest weight is the highest root; here the 0 weight space corresponds to a Cartan subalgebra. In turn, classical results such as Brauer's early method for decomposing tensor products (later extended by Klimyk) show that the highest weights $\nu$ of all irreducible summands lie below the sum of the two given highest weights in the usual partial ordering of weights. It follows immediately that the adjoint representation occurs as a summand of the tensor product only if $\lambda + \mu$ lies in the root lattice.

In some Lie types such as $G_2$, the root lattice equals the weight lattice. But here one sees readily that not all tensor products of two irreducibles have the 14-dimensional adjoint representation as a summand. I don't know of an easily stated sufficient condition in general.

As the comments suggest, there don't seem to be any easily stated necessary and sufficient conditions for the adjoint representation to occur as a summand of the tensor product of two irreducibles (say with highest weights $\lambda$ and $\mu$). Of course, one has to fix a simple system of roots to speak of "highest weight".

However, there is a simple necessary condition. Start with the classical fact that 0 is a weight of an irreducible representation precisely when the highest weight lies in the root lattice. In particular, this applies to the adjoint representation, whose highest weight is the highest root; here the 0 weight space corresponds to a Cartan subalgebra. In turn, classical results show that the highest weights $\nu$ of all irreducible summands lie below the sum of the two given highest weights in the usual partial ordering of weights. It follows immediately that the adjoint representation occurs as a summand of the tensor product only if $\lambda + \mu$ lies in the root lattice.

In some Lie types such as $G_2$, the root lattice equals the weight lattice. But here one sees readily that not all tensor products of two irreducibles have the 14-dimensional adjoint representation as a summand. I don't know of an easily stated sufficient condition in general.

Source Link

created Mar 28, 2017 at 16:11

52.9k
4
120
241

As the comments suggest, there don't seem to be any easily stated necessary and sufficient conditions for the adjoint representation to occur as a summand of the tensor product of two irreducibles (say with highest weights $\lambda$ and $\mu$). Of course, one has to fix a simple system of roots to speak of "highest weight".

However, there is a simple necessary condition. Start with the classical fact that 0 is a weight of an irreducible representation precisely when the highest weight lies in the root lattice. In particular, this applies to the adjoint representation, whose highest weight is the highest root; here the 0 weight space corresponds to a Cartan subalgebra. In turn, classical results such as Brauer's early method for decomposing tensor products (later extended by Klimyk) show that the highest weights $\nu$ of all irreducible summands lie below the sum of the two given highest weights in the usual partial ordering of weights. It follows immediately that the adjoint representation occurs as a summand of the tensor product only if $\lambda + \mu$ lies in the root lattice.

In some Lie types such as $G_2$, the root lattice equals the weight lattice. But here one sees readily that not all tensor products of two irreducibles have the 14-dimensional adjoint representation as a summand. I don't know of an easily stated sufficient condition in general.