Skip to main content
added 7 characters in body
Source Link
Johannes Ebert
  • 20.9k
  • 4
  • 74
  • 117

Let $V$ and $W$ be Hilbert spaces with irreducible unitary $G$-actions and let $T:V \to W$ be ana bounded intertwiner. Then the adjoint is an intertwiner as well and hence so are $T^\ast T$ and $TT^\ast$. Claim: These two are multiples of the identity. It follows that $T$ is zero or a multiple of an isometric isomorphism.

To prove the claim, it is enough to show that a selfadjoint intertwiner $A$ of an irreducible represenation is a multiple of the identity.

But the spectral projections of $A$ are intertwiners. By irreducibility, they are either 0 or 1. The spectrsl theorem then proves that $A$ is a multiple of the identity.

Let $V$ and $W$ be Hilbert spaces with irreducible unitary $G$-actions and let $T:V \to W$ be an intertwiner. Then the adjoint is an intertwiner as well and hence so are $T^\ast T$ and $TT^\ast$. Claim: These two are multiples of the identity. It follows that $T$ is zero or a multiple of an isometric isomorphism.

To prove the claim, it is enough to show that a selfadjoint intertwiner $A$ of an irreducible represenation is a multiple of the identity.

But the spectral projections of $A$ are intertwiners. By irreducibility, they are either 0 or 1. The spectrsl theorem then proves that $A$ is a multiple of the identity.

Let $V$ and $W$ be Hilbert spaces with irreducible unitary $G$-actions and let $T:V \to W$ be a bounded intertwiner. Then the adjoint is an intertwiner as well and hence so are $T^\ast T$ and $TT^\ast$. Claim: These two are multiples of the identity. It follows that $T$ is zero or a multiple of an isometric isomorphism.

To prove the claim, it is enough to show that a selfadjoint intertwiner $A$ of an irreducible represenation is a multiple of the identity.

But the spectral projections of $A$ are intertwiners. By irreducibility, they are either 0 or 1. The spectrsl theorem then proves that $A$ is a multiple of the identity.

Source Link
Johannes Ebert
  • 20.9k
  • 4
  • 74
  • 117

Let $V$ and $W$ be Hilbert spaces with irreducible unitary $G$-actions and let $T:V \to W$ be an intertwiner. Then the adjoint is an intertwiner as well and hence so are $T^\ast T$ and $TT^\ast$. Claim: These two are multiples of the identity. It follows that $T$ is zero or a multiple of an isometric isomorphism.

To prove the claim, it is enough to show that a selfadjoint intertwiner $A$ of an irreducible represenation is a multiple of the identity.

But the spectral projections of $A$ are intertwiners. By irreducibility, they are either 0 or 1. The spectrsl theorem then proves that $A$ is a multiple of the identity.