Skip to main content
added 26 characters in body
Source Link
Johannes Ebert
  • 20.9k
  • 4
  • 74
  • 117

Let $G$ be the group of all invertible operators on a Hilbert space $H$ that are of the form $1+K$, $K$ compact. Then the space of all invertible operators $GL(H)$ is a model for $EG$ and $BG$ is the identity component of the space of all units of the Calkin algebra $B(H)/K(H)$ (bounded modulo compact operators). By the way, $BG \simeq BU$, $G \simeq U$.

Let $G$ be the group of all invertible operators on a Hilbert space $H$ that are of the form $1+K$, $K$ compact. Then the space of all invertible operators $GL(H)$ is a model for $EG$ and $BG$ is the space of all units of the Calkin algebra $B(H)/K(H)$ (bounded modulo compact operators). By the way, $BG \simeq BU$, $G \simeq U$.

Let $G$ be the group of all invertible operators on a Hilbert space $H$ that are of the form $1+K$, $K$ compact. Then the space of all invertible operators $GL(H)$ is a model for $EG$ and $BG$ is the identity component of the space of all units of the Calkin algebra $B(H)/K(H)$ (bounded modulo compact operators). By the way, $BG \simeq BU$, $G \simeq U$.

Post Made Community Wiki by S. Carnahan
Source Link
Johannes Ebert
  • 20.9k
  • 4
  • 74
  • 117

Let $G$ be the group of all invertible operators on a Hilbert space $H$ that are of the form $1+K$, $K$ compact. Then the space of all invertible operators $GL(H)$ is a model for $EG$ and $BG$ is the space of all units of the Calkin algebra $B(H)/K(H)$ (bounded modulo compact operators). By the way, $BG \simeq BU$, $G \simeq U$.