Let $T^{2}=\mathcal{R}^{2}/2\pi\mathcal{Z}^{2}$ and consider the space $B(L^{2}(T^{2}))$, using the quotient topology, can we show that $B(L^{2}(T^{2}))$ is compact? Where B is the set of all bounded linear operators mapping elements of $L^{2}(T^{2})$ to itself.
Let $T^{2}=\mathcal{R}^{2}/2\pi\mathcal{Z}^{2}$ and consider the space $B(L^{2}(T^{2}))$, using the quotient topology, can we show that $B(L^{2}(T^{2}))$ is compact? Where B is the set of all bounded linear operators mapping elements of $L^{2}(T^{2})$ to itself.
The space $B(L^{2}(T^{2}))$ can never be compact in any topology, because it is unbounded. For example, the sequence (nI), where I is the identity operator, cannot have a convergent subsequence.
The unit ball of $B(L^{2}(T^{2}))$ is compact in the weak operator topology.