Compact?

**Mauritzvdworm** · August 22nd 2009, 01:22 PM

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.

**Opalg** · August 23rd 2009, 11:52 PM

Originally Posted by Mauritzvdworm

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.

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