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  "body": "\n## Definition\n\nLet $H$ be a complex [[Hilbert space|HM2HHE3J51]] with [[inner product|0NB0AWGJBW]] $\\langle \\cdot, \\cdot \\rangle$, and let $B(H)$ be the [[algebra of bounded operators|BMDG4JQPEH]] on $H$.\nThe *weak operator topology* on $B(H)$ is the coarsest topology such that, for all $x, y \\in H$, the map $B(H) \\to \\C$, $T \\mapsto \\langle T x, y \\rangle$, is continuous.\n\n## Remarks\n\nEquivalently, it is the [[topology generated by the seminorms|ZVFXQ7S6FY]] $T \\mapsto \\abs{\\langle T x, y \\rangle}$ for $x, y \\in H$.\nA net $(T_i)$ converges to $T$ in this topology if and only if $\\langle T_i x, y \\rangle \\to \\langle T x, y \\rangle$ for all $x, y \\in H$.\n\nExample: on $H = \\ell^2(\\N)$, let $S$ be the right shift, $S(x_0, x_1, \\dots) = (0, x_0, x_1, \\dots)$.\nThen $\\langle S^n x, y \\rangle = \\sum_{k \\ge 0} x_k \\overline{y_{k+n}}$, and by the Cauchy–Schwarz inequality its modulus is at most $\\|x\\| \\, (\\sum_{k \\ge n} \\abs{y_k}^2)^{1/2}$, which tends to $0$.\nSo $S^n \\to 0$ in the weak operator topology, while $\\|S^n\\| = 1$ for all $n$.\n\nNon-example: let $\\lambda \\in \\C$ with $\\abs{\\lambda} = 1$ and $\\lambda \\ne 1$, and let $T_n = \\lambda^n I$.\nFor $x = y \\ne 0$, the numbers $\\langle T_n x, x \\rangle = \\lambda^n \\|x\\|^2$ do not converge, so $(T_n)$ has no limit in this topology.\n\nThe weak operator topology is coarser than the strong operator topology, which is coarser than the norm topology.\nThe closed unit ball of $B(H)$ is compact in the weak operator topology.\nMultiplication is separately continuous but not jointly continuous when $H$ is infinite-dimensional, and $T \\mapsto T^*$ is continuous.\n",
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