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  "body": "\n## Definition\n\nIdentify $\\R^{2n}$ with $\\C^n$ by $(x, y) \\mapsto x + iy$, with the symplectic form $\\omega = \\sum_{j=1}^n dx_j \\wedge dy_j$.\nA linear subspace $\\Lambda \\subset \\R^{2n}$ is *Lagrangian* if $\\dim \\Lambda = n$ and $\\omega(u, v) = 0$ for all $u, v \\in \\Lambda$.\nLet $\\Lambda(n)$ be the set of Lagrangian subspaces, the *Lagrangian Grassmannian*.\n\nEvery $\\Lambda \\in \\Lambda(n)$ is $\\Lambda = U \\R^n$ for some $U$ in the [[unitary group|0R9R0Y1HDK]] $U(n)$.\nThis $U$ is unique up to right multiplication by an element of the [[orthogonal group|0TKYAEJ8N7]] $O(n)$.\nHence the map $\\Lambda \\mapsto \\det(U)^2 \\in S^1 \\subset \\C$ is well defined on $\\Lambda(n)$.\nLet $\\gamma \\colon [0, 1] \\to \\Lambda(n)$ be a continuous loop, so $\\gamma(0) = \\gamma(1)$.\nIts *Maslov index* is the [[winding number|0NQSV2EWDR]] about $0$ of the loop $t \\mapsto \\det(U(t))^2$, where $\\gamma(t) = U(t) \\R^n$.\n\n## Remarks\n\nExample: $n = 1$ and $\\gamma(t) = e^{i\\pi t} \\R \\subset \\C$ for $t \\in [0, 1]$, the line through $0$ turned by half a turn.\nHere $U(t) = e^{i\\pi t}$ and $\\det(U(t))^2 = e^{2\\pi i t}$ winds once around $0$, so the Maslov index is $1$.\n\nNon-example: in $\\R^4 = \\C^2$, the plane $P = \\C \\times \\set{0}$ has dimension $2$, but $\\omega$ pairs the vectors $(1, 0)$ and $(i, 0)$ of $P$ to $1 \\ne 0$.\nSo $P$ is not Lagrangian, and a loop of planes through $P$ is not a loop in $\\Lambda(2)$ and has no Maslov index.\n\nThe Maslov index depends only on the homotopy class of the loop, and it gives an isomorphism of the [[fundamental group|S1XK3HXHFX]] $\\pi_1(\\Lambda(n))$ with $\\Z$.\nFor a [[Lagrangian submanifold|H8ACPTMX08]] $L \\subset \\R^{2n}$, a loop in $L$ gives the loop of tangent spaces in $\\Lambda(n)$, and its Maslov index defines a class in $H^1(L; \\Z)$, the Maslov class.\n",
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