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  "body": "\n## Definition\n\nLet $X$ and $Y$ be Banach spaces, and let $T \\colon X \\to Y$ be a [[Fredholm operator|8HWGKWD6ZM]].\nThe *index* of $T$ is the integer $\\operatorname{ind} T = \\dim \\ker T - \\dim \\operatorname{coker} T$.\nHere $\\ker T$ is the [[kernel|0S23E6YE83]] of $T$.\nThe cokernel $\\operatorname{coker} T = Y / T(X)$ is the [[quotient|0BR3FY5Q8M]] of $Y$ by the [[image|0WSFKKT6FF]] of $T$.\n\n## Remarks\n\nExample: the right shift $S$ on $\\ell^2$ has $\\ker S = \\set{0}$ and $\\operatorname{coker} S$ of dimension $1$, so $\\operatorname{ind} S = -1$.\nThe left shift $S^*(x_1, x_2, \\ldots) = (x_2, x_3, \\ldots)$ is onto with a kernel of dimension $1$, so $\\operatorname{ind} S^* = 1$.\nNon-example: the diagonal operator $D(x_n) = (x_n / n)$ on $\\ell^2$ is not Fredholm, so it has no index.\nFor a linear map $T \\colon X \\to Y$ between finite-dimensional spaces, the rank–nullity theorem gives $\\operatorname{ind} T = \\dim X - \\dim Y$.\nFor Fredholm operators $T \\colon X \\to Y$ and $S \\colon Y \\to Z$, the product $ST$ is Fredholm and $\\operatorname{ind}(ST) = \\operatorname{ind} S + \\operatorname{ind} T$.\nIf $K \\colon X \\to Y$ is a [[compact operator|M2NEP1K5AE]], then $T + K$ is Fredholm and $\\operatorname{ind}(T + K) = \\operatorname{ind} T$.\nThe Fredholm operators form an open subset of the bounded operators, and the index is constant on its connected components.\n",
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