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  "body": "\n## Definition\n\nLet $D$ be a diagram of an oriented tame [[link|1A5WRCNTM3]] $L$, with crossings ordered $1, \\ldots, n$, of which $n_+$ are positive and $n_-$ are negative.\nLet $A = \\Z[x]/(x^2)$, graded by $\\deg 1 = 1$ and $\\deg x = -1$.\nFor $v \\in \\set{0,1}^n$, let $D_v$ be the diagram where crossing $i$ gets the $0$-smoothing (A-smoothing of the [[Kauffman bracket|JYKVYS4N32]]) if $v_i = 0$ and the $1$-smoothing otherwise.\nLet $k_v$ be the number of circles of $D_v$ and $r(v) = v_1 + \\cdots + v_n$.\nPut $C_v = A^{\\otimes k_v}$, with quantum grading shifted up by $r(v) + n_+ - 2n_-$.\nThe chain group in homological degree $i$ is $C^i(D) = \\bigoplus_{r(v) = i + n_-} C_v$.\nFor an edge $v \\to v'$ that changes one $0$ to a $1$, the map $C_v \\to C_{v'}$ is:\n\n1. $m \\colon A \\otimes A \\to A$ if two circles merge, with $m(1 \\otimes 1) = 1$, $m(1 \\otimes x) = m(x \\otimes 1) = x$, $m(x \\otimes x) = 0$;\n2. $\\Delta \\colon A \\to A \\otimes A$ if one circle splits, with $\\Delta(1) = 1 \\otimes x + x \\otimes 1$ and $\\Delta(x) = x \\otimes x$;\n\nand the identity on the other circles.\nThe differential $d \\colon C^i(D) \\to C^{i+1}(D)$ is the sum of these edge maps, each with a sign $(-1)^{s}$.\nHere $s$ is the number of $1$s in $v$ before the changed position.\nThe *Khovanov homology* of $L$ is $Kh^{i,j}(L) = H^i(C(D))$ in quantum degree $j$.\n\n## Remarks\n\nThe map $d$ satisfies $d^2 = 0$ and preserves the quantum grading.\nUp to isomorphism, $Kh^{i,j}(L)$ does not depend on the diagram $D$.\nExample: the round diagram of the [[unknot|1TQ06R4Q66]] has $n = 0$, so the complex is $A$ in degree $0$ and $Kh$ is $\\Z$ in bidegrees $(0, 1)$ and $(0, -1)$.\nIts graded Euler characteristic $q + q^{-1}$ is the unnormalized Jones polynomial of the unknot.\nIn general $\\sum_{i,j} (-1)^i q^j \\rank Kh^{i,j}(L) = (q + q^{-1}) V_L(t)$ at $t^{1/2} = -q$, where $V_L$ is the [[Jones polynomial|4SPC17FPXE]].\nNon-example: the homology of the complex $\\bigoplus_v C_v$ with the signs omitted is not defined, because without signs the faces of the cube commute and $d^2 \\ne 0$.\n",
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