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  "body": "\n## Definition\n\nLet $F$ be a field, let $V$ be a [[vector space|000AYPA9AJ]] over $F$, and let $Q$ be a [[quadratic form|0RBQS1JJCR]] on $V$.\nLet $T(V)$ be the [[tensor algebra|P4J5JZDW0B]] of $V$.\nLet $I_Q$ be the [[two-sided ideal|0CD87SFM2N]] of $T(V)$ generated by the elements $v \\otimes v - Q(v) \\cdot 1$ with $v \\in V$.\nThe *Clifford algebra* of $(V, Q)$ is the [[quotient ring|0076HWHGHQ]]\n$$\\mathrm{Cl}(V, Q) = T(V) / I_Q.$$\n\n## Remarks\n\nUniversal property: for every associative unital $F$-algebra $A$ and every linear map $f \\colon V \\to A$ with $f(v)^2 = Q(v) \\cdot 1_A$ for all $v \\in V$, there is exactly one unital algebra homomorphism $\\mathrm{Cl}(V, Q) \\to A$ that extends $f$.\nIf $\\dim V = n$, then $\\dim \\mathrm{Cl}(V, Q) = 2^n$, with basis the products $e_{i_1} \\cdots e_{i_k}$, $i_1 \u003c \\cdots \u003c i_k$, for a basis $e_1, \\ldots, e_n$ of $V$.\nExample: $V = \\R^2$ with $Q(x, y) = -(x^2 + y^2)$.\nThe relation $(e_1 + e_2)^2 = Q(e_1 + e_2) = -2$ with $e_1^2 = e_2^2 = -1$ gives $e_1 e_2 = -e_2 e_1$, and then $(e_1 e_2)^2 = -1$; so $\\mathrm{Cl}(V, Q)$ is the quaternion algebra with $i = e_1$, $j = e_2$, $k = e_1 e_2$.\nFor $Q = 0$, the Clifford algebra is the [[exterior algebra|0MDMZ965VE]] of $V$.\nNon-example: the tensor algebra $T(V)$ for $V \\ne 0$ is not a Clifford algebra of $(V, Q)$, because $v \\otimes v$ is not a scalar there.\nSome authors use the opposite sign convention $v^2 = -Q(v)$.\n",
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