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  "body": "\n## Definition\n\nLet $V$ be a finite-dimensional real vector space with a [[quadratic form|0RBQS1JJCR]] $q$.\nLet $\\mathrm{Cl}(V, q)$ be the [[Clifford algebra|A38B1QFK20]] of $(V, q)$, with the convention that $v \\cdot v = -q(v) \\cdot 1$ for $v \\in V$.\nLet $\\mathbb{K}$ be $\\R$ or $\\C$.\nA *Clifford module* over $\\mathrm{Cl}(V, q)$ is a $\\mathbb{K}$-vector space $W$ together with a homomorphism of real algebras $\\rho \\colon \\mathrm{Cl}(V, q) \\to \\End_{\\mathbb{K}}(W)$ that sends $1$ to $\\id_W$.\n\n## Remarks\n\nBy the universal property of the Clifford algebra, giving $\\rho$ is the same as giving a linear map $c \\colon V \\to \\End_{\\mathbb{K}}(W)$ with\n$$c(v)^2 = -q(v) \\, \\id_W \\quad \\text{for all } v \\in V.$$\nAuthors who use the convention $v \\cdot v = q(v) \\cdot 1$ get the same theory with $q$ replaced by $-q$.\nA Clifford module is irreducible if it has no proper nonzero invariant subspace; when $q$ is nondegenerate, every finite-dimensional Clifford module is a direct sum of irreducible ones.\n\nExample: let $V = \\R$ with $q(x) = x^2$, let $W = \\C$ as a real vector space and let $c(x)$ be multiplication by $ix$.\nThen $c(x)^2 = (ix)^2 = -x^2 = -q(x)$, so $W$ is a Clifford module, and in fact $\\mathrm{Cl}(V, q) \\cong \\C$.\n\nNon-example: with the same $V$ and $q$, let $W = \\R$ and $c(x) = x \\, \\id_W$.\nThen $c(1)^2 = 1 \\ne -1$, so this is not a Clifford module.\nIn fact no real vector space of dimension $1$ is a Clifford module here, since no real number squares to $-1$.\n\n## Explanations\n\n### L1 human\n\nA Clifford module is a vector space on which each vector $v$ acts by a linear map whose square is the scalar $-q(v)$.\n",
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