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  "body": "\n## Definition\n\nLet $A$ be a commutative complex Banach algebra, and let $\\Delta(A)$ be its [[character space|M7JVFJ3MMX]].\nFor $a \\in A$, the *Gelfand transform* of $a$ is the function\n$$\\hat a \\colon \\Delta(A) \\to \\C, \\qquad \\hat a(\\varphi) = \\varphi(a).$$\nThe *Gelfand transform* (or Gelfand representation) of $A$ is the map $\\Gamma \\colon A \\to C_0(\\Delta(A))$, $a \\mapsto \\hat a$, into the [[continuous functions vanishing at infinity|N59BW1QFD1]] on $\\Delta(A)$.\n\n## Remarks\n\nEach $\\hat a$ is continuous by the definition of the weak-* topology, and it vanishes at infinity.\nThe map $\\Gamma$ is an algebra homomorphism with $\\norm{\\hat a}_\\infty \\le \\norm{a}$.\nIf $A$ has a unit, then $\\hat a(\\Delta(A)) = \\sigma(a)$, the spectrum of $a$, so $\\norm{\\hat a}_\\infty$ is the spectral radius of $a$.\nThe kernel of $\\Gamma$ is the Jacobson radical of $A$.\n\nExample: let $A = \\ell^1(\\Z)$ with convolution $(a * b)_n = \\sum_k a_k b_{n-k}$.\nFor $z$ in the unit circle, $\\varphi_z(a) = \\sum_n a_n z^n$ is linear, nonzero, and multiplicative by the Cauchy product formula, so it is a character.\nEvery character has this form, and $z \\mapsto \\varphi_z$ identifies $\\Delta(A)$ with the circle.\nThe Gelfand transform of $a$ is the absolutely convergent Fourier series $\\hat a(z) = \\sum_n a_n z^n$.\n\nNon-example: the construction needs $A$ to be commutative.\nThe algebra $M_2(\\C)$ has no characters at all.\nA character vanishes on commutators, and the commutators span the trace-zero matrices.\nSo its kernel is a nonzero two-sided ideal of the simple algebra $M_2(\\C)$, hence everything.\nSo for $M_2(\\C)$ the \"Gelfand transform\" would map into functions on the empty set and lose all information.\n",
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