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  "body": "\n## Definition\n\nFor $n \\ge 0$, let $\\gamma_n$ be the [[universal $n$-plane bundle|1HDH716Z24]] over the [[classifying space|WSZ40H5E4T]] $BO(n)$, and let $\\varepsilon$ be the trivial line bundle.\nLet $(B_n, f_n, g_n)_{n \\ge 0}$ be a sequence of spaces $B_n$ with maps $f_n \\colon B_n \\to BO(n)$ and $g_n \\colon B_n \\to B_{n+1}$ such that $f_{n+1} \\circ g_n$ classifies the vector bundle $\\xi_n \\oplus \\varepsilon$, where $\\xi_n = f_n^* \\gamma_n$.\nWrite $\\operatorname{Th}(\\xi)$ for the [[Thom space|5MNYT3AFPH]] of a bundle $\\xi$.\nThe *Thom spectrum* of this sequence is the [[spectrum|1GHF4EKBXS]] $M$ with spaces $M_n = \\operatorname{Th}(\\xi_n)$.\nLet $\\Sigma$ be the reduced [[suspension|2VD7GR64JJ]].\nThe structure maps of $M$ are the composites $\\Sigma \\operatorname{Th}(\\xi_n) \\cong \\operatorname{Th}(\\xi_n \\oplus \\varepsilon) \\to \\operatorname{Th}(\\xi_{n+1})$.\nHere the second map is induced by the bundle map $\\xi_n \\oplus \\varepsilon \\to \\xi_{n+1}$ that covers $g_n$.\n\n## Remarks\n\nExample: $B_n = BO(n)$ with $f_n = \\id$ gives the spectrum $MO$.\nHere $g_n \\colon BO(n) \\to BO(n+1)$ classifies $\\gamma_n \\oplus \\varepsilon$, so the condition holds by construction.\nTaking $B_n = BSO(n)$ gives $MSO$.\nFor complex bundles one uses $BU(n)$ and the universal complex bundle, whose Thom space is placed in degree $2n$; this gives the spectrum $MU$ of complex cobordism.\n\nNon-example: keep the spaces $\\operatorname{Th}(\\gamma_n)$ but take all structure maps constant.\nThis is still a spectrum, but its structure maps are not induced by bundle maps, so it is not the Thom spectrum $MO$.\nIndeed, all its stable homotopy groups vanish, while $\\pi_0 MO \\cong \\Z/2$.\n\nMore generally one can form the Thom spectrum of a map $f \\colon B \\to BO$, taking for $B_n$ the preimage of $BO(n)$ under $f$, after replacing $f$ by a fibration.\n",
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