Formal group law of unoriented cobordism - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T18:10:50Z http://mathoverflow.net/feeds/question/74770 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/74770/formal-group-law-of-unoriented-cobordism Formal group law of unoriented cobordism Mark Grant 2011-09-07T16:29:38Z 2011-09-07T23:06:12Z <p>It is well known that the formal group law $F_U$ of complex cobordism, expressing the Euler class of a tensor product of complex line bundles, is universal.</p> <p>Also, the formal group law $F_O$ of unoriented cobordism, expressing the Euler class of a tensor product of real line bundles, is universal among formal group laws in characteristic 2 with the property that $F(X,X)=0$.</p> <p>There is a nice description of $F_U$ in terms of manifold generators, due to Buchstaber: $$F_U(X,Y) = \frac{\sum_{i,j\geq 0} [H_{ij}]\;X^iY^j}{\left(\sum_{r\geq 0}[\mathbb{C}P^r] X^r\right) \left(\sum_{s\geq 0}[\mathbb{C}P^s] Y^s\right)}$$ where the $H_{ij}$ are Milnor hypersurfaces. Here I am quoting <a href="http://www.map.him.uni-bonn.de/Formal_group_laws_and_genera#Formal_group_law_of_geometric_cobordisms" rel="nofollow">this page</a>.</p> <p>Is there a similar description of $F_O(X,Y)$?</p> http://mathoverflow.net/questions/74770/formal-group-law-of-unoriented-cobordism/74810#74810 Answer by Neil Strickland for Formal group law of unoriented cobordism Neil Strickland 2011-09-07T23:06:12Z 2011-09-07T23:06:12Z <p>I'm fairly sure you just get the same formula, with $\mathbb{C}P^k$ replaced by $\mathbb{R}P^k$, and $H_{ij}$ replaced by the corresponding real hypersurface in $\mathbb{R}P^i\times\mathbb{R}P^j$. The proof of the equivalent formula $$\left(\sum [\mathbb{R}P^r]\;X^r\right) \left(\sum [\mathbb{R}P^s]\;Y^s\right) F_O(X,Y) = \sum H_{ij} X^i Y^j$$ is quite direct and geometric. (I might come back and write more tomorrow.)</p>