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.

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 $F(X,X)=0$.

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 this page.

Is there a similar description of $F_O(X,Y)$?


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.)

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  • $\begingroup$ Hi Neil. You were right, of course. As I've just got round to checking, the proof on the page I linked goes through $\mathbb{R}$ replacing $\mathbb{C}$. $\endgroup$ – Mark Grant Feb 10 '12 at 21:17

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