Cobordism cohomology of Lie groups Are there any results about cobordism cohomology of Lie groups？For example, $\mathrm{MU}^*(\mathrm{SU}(n))$.
 A: Firstly, for any space $X$ we have an Atiyah-Hirzebruch spectral sequence
$$ H^p(X;MU^q) \Longrightarrow MU^{p+q}(X). $$
The differentials are always torsion-valued, essentially because the higher stable homotopy groups of spheres are torsion groups.  Now suppose that $X$ is a finite complex and that each group $H^p(X;\mathbb{Z})$ is free abelian.  Then the differentials must be zero, so the spectral sequence collapses.  This means that we can choose a homogeneous basis $e_1,\dotsc,e_m$ for $H^*(X;\mathbb{Z})$, then it will be possible to choose elements $f_i\in MU^*(X)$ that reduce to $e_i$ under the standard map $MU\to H$, and for any such choices, the list $f_1,\dotsc,f_m$ will be a basis for $MU^*(X)$ as a free module over $MU^*$.  This applies to many Lie groups, in particular to $U(n)$ and $SU(n)$ (but not $PSU(n)$).
One might want to have a canonical choice of generators.  In the case of $U(n)$ we can proceed as follows.  A theorem of Haynes Miller gives a canonical stable splitting of $\Sigma^\infty_+U(n)$, with the bottom piece being $S^0$ and the next piece being $\Sigma\mathbb{C}P^{n-1}_+$.  There is a canonical basis $a_0,\dotsc,a_{n-1}$ for $MU^*(\Sigma\mathbb{C}P^{n-1}_+)$ with $|a_i|=2i+1$.  The splitting allows us to regard these as elements of $MU^*(U(n))$, and the full ring $MU^*(U(n))$ is just the exterior algebra over $MU^*$ generated by these elements.
For groups $G$ such that $H^*(G;\mathbb{Z})$ has torsion, it will be easier to calculate the Morava $K$-theories $K(p,n)^*(G)$ than $MU^*(G)$.  Many problems that you might want to solve using $MU^*(G)$ can be solved using the whole family of rings $K(p,n)^*(G)$ instead.
