Let $\Omega U(n)$ be the loop space of $U(n)$. Is it true that the cohomology groups $H^\*(\Omega U(n); \mathbb{Z})$ are torsion-free? How can one calculate these groups?



The cohomology groups of $\Omega U(n)$
Given a one-dimensional subspace $L<\mathbb{C}^n$ and a number $z\in S^1\subset\mathbb{C}$ we can define $\rho(L)(z)\colon\mathbb{C}^n\to\mathbb{C}^n$ by $\rho(L)(z)=z$ on $L$ and $\rho(L)(z)=1$ on $L^\perp$. This defines $\rho(L)\colon S^1\to U(n)$ with $\rho(L)(1)=1$, or in other words $\rho(L)\in\Omega U(n)$. We thus have $\rho\colon\mathbb{C}P^{n-1}\to\Omega U(n)$, giving 
$$ \rho\_\*\colon H\_\*(\mathbb{C}P^{n-1}) = \mathbb{Z}\{b\_0,\dotsc,b\_{n-1}\} \to H\_\*(\Omega U(n)) $$
On the other hand, the space $\Omega U(n)=\Omega^2BU(n)$ is a double loop space and so is a homotopy commutative $H$-space, so $H\_\*(\Omega U(n))$ is a commutative ring, and it is not hard to see that the element $b\_0\in H\_0(\Omega U(n))$ is invertible. Thus, $\rho\_\*$ extends to give a ring map 
$$ \phi\colon\mathbb{Z}[b\_0,\dotsc,b\_{n-1}][b\_0^{-1}] \to H\_\*(\Omega U(n)). $$
It is a standard theorem that this map $\phi$ is an isomorphism. This can be proved using the Serre spectral sequence for the fibration $\Omega U(n)\to PU(n)\to U(n)$, for example. (Here $PU(n)$ is the space of based paths, which is contractible.) It is technically more convenient to study $SU(n)$ rather than $U(n)$, as $\Omega SU(n)$ is connected and of finite type, whereas $\Omega U(n)\simeq \mathbb{Z}\times\Omega SU(n)$. As the homology of $\Omega U(n)$ is free, the cohomology is just the dual and is torsion free. However, $H^\*(\Omega U(n))$ is not actually free; even in dimension $0$ we have $\prod\_{n\in\mathbb{Z}}\mathbb{Z}$, which is not a free abelian group.


