Let $ X $ be a $ n $ - dimentional oriented closed real manifold ( i.e : compact and without boundary ).


Can you tell me how to show that,
$$ \mathrm{CH}\_k (X) \otimes\_{ \mathbb{Z} } \mathbb{Q} \simeq \Omega\_k (X) \otimes\_{ \mathbb{Z} } \mathbb{Q} $$ where,
$ \mathrm{CH}\_k ( X ) $ is the Chow group that is the free abelian group on the set of real $ k $ - submanifolds $ Z \subset X $ of $ X $, with $ 0 \leq k \leq n $, modulo rational equivalence, and, $ \Omega\_k (X) $ is the oriented bordism group defined as the set of all isomorphism classes of $ k $ - manifolds $ M \to X $ modulo bordism, where $ M \to X $ is bordant to $ N \to X $ if there is a $ W $ with $ \partial W = M - N $. ( Here, $ M - N $ denotes the disjoint union of $ M $ and $ N $ with the orientation of $ N $ reversed ).


Here, I would like to point out that, the oriented bordism group $ \Omega\_k (X) $ is the one which is isomorphic to $ MSO\_k (X) := \displaystyle \lim\_ {\longrightarrow \\ \ \ k} \pi\_{n + k} (MSO(k) \wedge X\_+) $ if I'm not mistaken, and should not be confused with the unoriented bordism group $ \mathfrak {M} \_k (X) $ which is isomorphic to $ MO\_k(X) := \displaystyle \lim\_{\longrightarrow \\ \ \ k} \pi\_ {n + k} (MO(k) \wedge X\_+) $.


Thanks in advance for your help.



How to show that, $ \mathrm{CH}_k (X) \otimes_{ \mathbb{Z} } \mathbb{Q} \simeq \Omega_k (X) \otimes_{ \mathbb{Z} } \mathbb{Q} $?
I haven't checked the book by Levine and Morel, but there is a short arXiv article by Levine at <https://arxiv.org/pdf/math/0304206.pdf>. There he writes $\mathbb{L}$ for the Lazard ring, which is the same as $MU\_\*$, and is polynomial over $\mathbb{Z}$ on countably many generators. There is a natural map $\Omega\_\*(X)\otimes\_{\mathbb{L}}\mathbb{Z}\to CH\_\*(X)$, and Levine's Theorem 5.13 says that this becomes an isomorphism after tensoring with $\mathbb{Q}$. It is important here that we are tensoring over $\mathbb{L}$; as $\mathbb{Z}$ is much smaller than $\mathbb{L}$, we find that $\Omega\_\*(X)\otimes\_{\mathbb{L}}\mathbb{Z}$ is much smaller than $\Omega\_\*(X)$, consistent with the fact that $CH\_\*(X)$ is also fairly small. Note also that the claim is an isomorphism of graded rings and relies on the graded tensor product; there is no natural formulation that gives $CH\_k(X)$ for a single value of $k$.


Levine's article only gives a brief indication of a proof of Theorem 5.13; I'm afraid I can't help you with that. In the analogous topological situation, however, we can just recall that $(E\_\*X)\otimes\mathbb{Q}=H\_\*(X;\mathbb{Q})\otimes E\_\*$ for all spectra $E$ and $X$. This gives $MU\_\*(X)\otimes\mathbb{Q}=H\_\*(X;\mathbb{Q})\otimes\mathbb{L}$ and so $(MU\_\*(X)\otimes\_{\mathbb{L}}\mathbb{Z})\otimes\mathbb{Q}=H\_\*(X;\mathbb{Q})$.


