Any homology sphere is [stably parallelizable](https://mathoverflow.net/questions/242412/are-homology-spheres-stably-trivial), hence nullcobordant. However, rational homology spheres need not be nullcobordant, as the example of the Wu manifold shows, which generates $\text{torsion}({\Omega^{\text{SO}}\_{5}}) \cong \mathbb Z/2\mathbb Z$. This motivates the following question.


**Which classes in $\Omega^{\text{SO}}\_{\ast}$ can be represented by rational homology spheres?**


Of course, any such class is torsion, as all its composite Pontryagin numbers, as well as its signature, vanish.



Oriented cobordism classes represented by rational homology spheres
The necessary condition pointed out by Jens Reinhold is also sufficient: any torsion class $x = [M] \in \Omega^{SO}\_d$ admits a representative where $M$ is a rational homology sphere.


**EDIT**: This is Theorem 8.3 in [$\Lambda$-spheres](http://sites.mathdoc.fr/PMO/PDF/B_BARGE_61_7405.pdf) by Barge, Lannes, Latour, and Vogel. They also calculate the group of rational homology spheres up to rational h-cobordism, and more. I'll leave my argument below:


To prove this, we first dispense with low-dimensional cases: in any dimension $d < 5$ the only torsion class is $0 = [S^d]$. The high dimensional case follows from Claims 1 and 2 below.


I'll write $MX$ for the Thom spectrum of a map $X \to BO$ and $\Omega^X\_d \cong \pi\_d(MX)$ for the bordism group of smooth $d$-manifolds equipped with $X$-structure. Representatives are smooth closed $d$-manifolds $M$ with some extra structure, which includes a continuous map $f: M \to X$.


**Claim 1**: if $d \geq 5$ and $X$ is simply connected and rationally $\lfloor d/2 \rfloor$-connected, then any class in $\Omega^X\_d$ admits a representative where $M$ is a rational homology sphere.


**Claim 2**: There exists a simply connected space $X$ such that $\widetilde{H}\_\*(X;\mathbb{Z}[\frac12]) = 0$, and map $X \to BSO$ such that the image of the induced map $\Omega^X\_d = \pi\_d(MX) \to \pi\_d(MSO) = \Omega\_d^{SO}$ is precisely the torsion subgroup, for $d > 0$.


*Proof of Claim 1*: Starting from an arbitrary class in $\Omega^X\_d$ we can use surgery to improve the representative. Since $X$ is simply connected and $d > 3$ we can use connected sum and then surgery on embeddings $S^1 \times D^{d-1} \hookrightarrow M$ to make $M$ simply connected. Slightly better, such surgeries can be used to make the map $M \to X$ be 2-connected, meaning that its homotopy fibers are simply connected. From now on we need not worry about basepoints and will write $\pi\_{k+1}(X,M) = \pi\_k(\mathrm{hofib}(M \to X))$. These are abelian groups for all $k$.


If there exists a $k < \lfloor d/2\rfloor$ with $\widetilde{H}\_k(M;\mathbb{Q}) \neq 0$ we can choose $\lambda \in H\_k(M;\mathbb{Q})$ and $\mu \in H\_{d-k}(M;\mathbb{Q})$ with intersection number $\lambda \cdot \mu \neq 0$. If $d = 2k$ for even $k$ we can additionally assume $\lambda \cdot \lambda = 0$, since the signature of $M$ vanishes. The rational Hurewicz theorem implies that $\pi\_k(M) \otimes \mathbb{Q} \to H\_k(M;\mathbb{Q})$ is an isomorphism, and the long exact sequence implies that $\pi\_{k+1}(X,M) \otimes \mathbb{Q} \to \pi\_k(M)\otimes\mathbb{Q}$ is surjective. After replacing $\lambda$ by a non-zero multiple, we may therefore assume that it admits a lift to $\pi\_{k+1}(X,M)$. Such an element can be represented by an embedding $j: S^k \times D^{d-k} \hookrightarrow M$, together with a null homotopy of the composition of $j$ with $M \to X$. In the case $k < d/2$ this follows from Smale-Hirsh theory, in the case $d = 2k$ we must also use $\lambda \cdot \lambda = 0$ to cancel any self-intersections. (Actually there could also be obstructions to this in the case $d=2k$ for odd $k$, but those obstructions vanish after multiplying $\lambda$ by 2.) The embedding and the nullhomotopy gives the necessary data to perform surgery on $M$ and to promote the surgered manifold to a representative for the same class in $\Omega^X\_d$.


Performing the surgery gives a new manifold $M'$ where $H\_k(M';\mathbb{Q})$ has strictly smaller dimension than $H\_k(M;\mathbb{Q})$ and $\widetilde{H}\_\*(M';\mathbb{Q}) = 0$ for $\* < k$. This is seen in the same way as in [Kervaire-Milnor](https://www.jstor.org/stable/1970128). The case $d > 2k+1$ is easy, similar to their Lemma 5.2. In the case $d = 2k+1$ the diagram on page 515 shows that we can kill the homology class $j[S^k]$ and at worst create some new torsion in $H\_k(M')$. In the case $d = 2k$ the diagram on page 527 shows that we can kill the homology class $j[S^k]$ and at worst create some new torsion in $H\_{k-1}(M')$.


In finitely many steps we arrive at a representative where $\widetilde{H}\_k(M;\mathbb{Q}) = 0$ for all $k \leq \lfloor d/2\rfloor$. Poincaré duality then implies that $H\_\*(M;\mathbb{Q}) \cong H\_\*(S^d;\mathbb{Q})$. $\Box$.


*Proof of Claim 2*: Finiteness of the stable homotopy groups of spheres implies that $\pi\_d(MX)$ is a torsion group for $d > 0$ for any such $X$. Therefore we can never hit more than the torsion in $\pi\_d(MSO)$, all of which is exponent 2 by [Wall's theorem](http://dx.doi.org/10.2307/1970136). The difficult part is to construct an $X$ where all torsion is hit.


The non-trivial based map $S^1 \to BO$ factors through $\mathbb{R} P^\infty \to BO$, whose image in mod 2 homology generates the Pontryagin ring $H\_\*(BO;\mathbb{F}\_2)$. We can freely extend to double loop maps
$$\Omega^2 S^3 \to \Omega^2 \Sigma^2 \mathbb{R}P^\infty \to BO$$
where the second map then induces a surjection on mod 2 homology. Both $\Omega^2 \Sigma^2 \mathbb{R}P^\infty$ and $BO$ split as $\mathbb{R} P^\infty$ times their 1-connected cover, so the induced map of 1-connected covers $\tau\_{\geq 2}(\Omega^2 \Sigma^2 \mathbb{R}P^\infty) \to \tau\_{\geq 2}(BO) = BSO$ also induces a surjection on mod 2 homology.


Now let $X = \tau\_{\geq 2}(\Omega^2 \Sigma^2 \mathbb{R}P^\infty)$ with the map to $BSO$ constructed above. Take 1-connected covers of the double loop maps above, Thomify, 2-localize, and use the Hopkins-Mahowald theorem to get maps of $E\_2$ ring spectra
$$H \mathbb{Z} \_{(2)} \to MX\_{(2)} \to MSO\_{(2)}.$$
(See e.g. section 3 of [this paper](https://arxiv.org/abs/2004.06889).)
We can view $MX\_{(2)} \to MSO\_{(2)}$ as a map of $H\mathbb{Z}\_{(2)}$-module spectra, and hence $MX/2 \to MSO/2$ as a map of $H\mathbb{F}\_2$-module spectra. The induced map $H\_\*(MX/2;\mathbb{F}\_2) \to H\_\*(MSO/2;\mathbb{F}\_2)$ is still surjective (it looks like two copies of $H\_\*(X;\mathbb{F}\_2) \to H\_\*(BSO;\mathbb{F}\_2))$, and inherits the structure of a module map over the mod 2 dual Steenrod algebra $\mathcal{A}^\vee = H\_\*(H\mathbb{F}\_2;\mathbb{F}\_2)$. Both modules are free, because any $H\mathbb{F}\_2$-module spectrum splits as a wedge of suspensions of $H\mathbb{F}\_2$. In fact the Hurewicz homomorphism $\pi\_\*(MX/2) \to H\_\*(MX/2;\mathbb{F}\_2)$ induces an isomorphism
$$\mathcal{A}^\vee \otimes \pi\_\*(MX/2) \to H\_\*(MX/2;\mathbb{F}\_2),$$
and similarly for $MSO$. Therefore the map $\pi\_\*(MX/2) \to \pi\_\*(MSO/2)$ may be identified with the map obtained by applying $\mathbb{F}\_2 \otimes\_{\mathcal{A}^\vee} (-)$ to the map on homology, showing that the induced map $\pi\_\*(MX/2) \to \pi\_\*(MSO/2)$ is also surjective. Now any 2-torsion class $x \in \pi\_d(MSO)$ comes from $\pi\_{d+1}(MSO/2)$, hence from $\pi\_{d+1}(MX/2)$ and in particular from $\pi\_d(MX)$. $\Box$


