Explicit cobordism between Wu manifold and Dold manifold P(1,2)? The Wu manifold $SU(3)/SO(3)$ and the Dold manifold $P(1,2)$, the latter being defined as $(S^1\times \mathbb{C}P^2) / (p, x) \sim (-p, \overline{x})$, are cobordant because they are both generators of $Ω_{SO}^5$. As the Wu manifold is equipped with an $SU(3)$-action, and the Dold manifold $P(1,2)$ is a fiber bundle over $S^1$ with fiber equipped with an $SU(3)$-action, I am curious is there an explicit construction of this cobordism?
Probably a more well-defined question would be: as the Wu manifold and the Dold manifold are both equipped with $SO(3)$-actions, can there be a $6$-manifold equipped with $SO(3)$ action to give the cobordism?
 A: There is a cobordism relating the two manifolds of the simplest possible kind: it consists of the attachment of a single 6-dimensional 2-handle. Pick the Dold manifold $\bf D$, thicken it to a 6-manifold ${\bf D}\times [-1,1]$, and attach a 2-handle to ${\bf D} \times\{1\}$ along the unique loop (up to isotopy) that generates the fundamental group $\mathbb Z$. The new boundary that you get is diffeomorphic to the Wu manifold ${\bf W}$. That's all.
To prove this, we need to verify that if we replace $S^1 \times D^4 \subset {\bf D}$ with $D^2 \times S^3$ we get ${\bf W}$. (Here $S^1 \times D^4$ is a tubular neighbourhood of the loop generating the fundamental group of $\mathbf D$.) The first thing that we notice is that the result of this surgery is clearly simply connected, and that sounds promising.
 Exercise: The complement ${\bf D} \setminus (S^1 \times D^4)$ is diffeomorphic to a disc bundle over the non-orientable 3-manifold $M = S^2 \tilde \times S^1$ (this is the non-orientable 3-manifold fibering over $S^1$ with fiber $S^2$). We denote this disc bundle as $D^2 \tilde \times M$, or as $D^2 \tilde \times (S^2 \tilde \times S^1)$.
 Observation: The non-orientable 3-manifold $M$ is Poincaré dual to $w_2({\bf D})\in H^2({\bf D}, \mathbb Z_2)$. The fact that $M$ is non-orientable (and there is no way to find an orientable representative) is a manifestation of the fact that ${\bf D}$ has no ${\rm Spin}^c$ structure, that is $w_2$ has no integral lift.
We pick from the 1965 paper Simply connected five-manifolds of Barden a simple description of ${\bf W}$ as the union of two copies of the (unique) orientable non-trivial $D^3$-bundle over $S^2$, that we may indicate as $D^3 \tilde\times S^2$. The boundary of this bundle is the unique orientable non-trivial $S^2$-bundle over $S^2$, usually indicated as $S^2 \tilde\times S^2$.
One of the first exercises one proves in dimension 4 is that $S^2 \tilde \times S^2$ is diffeomorphic to the connected sum $\mathbb{CP}^2 \# \overline{\mathbb{CP}}^2$. The two copies of $D^3 \tilde \times S^2$ are glued to each other along a diffeomorphism of $\mathbb{CP}^2 \# \overline{\mathbb{CP}}^2$ that acts homologically like $(p,q) \mapsto (p,-q)$, that is it reverses the generator of the second factor in the connected sum. Consider the $S^3$ that separates the two factors $\mathbb{CP}^2$ and  $\overline{\mathbb{CP}}^2$. Its normal bundle is trivial, so it has a tubular neighbourhood $S^3\times D^2$.
 Exercise : The complement ${\bf W} \setminus (S^3 \times D^2)$ is again diffeomorphic to $D^2 \tilde \times M$.
Therefore ${\bf W}$ is obtained from ${\bf D}$ by surgery as stated above.
