TITLE: Cell structures of Dold manifold and Wu manifold
QUESTION [10 upvotes]: In Dold's 1956 paper Erzeugende der Thomschen Algebra N, Dold studied the Dold manifold $P(m,n)=(S^m\times\mathbb{CP}^n)/\tau$ where $\tau$ acts as $-1$ on $S^m$ and a complex conjugation on $\mathbb{CP}^n$.
In section B of that paper, Dold gave the cell structure of $P(m,n)$, the cells were denoted by $(C_i,D_j)$ where $C_i$ has dimension $i$ and $D_j$ has dimension $2j$, $(C_i,D_j)$ has dimension $i+2j$.
In equation (4) of that paper, Dold gave the boundary relations between the cells:
$$\partial(C_i,D_j)=(1+(-1)^{i+j})(C_{i-1},D_j),$$
$$\partial(C_0,D_j)=0$$
for $i=1,2,\dots,m$, $j=0,1,\dots,n$.
I determine the cell structure of the 5d Dold manifold $P(1,2)$,
following Dold's results, 
$$\partial(C_1,D_0)=\partial(C_1,D_2)=0,$$
$$\partial(C_1,D_1)=2(C_0,D_1),$$
$$\partial(C_0,D_j)=0$$ 
for $j=0,1,2$.
So there is exactly one cell in each dimension from 0 to 5 and there is only one nonzero boundary map from the 3-cell to the 2-cell.
Both Dold manifold $P(1,2)$ and Wu manifold $X=SU(3)/SO(3)$ generate $\Omega_5^{SO}$, actually they are bordant.
But $P(1,2)$ and $X$ have different cohomology groups 
$$H^*(P(1,2),\mathbb{Z}/2)=\mathbb{Z}/2[c,d]/(c^2,d^3)$$
where $|c|=1$, $|d|=2$.
$$H^*(X,\mathbb{Z}/2)=\Lambda(w_2(X),w_3(X)).$$
My question: What is the cell structure of Wu manifold? Is there a nonzero boundary map from the 2-cell to the 1-cell?
Thank you!

REPLY [5 votes]: The 1965 paper Simply connected five-manifolds of Barden contains a simple topological description of the Wu manifold. The Wu manifold decomposes into two copies of the (unique) orientable non-trivial $D^3$-bundle over $S^2$. Since a $D^3$-bundle over $S^2$ decomposes into one 0-handle and one 2-handle, we deduce that the Wu manifold decomposes into one 0-handle, one 2-handle, one 3-handle, and one 5-handle.
So there is a cell decomposition with one 0-cell, one 2-cell, one 3-cell, and one 5-cell.