When is the orientable double cover of a product of non-orientable surfaces spin? Let $M_{k,l}$ denote the orientable double cover of the non-orientable four-manifold $k\mathbb{RP}^2\times l\mathbb{RP}^2$; here $k\mathbb{RP}^2$ denotes the connected sum of $k$ copies of $\mathbb{RP}^2$.

For which $k$ and $l$ is $M_{k,l}$ a spin manifold?

If $\pi : M_{k,l} \to k\mathbb{RP}^2\times l\mathbb{RP}^2$ denotes covering map, note that $\pi^*T(k\mathbb{RP}^2\times l\mathbb{RP}^2) \cong TM_{k,l}$ so
\begin{align*}
w_2(M_{k,l}) &= \pi^*w_2(k\mathbb{RP}^2\times l\mathbb{RP}^2)\\ 
&= \pi^*(w_2(k\mathbb{RP}^2) + w_1(k\mathbb{RP}^2)w_1(l\mathbb{RP}^2) + w_2(l\mathbb{RP}^2)).
\end{align*}
I have only been able to determine the answer for one case: when $k = l = 1$, we have $M_{1,1} = \operatorname{Gr}(2, 4)$ which is not spin, see here for example.
Added Later: In this paper (English translation here), Habegger shows that the manifold $(S^2\times S^2)/\mathbb{Z}_2$, where $\mathbb{Z}_2$ acts on $S^2\times S^2$ by $(x, y) \mapsto (-x, -y)$, is not spin; note that $(S^2\times S^2)/\mathbb{Z}_2$ is precisely $M_{1,1}$. He does so by observing that the diagonal embedding $S^2 \hookrightarrow S^2\times S^2$ descends to an embedding $\mathbb{RP}^2 \hookrightarrow M_{1,1}$ which has self-intersection $1$. More generally, one can show that the diagonal embedding $\Sigma_{k-1} \hookrightarrow \Sigma_{k-1}\times\Sigma_{k-1}$ descends to an embedding $k\mathbb{RP}^2 \hookrightarrow M_{k,k}$ with self-intersection $\chi(k\mathbb{RP}^2) = 2 - k$. It follows that for $k$ odd, the manifold $M_{k,k}$ is not spin.
 A: The Gysin sequence associated to the double cover $\pi : M_{k,l} \to k\mathbb{RP}^2\times l\mathbb{RP}^2$ is given by
$$\dots \xrightarrow{\pi_*} H^1(k\mathbb{RP}^2\times l\mathbb{RP}^2; \mathbb{Z}_2) \xrightarrow{w_1(L)\cup} H^2(k\mathbb{RP}^2\times l\mathbb{RP}^2; \mathbb{Z}_2) \xrightarrow{\pi^*} H^2(M_{k,l}; \mathbb{Z}_2) \xrightarrow{\pi_*} \dots$$
where $L$ is the real line bundle determined by the double cover $\pi$. By exactness, we see that $0 = w_2(M_{k,l}) = \pi^*(w_2(k\mathbb{RP}^2\times l\mathbb{RP}^2))$ if and only if $w_2(k\mathbb{RP}^2\times l\mathbb{RP}^2)$ is of the form $w_1(L)\cup \eta$ for some $\eta \in H^1(k\mathbb{RP}^2\times l\mathbb{RP}^2; \mathbb{Z}_2)$.
As $\pi$ is the orientable double cover, $L$ is the determinant line bundle on $k\mathbb{RP}^2\times l\mathbb{RP}^2$, i.e. the line bundle with
$$w_1(L) = w_1(k\mathbb{RP}^2\times l\mathbb{RP}^2) = w_1(k\mathbb{RP}^2) + w_1(l\mathbb{RP}^2).$$
So $M_{k,l}$ is spin if and only if there is $\eta \in H^1(k\mathbb{RP}^2\times l\mathbb{RP}^2; \mathbb{Z}_2)$ such that
$$(w_1(k\mathbb{RP}^2) + w_1(l\mathbb{RP}^2))\cup\eta = w_2(k\mathbb{RP}^2) + w_1(k\mathbb{RP}^2)w_1(l\mathbb{RP}^2) + w_2(l\mathbb{RP}^2). \tag{1}$$
To determine when this equation has a solution, note that the $\mathbb{Z}_2$ cohomology ring of $k\mathbb{RP}^2\times l\mathbb{RP}^2$ is generated by degree one elements $a_1, \dots, a_k, b_1, \dots, b_l$ and degree two elements $a, b$ subject to the conditions $a_i^3 = a_ia_{i'} = b_jb_{j'} = b_j^3 = 0$, $a_i^2 = a$, and $b_j^2 = b$; here $i, i' \in \{1, \dots, k\}$ are distinct and $j, j' \in \{1, \dots, l\}$ are distinct. In terms of these generators, we have
\begin{align*}
w_1(k\mathbb{RP}^2) &= a_1 + \dots + a_k\\
w_1(l\mathbb{RP}^2) &= b_1 + \dots + b_k\\
w_2(k\mathbb{RP}^2) &= ka\\
w_2(l\mathbb{RP}^2) &= lb.
\end{align*}
An arbitrary element $\eta \in H^1(k\mathbb{RP}^2\times l\mathbb{RP}^2)$ takes the form $\eta = x_1a_1 + \dots + x_ka_k + y_1b_1 + \dots + y_lb_l$ for some $x_1, \dots, x_k, y_1, \dots, y_l \in \mathbb{Z}_2$. So we have
\begin{align*}
&\ (w_1(k\mathbb{RP}^2) + w_1(l\mathbb{RP}^2))\cup\eta\\ 
=&\ (a_1 + \dots + a_k + b_1 + \dots + b_k)\cup(x_1a_1 + \dots + x_ka_k + y_1b_1 + \dots + y_lb_l)\\
=&\ x_1a_1^2 + y_1a_1b_1 + \dots + y_la_1b_l + \dots + x_ka_k^2 + y_1a_kb_1 + \dots + y_la_kb_l\\
&\ + x_1a_1b_1 + \dots + x_ka_kb_1 + y_1b_1^2 + \dots + x_1a_1b_l + \dots + x_ka_kb_l + y_lb_l^2\\
=&\ (x_1 + \dots + x_k)a + \sum_{i=1}^k\sum_{j=1}^l(x_i + y_j)a_ib_j + (y_1 + \dots + y_l)b
\end{align*}
while
\begin{align*}
&\ w_2(k\mathbb{RP}^2) + w_1(k\mathbb{RP}^2)w_1(l\mathbb{RP}^2) + w_2(l\mathbb{RP}^2)\\
=&\ ka + (a_1 + \dots + a_k)(b_1 + \dots + b_l) + lb\\
=&\ ka + \sum_{i=1}^k\sum_{j=1}^la_ib_j + lb.
\end{align*}
Equating the coefficients of $a$, $b$, and $a_ib_j$ we obtain the following equations
\begin{align*}
x_1 + \dots + x_k &= k\\
y_1 + \dots + y_l &= l\\
x_i + y_j &= 1.
\end{align*}
Note that the equations $x_i + y_j = 1$ imply that $x_1 = \dots = x_k$ and $y_1 = \dots = y_l$ and that the two values are distinct. So there are two possible solutions:

*

*$x_1 = \dots = x_k = 0$ and $y_1 = \dots = y_l = 1$, or

*$x_1 = \dots = x_k = 1$ and $y_1 = \dots = y_l = 0$.

In the first case, the equation $y_1 + \dots + y_l = l$ is satisfied while $x_1 + \dots + x_k = k$ is satisfied if and only if $k$ is even. In the second case, the equation $x_1 + \dots + x_k = k$ is satisfied while $y_1 + \dots + y_l = l$ is satisfied if and only if $l$ is even.
In conclusion, equation $(1)$ has a solution if and only if $k$ is even, in which case $\eta = b_1 + \dots + b_l$, or $l$ is even, in which case $\eta = a_1 + \dots + a_k$. Hence, we have the following answer to the posed problem:

The manifold $M_{k,l}$ is spin if and only if $k$ or $l$ is even.

When $k$ or $l$ is even, it's not hard to see that $(1)$ has a solution. If $k$ is even, then $w_2(k\mathbb{RP}^2) = 0$. For any surface, its second Stiefel-Whitney class is equal to the square of its first Stiefel-Whitney class, so we have
\begin{align*}
w_2(k\mathbb{RP}^2) + w_1(k\mathbb{RP}^2)w_1(l\mathbb{RP}^2) + w_2(l\mathbb{RP}^2) &= w_1(k\mathbb{RP}^2)w_1(l\mathbb{RP}^2) + w_1(l\mathbb{RP}^2)^2\\
&= (w_1(k\mathbb{RP}^2) + w_1(l\mathbb{RP}^2))\cup w_1(l\mathbb{RP}^2).
\end{align*}
Note that this agrees with what we found above as in the $k$ even case we have $\eta = b_1 + \dots + b_l = w_1(l\mathbb{RP}^2)$. The case of $l$ even is completely analogous.

More generally, one could define $M_{k_1, \dots, k_d}$ to be the orientable double cover of the product of non-orientable surfaces $k_1\mathbb{RP}^2\times\dots\times k_d\mathbb{RP}^2$. The same argument can be used, however the outcome is not the same:

If $d > 2$, the manifold $M_{k_1, \dots, k_d}$ is never spin!

The issue is that there are now not only $x$ variables and $y$ variables, but variables which are coefficients of generators coming from the cohomology of the other factors. As before, all the coefficients of the generators from a single factor have to be equal, so we have a choice of an element of $\mathbb{Z}_2$ for each factor. However, each of these $d$ choices need to be different which is impossible if $d > 2$.
