Lens space $L^k(n) \equiv S^{2k-1}/\mathbb{Z}_n$: Stiefel-Whitney class and non/spin manifold

Question

Define the lens space $L^k(n) \equiv S^{2k-1}/\mathbb{Z}_n$.

1. What is the property of Stiefel-Whitney class $w_1(TM)$ and $w_2(TM)$ for $M= L^k(n)$?

2. What is the spin or nonspin manifold property? Is that true that we can show:

• odd $k$, odd $n$: spin?

• odd $k$, even $n$: nonspin?

• even $k$, odd $n$: spin?

• even $k$, even $n$: spin?

Answer 1

Since $M$ is orientable, we have $w_1(TM)=0$ for all $k,n$.

We know that $w_2(TM)\in H^2(M;\Bbb Z_2)$, and explicit computation shows that when $n$ is odd, we have $H^i(M;\Bbb Z_2)\cong 0$ for $1\le i\le 2k-2$. This means that $w_1(TM)=0=w_2(TM)$, so $M$ admits a spin structure.

Now, let's suppose $n=2m$ is even and $k\ge 2$ (as $k=1$ needs to be dealt separately). In this case, we have $H^i(M;\Bbb Z_2)\cong \Bbb Z_2$ for all $i$. We will use the ring homomorphism $\phi_\ast:H^\ast(M;\Bbb Z_{2m})\to H^\ast(M;\Bbb Z_2)$ induced by coefficient reduction to derive the mod-$2$ cup product structure and the action of Steenrod squares, and then we can compute the Wu classes and hence deduce all Stiefel-Whitney classes using the relation $$w_j=\sum_{i=0}^jSq^i(v_{j-i})\tag{$\ast$}$$ where $v_j$ is the $j$-th Wu class.

• ($m$ odd) We can deduce that $H^\ast (M;\Bbb Z_2)$ is generated by $\bar x=\phi_\ast(x)\in H^1(M;\Bbb Z_2)$ since $x^2=my$ gets reduced to the generator of $H^2(M;\Bbb Z_2)$, where $y$ generates $H^2(M;\Bbb Z_{2m})$. The cohomology ring is thus the same as that of real projective space. The Steenrod squares are given by $$Sq^i(\bar{x}^j)=\binom{j}{i}\bar x^{i+j}$$ Using this information, we can compute the Wu classes $$v_i\bar x^{2k-1-i}=\binom{2k-1-i}{i}\bar x^{2k-1}\implies v_i=\binom{2k-1-i}{i}\bar x^i$$ Finally we apply $(\ast)$ for $j=2$ to get $$w_2(TM)=\binom{2k-3}{2}\bar x^2+(2k-2)\bar x^2\equiv k(2k-1)\bar x^2$$ which implies that $$w_2(TM)=0\Leftrightarrow k\equiv 0\pmod{2}$$

• ($m$ even) $H^\ast(M;\Bbb Z_2)$ is generated by $\bar x\in H^1$ and $\bar y\in H^2$ because $x^2=my$ gets reduced to $0$ by the induced homomorphism. We also know that $\bar y^i$ generates $H^{2i}$ and $\bar x\bar y^i$ generates $H^{2i+1}$. To compute $w_2(TM)$, we just need $v_1$ and $v_2$: $$v_1\bar y^{k-1}=Sq^1(\bar y^{k-1})=\beta(\bar y^{k-1})=0\implies v_1=0$$ since the bockstein $\beta$ is a composition $H^{2k-2}(M;\Bbb Z_2)\to H^{2k-1}(M;\Bbb Z)\to H^{2k-1}(M;\Bbb Z_2)$, and the group in the middle is isomorphic to $\Bbb Z$. $$v_2\bar x\bar y^{k-2}=Sq^2(\bar x\bar y^{k-2})=Sq^0(\bar x)Sq^2(\bar y^{k-2})=\binom{2k-4}{2}\bar x\bar y^{k-1}\equiv k(2k-1)\bar x\bar y^{k-1}$$ which implies that $v_2\equiv k(2k-1)\bar y$. Substituting into $(\ast)$, we get $$w_2(TM)=k(2k-1)\bar y=0\Leftrightarrow k\equiv 0\pmod{2}$$

If $k=1$, then we have a rather simple space $S^1/{\Bbb Z_n}$. The second cohomology group vanishes since the space is 1-dimensional. Therefore, $w_2=0$.

In summary,

• If $n$ is odd, then $M$ admits a spin structure (the parity of $k$ doesn't matter in this case)
• If $n$ is even, then $M$ admits a spin structure if and only if $k$ is even or $k=1$. (Thanks to wonderich for pointing out the last special case in the comment.)