Obstruction to a Spin structure on a bundle ξ, and ξ ⊕ $n$ det ξ

Question

In Ref, it says that:

The obstruction to putting a Spin structure on a bundle $ξ (= Rn → E → B)$ is $w_2(ξ) \in H^2(B;Z/2Z)$.

Pin± structures is that

• Pin− structures on ξ correspond to Spin structures on ξ ⊕ det ξ.

• Pin+ to Spin structures on ξ ⊕ 3 det ξ where det ξ is the determinant line bundle.

Question

1. Why do we have det ξ and 3 det ξ instead of other numbers? I suppose the 2 det ξ is related to the $w_1(ξ)^2$?

2. How to show: a variety of “descent” theorems of the type: a Pin± structure on ξ⊕η descends to a Pin+, Pin−, or Spin structure on ξ when dim η=1 or 2 and various conditions on η are satisfied.

Answer 1

Your questions are addressed in Lemma 1.7 of the paper you cite, but I shall give an explanation of this from the perspective of classifying spaces.

Let me answer your questions in reverse order. First, a general setting for the "descent theorems" you are looking for can be encapsulated as the existence of pullback squares of the form \begin{equation} \tag{1} \begin{array}{ccc} BG_1 & \rightarrow & BG_2 \\ \downarrow & & \downarrow \\ BO(n) & \xrightarrow{\oplus \eta} & BO(n+r), \end{array} \end{equation} which should be read as: "a $G_1$-structure on a $n$-plane bundle $\xi$ corresponds to a $G_2$-structure on $\xi \oplus \eta$". Any time you can produce such a square will give you a descent theorem.

Where might these pullback squares come from? For the rest of this answer, we focus primarily on the case $G_1, G_2 \in \{\mathrm{Spin}(n), \mathrm{Pin}^\pm(n)\}$ and $\eta = r \det \xi$ for some integer $r$.

Suppose we had a diagram \begin{equation} \tag{2} \begin{array}{ccc} BO(n) & \xrightarrow{\oplus r \det} & BO(n+r) \\ \downarrow & & \downarrow \\ K(\mathbb{Z}/2, 2) & \xrightarrow{\sim} & K(\mathbb{Z}/2, 2). \end{array} \end{equation}

From the cohomology of $BO(n)$, we know that the vertical maps are spanned by $w_2$ and $w_1^2$. Suppose they are given by $w_2 + a_1 w_1^2$ and $w_2 + a_2 w_1^2$ respectively, where $a_1, a_2 \in \mathbb{Z}/2$. Then, we can determine when this square commutes up to homotopy based on $r$, $a_1$, and $a_2$ by solving the equation $$(w_2 + a_1 w_1^2)(\xi) = (w_2 + a_2 w_1^2)(\xi \oplus r \det \xi).$$

We summarize the results in the following table: $$ \begin{array}{|ccc|} \hline a_1 & a_2 & r \\ \hline 0 & 0 & 0, 3 \pmod{4} \\ 0 & 1 & 2, 3 \pmod{4} \\ 1 & 0 & 1, 2 \pmod{4} \\ 1 & 1 & 0, 1 \pmod{4} \\ \hline \end{array} $$

Once we have a commuting square (2), we can take the vertical fibers of the diagram to get a diagram that looks like (1), where $G_i = \mathrm{Pin}^+$ (resp. $\mathrm{Pin}^-$) if $a_i = 0$ (resp. $a_i = 1$).

It remains to show that this resulting diagram (1) is cartesian. To see why, note that we may back up the fibration one more time to get a map $K(\mathbb{Z}/2, 1) \to K(\mathbb{Z}/2, 1)$ which is an equivalence iff (1) is a pullback. But this map is the loop of the map $K(\mathbb{Z}/2, 2) \xrightarrow{\sim} K(\mathbb{Z}/2, 2)$, which is an equivalence by construction.

There is one more observation to make. If $r$ is odd, then the map $BO(n) \xrightarrow{\oplus r \det} BO(n+r)$ factors through $BSO(n+r)$, and we may replace $G_2$ with $\mathrm{Spin}(n+r)$. Combining this with the table above, we obtain:

• $r \equiv 0 \pmod{4}$: a $\mathrm{Spin}$- or $\mathrm{Pin}^+$- or $\mathrm{Pin}^-$-structure on $\xi$ corresponds to the same on $\xi \oplus r \det \xi$.
• $r \equiv 1 \pmod{4}$: a $\mathrm{Pin}^-$-structure on $\xi$ corresponds to a $\mathrm{Spin}$-structure on $\xi \oplus r \det \xi$.
• $r \equiv 2 \pmod{4}$: a $\mathrm{Pin}^+$-structure on $\xi$ corresponds to a $\mathrm{Pin}^-$-structure on $\xi \oplus r \det \xi$, and vice-versa.
• $r \equiv 3 \pmod{4}$: a $\mathrm{Pin}^+$-structure on $\xi$ corresponds to a $\mathrm{Spin}$-structure on $\xi \oplus r \det \xi$.

This proves Lemma 1.7 in your reference.