The existence of fields of tangent $2$-planes on real projective spaces

Question

I'm interested in the following problem:

When does $\Bbb RP^n$ admit a field of tangent $2$-planes? (Please assume $n>2$.)

A field of tangent $k$-planes is a subbundle of $T\Bbb RP^n$ of rank $k$.

This is essentially the second part of problem [4-C] in Milnor-Stasheff, but instead of focusing $\Bbb RP^4$ and $\Bbb RP^6$, I would like to know exactly which real projective spaces have fields of tangent $2$-planes.

---

My attempt:

• Note that when $n$ is even $T\Bbb RP^n$ contains no (proper) subbundles. This is because any subbundle $\xi$ can be pulled back to $S^n$ to split $TS^n$ as $\pi^*(\xi)\oplus\pi^*(\xi^\perp)$. Since both summands are orientable, we can consider the Euler class, which leads to a contradiction as we would get $2=\chi(S^n)=\langle e(TS^n),[S^n]\rangle=\langle e(\pi^*\xi)e(\pi^*\xi^\perp),[S^n]\rangle=0$. This rules out all even dimensional projective spaces except $\Bbb RP^2$, whose tangent bundle is a field of tangent $2$-planes.

• If $n=4m+3$, then $S^n$ embeds in $\Bbb H^{m+1}$. Define the map $f_1:S^n\to TS^n,\ v\mapsto iv$ and similarly $f_2$ using left multiplication by $j$. By antipodal invariance, these maps give rise to two linearly independent sections of $T\Bbb RP^n$, which suggests the existence of a (trivial) rank $2$ subbundle. With $f_3:v\mapsto kv$, we can actually show that there is a trivial rank $3$ subbundle.

• Now suppose that $n=4m+1$ and that there is a rank $2$ subbundle $\xi$, then we have a splitting $\xi\oplus\xi^\perp=T\Bbb RP^n$. We note that if $w(\xi)=1$, then $w(\xi^\perp)=(1+a)^{4m+2}$ and we deduce that $$w_{4m}(\xi^\perp)= a^{4m}$$ but $\xi^\perp$ has rank $4m-1$, which is a contradiction. If instead we have $w(\xi)=1+a$, then $w(\xi^\perp)=(1+a)^{4m+1}$, which again implies that $w_{4m}(\xi^\perp)\neq 0$, contradiction. So we have either $w(\xi)=1+a^2$ or $w(\xi)=1+a+a^2$. In the first case, we have \begin{align}&w(\xi^\perp)=(1+a^2+a^4+a^6+\ldots)(1+a)^{4m+2}\\ &\implies w_{4m}(\xi^\perp)=\sum_{j=0}^{2m}\binom{4m+2}{2j}a^{4m}=(2^{4m+1}-1)a^{4m}\neq 0\end{align} This is impossible by considering the rank of $\xi^\perp$. In the second case, $$w(\xi^\perp)=(1+a+a^2)^{-1}(1+a)^{4m+2}=(1-a+a^3-a^4+a^6-a^7+\ldots)(1+a)^{4m+2}$$ If $m\equiv 0,2\pmod{3}$, i.e., $n\equiv 1,9\pmod{12}$, then $w_{4m+1}(\xi^\perp)\neq 0$, which is a contradiction. However, if $m\equiv 1\pmod{3}$, i.e., $n\equiv 5\pmod{12}$, a problem arises because the $4m$th and $(4m+1)$th Stiefel-Whitney classes both vanish. $(*)$

Up to this point, we have deduced that (for $n>2$)

1. $\Bbb RP^n$ admits a field of tangent $2$-planes if $n\equiv 3\pmod{4}$.
2. $\Bbb RP^n$ does not admit a field of tangent $2$-planes if $n\equiv 1,9\pmod{12}$ or $n\equiv 0\pmod{2}$.

So here is my question:

Suppose $n\equiv 5\pmod{12}$, is it possible to construct a rank $2$ vector bundle over $\Bbb RP^n$ such that $w(\xi)=1+a+a^2$? If yes, can it be realized as a rank $2$ subbundle of $T\Bbb RP^n$?

Thanks in advance for your time and effort.

---

$(*)$ I decided to include the computation of $w(\xi^\perp)=(1+a+a^2)^{-1}(1+a)^{4m+2}$ here for reference in case this mess is due to an arithmetic mistake. We can ignore the sign because we're working in $\Bbb Z_2$, which leads to $$w(\xi^\perp)=(1+a+a^3+a^4+\ldots)(1+a)^{4m+2}$$

Note that the degree $j$ term vanishes precisely when $j\equiv 2\pmod{3}$. With some algebra, we end up with $$w_{4m+1}(\xi^\perp)\equiv \sum_{j\ge 0}\left(\binom{4m+2}{1+3j}+\binom{4m+2}{2+3j}\right)a^{4m+1}\equiv \sum_{j\ge 0}\binom{4m+2}{3j}a^{4m+1}$$ $$w_{4m}(\xi^\perp)\equiv \sum_{j\ge 0}\left(\binom{4m+2}{2+3j}+\binom{4m+2}{3+3j}\right)a^{4m}\equiv \left(1+\sum_{j\ge 0}\binom{4m+2}{1+3j}\right)a^{4m}$$ These sums can be evaluated, and the closed forms depend on $m\pmod3$. (Reference)

Answer 1

Let $n \geq 3$ and let $\pi : S^n \to \mathbb{RP}^n$ be the double cover.

If $E$ is a rank two subbundle of $T\mathbb{RP}^n$ then $\pi^*E$ is isomorphic to a rank two subbundle of $TS^n$. Since $S^n$ is simply connected, the bundle $\pi^*E$ is orientable. As orientable rank two bundles are determined up to isomorphism by their Euler class and $H^2(S^n; \mathbb{Z}) = 0$, we see that $\pi^*E$ is trivial and hence $S^n$ admits two pointwise linearly independent vector fields.

The maximal number of pointwise linearly independent vector fields a smooth manifold $M$ admits is called the span of the manifold, and is denoted $\operatorname{span}(M)$. By the above, if $T\mathbb{RP}^n$ admits a rank two subbundle, then $\operatorname{span}(S^n) \geq 2$. The span of $S^n$ is known by work of Adams, it is equal to $\rho(n+1) - 1$ where $\rho(n)$ denotes the $n^{\text{th}}$ Radon-Hurwitz number which is defined as follows: if $n = 2^{4a + b}c$ where $a, b, c$ are non-negative integers, $0 \leq b \leq 3$ and $c$ is odd, then $\rho(n) = 8a + 2^b$.

If $n = 4m + 1$, then $n + 1 = 4m + 2 = 2(2m+1) = 2^{4(0)+1}(2m+1)$, so

$$\operatorname{span}(S^n) = \rho(n+1) - 1 = 8(0) + 2^1 - 1 = 1.$$

Therefore $\mathbb{RP}^n$ does not admit a field of tangent $2$-planes for $n \equiv 1 \bmod 4$. In particular, when $n \equiv 5 \bmod{12}$, there is no rank two subbundle $\xi$ of $T\mathbb{RP}^n$ with $w(\xi) = 1 + a + a^2$.

Note that for $n = 4m + 3$, not only does $T\mathbb{RP}^n$ admit a rank two subbundle, it admits a trivial rank two subbundle. To see this, first note that $\operatorname{span}(\mathbb{RP}^n) = \operatorname{span}(S^n)$, see this question. Now $n + 1 = 4m + 4 = 4(m + 1) = 2^2(m+1)$, and we can write $m + 1 = 2^kq$ where $q$ is odd, so $n + 1 = 2^{k+2}q$. If $k + 2 = 4a + b$ with $0 \leq b \leq 3$, then there are two possibilities: $a > 0$ and $a = 0$.

• If $a > 0$, then $\rho(n + 1) - 1 = 8a + 2^b - 1 \geq 8 + 1 - 1 = 8$.
• If $a = 0$, then $b = 2$ or $3$, so $\rho(n+1) - 1 = 8a + 2^b - 1 \geq 2^2 - 1 = 3$.

Combining, we see that if $n \equiv 3 \bmod 4$, then $\operatorname{span}(\mathbb{RP}^n) \geq 3$ and hence $T\mathbb{RP}^n$ has a trivial rank two subbundle.