2

See the related question here. This is the second part of question 4-C in Milnor and Stasheff's book on characteristic classes. In the solution to the first part, we rely on the fact that having a field of tangent $1$-planes allows us to split the tangent bundle into a direct sum. It doesn't seem like this should be true for a field of tangent $2$-planes (otherwise the question wouldn't just be asking about $P^4$ and $P^6$), so I'm not sure exactly where to start.

edit: actually, since $P^n$ can be given a Riemannian metric, any subbundle should be a Whitney summand; this doesn't answer the question as to why they only ask for $P^4$ and $P^6$ though, so I'm still confused.

Community
  • 1
cats
  • 4,298

1 Answers1

3

To simplify notation, all cohomology here is over $\mathbb{Z}_2$. As you said in your edit, such a bundle $\xi^2 \to \mathbb{RP}^n$ induces a Whitney sum $T\mathbb{RP}^n = \xi \oplus \xi^\perp$. The Stiefel-Whitney class $w(\mathbb{RP^n}) = (1 + x)^{n+1}\in H^*(\mathbb{RP}^n) = \mathbb{Z}_2[x]/(x^{n+1})$ with $\deg x = 1$.

In dimension $n = 4$, we have $$w(\xi) w(\xi^\perp) = (1 + x)^5 = 1 + x + x^4.$$ Since $w_i(\xi), w_i(\xi^\perp) = 0$ for $i > 2$, it follows that $w(\xi) = 1 + ax + x^2$ for some $a\in \mathbb{Z_2}$. But we can check that $(1 + ax + x^2)^{-1}(1 + x^4)\in H^*(\mathbb{RP}^n)$ is not a polynomial of degree $2$ in $x$, implying that a splitting $T\mathbb{RP^4} = \xi \oplus \xi^\perp$ can't exist. The case $n = 6$ is completely analogous, except there you're looking for a degree $4$ polynomial $w(\xi^\perp)$ with $$w(\xi)w(\xi^\perp) = (1 + ax + x^2)w(\xi^\perp) = (1 + x)^7 = 1 + x + \cdots + x^6.$$ (An analogous polynomial does exist for $n = 8$, to answer your question in the edit.)

anomaly
  • 25,364
  • 5
  • 44
  • 85
  • Thanks! I see the mistake I was making before - I didn't check that $(1+ax+x^2)^{-1}(1+x)^{n+1}$ could have degree less than $n-1$ in larger cases. Quite silly of me – cats May 02 '15 at 03:46
  • Yeah, it sounds like you were on the right track after the first part of the question, and it was just a matter of going through the computation. – anomaly May 02 '15 at 03:47
  • Why is $w_2(\xi) = 1$? – Michael Albanese Aug 17 '15 at 17:48
  • @MichaelAlbanese: Just a direct computation. The only possible such factor is $x + 1$. – anomaly Aug 17 '15 at 17:55
  • I see. There is an $x^4$ term in $w(T\mathbb{RP}^4)$, so there has to be an $x^2$ term in $w(\xi)$ (as $w(\xi^{\perp})$ is at most quadratic). Silly me. – Michael Albanese Aug 17 '15 at 17:58
  • No problem; it took me a minute or two to remember exactly what I was doing in the original response. :) – anomaly Aug 17 '15 at 17:58