# Characteristic classes over the real projective spaces ${\mathbb{P}}_d({\mathbb{R}})$ [closed]

Inspired by other users --- this and others, also here: We may explore the characteristic classes of real projective spaces ${\mathbb{P}}_d({\mathbb{R}})$, in some sense, it is related to some sort of "sphere" when we consider the most simple case ${\mathbb{P}}_1({\mathbb{R}}) \simeq S^1$. We however like to know general $d$.

It is still worthwhile to know the basic characteristic class data of them:

1. Stiefel–Whitney class $w_i$: ${\mathbb{P}}_d({\mathbb{R}})$ may be non-orientable and non-spin. Note $$w_1({\mathbb{P}}_d({\mathbb{R}})) =d+1 \mod 2\neq 0, \quad \text{ if d=0 \mod 2}$$ $$w_2({\mathbb{P}}_d({\mathbb{R}}))=\frac{(d+1)d}{2} \mod 2, \text{ if d \geq 2}$$ Thus $w_2({\mathbb{P}}_1({\mathbb{R}}))=0$, and for $d \geq 2$, we have $w_2({\mathbb{P}}_d({\mathbb{R}}))=0$ for $d=3$ or $0$ mod $4$.

More generally, we have for other $w_i$: $$w({\mathbb{P}}_d({\mathbb{R}}))={(1+a)}^{d+1},$$ where $H^*({\mathbb{P}}_d({\mathbb{R}}),\mathbb{Z}_2)=\frac{\mathbb{Z}_2[a]}{a^{d+1}}$

1. Euler class: $$e({\mathbb{P}}_d({\mathbb{R}}))=?$$

2. Wu class $u_i$: is related to the Stiefel–Whitney class $w_i$ through Stenrod square, so $$u_i({\mathbb{P}}_d({\mathbb{R}}))=?$$

3. Pontryagin class $p_i$: $$p_i({\mathbb{P}}_d({\mathbb{R}}))=? \in H^{4i}({\mathbb{P}}_d({\mathbb{R}}), \mathbb{Z})$$ We can consider all the $d=0 \pmod 4$ dimensions. e.g. $p_1(T{\mathbb{P}}_4({\mathbb{R}}))=0$ (yes?) and the frame bundle $p_1(F{\mathbb{P}}_d({\mathbb{R}}))=?$.

4. Are there other helpful / useful Characteristic classes for real projective spaces ${\mathbb{P}}_d({\mathbb{R}})$ that are nontrivial (or trivial)? A list of such info is greatly appreciated.

Edit. Chern class $c_i$ removed.

## closed as too broad by Balarka Sen, Namaste, Leucippus, user99914, Nils MatthesJul 12 '18 at 13:18

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• $T\Bbb{RP}^d$ is not a complex vector bundle. It does not make sense to take its Chern classes. – user98602 Jul 11 '18 at 0:23
• when $d$ is even, is it possible (as said above)? – wonderich Jul 11 '18 at 0:26
• You can take the Chern classes of a complex vector bundle. $d$ even does not make that into a complex vector bundle; it's not even oriented, so it can't be made into one. (Complex vector bundles come equipped with a canonical orientation.) – user98602 Jul 11 '18 at 0:29
• That is fine - thanks - you can still give a negative answer (No) for that case - I am just asking - it is a legal question. Just a negative answer – wonderich Jul 11 '18 at 0:30
• @Mike Miller, Chern class removed. (Why) Did you vote down? – wonderich Jul 11 '18 at 3:31