Good time of day. I try to find Pontrjagin number for such task
$\langle p_1(r\gamma_H^1), [S^4] \rangle=\oint_{S^4}{p_1(r\gamma_H^1)}$
where $\gamma_H^1=(E\rightarrow \Bbb HP^1 \simeq S^4)$ - universal bundle over quaternionic projective line $\Bbb HP^1$
$r$-the operation of realification
My attempt is the following
I know from Milnor and Stasheff's Characteristic Classes book Pontryagin class is equal
$p_1(\gamma_r)=-2u$ where $u \in H^4(S^4)$
I hope it is true but I'm not sure. I don't understand how to compute Pontrjagin number. Can you help me please in more detail. And How is non-commutativity of quaternions affect on computations?
Thank you
Signs for the Chern and Pontryagin classes are a little confusing to me, so I'd recommend taking my final answer up to a sign.
Let's say $V=\gamma_H ^1$. By definition, $p_1(V)=c_2(V \otimes C)$. Since $V$ has a complex structure coming from the fact that the complex numbers sit inside the quaternions, we know that $V \otimes C$ is isomorphic to $V \oplus \overline V$. Using the Cartan formula and the fact that $c_i (\overline V) = (-1)^i c_i(V)$, we deduce that $c_2(V \oplus \overline V)=2c_2(V)$.
So we are left to calculate the second Chern class of $V$. Since this is the top class of this vector bundle, it coincides with the Euler class. Recall that the Euler class is defined as the pullback of the Thom class of $V$ via the zero section. The Thom class of $V$ lies in the relative cohomology $H^4 (V, V-\mathbb{H}P^1)$. This latter cohomology can be modeled as $\bar H^*(V^+ )$, where $V^+$ denotes the one point compactification of V. Under this identification, we simply pullback by the inclusion of $\mathbb{H}P^1$ into the compactification.
The one point compactification of $\gamma^n _H$ is homeomorphic to $\mathbb{H}P^{n+1}$. In this case, the Thom class is represented by the dual of the 4 cell in the standard cell structure on $\mathbb{H}P^2$ and we are pulling this back to $\mathbb{H}P^1 =S^4$ via the inclusion of this cell. Hence, the Euler class is $[S^4]^c$. If this is not obvious to you, it is also possible to argue that the Euler class is $\langle[\mathbb{H}P^1 ]^* \cup[\mathbb{H}P^1]^*,[\mathbb{H}P^2]\rangle[S^4]^c$, here $*$ denotes Poincare dual and $c$ means the cohomological fundamental class (the one that evaluates to 1 on the fundamental class). This follows from the formulation of the Euler class as counting the number of zeroes in a generic section.
Hence, $c_2(V)=[S^4]^c$ and $p_1(V)=2[S^4]^c$. By definition the Pontryagin number of $V$ for $p_1$ is $\langle p_1(V), [S^4]\rangle$ which in our case is $\langle 2[S^4]^c, [S^4]\rangle$ which is just 2 since $[S^4]^c$ evaluates to 1 on $[S^4]$.