ABBV with Positive-Dimensional Fixed Components

The first-factor action fixes exactly $p_0=[1:0]$ and $p_{\mathrm{\infty }}=[0:1]$, while it fixes every point of the second factor. Thus $M^T=F_0\bigsqcup F_{\mathrm{\infty }}$, with each component a copy of ${\mathbb{C}\mathbb{P}}^1$. The normal bundle is the tangent line of the first factor at its fixed point; in the coordinates $z_1\mathrm{/}{z}_0$ and $z_0\mathrm{/}{z}_1$ its weights are $+1$ on $F_0$ and $-1$ on $F_{\mathrm{\infty }}$.

The normal complex line is the first factor and the fixed-component tangent space is the second. Moving the real two-dimensional first-factor orientation past the real two-dimensional second-factor orientation has sign $(-1{)}^{2\cdot 2}=+1$, so the induced component orientations agree with the product orientations.

The dual tautological fiber at $p_0$ has weight $0$ and at $p_{\mathrm{\infty }}$ has weight $-1$, so the first-factor class restricts to $0$ and $-u$. Thus the contribution from $F_0$ is $0\mathrm{/}u=0$; on $F_{\mathrm{\infty }}$, the restricted class is $(-u)\eta$ and the normal Euler class is $-u$, leaving $\eta$, whose integral is $1$. The fixed-component sum is therefore $1$.

Product fiber integration factors over the two compact oriented factors. On the first ${\mathbb{C}\mathbb{P}}^1$, the section of $\mathcal{O}(1)$ induced by $\mathrm{\ell }(z_0,z_1)=z_0$ has one positive transverse zero at $p_{\mathrm{\infty }}$, so its Euler integral is $1$ by [F3]; the second-factor integral of $\eta$ is $1$ by definition. Hence ${\int }_M\alpha =1\cdot 1=1$, agreeing with step 3.1.