A free circle orbit has zero localized Borel cohomology

Translation has no fixed point: translation by $[1\mathrm{/}2]$ moves every $x\in T$. Define $\mathrm{\Phi }([e,x])=e\cdot x$. If $(e,x)$ is replaced by $(e\cdot h,h^{-1}x)$, then $(e\cdot h)\cdot (h^{-1}x)=e\cdot x$, so $\mathrm{\Phi }$ is well-defined. Its inverse is $e\mapsto [e,0]$, where $0$ is the identity of the additive circle: $\mathrm{\Phi }([e,0])=e$, and the quotient relation gives $[e,x]=[e\cdot x,0]$. Both maps are continuous by the quotient topology, hence $ET{\times }_{T}T\cong ET$. Contractibility of $ET$ and [F2] give cohomology $k$ in degree zero and zero in positive degrees.

The projection $ET\to BT$ is null-homotopic because $ET$ is contractible, so every positive-degree element of $A_T(k)$ pulls back to zero and acts by zero on $H^{\ast }(ET;k)$. In particular the invertible denominator $u\in {\mathcal{S}}_T(k)$ annihilates every class; for any $a$, $a\mathrm{/}1=(ua)\mathrm{/}u=0$ in the localization. The circle is a compact smooth manifold without boundary and translation is jointly smooth, so [F4] applies under [A1] and agrees with the zero cohomology of the empty fixed set.