Rational coefficient ring and characters of a product torus classifying space

The identity point is a zero-cell of the finite CW circle, so its inclusion is a relative CW subcomplex inclusion and is a cofibration by [F2]. Together with compact metrizability from [F1], this makes $S^1$ a well-pointed group of CW type in the sense of [F2]. Hence [F3] applies.

On one ${\mathbb{C}\mathbb{P}}^{\mathrm{\infty }}$ factor, apply [F5] with $G=\mathbb{Q}$. By [F4], each integral homology group is free, so the Ext term is zero and evaluation identifies $H^n({\mathbb{C}\mathbb{P}}^{\mathrm{\infty }};\mathbb{Q})$ with $\mathrm{Hom}(H_n({\mathbb{C}\mathbb{P}}^{\mathrm{\infty }};\mathbb{Z}),\mathbb{Q})$. The natural coefficient map sends each integral generator $u^k$ to the nonzero generator of the corresponding rational group. Thus the rational ring is $\mathbb{Q}[u_{\mathbb{Q}}]$, with $\mathrm{\mid }{u}_{\mathbb{Q}}\mathrm{\mid }=2$.

Apply [F6] to the integral singular chain complex with $G=\mathbb{Q}$. Each integral homology group in [F4] is free, so the Tor term vanishes and $H_n({\mathbb{C}\mathbb{P}}^{\mathrm{\infty }};\mathbb{Q})$ is $\mathbb{Q}$ for even $n$ and zero for odd $n$; it is finite free in every degree.

Let $\chi :T\to S^1$ be a smooth character and restrict it to each coordinate one-parameter subgroup ${\gamma }_i(s)=[0,\dots ,s,\dots ,0]$. Write $f_i(s)=\chi ({\gamma }_i(s))$. The homomorphism law gives $f_i(s+h)=f_i(s)f_i(h)$; differentiating in $h$ at zero yields $f_i^{\mathrm{\prime }}(s)=f_i(s)f_i^{\mathrm{\prime }}(0)$. Since $\mathrm{\mid }{f}_i(s)\mathrm{\mid }=1$, differentiation of $\mathrm{\mid }{f}_i(s){\mathrm{\mid }}^2=1$ at zero shows $f_i^{\mathrm{\prime }}(0)=2\pi i{a}_i$ for a real $a_i$. Put $g_i(s)=e^{-2\pi i{a}_{i}s}{f}_i(s)$. Under $\mathbb{C}\cong {\mathbb{R}}^2$, this is a differentiable map of the real variable $s$, and the product rule and [F9] give $g_i^{\mathrm{\prime }}(s)=e^{-2\pi i{a}_{i}s}(f_i^{\mathrm{\prime }}(s)-2\pi i{a}_i{f}_i(s))=0$ for all $s\in \mathbb{R}$. The real-linear coordinate projections commute with this derivative (apply them to the difference quotient), so $\mathrm{Re}{g}_i$ and $\mathrm{Im}{g}_i$ are differentiable real functions on the interval $\mathbb{R}$, hence continuous there, with zero derivative everywhere. Applying [F10] separately to these two functions makes both constant. Since $f_i(0)=\chi (e)=1$, we have $g_i(0)=1$, so their constant values are $1$ and $0$, respectively. Thus $g_i(s)=1$ and $f_i(s)=e^{2\pi i{a}_{i}s}$ for every real $s$. Since ${\gamma }_i(1)$ is the identity, $f_i(1)=1$, and [F12] gives $a_i=n_i\in \mathbb{Z}$. The product decomposition of $T$ now gives $\chi ([t])=e^{2\pi i{\sum }_i{n}_i{t}_i}$. Uniqueness follows by restricting again to each ${\gamma }_i$.

The product $E_T=(E_{S^1}{)}^r$ is free for the coordinatewise right $T$-action, since a point fixed by $t$ has every coordinate fixed and each factor action is free. It is contractible because it is a finite product of contractible spaces. Each Milnor quotient map is open: the saturation of an open set is the union of its translates. The finite product quotient map is therefore open and its fibers are exactly the $T$-orbits, so it induces $BT\cong (B{S}^1{)}^r\cong ({\mathbb{C}\mathbb{P}}^{\mathrm{\infty }}{)}^r$ by [F3]. For $r=0$ this is the one-point product.

Iterating the ring cross product [F7] over the finite product $({\mathbb{C}\mathbb{P}}^{\mathrm{\infty }}{)}^r$ identifies its rational cohomology ring with the graded tensor product of $r$ copies of $\mathbb{Q}[u_{\mathbb{Q}}]$, using the finite-free homology in step 1.3. All generators have even degree, so the graded tensor multiplication has no sign changes and is $\mathbb{Q}[u_1,\dots ,u_r]$.

For $k=\mathbb{Q}$ or $\mathbb{R}$, [F6] applied to the same integral chain complex and $G=k$ gives $H_n({\mathbb{C}\mathbb{P}}^{\mathrm{\infty }};k)=k$ for even $n$ and zero for odd $n$, since the integral groups in [F4] are free. Iterating [F11] over the finite product in step 2.1 identifies $H_n(BT;k)$ with a direct sum over the finitely many $r$-tuples of nonnegative degrees summing to $n$. Every summand is either zero or one-dimensional over $k$, so $H_n(BT;k)$ is finite-dimensional, and therefore finite-free, for every $n$. For $r=0$, only the degree-zero group $k$ occurs.

For the coordinate character ${\chi }_i$, write a point of the $N$th Milnor join in the $i$th circle factor as $e={\sum }_{j=0}^N{t}_j{z}_j$, with $t_j\ge 0$, ${\sum }_j{t}_j=1$, and $z_j\in S^1$ ignored whenever $t_j=0$. By [F3], it determines the unit vector $v_N(e)={\mathrm{\Phi }}_N(e)=(\sqrt{t_0}{z}_0,\dots ,\sqrt{t_N}{z}_N)\in S^{2N+1}$. The inverse of ${\mathrm{\Phi }}_N$ recovers $t_j=\mathrm{\mid }{v}_j{\mathrm{\mid }}^2$ and $z_j=v_j\mathrm{/}\mathrm{\mid }{v}_j\mathrm{\mid }$ when $v_j\mathrm{\ne }0$. The right circle action sends $v_N$ to $\lambda v_N$, and its orbit is the complex line $\mathbb{C}{v}_N$ represented by $[v_N]\in {\mathbb{C}\mathbb{P}}^N$. For the product model $ET=(E_{S^1}{)}^r$, the associated-line map on this stage is $[e,z]\mapsto z{v}_N(e)$ in the $i$th factor. It is well defined because $(e,z)\sim (e\cdot \lambda ,{\lambda }^{-1}z)$ gives $({\lambda }^{-1}z)(\lambda v_N(e))=z{v}_N(e)$. On each fiber it is a complex-linear isomorphism onto $\mathbb{C}{v}_N(e)$. On each projective coordinate chart choose a continuous unit representative $v$; the inverse coefficient $w\mapsto {\sum }_j\overline{v_j}{w}_j$ is continuous there, and replacing $v$ by $\lambda v$ replaces the coefficient by ${\lambda }^{-1}$ times itself, which gives the same associated-bundle class. Thus the bundle map is a homeomorphism onto the tautological line over ${\mathbb{C}\mathbb{P}}^N$. These maps commute with the finite-stage inclusions, and hence identify $L_i$ with the tautological line over the selected ${\mathbb{C}\mathbb{P}}^{\mathrm{\infty }}$ quotient. By [F4] and [F8], $u_i=c_1(L_i)$ is a rational degree-two generator. For $\mathbf{n}\in {\mathbb{Z}}^r$, the associated line is the tensor product of $L_i^{\otimes n_i}$, interpreting negative powers as dual lines; additivity and duality in [F8] give $c_1(L_{{\chi }_{\mathbf{n}}})={\sum }_i{n}_i{u}_i$. This is nonzero exactly when $\mathbf{n}\mathrm{\ne }0$, by the polynomial basis in step 2.2.

For $r=0$, $T$ and $BT$ are points, the cohomology is $\mathbb{Q}$ in degree zero, and there are no nontrivial characters. The finite product, universal-coefficient, and ring arguments above add no additional restriction beyond AC.