Contents

# Contents

## Idea

Complex projective space $\mathbb{C}P^n$ is the projective space $\mathbb{A}P^n$ for $\mathbb{A} = \mathbb{C}$ being the complex numbers (and for $n \in \mathbb{N}$), a complex manifold of complex dimension $n$ (real dimension $2n$). Equivalently, this is the complex Grassmannian $Gr_1(\mathbb{C}^{n+1})$. For the special case $n = 1$ then $\mathbb{C}P^1 \simeq S^2$ is the Riemann sphere.

As $n$ ranges, there are natural inclusions

$\ast = \mathbb{C}P^0 \hookrightarrow \mathbb{C}P^1 \hookrightarrow \mathbb{C}P^2 \hookrightarrow \mathbb{C}P^3 \hookrightarrow \cdots \,.$

The sequential colimit over this sequence is the infinite complex projective space $\mathbb{C}P^\infty$. This is a model for the classifying space $B U(1)$ of circle principal bundles/complex line bundles (an Eilenberg-MacLane space $K(\mathbb{Z},2)$).

## Definition

###### Definition

For $n \in \mathbb{N}$, then complex $n$-dimensional complex projective space is the complex manifold (often just regarded as its underlying topological space) defined as the quotient

$\mathbb{C}P^n \coloneqq (\mathbb{C}^{n+1}-\{0\})/_\sim$

of the Cartesian product of $(n+1)$-copies of the complex plane, with the origin removed, by the equivalence relation

$(z \sim w) \Leftrightarrow (z = \kappa \cdot w)$

for some $\kappa \in \mathbb{C} - \{0\}$ and using the canonical multiplicative action of $\mathbb{C}$ on $\mathbb{C}^{n+1}$.

The canonical inclusions

$\mathbb{C}^{n+1} \hookrightarrow \mathbb{C}^{n+2}$

induce canonical inclusions

$\mathbb{C}P^n \hookrightarrow \mathbb{C}P^{n+1} \,.$

The sequential colimit over this sequence of inclusions is the infinite complex projective space

$\mathbb{C}P^\infty \coloneqq \underset{\longleftarrow}{\lim}_n \mathbb{C}P^n \,.$

The following equivalent characterizations are immediate but use#ful:

###### Proposition

For $n \in \mathbb{N}$ then complex projective space, def. , is equivalently the complex Grassmannian

$\mathbb{C}P^n \simeq Gr_1(\mathbb{C}^{n+1}) \,.$
###### Proposition

For $n \in \mathbb{N}$ then complex projective space, def. , is equivalently

1. the coset

$\mathbb{C}P^n \simeq U(n+1)/(U(n) \times U(1)) \,,$
2. the quotient of the (2n+1)-sphere by the circle group $S^1 \simeq \{ \kappa \in \mathbb{C}| {\vert \kappa \vert} = 1\}$

$\mathbb{C}P^n \simeq S^{2n+1}/S^1 \,.$
###### Proof

To see the second characterization from def. :

With ${\vert -\vert} \colon \mathbb{C}^{n} \longrightarrow \mathbb{R}$ the standard norm, then every element $\vec z \in \mathbb{C}^{n+1}$ is identified under the defining equivalence relation with

$\frac{1}{\vert \vec z\vert}\vec z \in S^{2n+1} \hookrightarrow \mathbb{C}^{n+1}$

lying on the unit $(2n+1)$-sphere. This fixes the action of $\mathbb{C}-0$ up to a remaining action of complex numbers of unit absolute value. These form the circle group $S^1$. This shows that we have a commuting diagram of functions of underlying sets of the form

$\array{ S^{2n+1} &\hookrightarrow& \mathbb{C}^{n+1} \setminus \{0\} \\ {}^{\mathllap{q_{S^{2n+1}}}}\downarrow &\searrow^{\mathrlap{f}}& \downarrow^{\mathrlap{q_{\mathbb{C}^{n+1}}}} \\ S^{2n+1}/S^1 &\longrightarrow& \mathbb{C}P^n }$

where the top horizontal and the two vertical functions are continuous, and where the bottom function is is a bijection. Since the diagonal composite is also continuous, the nature of the quotient space topology implies that the bottom function is also continuous. To see that it is a homeomorphism it hence remains to see that it is an open map (by this prop.).

So let $U \subset S^{2n+1}/S^1$ be an open set, which means that $q_{S^{2n+1}}^{-1}(U) \subset S^{2n+1}$ is an open set. We need to see that $f(q_{S^{2n+1}}^{-1}(U)) \subset \mathbb{C}P^{n}$ is open, hence that $q_{\mathbb{C}^{n+1}}^{-1}(f(q_{S^{2n+1}}^{-1}(U))) \subset \mathbb{C}^{n+1}$ is open. Now by the nature of the Euclidean metric topology, the open subset $q_{S^{2n+1}}^{-1}(U)$ is a union of open balls $B^\circ_x(\epsilon)$ in $\mathbb{C}^{n+1}$ intersected with $S^{2n+1}$. But then $q_{\mathbb{C}^{n+1}}^{-1}(f(B^\circ_x(\epsilon)\vert_{S^{2n+1}}))$ is their orbit under the multiplicative action by $\mathbb{C} \setminus \{0\}$, hence is a cylinder $B^\circ_x(\epsilon)\vert_{S^{2n+1}} \times (\mathbb{C} \setminus \{0\})$. This is clearly open.

The first characterization follows via prop. from the general discusion at Grassmannian. With this the second characterization follows also with the coset identification of the $(2n+1)$-sphere: $S^{2n+1} \simeq U(n+1)/U(n)$ (exmpl.).

## Properties

### Cell structure

###### Proposition

There is a CW-complex structure on complex projective space $\mathbb{C}P^n$ (def. ) for $n \in \mathbb{N}$, given by induction, where $\mathbb{C}P^{n+1}$ arises from $\mathbb{C}P^n$ by attaching a single cell of dimension $2(n+1)$ with attaching map the projection $S^{2n+1} \longrightarrow \mathbb{C}P^n$ from prop. :

$\array{ S^{2n+1} &\longrightarrow& S^{2n+1}/S^1 \simeq \mathbb{C}P^n \\ {}^{\mathllap{\iota_{2n+2}}}\downarrow^{\mathrlap{i_n}} &(po)& \downarrow \\ D^{2n+2} &\underset{q}{\longrightarrow}& \mathbb{C}P^{n+1} } \,.$
###### Proof

Given homogeneous coordinates $(z_0 , z_1 , \cdots , z_n , z_{n+1} , z_{n+2}) \in \mathbb{C}^{n+2}$ for $\mathbb{C}P^{n+1}$, let

$\phi \coloneqq -arg(z_{n+2})$

be the phase of $z_{n+2}$. Then under the equivalence relation defining $\mathbb{C}P^{n+1}$ these coordinates represent the same element as

$\frac{1}{\vert \vec z\vert}(e^{i \phi} z_0, e^{i \phi}z_1,\cdots, e^{i \phi}z_{n+1}, r) \,,$

where

$r = {\vert z_{n+2}\vert}\in [0,1) \subset \mathbb{C}$

is the absolute value of $z_{n+2}$. Representatives $\vec z'$ of this form (${\vert \vec z' \vert = 1}$ and $z'_{n+2} \in [0,1]$) parameterize the 2n+2-disk $D^{2n+2}$ with boundary the $(2n+1)$-sphere at $r = 0$.

The resulting function $q \colon D^{2n+2} \to \mathbb{C}P^{n+1}$ is continuous: It may be factored as

$\array{ q_{D^{2n+2}} \colon & D^{2n+2} &\overset{\phantom{AAA}}{\hookrightarrow}& \mathbb{C}^{n+2} \setminus \{0\} &\overset{q_{\mathbb{C}^{n+2}}}{\longrightarrow}& \mathbb{C}P^{n+1} \\ & (Re(z_1), Im(z_1), \cdots, Re(z_{n+1}), Im(z_{n+1}), r) &\mapsto& (z_1, \cdots, z_{n+1}, r) &\mapsto& [ z_1 : \cdots : z_{n+1} : r ] }$

and here the first map is the embedding of the disk $D^{2n+2}$ as a hemisphere in $\mathbb{R}^{2n+1} \hookrightarow \mathbb{R}^{2n+2} \simeq \mathbb{C}^{2n+2}$, while the second is the defining quotient space projection. Both of these are continuous, and hence so is their composite.

The only remaining part of the action of $\mathbb{C}-\{0\}$ which fixes the conditions ${\vert z'\vert} = 0$ and $z'_{n+2}$ is $S^1 \subset \mathbb{C} \setminus \{0\}$ acting on the elements with $r = \{z'_{n+2}\} = 0$ by phase shifts on the $z_0, \cdots, z_{n+1}$. The quotient of this remaining action on $D^{2(n+1)}$ identifies its boundary $S^{2n+1}$-sphere with $\mathbb{C}P^{n}$, by prop. .

This shows that the above square is a pushout diagram of underlying sets.

By the nature of colimits in Top (this prop.) it remains to see that the topology on $\mathbb{C}P^{n+1}$ is the final topology induced by the functions $D^{2n+2} \to \mathbb{C}P^{n+1}$ and $\mathbb{C}P^n \to \mathbb{C}P^{n+1}$, hence that a subset of $\mathbb{C}P^{n+1}$ is open precisely if its pre-images under these two functions are open.

We saw above that $q_{D^{2n+2}}$ is continuous. Moreover, also the function $i_n \colon \mathbb{C}P^n \to \mathbb{C}P^{n+1}$ is continuous (by this lemma).

This shows that if a subset of $\mathbb{C}P^{n+1}$ is open, then its pre-images under these functions are open. It remains to see that if $S \subset \mathbb{C}P^{n+1}$ is a subset with $q_{S^{2n+2}}^{-1}(S) \subset D^{2n+2}$ open and $i_n^{-1}(S) \subset \mathbb{C}P^n$ open, then $S \subset \mathbb{C}P^{n+1}$ is open.

Notice that $q_{\mathbb{C}^{n+2}}^{-1}(S)$ contains with every point also its orbit under the action of $\mathbb{C} \setminus \{0\}$, and that every open subset of $D^{2n+2}$ is a unions of open balls. By the above factorization of $q_{D^{2n+2}}$ this means that if $q_{D^{2n+2}}^{-1}(S)$ is open, then $q_{\mathbb{C}^{n+2}}^{-1}(S)$ is a union of open cyclinders, hence is open. By the nature of the quotient topology, this means that $S \subset \mathbb{C}P^n$ is open.

### Homotopy

Write $\Sigma^\infty (\mathbb{C}P^\infty)_+ \in Ho(Spectra)$ for the H-group ring spectrum of $\mathbb{C}P^\infty \simeq B U(1)$ (see there for details).

For $X \in Ho(Top)$ the homotopy type of a topological space in the classical homotopy category, write

$[\Sigma^\infty X_+ , \Sigma^\infty, \mathbb{C}P^\infty_+] \simeq [X, \Omega^\infty \Sigma^\infty \mathbb{C}P^\infty_+] \in Ab$

for the hom-group in the stable homotopy category, which, by adjunction, is equivalently computed in the classical homotopy category as shown on the right.

Write

$i \;\colon\; \mathbb{C}P^\infty \simeq B U(1) \simeq B U(1) \times \{1\} \hookrightarrow B U \times \mathbb{Z}$

for the inclusion into the classifying space for complex topological K-theory which classifies the inlusion of complex line bundles $E$ as virtual vector bundles $[E] - 0$.

###### Proposition

For any $X \in Ho(Top)$ the ring homomorphism

$i_\ast \;\colon\; [\Sigma^\infty X_+, \Sigma^\infty \mathbb{C}P_+] \longrightarrow K_{\mathbb{C}}(X)$

This is due to (Segal 73, prop. 1).

Prop. is sharpened by Snaith's theorem. See there for more. The version for real projective space is called the Kahn-Priddy theorem. #

### Homology and Cohomology

#### Ordinary

###### Proposition

For $A \in$ Ab any abelian group, then the ordinary homology groups of complex projective space $\mathbb{C}P^n$ with coefficients in $A$ are

$H_k(\mathbb{C}P^n,A)\simeq \left\{ \array{ A & for \; k \;even\; and \; k \leq 2n \\ 0 & otherwise } \right. \,.$

Similarly the ordinary cohomology groups of $\mathbb{C}P^n$ is

$H^k(\mathbb{C}P^n,A) \simeq \left\{ \array{ A & for \; k \;even\; and \; k \leq 2n \\ 0 & otherwise } \right. \,.$

Moreover, if $A$ carries the structure of a ring $R = (A, \cdot)$, then under the cup product the cohomology ring of $\mathbb{C}P^n$ is the the graded ring

$H^\bullet(\mathbb{C}P^n, R) \simeq R[c_1] / (c_1^{n+1})$

which is the quotient of the polynomial ring on a single generator $c_1$ in degree 2, by the relation that identifies cup products of more than $n$-copies of the generator $c_1$ with zero.

Finally, the cohomology ring of the infinite-dimensional complex projective space is the formal power series ring in one generator:

$H^\bullet(\mathbb{C}P^\infty, R) \simeq R[ [ c_1 ] ] \,.$

(Or else the polynomial ring $R[c_1]$, depending on how one chooses to extract a ring from a graded ring, see remark .)

###### Proof

First consider the case that the coefficients are the integers $A = \mathbb{Z}$.

Since $\mathbb{C}P^n$ admits the structure of a CW-complex by prop. , we may compute its ordinary homology equivalently as its cellular homology (thm.). By definition (defn.) this is the chain homology of the chain complex of relative homology groups

$\cdots \overset{\partial_{cell}}{\longrightarrow} H_{q+2}((\mathbb{C}P^n)_{q+2}, (\mathbb{C}P^n)_{q+1}) \overset{\partial_{cell}}{\longrightarrow} H_{q+1}((\mathbb{C}P^n)_{q+1}, (\mathbb{C}P^n)_{q}) \overset{\partial_{cell}}{\longrightarrow} H_{q}((\mathbb{C}P^n)_{q}, (\mathbb{C}P^n)_{q-1}) \overset{\partial_{cell}}{\longrightarrow} \cdots \,,$

where $(-)_q$ denotes the $q$th stage of the CW-complex-structure. Using the CW-complex structure provided by prop. , then there are cells only in every second degree, so that

$(\mathbb{C}P^n)_{2k+1} = (\mathbb{C}P)_{2k}$

for all $k \in \mathbb{N}$. It follows that the cellular chain complex has a zero group in every second degree, so that all differentials vanish. Finally, since prop. says that $(\mathbb{C}P^n)_{2k+2}$ arises from $(\mathbb{C}P^n)_{2k+1} = (\mathbb{C}P^n)_{2k}$ by attaching a single $2k+2$-cell it follows that (by passage to reduced homology)

$H_{2k}(\mathbb{C}P^n, \mathbb{Z}) \simeq \tilde H_{2k}(S^{2k})((\mathbb{C}P^n)_{2k}/(\mathbb{C}P^n)_{2k-1}) \simeq \tilde H_{2k}(S^{2k}) \simeq \mathbb{Z} \,.$

This establishes the claim for ordinary homology with integer coefficients.

In particular this means that $H_q(\mathbb{C}P^n, \mathbb{Z})$ is a free abelian group for all $q$. Since free abelian groups are the projective objects in Ab (prop.) it follows (with the discussion at derived functors in homological algebra) that the Ext-groups vanishe:

$Ext^1(H_q(\mathbb{C}P^n, \mathbb{Z}),A) = 0$

and the Tor-groups vanishes:

$Tor_1(H_q(\mathbb{C}P^n), A) = 0 \,.$

With this, the statement about homology and cohomology groups with general coefficients follows with the universal coefficient theorem for ordinary homology (thm.) and for ordinary cohomology (thm.).

Finally to see the action of the cup product: by definition this is the composite

$\cup \;\colon\; H^\bullet(\mathbb{C}P^n, R) \otimes H^\bullet(\mathbb{C}P^n, R) \longrightarrow H^\bullet(\mathbb{C}P^n \times \mathbb{C}P^n , R) \overset{\Delta^\ast}{\longrightarrow} H^\bullet(\mathbb{C}P^n,R)$

of the “cross-product” map that appears in the Kunneth theorem, and the pullback along the diagonal $\Delta\colon \mathbb{C}P^n \to \mathbb{C}P^n \times \mathbb{C}P^n$.

Since, by the above, the groups $H^{2k}(\mathbb{C}P^n,R) \simeq R[2k]$ and $H^{2k+1}(\mathbb{C}P^n,R) = 0$ are free and finitely generated, the Kunneth theorem in ordinary cohomology applies (prop.) and says that the cross-product map above is an isomorphism. This shows that under cup product pairs of generators are sent to a generator, and so the statement $H^\bullet(\mathbb{C}P^n , R)\simeq R[c_1](c_1^{n+1})$ follows.

This also implies that the projection maps

$H^\bullet((\mathbb{C}P^\infty)_{2n+2}, R) = H^\bullet(\mathbb{C}P^{n+1}, R) \to H^\bullet(\mathbb{C}P^{n+}, R) = H^\bullet((\mathbb{C}P^\infty)_{2n}, R)$

are all epimorphisms. Therefore this sequence satisfies the Mittag-Leffler condition (def., exmpl.) and therefore the Milnor exact sequence for cohomology (prop.) implies the last claim to be proven:

\begin{aligned} H^\bullet(\mathbb{C}P^\infty, R) \\ & \simeq H^\bullet( \underset{\longleftarrow}{\lim}_n \mathbb{C}P^n , R) \\ &\simeq \underset{\longrightarrow}{\lim}_n H^\bullet(\mathbb{C}P^n, R) \\ &\simeq \underset{\longrightarrow}{\lim}_n ( R [c_1^E] / ((c_1)^{n+1}) ) \\ & \simeq R[ [ c_1 ] ] \,, \end{aligned}

where the last step is this prop..

###### Remark

There is in general a choice to be made in interpreting the cohomology groups of a multiplicative cohomology theory $E$ as a ring:

a priori $E^\bullet(X)$ is a sequence

$\{E^n(X)\}_{n \in \mathbb{Z}}$

of abelian groups, together with a system of group homomorphisms

$E^{n_1}(X) \otimes E^{n_2}(X) \longrightarrow E^{n_1 + n_2}(X) \,,$

one for each pair $(n_1,n_2) \in \mathbb{Z}\times\mathbb{Z}$.

In turning this into a single ring by forming formal sums of elements in the groups $E^n(X)$, there is in general the choice of whether allowing formal sums of only finitely many elements, or allowing arbitrary formal sums.

In the former case the ring obtained is the direct sum

$\oplus_{n \in \mathbb{N}} E^n(X)$

while in the latter case it is the Cartesian product

$\prod_{n \in \mathbb{N}} E^n (X) \,.$

These differ in general. For instance if $E$ is ordinary cohomology with integer coefficients and $X$ is complex projective space $\mathbb{C}P^\infty$, then (prop. )

$E^n(X) = \left\{ \array{ \mathbb{Z} & n \; even \\ 0 & otherwise } \right.$

and the product operation is given by

$E^{2{n_1}}(X)\otimes E^{2 n_2}(X) \longrightarrow E^{2(n_1 + n_2)}(X)$

for all $n_1, n_2$ (and zero in odd degrees, necessarily). Now taking the direct sum of these, this is the polynomial ring on one generator (in degree 2)

$\oplus_{n \in \mathbb{N}} E^n(X) \;\simeq\; \mathbb{Z}[c_1] \,.$

But taking the Cartesian product, then this is the formal power series ring

$\prod_{n \in \mathbb{N}} E^n(X) \;\simeq\; \mathbb{Z} [ [ c_1 ] ] \,.$

A priori both of these are sensible choices. The former is the usual choice in traditional algebraic topology. However, from the point of view of regarding ordinary cohomology theory as a multiplicative cohomology theory right away, then the second perspective tends to be more natural:

The cohomology of $\mathbb{C}P^\infty$ is naturally computed as the inverse limit of the cohomolgies of the $\mathbb{C}P^n$, each of which unambiguously has the ring structure $\mathbb{Z}[c_1]/((c_1)^{n+1})$. So we may naturally take the limit in the category of commutative rings right away, instead of first taking it in $\mathbb{Z}$-indexed sequences of abelian groups, and then looking for ring structure on the result. But the limit taken in the category of rings gives the formal power series ring (see here).

Incidentally, this is the default choice of ring structure for generalized multiplicative cohomology theories evaluated on $\mathbb{C}P^\infty$. In particular in complex oriented cohomology (see there) this choice is of paramount importance.

See also for instance remark 1.1. in Jacob Lurie: A Survey of Elliptic Cohomology.

#### Complex-oriented

###### Proposition

Given a complex oriented cohomology theory $(E^\bullet, c^E_1)$ (defn.), then there is an isomorphism of graded rings

$E^\bullet(\mathbb{C}P^\infty) \simeq E^\bullet(\ast)[ [ c_1^E ] ]$

between the $E$-cohomology ring of infinite-dimensional complex projective space and the formal power series in one generator of even degree over the $E$-cohomology ring of the point.

###### Proof

Using the CW-complex-structure on $\mathbb{C}P^\infty$ from prop. , given by inductively identifying $\mathbb{C}P^{n+1}$ with the result of attaching a single $2n$-cell to $\mathbb{C}P^n$. With this structure, the unique 2-cell inclusion $i \;\colon\; S^2 \hookrightarrow \mathbb{C}P^\infty$ is identified with the canonical map $S^2 \to B U(1)$.

Then consider the Atiyah-Hirzebruch spectral sequence for the $E$-cohomology of $\mathbb{C}P^n$.

$H^\bullet(\mathbb{C}P^n, E^\bullet(\ast)) \;\Rightarrow\; E^\bullet(\mathbb{C}P^n) \,.$

Since, by prop. , the ordinary cohomology with integer coefficients of projective space is

$H^\bullet(\mathbb{C}P^n, \mathbb{Z}) \simeq \mathbb{Z}[c_1]/((c_1)^{n+1}) \,,$

where $c_1$ represents a unit in $H^2(S^2, \mathbb{Z})\simeq \mathbb{Z}$, and since similarly the ordinary homology of $\mathbb{C}P^n$ is a free abelian group, hence a projective object in abelian groups (prop.), the Ext-group vanishes in each degree ($Ext^1(H_n(\mathbb{C}P^n), E^\bullet(\ast)) = 0$) and so the universal coefficient theorem (prop.) gives that the second page of the spectral sequence is

$H^\bullet(\mathbb{C}P^n, E^\bullet(\ast)) \simeq E^\bullet(\ast)[ c_1 ] / (c_1^{n+1}) \,.$

By the standard construction of the Atiyah-Hirzebruch spectral sequence (here) in this identification the element $c_1$ is identified with a generator of the relative cohomology

$E^2((\mathbb{C}P^n)_2, (\mathbb{C}P^n)_1) \simeq \tilde E^2(S^2)$

(using, by the above, that this $S^2$ is the unique 2-cell of $\mathbb{C}P^n$ in the standard cell model).

This means that $c_1$ is a permanent cocycle of the spectral sequence (in the kernel of all differentials) precisely if it arises via restriction from an element in $E^2(\mathbb{C}P^n)$ and hence precisely if there exists a complex orientation $c_1^E$ on $E$. Since this is the case by assumption on $E$, $c_1$ is a permanent cocycle. (For the fully detailed argument see (Pedrotti 16).)

The same argument applied to all elements in $E^\bullet(\ast)[c]$, or else the $E^\bullet(\ast)$-linearity of the differentials (prop.), implies that all these elements are permanent cocycles.

Since the AHSS of a multiplicative cohomology theory is a multiplicative spectral sequence (prop.) this implies that the differentials in fact vanish on all elements of $E^\bullet(\ast) [c_1] / (c_1^{n+1})$, hence that the given AHSS collapses on the second page to give

$\mathcal{E}_\infty^{\bullet,\bullet} \simeq E^\bullet(\ast)[ c_1^{E} ] / ((c_1^E)^{n+1})$

or in more detail:

$\mathcal{E}_\infty^{p,\bullet} \simeq \left\{ \array{ E^\bullet(\ast) & \text{if}\; p \leq 2n \; and\; even \\ 0 & otherwise } \right. \,.$

Moreover, since therefore all $\mathcal{E}_\infty^{p,\bullet}$ are free modules over $E^\bullet(\ast)$, and since the filter stage inclusions $F^{p+1} E^\bullet(X) \hookrightarrow F^{p}E^\bullet(X)$ are $E^\bullet(\ast)$-module homomorphisms (prop.) the extension problem trivializes, in that all the short exact sequences

$0 \to F^{p+1}E^{p+\bullet}(X) \longrightarrow F^{p}E^{p+\bullet}(X) \longrightarrow \mathcal{E}_\infty^{p,\bullet} \to 0$

split (since the Ext-group $Ext^1_{E^\bullet(\ast)}(\mathcal{E}_\infty^{p,\bullet},-) = 0$ vanishes on the free module, hence projective module $\mathcal{E}_\infty^{p,\bullet}$).

In conclusion, this gives an isomorphism of graded rings

$E^\bullet(\mathbb{C}P^n) \simeq \underset{p}{\oplus} \mathcal{E}_\infty^{p,\bullet} \simeq E^\bullet(\ast)[ c_1 ] / ((c_1^{E})^{n+1}) \,.$

A first consequence is that the projection maps

$E^\bullet((\mathbb{C}P^\infty)_{2n+2}) = E^\bullet(\mathbb{C}P^{n+1}) \to E^\bullet(\mathbb{C}P^{n+}) = E^\bullet((\mathbb{C}P^\infty)_{2n})$

are all epimorphisms. Therefore this sequence satisfies the Mittag-Leffler condition (def., exmpl.) and therefore the Milnor exact sequence for generalized cohomology (prop.) finally implies the claim:

\begin{aligned} E^\bullet(B U(1)) & \simeq E^\bullet(\mathbb{C}P^\infty) \\ & \simeq E^\bullet( \underset{\longleftarrow}{\lim}_n \mathbb{C}P^n ) \\ &\simeq \underset{\longrightarrow}{\lim}_n E^\bullet(\mathbb{C}P^n) \\ &\simeq \underset{\longrightarrow}{\lim}_n ( E^\bullet(\ast) [c_1^E] / ((c_1^E)^{n+1}) ) \\ & \simeq E^\bullet(\ast)[ [ c_1^E ] ] \,, \end{aligned}

where the last step is this prop..

## References

Computation of the stable homotopy groups of $\mathbb{C}P^\infty$ is due to

• R. E. Mosher, Some stable homotopy of complex projective space, Topology 7 (1968), 179-193 (pdf)

• Graeme Segal, The stable homotopy of complex of projective space, The quarterly journal of mathematics (1973) 24 (1): 1-5. (pdf, doi:10.1093/qmath/24.1.1)