# Obstructions to almost complex structures following Massey

Michael Albanese  University of Waterloo, Faculty of Mathematics, 200
University Ave W, Waterloo, N2L 3G1, Ontario<EMAIL_ADDRESS><EMAIL_ADDRESS>and  Aleksandar Milivojević

###### Abstract.

We provide proofs of two theorems stated by Massey in 1961, concerning the
obstructions to finding complex structures on real vector bundles. In
addition, we determine the second obstruction to a complex structure on a rank
six orientable real vector bundle. The obstructions are fractional parts of
integral Stiefel–Whitney classes, and a fourth of an appropriate combination
of Pontryagin, Chern, and Euler classes.

###### Key words and phrases:

Almost complex structures, obstruction theory

###### 2020 Mathematics Subject Classification:

32Q60, 55S35, 53C15

##  1\. Introduction

In a 1961 paper, Massey [M61] studied the obstructions to reducing the
structure group of an oriented real vector bundle over a CW complex to the
unitary group. This includes as a special case of interest the obstructions to
finding an almost complex structure on a smooth oriented $2n$–manifold. The
problem of reducing the structure group can be phrased as finding a section of
the natural associated bundle with fiber $SO(2n)/U(n)$.

By the time of Massey’s paper, it was understood how to relate the
obstructions arising in the problem of lifting through the map $BU(n)\to
BSO(2n)$ to the cohomology classes classifying the stages of its
Moore–Postnikov system. This approach was developed by Hermann [He59, He60],
whose technique Massey employed [M61, Section 3]. Finding a section of the
$SO(2n)/U(n)$ bundle over a certain skeleton of the CW complex is tantamount
to lifting the classifying map of the tangent bundle to a certain stage of the
Moore–Postnikov system of the map $BU(n)\to BSO(2n)$.

On a complex vector bundle, one observes certain relations that must hold on
the level of characteristic classes. First of all, since the Chern classes
$c_{i}$ reduce modulo two to the even-indexed Stiefel–Whitney classes
$w_{2i}$, one sees that the odd–indexed integral Stiefel–Whitney classes
$W_{2i+1}$ must vanish, being the integral Bockstein of $w_{2i}$. Furthermore,
the Pontryagin classes are related to the Chern classes by the formula

$1-p_{1}+p_{2}-\cdots=(1-c_{1}+c_{2}-\cdots)(1+c_{1}+c_{2}+\cdots),$

yielding

$(-1)^{k}p_{k}=\sum_{i+j=2k}(-1)^{i}c_{i}c_{j}.$

Lastly, the top degree Chern class $c_{n}$ on a complex rank $n$ vector bundle
coincides with its Euler class.

Massey observed that even though the above are necessary conditions for having
an almost complex structure, the actual obstructions in the relevant degrees
are generally classes such that the vanishing of some non-trivial multiple of
them corresponds to the above relations. More precisely:

######  Theorem.

[M61, Theorems I and II] Let $\xi\to X$ be an orientable real vector bundle of
rank $2n$, whose structure group we assume has been reduced to $SO(2n)$, over
a CW complex $X$. Consider the associated $SO(2n)/U(n)$ bundle $\theta$,
sections of which correspond to reductions of the structure group of $\xi$ to
$U(n)$. If $s$ is a section of $\theta$ over the $q$–skeleton of $X$, then the
obstruction to extending over the $(q+1)$–skeleton is a cohomology class
$\mathfrak{o}_{q+1}\in H^{q+1}(X;\pi_{q}SO(2n)/U(n))$. We have the following:

  1. (1)

If $q=4k+2$ and $q<2n-1$, then

$W_{4k+3}(\xi)=\begin{cases}(2k)!\,\mathfrak{o}_{4k+3}\ \ \ \
&k\,\mathrm{even},\\\ \tfrac{1}{2}(2k)!\,\mathfrak{o}_{4k+3}\ \ \ \
&\,k\,\mathrm{odd}.\end{cases}$

  2. (2)

Assume $n=2k$ and $q=2n-1=4k-1$. The obstruction
$\mathfrak{o}_{2n}=\mathfrak{o}_{4k}$ is an integral class if $k$ is odd,
while if $k$ is even, it is a pair consisting of an integral class and a mod 2
class. If $k$ is odd, then

$\sum_{i+j=2k}(-1)^{i}c_{i}c_{j}-(-1)^{k}p_{k}=4\mathfrak{o}_{2n},$

while for $k$ even, this same formula holds with $\mathfrak{o}_{2n}$ replaced
by its integral component. In the formula, $c_{0},\ldots,c_{n-1}$ are the
Chern classes of the induced $U(n)$ bundle over the $(2n-1)$–skeleton of $X$,
while $c_{n}$ denotes the Euler class of $\xi$.

Given that all the odd-indexed integral Stiefel–Whitney classes vanish on an
almost complex manifold, one might wonder whether the classes $W_{4k+1}$
appear in certain obstructions. However, as we will see in Proposition 1,
their vanishing is ensured by the vanishing of the classes $W_{4k+3}$.

As an example of statement (1) above, the obstruction to extending an almost
complex structure on the 10–skeleton of a 12–manifold over its 11–skeleton is
not $W_{11}$, but rather a class $\mathfrak{o}_{11}$ satisfying
$W_{11}=24\mathfrak{o}_{11}$. Likewise, as an example of statement (2), the
obstruction to extending an almost complex structure on the three–skeleton of
a four–manifold over the four–skeleton (i.e., the whole manifold) is
$\tfrac{1}{4}(p_{1}-c_{1}^{2}+2e)$, where $e$ is the Euler class. Of course,
on an orientable four–manifold, this class vanishes if and only if
$p_{1}-c_{1}^{2}+2e$ vanishes, but on a general CW complex this need not be
the case.

Though Massey included remarks in [M61] on the ingredients of the proofs of
the above results, the arguments are not made explicit, and we take this
opportunity to fill in the details. Massey also addressed a particular
obstruction not covered by the above, namely the degree eight obstruction to
extending an almost complex structure on the seven–skeleton over the
eight–skeleton, in the case of $2n\geq 10$ [M61, Theorem III]. This mod 2
obstruction to a stable almost complex structure, in the case of closed
manifolds, was later studied in detail by Thomas [T67, Theorem 1.2] and Heaps
[H60, §1], and we will touch upon it only briefly in the last section.

It is wholly possible that some of the arguments we write out here for proving
[M61, Theorems I and II] were not the ones Massey had in mind, particularly
since we do not make use of any of the general statements (a)–(f) in [M61,
Section 3] listed as ingredients of the proofs of [M61, Theorems I–III],
though perhaps these were mostly intended for the proof of [M61, Theorem III].
Our method, which only employs technology known at the time of [M61], is
inspired by an argument of Kervaire used in [K59, Lemma 1.1], which is
mentioned in [M61, Remark 3].

### Acknowledgements

We would like to thank Mark Grant for providing helpful references.

##  2\. Moore–Postnikov systems

We begin with some general topological preliminaries we will make use of
heavily, namely Postnikov systems of spaces, and their relative versions which
we refer to as Moore–Postnikov systems.

###  2.1. The Postnikov system of a space

For the content of this subsection, we refer the reader to e.g. [Wh, Chapter
IX]. Given a simply connected CW complex $F$, there is a sequence of spaces
$F_{n}$, $n\geq 2$, maps $F\xrightarrow{g_{n}}F_{n}$, and maps
$F_{n}\xrightarrow{f_{n}}F_{n-1}$, such that:

  * •

$\pi_{>n}F_{n}=0$.

  * •

The map $F\xrightarrow{g_{n}}F_{n}$ induces an isomorphism on $\pi_{\leq n}$.

  * •

The map $F_{n}\xrightarrow{f_{n}}F_{n-1}$ is a principal fibration with fiber
$K(\pi_{n}F,n)$, classified by a map $F_{n-1}\to K(\pi_{n}F,n+1)$. The
corresponding element in $H^{n+1}(F_{n-1};\pi_{n})$ is called the
$k$–invariant in degree $n+1$.

  * •

$f_{n}\circ g_{n}$ is homotopic to $g_{n-1}$.

${F}$${F_{n}}$${\vdots}$${{K(\pi_{4}F,4)}}$${F_{4}}$${{K(\pi_{3}F,3)}}$${F_{3}}$${{F_{2}=K(\pi_{2}F,2)}}$$\scriptstyle{g_{n}}$$\scriptstyle{f_{n}}$$\scriptstyle{f_{5}}$$\scriptstyle{f_{4}}$$\scriptstyle{f_{3}}$

This data is referred to as a _Postnikov system_ for the space $F$. We refer
to $F\xrightarrow{g_{n}}F_{n}$ as the $n^{\mathrm{th}}$ Postnikov map of $F$.
Denoting the homotopy fiber of $F\xrightarrow{g_{n}}F_{n}$ by
$F_{n}^{\prime}$, we have $\pi_{\leq n}F_{n}^{\prime}=0$ and the map
$F_{n}^{\prime}\to F$ induces an isomorphism on $\pi_{>n}$. A map $X\to
F_{n-1}$ lifts through $f_{n}$ if and only if the pullback of the
$k$-invariant in $H^{n+1}(F_{n-1};\pi_{n})$ to $H^{n+1}(X;\pi_{n})$ is the
zero class.

###  2.2. The Moore–Postnikov system of a map

For the content of this subsection, we refer the reader to [He60], in
particular Section 7, and to [May, Chapter II.8], [Wh, Chapter IX.6], [Ba,
Chapter 5.3].

Consider a fibration consisting of simply connected CW complexes $F\to
E\xrightarrow{p}B$. Define $E_{1}=B$. There are spaces $E_{n}$, $n\geq 2$, and
fibrations $F_{n}\to E_{n}\xrightarrow{p_{n}}B$, $F_{n}^{\prime}\to
E\xrightarrow{h_{n}}E_{n}$, and $K(\pi_{n}F,n)\to
E_{n}\xrightarrow{e_{n}}E_{n-1}$ such that

  * •

the diagram

(1)
${E}$${E_{n}}$${B}$${B}$$\scriptstyle{p}$$\scriptstyle{h_{n}}$$\scriptstyle{p_{n}}$$\scriptstyle{id}$

commutes (up to homotopy), and the induced map of fibers $F\to F_{n}$ is an
$n^{\mathrm{th}}$ Postnikov map for $F$ as described in Section 2.1. In
particular, the Moore–Postnikov system of a constant map to a point induces a
Postnikov system for its domain.

  * •

$E_{n}\xrightarrow{e_{n}}E_{n-1}$ is a principal $K(\pi_{n}F,n)$ fibration
such that the diagram

${F_{n}}$${F_{n-1}}$${E_{n}}$${E_{n-1}}$${B}$${B}$$\scriptstyle{f_{n}}$$\scriptstyle{p_{n}}$$\scriptstyle{e_{n}}$$\scriptstyle{p_{n-1}}$$\scriptstyle{id}$

is homotopy commutative, where $f_{n}$ is the induced map of homotopy fibers.
Note that if $\pi_{n}F=0$, then $E_{n}\xrightarrow{e_{n}}E_{n-1}$ is a
homotopy equivalence.

We can complete diagram (1) to a three–by–three diagram with the rows and
columns being fibrations, see e.g. [N10, Section 3.2], and from the long exact
sequences in homotopy groups we see that the induced map from the homotopy
fiber of $g_{n}$ (as in Section 2.1) to the homotopy fiber $F_{n}^{\prime}$ of
$h_{n}$ is a homotopy equivalence, justifying the notation $F_{n}^{\prime}$
for the homotopy fiber of $h_{n}$.

${F_{n}^{\prime}}$${F}$${F_{n}}$${F_{n}^{\prime}}$${E}$${E_{n}}$${\ast}$${B}$${B}$$\scriptstyle{\sim}$$\scriptstyle{g_{n}}$$\scriptstyle{p}$$\scriptstyle{h_{n}}$$\scriptstyle{p_{n}}$$\scriptstyle{id}$

Recall, the homotopy groups of $F_{n}^{\prime}$ are trivial in degrees $\leq
n$, while they are isomorphic to those of $F$ in degrees $>n$. In particular,
the map $E\xrightarrow{h_{n}}E_{n}$ induces an isomorphism on $\pi_{\leq n}$
and a surjection on $\pi_{n+1}$, so $h_{n}^{*}$ is an isomorphism on
cohomology in degrees $\leq n$ and injective in degree $n+1$.

The data of the fibrations $F_{n}\to E_{n}\xrightarrow{p_{n}}B$,
$F_{n}^{\prime}\to E\xrightarrow{h_{n}}E_{n}$, and $K(\pi_{n}F,n)\to
E_{n}\xrightarrow{e_{n}}E_{n-1}$ satisfying the above conditions is called a
_Moore–Postnikov system_ for $F\to E\xrightarrow{p}B$. The principal
$K(\pi_{n}F,n)$ fibration $E_{n}\xrightarrow{e_{n}}E_{n-1}$ is classified by a
map $E_{n-1}\to K(\pi_{n}F,n+1)$, with the corresponding cohomology class
$\kappa$ of $E_{n-1}$ referred to as the $k$-invariant in degree $n+1$ of the
Moore–Postnikov system. The $k$-invariant pulls back to that of the Postnikov
system of $F$, i.e. the diagram

${F_{n}}$${E_{n}}$${F_{n-1}}$${E_{n-1}}$${{K(\pi_{n}F,n+1)}}$${{K(\pi_{n}F,n+1)}}$$\scriptstyle{f_{n}}$$\scriptstyle{e_{n}}$

is homotopy commutative.

Since $E_{n}$ is the homotopy fiber of the map $E_{n-1}\to K(\pi_{n}F,n+1)$,
we have $e_{n}^{*}(\kappa)=0$. In fact, from the Serre long exact sequence we
have that $\kappa$ generates the kernel of
$H^{n+1}(E_{n-1};\pi_{n}F)\xrightarrow{e_{n}^{*}}H^{n+1}(E_{n};\pi_{n}F).$

###  2.3. Naturality of the Moore–Postnikov system

Naturality of the ordinary Postnikov system (i.e. the Moore–Postnikov system
of the map to a point) is discussed in detail in [Ka63]. For an explicit
treatment of naturality of the relative case, we point the reader to [Ba,
(5.3.5)]; see also [He60, Section 3].

Suppose we have a fibration $F^{1}\to E^{1}\xrightarrow{p_{1}}B^{1}$ mapping
to another fibration $F^{2}\to E^{2}\xrightarrow{p_{2}}B^{2}$, i.e. a homotopy
commutative diagram

${F^{1}}$${F^{2}}$${E^{1}}$${E^{2}}$${B^{1}}$${B^{2}}$$\scriptstyle{p_{1}}$$\scriptstyle{p_{2}}$

Then, we can find Moore–Postnikov systems $\\{E^{1}_{n}\\}$ for $E^{1}\to
B^{1}$ and $\\{E^{2}_{n}\\}$ for $E^{2}\to B^{2}$ and a homotopy commutative
diagram

${F^{1}}$${F^{2}}$${\vdots}$${E^{1}}$${\vdots}$${E^{2}}$${\vdots}$${\vdots}$${F_{3}^{1}}$${F_{3}^{2}}$${E_{3}^{1}}$${E_{3}^{2}}$${{K(\pi_{4}F^{1},5)}}$${{K(\pi_{4}F^{1},5)}}$${F_{2}^{1}}$${F_{2}^{2}}$${E_{2}^{1}}$${E_{2}^{2}}$${{K(\pi_{3}F^{1},4)}}$${{K(\pi_{3}F^{2},4)}}$${B^{1}}$${B^{2}}$${{K(\pi_{2}F^{1},3)}}$${{K(\pi_{2}F^{1},3)}}$

where the columns in the back are the induced Postnikov systems for $F^{1}$
and $F^{2}$.

##  3\. More preliminaries, and a proof of Ehresmann’s theorem

Let $X$ be a CW complex, and let $\xi$ be an oriented real rank $2n$ vector
bundle over $X$, whose structure group we have reduced to $SO(2n)$ without
loss of generality. Let $X\xrightarrow{f}BSO(2n)$ be the homotopy class of
maps classifying this bundle up to isomorphism. The problem of determining
whether $\xi$ admits an almost complex structure is equivalent to the problem
of lifting $f$ through the natural map $BU(n)\to BSO(2n)$.

${BU(n)}$${X}$${BSO(2n)}$$\scriptstyle{f}$

This is an obstruction-theoretic problem, which we address by decomposing
$BU(n)\to BSO(2n)$ into a Moore–Postnikov system and determining the
obstructions to lifting stage-by-stage.

###  3.1. Some relevant homotopy groups

We list the relevant homotopy groups of $SO(2n)/U(n)$ for our lifting problem
here. First of all, since $U(n+1)$ acts transitively on $S^{2n+1}$ with
isotropy $U(n)$, we have the fiber bundle $U(n)\to U(n+1)\to S^{2n+1}$, giving
us a fibration $S^{2n+1}\to BU(n)\to BU(n+1)$. From here one sees that the
natural map $BU(n)\to BU(n+1)$ induces an isomorphism on $\pi_{\leq 2n}$.
Hence the homotopy groups $\pi_{\leq 2n}BU(n)$ are stable, and in particular
isomorphic, via the map induced by the inclusion, to $\pi_{\leq 2n}BU$. The
homotopy groups of $BU$ are known by Bott’s periodicity theorem: concretely,
$\pi_{2k}BU\cong\mathbb{Z}$ and $\pi_{2k+1}BU=0$ for all $k$. Similarly, one
obtains that the groups $\pi_{\leq 2n-1}BSO(2n)$ are stable. The homotopy
groups of the stable space $BO$ are eightfold-periodic: starting from
$\pi_{1}BO$ they read
$\mathbb{Z}_{2},\mathbb{Z}_{2},0,\mathbb{Z},0,0,0,\mathbb{Z}$. The homotopy
groups of $BSO$ differ only in $\pi_{1}$, where we have $\pi_{1}BSO=0$.

From the map of fibrations

${SO(2n)/U(n)}$${SO/U}$${BU(n)}$${BU}$${BSO(2n)}$${BSO}$

and the induced map of long exact sequences in homotopy groups, we see that
the groups $\pi_{\leq 2n-2}SO(2n)/U(n)$ are stable using the five lemma, i.e.
they are isomorphic via the natural map to those of $SO/U$. A part of Bott’s
periodicity theorem also gives us that $SO/U$ has the homotopy type of a
connected component of the based loop space $\Omega O$. Hence its first
several homotopy groups, starting with $\pi_{1}$, are given by

$0,\mathbb{Z},0,0,0,\mathbb{Z},\mathbb{Z}_{2},\mathbb{Z}_{2},0,\mathbb{Z},0,0,0,\mathbb{Z},\ldots$

In particular, we have

$\pi_{4k+2}SO/U\cong\mathbb{Z}.$

The first unstable homotopy group of $SO(2n)/U(n)$, namely
$\pi_{2n-1}SO(2n)/U(n)$, depends on the mod 4 class of $n$, and is given by

(2)
$\pi_{2n-1}(SO(2n)/U(n))=\begin{cases}\mathbb{Z}\oplus\mathbb{Z}_{2}&n\equiv
0\bmod 4,\\\ \mathbb{Z}_{(n-1)!}&n\equiv 1\bmod 4,\\\ \mathbb{Z}&n\equiv
2\bmod 4,\\\ \mathbb{Z}_{\tfrac{1}{2}(n-1)!}&n\equiv 3\bmod 4.\end{cases}$

This is summarized in [M61, Lemma 1], and the reference to (then forthcoming)
work of Bruno Harris therein is to [Ha63].

###  3.2. The first obstruction to an almost complex structure

Since $SO(2)/U(1)$ is a point, there are no obstructions to reducing the
structure group. From now on we suppose $n\geq 2$. Then $\pi_{2}SO(2n)/U(n)$
is stable, and hence isomorphic to $\mathbb{Z}$. Therefore, the first
$k$-invariant in the Moore–Postnikov system for $BU(n)\to BSO(2n)$ is a
cohomology class in $H^{3}(BSO(2n);\mathbb{Z})$.

Since $BSO(2n)$ is simply connected, by the Hurewicz theorem we have that

$H_{2}(BSO(2n);\mathbb{Z})\cong\pi_{2}(BSO(2n);\mathbb{Z})\cong\mathbb{Z}_{2},$

and that $\pi_{3}(BSO(2n))=0$ surjects onto $H_{3}(BSO(2n);\mathbb{Z})$, i.e.
the latter group is trivial. By the universal coefficient theorem, we have
$H^{3}(BSO(2n);\mathbb{Z})\cong\mathbb{Z}_{2}$.

The non-zero element in this group is the third integral Stiefel–Whitney class
$W_{3}=\beta w_{2}$. Denote by $E_{2}$ the second stage of the Moore–Postnikov
system for $BU(n)\to BSO(2n)$, as in Section 2.2. Since $BU(n)\to E_{2}$ has
two–connected fiber, it induces an injection on $H^{3}(-;\mathbb{Z})$. As
$H^{3}(BU(n);\mathbb{Z})=0$, we have $H^{3}(E_{2};{\mathbb{Z}})=0$. The
$k$-invariant for the fibration $E_{2}\to BSO(2n)$ generates the kernel of the
induced map on $H^{3}(-;{\mathbb{Z}})$, and therefore it is the class $W_{3}$.
We note that $W_{3}$ of an orientable bundle vanishes precisely when $w_{2}$
has an integral lift, i.e. when $w_{2}$ is the mod 2 reduction of an integral
cohomology class, which would be the first Chern class of the almost complex
structure we would obtain if we could lift to $BU(n)$. The homotopy fiber of
$BSO(2n)\xrightarrow{W_{3}}K(\mathbb{Z},3)$ can be identified with
$BSpin^{c}(2n)$, the classifying space of the Lie group $Spin^{c}(2n)$, though
we will not make use of this here.

###  3.3. The second obstruction to an almost complex structure

Supposing $W_{3}$ of the bundle is $0$, we now have to distinguish three
cases: $n=2$, $n=3$, and $n\geq 4$. If $n=2$, then the second (and, in the
case of $X$ having dimension $\leq 4$, last) obstruction to lifting to $BU(2)$
lies in $H^{4}(X;\pi_{3}SO(4)/U(2))\cong H^{4}(X;\mathbb{Z})$. Since $W_{3}=0$
we can choose a lift $X\to E_{2}$ of the map $\tau$, which yields an integral
lift $c$ of $w_{2}$ on $X$. The second obstruction is then obtained by pulling
back the $k$-invariant $E_{2}\to K(\mathbb{Z},4)$ to $X$ via this lift (see
e.g. [Wh, Chapter IX.7]). Different choices of lift will generally give
different classes in $H^{4}(X;\mathbb{Z})$. A theorem of Wu [W52, Théorème
IV.10] tells us that, on a closed orientable four–manifold, this obstruction
class vanishes if $c^{2}-2e-p_{1}$ vanishes. For an $SO(4)$ bundle over a
general CW complex, $c^{2}-2e-p_{1}$ is four times this second obstruction;
see 6.

For $n=3$, we have $SO(6)/U(3)=\mathbb{CP}^{3}$, so the second obstruction
lies in $H^{8}(X;\pi_{7}\mathbb{CP}^{3})\cong H^{8}(X;{\mathbb{Z}})$, and we
will address this case later, in Proposition 9, after proving the main
results. Let us focus on the case $n\geq 4$, where the second obstruction to
lifting to $BU(n)$ is stable, and lies in $H^{7}(X;\pi_{6}SO(2n)/U(n))\cong
H^{7}(X;\mathbb{Z})$. By the obstruction being stable, we mean that it is
obtained by pulling back the corresponding obstruction for lifting the
stabilized real bundle $X\to BSO$ through $BU\to BSO$, where one has the
commutative diagram

${SO(2n)/U(n)}$${SO/U}$${BU(n)}$${BU}$${BSO(2n)}$${BSO}$

We will not make use of this observation so we will not go into the details.

As with $W_{3}$, we see that since $W_{7}$ pulls back to the zero class in
$BU(n)$, we must have $W_{7}=0$ for a complex vector bundle (since $c_{3}$
will be an integral lift of $w_{6}$). However, it may be the case that when
pulling back $W_{7}$ to $E_{2}$, the class becomes divisible by an integer
$\geq 2$, and the relevant $k$-invariant is in fact a “fractional part” of
$W_{7}$. Even though this does not happen in the case of $W_{7}$ as we will
now see, later on we will observe that this phenomenon does occur in higher
degrees, for $W_{11},W_{15},\ldots$ One might also find it curious that
$W_{5}$ has been skipped; as soon as $W_{3}=0$, we have that $W_{5}=0$. Let us
record a proof of this phenomenon here; the method will also become relevant
again in Section 5.

######  Proposition 1.

Let $\xi\to X$ be an orientable real vector bundle such that $W_{4k-1}=0$ for
$k\leq m$. Then $W_{4m+1}=0$.

###### Proof.

We will use the following formula [T60, Theorem C] attributed to Wu111In the
notation of loc. cit., $W_{i}$ denotes the mod 2 Stiefel–Whitney class
$w_{i}$., involving the Pontryagin square operation
$H^{2m}(X;\mathbb{Z}_{2})\xrightarrow{\mathfrak{P}}H^{4m}(X;\mathbb{Z}_{4})$,
which for orientable bundles (i.e. $w_{1}=0$) becomes

$\mathfrak{P}(w_{2m})=\rho_{4}(p_{m})+\theta_{2}\left(\sum_{j=0}^{m-1}w_{2j}w_{4m-2j}\right).$

Here, $\rho_{4}$ denotes mod 4 reduction, and $\theta_{2}$ denotes the map on
cohomology $H^{4m}(X;\mathbb{Z}_{2})\to H^{4m}(X;\mathbb{Z}_{4})$ induced by
the inclusion $\mathbb{Z}_{2}\hookrightarrow\mathbb{Z}_{4}$. Since the classes
$W_{\leq 4m-1}$ vanish, the classes $w_{\leq 4m-2}$ are the mod 2 reductions
of integral classes. Let us denote by $c_{k}$ some fixed integral lift of
$w_{2k}$, for $k\leq 2m$. Since $c_{m}$ is an integral lift of $w_{2m}$, a
general property of the Pontryagin square gives us that
$\mathfrak{P}(w_{2m})=\rho_{4}(c_{m}^{2})$:

######  Lemma 2.

Let $\tilde{u}$ be an integral lift of a mod 2 cohomology class $u$. Then
$\mathfrak{P}(u)=\rho_{4}(\tilde{u}^{2})$.

###### Proof.

By construction, the Pontryagin square satisfies

$\mathfrak{P}(\rho_{2}^{4}(x))=x^{2}$

for any mod 4 class $x$, where $\rho_{2}^{4}$ denotes the mod 2 map
$H^{*}(-;{\mathbb{Z}}_{4})\to H^{*}(-;{\mathbb{Z}}_{2})$; see e.g. [BT62,
(1.3)]. Note that $\rho_{2}^{4}(\rho_{4}(\tilde{u}))=\rho_{2}(\tilde{u})=u$.
Therefore

$\mathfrak{P}(u)=\mathfrak{P}(\rho_{2}^{4}(\rho_{4}(\tilde{u})))=\rho_{4}(\tilde{u})^{2}=\rho_{4}(\tilde{u}^{2}).$

∎

Now, consider the following commutative diagram of short exact sequences of
abelian groups

${0}$${{\mathbb{Z}}}$${{\mathbb{Z}}}$${{\mathbb{Z}}_{2}}$${0}$${0}$${{\mathbb{Z}}_{2}}$${{\mathbb{Z}}_{4}}$${{\mathbb{Z}}_{2}}$${0}$$\scriptstyle{\mathrm{mod}\,2}$$\scriptstyle{\mathrm{mod}\,4}$$\scriptstyle{=}$

and the following commutative diagram coming from the induced map of long
exact sequences in cohomology

${H^{4m-1}(X;{\mathbb{Z}}_{2})}$${H^{4m}(X;{\mathbb{Z}})}$${H^{4m}(X;{\mathbb{Z}})}$${H^{4m}(X;{\mathbb{Z}}_{2})}$${H^{4m-1}(X;{\mathbb{Z}}_{2})}$${H^{4m}(X;{\mathbb{Z}}_{2})}$${H^{4m}(X;{\mathbb{Z}}_{4})}$${H^{4m}(X;{\mathbb{Z}}_{2})}$$\scriptstyle{\beta}$$\scriptstyle{=}$$\scriptstyle{\rho_{2}}$$\scriptstyle{\times
2}$$\scriptstyle{\rho_{4}}$$\scriptstyle{\rho_{2}}$$\scriptstyle{=}$$\scriptstyle{Sq^{1}}$$\scriptstyle{\theta_{2}}$$\scriptstyle{\rho_{2}^{4}}$

In particular, $\theta_{2}(\rho_{2}(x))=\rho_{4}(2x)$ for any class $x$, from
the commutativity of the middle square. Now, since
$\sum_{j=1}^{m-1}w_{2j}w_{4m-2j}=\rho_{2}\left(\sum_{j=1}^{m-1}c_{j}c_{2m-j}\right)$,
we conclude that

$\theta_{2}\left(\sum_{j=1}^{m-1}w_{2j}w_{4m-2j}\right)=\rho_{4}\left(2\sum_{j=1}^{m-1}c_{j}c_{2m-j}\right).$

Recall $\mathfrak{P}(w_{2m})=\rho_{4}(c_{m}^{2})$. Since
$\theta_{2}(w_{4m})=\mathfrak{P}(w_{2m})-\rho_{4}(p_{m})-\theta_{2}\left(\sum_{j=1}^{m-1}w_{2j}w_{4m-2j}\right),$
we therefore have

$\theta_{2}(w_{4m})=\rho_{4}\left(c_{m}^{2}-p_{m}-2\sum_{j=1}^{m-1}c_{j}c_{2m-j}\right).$

Since $\rho_{2}^{4}(\theta_{2}(w_{4m}))=0$ by exactness, we have

$\rho_{2}\left(c_{m}^{2}-p_{m}-2\sum_{j=1}^{m-1}c_{j}c_{2m-j})\right)=0$

using the right-most square of the diagram. Therefore, there is a class $x\in
H^{4m}(X;{\mathbb{Z}})$ such that

$2x=c_{m}^{2}-p_{m}-2\sum_{j=1}^{m-1}c_{j}c_{2m-j}.$

Then $\theta_{2}(\rho_{2}(x))=\rho_{4}(2x)=\theta_{2}(w_{4m})$. Therefore,
$\rho_{2}(x)+w_{4m}\in\ker(\theta_{2})=\mathrm{im}(Sq^{1})$. Hence, there is
some $y\in H^{4m-1}(X;{\mathbb{Z}}_{2})$ such that
$Sq^{1}(y)=\rho_{2}(x)+w_{4m}$. Since $Sq^{1}=\rho_{2}\circ\beta$, we have

$w_{4m}=\rho_{2}(x+\beta(y)).$

That is, $x+\beta(y)$ is an integral lift of $w_{4m}$, and hence $W_{4m+1}=0$.
∎

We return to the second obstruction to lifting through $BU(n)\to BSO(2n)$, for
$n\geq 4$. We will identify it as $W_{7}$ in another way in Section 4, as part
of proving the more general 4. However, here we address it directly, as Massey
points out [M61, Remark 1], this obstruction was known earlier, yet stated
without proof, by Ehresmann [E50].

Denote by $p$ the map $E_{2}\to BSO(2n)$. To show that the $k$-invariant
$E_{2}\to K(\mathbb{Z},7)$ is $p^{*}W_{7}$, we calculate the relevant
cohomology of $E_{2}$ from the Serre spectral sequence associated to the
fibration $K(\mathbb{Z},2)\to E_{2}\to BSO(2n)$. For the integral cohomology
of $BSO(2n)$, we point the reader to [B82, Theorem 1.6] for a detailed
treatment. Below we have the relevant terms of the second page of the spectral
sequence for $n\geq 5$. The only difference in the case of $n=4$ is that the
Euler class appears in degree eight. However, it does not affect our
computation, and so our conclusion will hold in this case as well.

${6}$${\mathbb{Z}\langle\alpha^{3}\rangle}$${5}$${4}$${\large\mathbb{Z}\langle\alpha^{2}\rangle}$${\mathbb{Z}_{2}\langle
W_{3}\alpha^{2}\rangle}$${\mathbb{Z}\langle
p_{1}\alpha^{2}\rangle}$${3}$${2}$${\mathbb{Z}\langle\alpha\rangle}$${\mathbb{Z}_{2}\langle
W_{3}\alpha\rangle}$${\mathbb{Z}\langle
p_{1}\alpha\rangle}$${\mathbb{Z}_{2}\langle
W_{5}\alpha\rangle}$${1}$${0}$${\mathbb{Z}\langle
1\rangle}$${\mathbb{Z}_{2}\langle W_{3}\rangle}$${\mathbb{Z}\langle
p_{1}\rangle}$${\mathbb{Z}_{2}\langle W_{5}\rangle}$${\mathbb{Z}_{2}\langle
W_{3}^{2}\rangle}$${{\mathbb{Z}_{2}\langle
W_{3}p_{1},W_{7},\beta(w_{2}w_{4})\rangle}}$${\mathbb{Z}_{2}\langle
W_{3}W_{5}\rangle\oplus\mathbb{Z}\langle
p_{1}^{2},p_{2}\rangle}$${0}$${1}$${2}$${3}$${4}$${5}$${6}$${7}$${8}$

Note that, since the cohomology of the fiber is concentrated in even degrees,
only the odd-indexed pages of the spectral sequence can have non-trivial
differentials. On the third page, since $p^{*}W_{3}=0$, we must have
$d_{3}(\alpha)=W_{3}$. Note that then $d_{3}(\alpha^{2})=2\alpha W_{3}=0$, so
$\alpha^{2}$ together with $2\alpha$ survives to the fourth, and hence fifth,
page. We have
$d_{3}(\alpha^{3})=3W_{3}\alpha^{2}=W_{3}\alpha^{2},\,\,\,\,d_{3}(W_{3}\alpha)=W_{3}^{2}$,
and $d_{3}(W_{5}\alpha)=W_{3}W_{5}$. Moving on to the fifth page of the
spectral sequence, we have $d_{5}(\alpha^{2})=W_{5}$:

${6}$${\mathbb{Z}\langle
2\alpha^{3}\rangle}$${5}$${4}$${\mathbb{Z}\langle\alpha^{2}\rangle}$${\mathbb{Z}\langle
p_{1}\alpha^{2}\rangle}$${3}$${2}$${\mathbb{Z}\langle
2\alpha\rangle}$${\mathbb{Z}\langle
2p_{1}\alpha\rangle}$${1}$${0}$${\mathbb{Z}\langle
1\rangle}$${\mathbb{Z}_{2}\langle W_{3}\rangle}$${\mathbb{Z}\langle
p_{1}\rangle}$${\mathbb{Z}_{2}\langle W_{5}\rangle}$${{\mathbb{Z}_{2}\langle
W_{7},\beta(w_{2}w_{4})\rangle}}$${0}$${1}$${2}$${3}$${4}$${5}$${6}$${7}$

Indeed, first of all, since the third, fourth, and fifth homotopy groups of
$SO(2n)/U(n)$ are trivial, we have that
$E_{5}\xrightarrow{e_{5}}E_{4}\xrightarrow{e_{4}}E_{3}\xrightarrow{e_{3}}E_{2}$
is a composition of homotopy equivalences. Therefore, $BU(n)\to E_{2}\simeq
E_{5}$ is an isomorphism on $H^{\leq 5}(-;\mathbb{Z})$. Since
$H^{5}(BU(n);\mathbb{Z})=0$, we have $p^{*}W_{5}=0$ and hence it must be at
this page that $W_{5}$ is killed, i.e. $d_{5}(\alpha^{2})=p^{*}W_{5}$. On the
seventh page, we have $d_{7}(2\alpha^{3})=p^{*}\beta(w_{2}w_{4})$.

${6}$${\mathbb{Z}\langle 2\alpha^{3}\rangle}$${5}$${4}$${\mathbb{Z}\langle
2\alpha^{2}\rangle}$${\mathbb{Z}\langle
p_{1}\alpha^{2}\rangle}$${3}$${2}$${\mathbb{Z}\langle
2\alpha\rangle}$${\mathbb{Z}\langle
2p_{1}\alpha\rangle}$${1}$${0}$${\mathbb{Z}\langle
1\rangle}$${\mathbb{Z}_{2}\langle W_{3}\rangle}$${\mathbb{Z}\langle
p_{1}\rangle}$${\mathbb{Z}_{2}\langle W_{5}\rangle}$${{\mathbb{Z}_{2}\langle
W_{7},\beta(w_{2}w_{4})\rangle}}$${0}$${1}$${2}$${3}$${4}$${5}$${6}$${7}$

Indeed, since $p^{*}W_{3}=p^{*}W_{5}=0$, the classes $p^{*}w_{2}$ and
$p^{*}w_{4}$ have integral lifts in $H^{*}(E_{2};\mathbb{Z})$. Hence
$p^{*}(w_{2}w_{4})$ has an integral lift as well, so
$p^{*}\beta(w_{2}w_{4})=\beta(p^{*}(w_{2}w_{4}))=0$. The only possible
differential that could kill $\beta(w_{2}w_{4})$ is the one above on the
seventh page.

We conclude that $H^{7}(E_{2};\mathbb{Z})=\\{0,p^{*}W_{7}\\}$. Since $BU(n)\to
E_{6}$ induces an injection on $H^{7}(-;\mathbb{Z})$, it must be that
$p^{*}W_{7}$ pulls back to be zero in $E_{6}$, and $W_{7}$ vanishes when
pulled back to $BU(n)$. We conclude that the $k$-invariant $E_{2}\to
K(\mathbb{Z},7)$ is $p^{*}W_{7}$:

######  Theorem 3.

(Ehresmann, [E50, Section 6]) The second obstruction to reducing the structure
group of an $SO(2n)$ bundle to $U(n)$, where $n\geq 4$, is $W_{7}$.

Note that if $n=3$, then $W_{7}$ automatically vanishes, as the Euler class is
an integral lift of $w_{6}$.

##  4\. Massey’s Theorem I

We now provide the details for Massey’s generalization of the above result on
$W_{7}$:

######  Theorem 4.

[M61, Theorem I] Let $\theta\to X$ be the $SO(2n)/U(n)$ bundle associated to
an oriented rank $2n$ real vector bundle $\xi$ over $X$, and let $s$ be a
section of the bundle over the $(4k+2)$–skeleton of $X$. If $4k+3<2n$, then

$W_{4k+3}(\xi)=\begin{cases}(2k)!\,\mathfrak{o}_{4k+3}\ \ \ \
&k\,\mathrm{even},\\\ \tfrac{1}{2}(2k)!\,\mathfrak{o}_{4k+3}\ \ \ \
&k\,\mathrm{odd},\end{cases}$

where $\mathfrak{o}_{4k+3}$ denotes the obstruction222Note that the
obstruction is only well-defined up to sign, and we make a particular choice
in the above statement. to extending $s$ over the $(4k+3)$–skeleton of $X$.

We want to relate the obstruction $\mathfrak{o}=\mathfrak{o}_{4k+3}$, which is
the pullback of a $k$–invariant in the Moore–Postnikov system for $BU(n)\to
BSO(2n)$, to the integral Stiefel–Whitney class $W_{4k+3}$. The latter
naturally appears as a $k$–invariant in the Moore–Postnikov system for
$BSO(4k+2)\to BSO(2n)$ as the primary obstruction to finding $2n-4k-2$
linearly independent sections of a rank $2n$ oriented real bundle [St, §38].
Now, there is no natural map between $U(n)$ and $SO(4k+2)$ in either
direction, so we consider an intermediate space, the intersection of $U(n)$
and $SO(4k+2)$, namely $U(2k+1)$. The inclusion of $U(2k+1)$ into $U(n)$ and
$SO(4k+2)$ gives us the following commutative diagram:

(3)
${SO(2n)/U(n)}$${SO(2n)/SO(4k+2)}$${BU(n)}$${SO(2n)/U(2k+1)}$${BSO(4k+2)}$${BSO(2n)}$${BU(2k+1)}$${BSO(2n)}$${BSO(2n)}$$\scriptstyle{r}$$\scriptstyle{q}$$\scriptstyle{id}$$\scriptstyle{id}$

We focus for the moment on the induced maps $q$ and $r$ on fibers.

######  Lemma 5.

We have the following:

  1. (a)

The map $q$ induces an isomorphism on $\pi_{\leq 4k+2}$.

  2. (b)

$\pi_{4k+2}$ of each of $SO(2n)/U(n),SO(2n)/U(2k+1)$, and $SO(2n)/SO(4k+2)$ is
isomorphic to $\mathbb{Z}$.

  3. (c)

$SO(2n)/SO(4k+2)$ is $(4k+1)$–connected, and $r$ induces multiplication by
$\ell$ on $\pi_{4k+2}$, up to sign, where $\ell=(2k)!$ if $k$ is even, and
$\ell=\tfrac{1}{2}(2k)!$ if $k$ is odd.

###### Proof.

(a) We have the fiber bundle

$U(n)/U(2k+1)\to SO(2n)/U(2k+1)\xrightarrow{q}SO(2n)/U(n).$

The complex Stiefel manifold $U(n)/U(2k+1)$ is $(4k+2)$–connected, so $q$
induces an isomorphism on $\pi_{\leq 4k+2}$.

(b) Since $4k+2\leq 2n-2$, we have
$\pi_{4k+2}SO(2n)/U(n)\cong\pi_{4k+2}SO/U\cong\mathbb{Z}$. From part (a) we
conclude that $\pi_{4k+2}SO(2n)/U(2k+1)\cong\mathbb{Z}$. Finally, the Stiefel
manifold $SO(2n)/SO(4k+2)$ is well known to be $(4k+1)$–connected with
$\pi_{4k+2}SO(2n)/SO(4k+2)\cong\mathbb{Z}$.

(c) We have the fiber bundle

(4) $SO(4k+2)/U(2k+1)\to SO(2n)/U(2k+1)\xrightarrow{r}SO(2n)/SO(4k+2)$

and the following part of the associated long exact sequence in homotopy
groups:

$\pi_{4k+2}SO(2n)/U(2k+1)\xrightarrow{r_{*}}\pi_{4k+2}SO(2n)/SO(4k+2)\to\pi_{4k+1}SO(4k+2)/U(2k+1)\to\pi_{4k+1}SO(2n)/U(2k+1)$

By part (b), the first two groups are isomorphic to $\mathbb{Z}$. The third
group is the first unstable homotopy group of $SO(4k+2)/U(2k+1)$ listed in
(2). Namely, it is isomorphic to $\mathbb{Z}_{\ell}$, where $\ell=(2k)!$ if
$k$ is even, and $\ell=\tfrac{1}{2}(2k)!$ if $k$ is odd. The fourth group is,
by part (a), isomorphic to $\pi_{4k+1}SO(2n)/U(n)\cong\pi_{4k+1}SO/U=0$, as
$4k+1<2n-2$. Therefore, the above exact sequence is given by

$\mathbb{Z}\xrightarrow{r_{*}}\mathbb{Z}\to\mathbb{Z}_{\ell}\to 0,$

and so the induced map $\mathbb{Z}\xrightarrow{r_{*}}\mathbb{Z}$ on
$\pi_{4k+2}$ is, up to sign, multiplication by $\ell$. ∎

Consider now the following subdiagrams of diagram (3):

(5) ${BU(2k+1)}$${BU(n)}$${BSO(2n)}$${BSO(2n)}$$\scriptstyle{id}$

inducing the map $SO(2n)/U(2k+1)\xrightarrow{q}SO(2n)/U(n)$ on fibers, and

(6) ${BU(2k+1)}$${BSO(4k+2)}$${BSO(2n)}$${BSO(2n)}$$\scriptstyle{id}$

inducing the map $SO(2n)/U(2k+1)\xrightarrow{r}SO(2n)/SO(4k+2)$ on fibers. Let

$F^{1}=SO(2n)/U(2k+1),\,\,\,\,\,F^{2}=SO(2n)/U(n),\,\,\,\,\,F^{3}=SO(2n)/SO(4k+2)$

so that $F^{1}\xrightarrow{q}F^{2}$ and $F^{1}\xrightarrow{r}F^{3}$.

### Induced lift and effect on the $k$-invariant in diagram (5)

Consider the induced map of Moore–Postnikov systems corresponding to diagram
(5):

${F_{2}^{1}}$${\vdots}$${F_{2}^{2}}$${\vdots}$${E_{2}^{1}}$${E_{2}^{2}}$${BSO(2n)}$${BSO(2n)}$

Since $F^{1}\xrightarrow{q}F^{2}$ induces isomorphisms on $\pi_{\leq 4k+2}$ by
Lemma 5(a), from the commutative diagram

${F^{1}}$${F^{2}}$${F^{1}_{s}}$${F^{2}_{s}}$$\scriptstyle{g^{1}_{s}}$$\scriptstyle{q}$$\scriptstyle{g^{2}_{s}}$

and from $\pi_{>s}F^{i}_{s}=0$, $i=1,2$ (recall Section 2.1), we see that the
maps $F^{1}_{s}\to F^{2}_{s}$ are weak homotopy equivalences for $s\leq 4k+2$.
From the induced map of long exact sequences in homotopy groups associated to
the diagram

${F_{s}^{1}}$${F_{s}^{2}}$${E_{s}^{1}}$${E_{s}^{2}}$${E_{s-1}^{1}}$${E_{s-1}^{2}}$

repeated application of the five lemma shows that the map $E_{s}^{1}\to
E_{s}^{2}$ is a weak homotopy equivalence for $s\leq 4k+2$.

By assumption, we have a lift of the map $X\to BSO(2n)$ up to $E_{4k+1}^{2}$.
Choosing a homotopy inverse of $E_{4k+1}^{1}\to E_{4k+1}^{2}$ then gives us a
lift to $E^{1}_{4k+1}$ in the left hand Moore–Postnikov system.

Our obstruction $\mathfrak{o}$ in the statement of 4 is (the pull back to $X$
of) the $k$–invariant, which takes the form $E_{4k+1}^{2}\to
K({\mathbb{Z}},4k+3)$ by Lemma 5(b). Since the homotopy fiber of a
$k$–invariant in a Moore–Postnikov system is the next stage in the system,
from the induced map on long exact sequences in homotopy groups for the
diagram

${E_{4k+2}^{1}}$${E_{4k+2}^{2}}$${E_{4k+1}^{1}}$${E_{4k+1}^{2}}$${{K(\mathbb{Z},4k+3)}}$${{K(\mathbb{Z},4k+3)}}$$\scriptstyle{\mathfrak{o}}$

we have that the bottom map $K({\mathbb{Z}},4k+3)\to K({\mathbb{Z}},4k+3)$ is
a homotopy equivalence by the five lemma. Therefore, it induces an isomorphism
$H^{4k+3}(X;{\mathbb{Z}})\to H^{4k+3}(X;{\mathbb{Z}})$; let us denote the
preimage of $\mathfrak{o}$ under this isomorphism simply by $\mathfrak{o}$.

### Induced lift and effect on the $k$-invariant in diagram (6)

Now we turn to the map of Moore–Postnikov systems induced by the diagram (6).
This time, as opposed to diagram (5), the induced map of homotopy fibers is
not highly connected. Regardless, our chosen lift of $X\to BSO(2n)$ through
$E_{4k+1}^{1}$ will induce a lift of $X\to BSO(2n)$ through $E_{4k+1}^{3}$ by
post-composition. (In fact, this holds for an even more immediate reason,
namely that $E_{4k+1}^{3}\to BSO(2n)$ is a homotopy equivalence. For
uniformity of notation we will put this observation on hold for now.)

${F_{4k+2}^{1}}$${F_{4k+2}^{3}}$${E_{4k+2}^{1}}$${E_{4k+2}^{3}}$${E_{4k+1}^{1}}$${E_{4k+1}^{3}}$${{K(\mathbb{Z},4k+3)}}$${{K(\mathbb{Z},4k+3)}}$${\vdots}$${\vdots}$${BSO(2n)}$${BSO(2n)}$

In the diagram

${F^{1}}$${F^{3}}$${F_{4k+2}^{1}}$${F_{4k+2}^{3}}$$\scriptstyle{g^{1}_{4k+2}}$$\scriptstyle{r}$$\scriptstyle{g^{3}_{4k+2}}$

the vertical arrows induce isomorphisms on $\pi_{\leq 4k+2}$ and $r$ induces
the map $\mathbb{Z}\xrightarrow{\times\ell}\mathbb{Z}$ on $\pi_{4k+2}$ by
Lemma 5(c). Therefore, the map $F_{4k+2}^{1}\to F_{4k+2}^{3}$ induces
$\mathbb{Z}\xrightarrow{\times\ell}\mathbb{Z}$ on $\pi_{4k+2}$, up to sign.

(7)
${{K(\mathbb{Z},4k+2)}}$${F_{4k+2}^{1}}$${{K(\mathbb{Z},4k+2)}}$${F_{4k+2}^{3}}$${E_{4k+2}^{1}}$${E_{4k+2}^{1}}$${E_{4k+2}^{3}}$${E_{4k+2}^{3}}$${E_{4k+1}^{1}}$${BSO(2n)}$${E_{4k+1}^{3}}$${BSO(2n)}$$\scriptstyle{\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
id}$$\scriptstyle{\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!id}$$\scriptstyle{id}$

Since the fibers $F_{4k+1}^{1}$ and $F_{4k+1}^{3}$ of the maps
$E_{4k+1}^{1}\to BSO(2n)$ and $E_{4k+1}^{3}\to BSO(2n)$ have $\pi_{>4k+1}=0$,
we see that the morphisms $E_{4k+1}^{i}\to BSO(2n)$, for $i=1,3$ induce
injections on $\pi_{4k+2}$ and isomorphisms on $\pi_{\geq 4k+3}$. Now consider
the following part of the map between long exact sequences in homotopy induced
either by the back face (for $i=1$) or the front face (for $i=3$) of the above
diagram:

${\pi_{4k+3}E_{4k+2}^{i}}$${\pi_{4k+3}E_{4k+2}^{i}}$${\pi_{4k+3}E_{4k+1}^{i}}$${\pi_{4k+3}BSO(2n)}$${{\pi_{4k+2}K(\mathbb{Z},4k+2)}}$${\pi_{4k+2}F_{4k+2}^{i}}$${\pi_{4k+2}E_{4k+2}^{i}}$${\pi_{4k+2}E_{4k+2}^{i}}$${\pi_{4k+2}E_{4k+1}^{i}}$${\pi_{4k+2}BSO(2n)}$$\scriptstyle{\sim}$$\scriptstyle{\sim}$$\scriptstyle{\sim}$

From the five lemma we conclude that the induced maps
$\pi_{4k+2}K({\mathbb{Z}},4k+2)\to\pi_{4k+2}F_{4k+2}^{i}$, for $i=1,3$, are
isomorphisms ${\mathbb{Z}}\to{\mathbb{Z}}$. From the top face of diagram (7),
we conclude that the induced map
$\pi_{4k+2}K({\mathbb{Z}},4k+2)\to\pi_{4k+2}K({\mathbb{Z}},4k+2)$ is the map
${\mathbb{Z}}\xrightarrow{\times\ell}{\mathbb{Z}}$ given by multiplication by
$\ell$, up to sign.

Now, recall that the fiber of the map $BSO(4k+2)\to BSO(2n)$, namely the
Stiefel manifold $SO(2n)/SO(4k+2)$, is $(4k+1)$–connected, and
$\pi_{4k+2}SO(2n)/SO(4k+2)\cong\mathbb{Z}$. Therefore $E_{4k+1}^{3}\to
BSO(2n)$ is a homotopy equivalence, and since $BSO(4k+2)\to E_{4k+2}^{3}$
induces an injection on $H^{4k+3}(-;{\mathbb{Z}})$, we see that the
$k$-invariant $E_{4k+1}^{3}\to K({\mathbb{Z}},4k+3)$ is the integral
Stiefel–Whitney class $W_{4k+3}$. Recall, from the Serre long exact sequence
for the fibration

$E^{3}_{4k+2}\xrightarrow{e_{4k+2}}E^{3}_{4k+1}\xrightarrow{\kappa}K({\mathbb{Z}},4k+3)$

we have that the kernel of $e_{4k+2}^{*}$ in degree $4k+3$ consists of
multiples of the $k$-invariant $\kappa$. Therefore, $W_{4k+3}=m\kappa$ for
some integer $m$. Since $W_{4k+3}$ has non-zero mod 2 reduction, namely
$w_{4k+3}$, we see that $m$ is odd. Now note that $m(\kappa-W_{4k+3})=0$ since
$W_{4k+3}$ is two-torsion. As the integral cohomology of $BSO(2n)$ contains no
odd torsion [B82], we see that $\kappa=W_{4k+3}$.

Consider now at last the map of fibrations

${{K(\mathbb{Z},4k+2)}}$${{K(\mathbb{Z},4k+2)}}$${E_{4k+2}^{1}}$${E_{4k+2}^{3}}$${E_{4k+1}^{1}}$${E_{4k+1}^{3}}$$\scriptstyle{\varphi_{4k+1}}$

where $\varphi_{4k+1}$ is the induced map between the stages $E_{4k+1}^{1}$
and $E_{4k+1}^{3}$ of the Moore–Postnikov systems. The $k$–invariants are
given by the transgression $\tau$ of the corresponding fiber generator
$\iota\in H^{4k+2}(K({\mathbb{Z}},4k+2),4k+2)$ in the Serre spectral sequence.
Since the map $K({\mathbb{Z}},4k+2)\to K({\mathbb{Z}},4k+2)$ on fibers induces
multiplication by $\pm\ell$ on $H^{4k+2}(-;{\mathbb{Z}})$, by naturality of
the transgression we have

$\varphi_{4k+1}^{*}W_{4k+3}=\varphi_{4k+1}^{*}(\tau(\iota))=\tau(\pm\ell\iota)=\pm\ell\tau(\iota)=\pm\ell\mathfrak{o}.$

Recall that $\ell=(2k)!$ if $k$ is even, and $\ell=\tfrac{1}{2}(2k)!$ if $k$
is odd. Since the obstruction $\mathfrak{o}$ is only well-defined up to sign,
due to the non-trivial automorphism of $K({\mathbb{Z}},4k+3)$ corresponding to
multiplication by $-1$, this proves 4.

##  5\. Proof of Massey’s Theorem II

Suppose $\xi\to X$ is an $SO(4k)$-bundle with a lift to $BU(2k)$ over the
$(4k-1)$–skeleton of $X$. The obstruction to extending the section over the
$4k$–skeleton is the pullback of the $k$-invariant $E_{4k-2}\to
K(\pi_{4k-1}SO(4k)/U(2k),4k)$ in the Moore–Postnikov system of $BU(2k)\to
BSO(4k)$. If $k$ is odd, then $\pi_{4k-1}SO(4k)/U(2k)\cong{\mathbb{Z}}$, and
if $k$ is even,
$\pi_{4k-1}SO(4k)/U(2k)\cong{\mathbb{Z}}\oplus{\mathbb{Z}}_{2}$.

As a warm up, we recover the following refinement of Wu’s theorem [W52,
Théorème IV.10], proved also in [HiHo58, 4.6], which is the $k=1$ case of
[M61, Theorem II]:

######  Theorem 6.

Suppose $\xi\to X$ is an $SO(4)$ bundle, and that we have reduced the
structure group over the three–skeleton of $X$ to $U(2)$. The second
obstruction $\mathfrak{o}$ to reducing the structure group to $U(2)$ satisfies

$p_{1}-c_{1}^{2}+2e=4\mathfrak{o}.$

###### Proof.

Consider the beginning of the Moore–Postnikov system for the map $BU(2)\to
BSO(4)$,

${BU(2)}$${\vdots}$${E_{3}}$${{K(\mathbb{Z},2)}}$${E_{2}}$${{K(\mathbb{Z},4)}}$${BSO(4)}$${{K(\mathbb{Z},3)}}$$\scriptstyle{\kappa}$$\scriptstyle{W_{3}}$

We compute $H^{4}(E_{2};\mathbb{Z})$ using the Serre spectral sequence.
(Recall $E_{2}=BSpin^{c}(4)$; for a presentation of the cohomology of
$BSpin^{c}(n)$ for all $n$, we refer the reader to [D18].) The third page of
the spectral sequence in integral cohomology associated to
$K({\mathbb{Z}},2)\to E_{2}\to BSO(4)$ is given by

${5}$${4}$${\large\mathbb{Z}\langle\alpha^{2}\rangle}$${3}$${2}$${\large\mathbb{Z}\langle\alpha\rangle}$${\large\mathbb{Z}_{2}\langle
W_{3}\alpha\rangle}$${1}$${0}$${\large\mathbb{Z}\langle
1\rangle}$${\large\mathbb{Z}_{2}\langle
W_{3}\rangle}$${{\large\mathbb{Z}\langle
p_{1},e\rangle}}$${0}$${1}$${2}$${3}$${4}$${5}$

We have $d_{3}(\alpha)=W_{3}$, and so $d_{3}(2\alpha)=0$ and
$d_{3}(\alpha^{2})=2\alpha W_{3}=0$.

Denote the inclusion $K({\mathbb{Z}},2)\to E_{2}$ by $i$. We conclude now from
the spectral sequence that there exists a unique class $\gamma\in
H^{2}(E_{2};{\mathbb{Z}})$ such that $i^{*}\gamma=2\alpha$, and that there
exists a class $\beta\in H^{4}(E_{2};{\mathbb{Z}})$ such that
$i^{*}\beta=\alpha^{2}$ and $\\{p_{1},e,\beta\\}$ is a ${\mathbb{Z}}$-basis
for $H^{4}(E_{2};{\mathbb{Z}})\cong{\mathbb{Z}}^{3}$. Now, $\gamma^{2}$ is a
linear combination of $p_{1}$, $e$, and $\beta$, but since
$i^{*}\gamma^{2}=4\alpha^{2}=4i^{*}\beta$ and $i^{*}p_{1}=i^{*}e=0$, we see
that

$\gamma^{2}=ap_{1}+be+4\beta$

for some integers $a,b$.

Since $BU(2)\xrightarrow{h_{2}}E_{2}$ induces an isomorphism on
$H^{2}(-;{\mathbb{Z}})$ and $H^{2}(-;{\mathbb{Z}}_{2})$, we have
$h_{2}^{*}\gamma=\pm c_{1}$ and hence $\rho_{2}(\gamma)=w_{2}$. The Pontryagin
square formula we used earlier reads

$\mathfrak{P}(w_{2})=\rho_{4}(p_{1})+\theta_{2}(w_{4}),$

which, using (and its proof), becomes

$\rho_{4}(\gamma^{2})=\rho_{4}(p_{1})+\rho_{4}(2e).$

Therefore $\gamma^{2}-p_{1}-2e=(a-1)p_{1}+(b-2)e+4\beta$ is divisible by four,
so we have that $a-1$ and $b-2$ are divisible by four. By a change of basis
preserving $p_{1}$ and $e$ and relabelling
$\beta+\tfrac{a-1}{4}p_{1}+\tfrac{b-2}{4}e$ to $\beta$, we have

$\gamma^{2}=p_{1}+2e+4\beta.$

Now, the $k$-invariant $\kappa$ equals $kp_{1}+\ell e+m\beta$, for some
integers $k,\ell,m$. Pulling this class back by the map
$BU(2)\xrightarrow{h_{2}}E_{2}$ must yield zero. Now, we have
$4\kappa=4kp_{1}+4\ell e+m(\gamma^{2}-p_{1}-2e)$, and so

$0=4h_{2}^{*}\kappa=4k(c_{1}^{2}-2c_{2})+4\ell
c_{2}+m(c_{1}^{2}-(c_{1}^{2}-2c_{2})-2c_{2})=4kc_{1}^{2}+(4\ell-8k)c_{2}.$

Therefore, since $\\{c_{1}^{2},c_{2}\\}$ is a ${\mathbb{Z}}$-basis for
$H^{4}(BU(2);{\mathbb{Z}})\cong{\mathbb{Z}}^{2}$, it follows that $k=\ell=0$.
We conclude $\kappa=m\beta$, and so $m\beta$ generates the kernel of the map
$H^{4}(E_{2};{\mathbb{Z}})\xrightarrow{e_{3}^{*}}H^{4}(E_{3};{\mathbb{Z}})$.
The map $BU(2)\to E_{3}$ induces an injection on $H^{4}(-;{\mathbb{Z}})$. If
$m=0$, then $H^{4}(E_{3};\mathbb{Z}$ would be isomorphic to $\mathbb{Z}^{3}$,
which cannot inject into $H^{4}(BU(2);\mathbb{Z})\cong\mathbb{Z}^{2}$. If
$|m|>1$, $\beta$ would pull back to be a torsion class in the torsion-free
group $H^{4}(BU(2);{\mathbb{Z}})$. Therefore $\kappa=\pm\beta$. That is, the
$k$-invariant $\kappa$ satisfies (up to sign, as it is only well-defined up to
sign)

$4\kappa=\gamma^{2}-p_{1}-2e.$

Pulled back to $X$, this becomes $4\mathfrak{o}=c_{1}^{2}-p_{1}-2e$, as we
have a lift all the way to $BU(2)$ over the three–skeleton of $X$, and
$h_{2}^{*}\gamma=\pm c_{1}$. ∎

######  Example 7.

Consider the complex projective plane $\overline{\mathbb{CP}^{2}}$ with its
opposite orientation. Choose a generator $\alpha$ for
$H^{2}(\overline{\mathbb{CP}^{2}};{\mathbb{Z}})\cong{\mathbb{Z}}$. Then
$-\alpha^{2}$ is the oriented generator for
$H^{4}(\overline{\mathbb{CP}^{2}};{\mathbb{Z}})\cong{\mathbb{Z}}$, i.e.
$\langle-\alpha^{2},[\overline{\mathbb{CP}^{2}}]\rangle=1$. Suppose we have a
reduction of structure group for the tangent bundle to $U(2)$ over the
3–skeleton. We have $p_{1}=3\alpha^{2}$; furthermore,
$w_{2}(\overline{\mathbb{CP}^{2}})=\rho_{2}(\alpha)$, so $c_{1}=(2m+1)\alpha$
for some integer $m$. The Euler class is given by $e=-3\alpha^{2}$. Then

$2e-c_{1}^{2}+p_{1}=(-6-(2m+1)^{2}+3)\alpha^{2}=-4(m^{2}+m+1)$

which is divisible by four, but non-zero for all $m$.

Generally, for any $k$, consider the class
$\sum_{i=0}^{2k}(-1)^{i}c_{i}c_{2k-i}-(-1)^{k}p_{k}$. Here, $c_{i}$, for
$0\leq i\leq 2k-1$, denotes the Chern classes of the $U(n)$ bundle induced
over the $(4k-1)$–skeleton of $X$, while $c_{2k}$ denotes the Euler class of
$\xi$.

######  Theorem 8.

[M61, Theorem II] Suppose $n=2k$. Let $\theta\to X$ be the $SO(2n)/U(n)$
bundle associated to an oriented rank $2n$ real vector bundle $\xi$ over $X$,
and let $s$ be a section of the bundle over the $(2n-1)$–skeleton of $X$. The
obstruction $\mathfrak{o}_{2n}=\mathfrak{o}_{4k}$ is an integral class if $k$
is odd, while if $k$ is even, it is a pair consisting of an integral class and
a mod 2 class. If $k$ is odd, then

$\sum_{i+j=2k}(-1)^{i}c_{i}c_{j}-(-1)^{k}p_{k}=4\mathfrak{o}_{2n},$

while for $k$ even, this same formula holds with $\mathfrak{o}_{2n}$ replaced
by its integral component333The integral component is only well-defined once a
choice of isomorphism between $\pi_{2n-1}SO(2n)/U(n)$ and
${\mathbb{Z}}\oplus{\mathbb{Z}}_{2}$ is made. However, up to sign as usual,
$4\mathfrak{o}_{2n}$ does not depend on this choice.. In the formula,
$c_{0},\ldots,c_{n-1}$ are the Chern classes of the induced $U(n)$ bundle over
the $(2n-1)$–skeleton of $X$, while $c_{n}$ denotes the Euler class of $\xi$.

###### Proof.

Recall the following formula involving the Pontryagin square that we used in
Proposition 1:

$\mathfrak{P}(w_{2k})=\rho_{4}(p_{k})+\theta_{2}\left(\sum_{i=0}^{k-1}w_{2i}w_{4k-2i}\right).$

As argued in Proposition 1, we have that
$\theta_{2}(\sum_{i=0}^{k-1}w_{2i}w_{4k-2i})$ is the mod 4 reduction of
$2\sum_{i=0}^{k-1}c_{i}c_{2k-i}$, or equivalently, the mod 4 reduction of
$\sum_{i=0}^{2k}(-1)^{i}c_{i}c_{2k-i}-(-1)^{k}c_{k}^{2}$. Since
$\mathfrak{P}(w_{2k})=\rho_{4}(c_{k}^{2})$ by Lemma 2, we conclude that

$\displaystyle\rho_{4}\left(\sum_{i=0}^{2k}(-1)^{i}c_{i}c_{2k-i}-(-1)^{k}p_{k}\right)$
$\displaystyle=\rho_{4}\left(\sum_{i=0}^{2k}(-1)^{i}c_{i}c_{2k-i}-(-1)^{k}c_{k}^{2}\right)+\rho_{4}\left((-1)^{k}c_{k}^{2}-(-1)^{k}p_{k}\right)$
$\displaystyle=\theta_{2}\left(\sum_{i=0}^{k-1}w_{2i}w_{4k-2i}\right)+(-1)^{k}\mathfrak{P}(w_{2k})-(-1)^{k}\rho_{4}(p_{k})$
$\displaystyle=(-1)^{k}\left[\mathfrak{P}(w_{2k})-\rho_{4}(p_{k})+(-1)^{k}\theta_{2}\left(\sum_{i=0}^{k-1}w_{2i}w_{4k-2i}\right)\right].$

Since $\theta_{2}$ is the map on cohomology induced by the map
${\mathbb{Z}}_{2}\to{\mathbb{Z}}_{4}$ sending $[1]$ to $[2]$, we have
$\theta_{2}=-\theta_{2}$. Therefore the righthand side in the above equation
vanishes regardless of the parity of $k$.

Now, the $k$-invariant
$E_{4k-2}\xrightarrow{\kappa}K(\pi_{4k-1}SO(4k)/U(2k),4k)$ is either an
integral cohomology class, or a pair consisting of an integral cohomology
class and a mod 2 cohomology class. In either case, let $\mathfrak{o}$ denote
the integral component. Consider the Serre long exact sequence in integral
cohomology for the fibration
$E_{4k-1}\xrightarrow{e_{4k-2}}E_{4k-2}\xrightarrow{\kappa}K(\pi_{4k-1}SO(4k)/U(2k),4k)$,
namely,

$H^{4k}(K(\pi_{4k-1}SO(4k)/U(2k),4k);{\mathbb{Z}})\xrightarrow{\kappa^{*}}H^{4k}(E_{4k-2};{\mathbb{Z}})\xrightarrow{e_{4k-2}^{*}}H^{4k}(E_{4k-1};{\mathbb{Z}}).$

Note that $H^{4k}(K({\mathbb{Z}}_{2},4k);{\mathbb{Z}})=0$, and hence
regardless of the parity of $k$, we have

$H^{4k}(K(\pi_{4k-1}SO(4k)/U(2k),4k);{\mathbb{Z}})\cong
H^{4k}(K({\mathbb{Z}},4k);{\mathbb{Z}})\cong{\mathbb{Z}},$

generated by the class $\iota_{4k}$ generating
$H^{4k}(K({\mathbb{Z}},4k);{\mathbb{Z}})$. From the long exact sequence we
thus see that the kernel of $e_{4k-2}^{*}$ on $H^{4k}(-;{\mathbb{Z}})$ is the
subgroup of $H^{4k}(E_{4k-2};{\mathbb{Z}})$ generated by
$\kappa^{*}\iota_{4k}$, which we denote by $\mathfrak{o}$ (this class pulls
back to $\mathfrak{o}_{4k}$ on $X$).

Since the map $BU(2k)\to E_{4k-1}$ induces an injection on
$H^{4k}(-;{\mathbb{Z}})$, and the class
$\sum_{i=0}^{2k}(-1)^{i}c_{i}c_{2k-i}-(-1)^{k}p_{k}\in
H^{4k}(E_{4k-2};{\mathbb{Z}})$ pulls back to be 0 in
$H^{4k}(BU(2k);{\mathbb{Z}})$, we have

$\sum_{i=0}^{2k}(-1)^{i}c_{i}c_{2k-i}-(-1)^{k}p_{k}\in\ker(e_{4k-2}^{*}),$

and hence there is some integer $\ell$ such that

$\sum_{i=0}^{2k}(-1)^{i}c_{i}c_{2k-i}-(-1)^{k}p_{k}=\ell\mathfrak{o}.$

Before proceeding, we show that $\mathfrak{o}$ generates a ${\mathbb{Z}}$
summand in $H^{4k}(E_{4k-2};{\mathbb{Z}})$. First, if there were a non-zero
integer $m$ such that $m\mathfrak{o}=0$, then we would have
$m(\sum_{i=0}^{2k}(-1)^{i}c_{i}c_{2k-i}-(-1)^{k}p_{k})=0\in
H^{4k}(E_{4k-2};{\mathbb{Z}})$. Now consider the classifying map for the
tangent bundle of the sphere $S^{4k}\to BSO(4k)$. Since $S^{4k}$ can be built
with one 0–cell and one $4k$–cell, this map lifts to $E_{4k-2}$. Pulling back
$\sum_{i=0}^{2k}(-1)^{i}c_{i}c_{2k-i}-(-1)^{k}p_{k}$ to $S^{4k}$ yields twice
the Euler class ($p_{k}$ is trivial as the sphere is stably parallelizable),
i.e. the element $4\in H^{4k}(S^{4k};{\mathbb{Z}})\cong{\mathbb{Z}}$. If
$m((\sum_{i=0}^{2k}(-1)^{i}c_{i}c_{2k-i}-(-1)^{k}p_{k})=0$, then $4m=0$, a
contradiction. Secondly, we observe that $\mathfrak{o}$ is a primitive class.
Suppose that $\mathfrak{o}=m\alpha$, for some $\alpha\in
H^{4k}(E_{4k-2};{\mathbb{Z}})$. Then, as $\mathfrak{o}$ is non-zero by the
previous argument, $\alpha$ would pull back to be a non-zero class of order
$m$ in $H^{4k}(BU(2k);{\mathbb{Z}})$. Since the integral cohomology of
$BU(2k)$ is torsion-free, we conclude that $m=\pm 1$.

Since the mod 4 reduction of
$\sum_{i=0}^{2k}(-1)^{i}c_{i}c_{2k-i}-(-1)^{k}p_{k}$ is zero, we conclude that

$\sum_{i=0}^{2k}(-1)^{i}c_{i}c_{2k-i}-(-1)^{k}p_{k}=4m\mathfrak{o}$

for some integer $m$. Pulling back again to the sphere $S^{4k}$ we see that
$4m\mathfrak{o}_{4k}=4\in H^{4k}(S^{4k};{\mathbb{Z}})$, so $m=\pm 1$ (as
$\mathfrak{o}_{4k}$ is only defined up to sign). ∎

We can now address the second obstruction to reducing the structure group of
an $SO(6)$ bundle to $U(3)$.

######  Proposition 9.

The second obstruction $\mathfrak{o}$ to reducing the structure group of an
$SO(6)$ bundle to $U(3)$ satisfies

$-2c_{1}c_{3}+c_{2}^{2}-p_{2}=4\mathfrak{o}.$

###### Proof.

First of all, from the well known identification of $SO(6)/U(3)$ with
$\mathbb{CP}^{3}$, we see that after $\pi_{2}$, the first non-zero homotopy
group is $\pi_{7}SO(6)/U(3)\cong\mathbb{Z}$.

Next, consider the fiber bundle $SO(6)/U(3)\to SO(8)/U(4)\to S^{6}$. Since
$S^{6}$ admits an almost complex structure, this bundle has a section, and
hence the long exact sequence in homotopy groups splits. Therefore,
$\pi_{7}SO(6)/U(3)\cong{\mathbb{Z}}$ maps isomorphically onto the
${\mathbb{Z}}$ summand of
$\pi_{7}SO(8)/U(4)\cong{\mathbb{Z}}\oplus{\mathbb{Z}}_{2}$.

Now we look at the map of Moore–Postnikov systems induced by the map of
fibrations

${SO(6)/U(3)}$${SO(8)/U(4)}$${BU(3)}$${BU(4)}$${BSO(6)}$${BSO(8)}$

The relevant part for us will be the following map of fibrations:

${{K(\mathbb{Z},7)}}$${{K(\mathbb{Z}\oplus\mathbb{Z}_{2},7)}}$${E_{7}^{1}}$${E_{7}^{2}}$${E_{6}^{1}}$${E_{6}^{2}}$

Arguing as in the last part of Section 4, involving diagram (6), we see that
the map $K({\mathbb{Z}},7)\to K({\mathbb{Z}}\oplus{\mathbb{Z}}_{2},7)$ induces
the inclusion of the ${\mathbb{Z}}$ summand on $\pi_{7}$, as $SO(6)/U(3)\to
SO(8)/U(4)$ does. Therefore, by naturality of the transgression for the Serre
spectral sequence with ${\mathbb{Z}}$ coefficients again, and since
$H^{8}(K({\mathbb{Z}},8);{\mathbb{Z}})\cong
H^{8}(K({\mathbb{Z}}\oplus{\mathbb{Z}}_{2},8);{\mathbb{Z}})$ induced by the
inclusion ${\mathbb{Z}}\hookrightarrow{\mathbb{Z}}\oplus{\mathbb{Z}}_{2}$ of
the left summand, the $k$-invariant $E_{6}\to K({\mathbb{Z}},8)$ is the
pullback of the class $\mathfrak{o}_{8}$ in 8. In particular, four times the
$k$-invariant is the pullback of $2e-2c_{1}c_{3}+c_{2}^{2}-p_{2}$. Since the
Euler class $e$ of $BSO(8)$ pulls back to zero in $BSO(6)$, we have that four
times the $k$-invariant (i.e. four times the second obstruction
$\mathfrak{o}$) is $-2c_{1}c_{3}+c_{2}^{2}-p_{2}$. ∎

##  6\. Remarks on Massey’s Theorem III

In [M61, Theorem III], the degree eight obstruction $\mathfrak{o}_{8}\in
H^{8}(-;\mathbb{Z}_{2})$ to a stable almost complex structure is discussed.
Concretely, suppose we have an $SO(2n)$-bundle over a CW complex $X$, such
that $n\geq 5$, with a section $s$ of the associated $SO(2n)/U(n)$-bundle over
the 7–skeleton of $X$. Denoting the total space of the $SO(2n)/U(n)$-bundle by
$E$, there are classes $u\in H^{2}(E;\mathbb{Z})$ and $v\in
H^{6}(E;\mathbb{Z})$ such that $s^{*}u=0$, $s^{*}v=0$, and $u,v$ restrict to
generators $x,y$ of the degree $\leq 8$ cohomology of $SO(2n)/U(n)$. The
obstruction $\mathfrak{o}_{8}$ is then claimed to be a certain class $b_{8}$
pulled back from $X$ that appears in the expansion of $Sq^{2}\rho_{2}(v)$ in
terms of the cohomology of $X$ and $SO(2n)/U(n)$. Namely,

$Sq^{2}\rho_{2}(v)=p^{*}b_{8}+(p^{*}b_{6})\rho_{2}(u)+(p^{*}b_{4})\rho_{2}(u)^{2}+(p^{*}b_{2})\rho_{2}(u)^{3}+(p^{*}b_{2}^{\prime})\rho_{2}(v)+\rho_{2}(u)^{4}.$

Such an equation holds since $x$ and $y$ transgress to $W_{3}$ and $W_{7}$,
which are assumed to vanish, and hence the spectral sequence collapses in
sufficiently low degree.

We have the relation $Sq^{2}\rho_{2}(y)=\rho_{2}(x)^{4}$, which can be seen by
combining [MT, Theorem 6.11] and [MT, Theorem 6.6]. It is clear that the
vanishing of the obstruction $\mathfrak{o}_{8}$ implies the vanishing of
$b_{8}$. Indeed, if the obstruction vanishes, then one can extend $s$ over the
eight–skeleton of $X$, in which case we can apply the pull back by this
extended section to the above equation, obtaining $b_{8}=0$ since
$s^{*}u=s^{*}v=0$.

In the cases of most interest, namely stable almost complex structures on
closed eight–manifolds, this obstruction was addressed by Heaps [H60] and
Thomas [T67]. In [H60], the obstruction itself is not explicitly identified,
but it is shown (making use of another theorem of Massey that $W_{7}=0$ for
the tangent bundle of a closed orientable eight–manifold) that a closed
orientable eight–manifold admits a stable almost complex structure if and only
if $W_{3}=0$ and $w_{8}$ is in the image of $Sq^{2}\rho_{2}$ [H60, Corollary
1.3]. Thomas [T67, Theorem 1.2] showed that a general real orientable vector
bundle $\xi$ over a CW complex admits a stable almost complex structure if and
only if $W_{3}(\xi)=0,W_{7}(\xi)=0$, and a certain secondary cohomology
operation $\Omega$ vanishes on
$w_{8}(\xi)+w_{4}(\xi)^{2}+w_{2}(\xi)^{2}w_{4}(\xi)$.

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