6
$\begingroup$

Let $X$ be CW complex having only cells up to dimension $n$.

I want to prove that every $m$-dimensional vector bundle $E$ over $X$ decomposes as a sum $E\cong A\oplus B$ where $A$ is a $n$-dimensional (perhaps $n+1$?) vector bundle and $B$ is a trivial bundle $X\times \mathbb{R}^{m-n}$.

I remember vaguely that I've seen such a statement somewhere but I have no idea how to prove it. My first thought was to use the cellular approximation theorem an show that every map $X\to Gr(m,\infty)$ is homotopic to a map with image in $Gr(n,\infty)$ but I have doubts now that the cell structure on $Gr(n,\infty)$ can be choosen in such a way.

  • 1
    I still don't get it. To use cellular approximation for the map $X\to Gr(m,\infty)$, $Gr(n,\infty)$ would have to be the $n$-skeleton of $Gr(m,\infty)$ but this isn't given by theorem 6.4, is it?2012-02-14

1 Answers 1

2

Let $E$ be a rank $m$ vector bundle on $X$, and let $f : X \to BO(m)$ be a classifying map for $E$.

The inclusion $O(m-1) \to O(m)$ induces a map $BO(m-1) \to BO(m)$. The map $f$ lifts to $\tilde{f} : X \to BO(m-1)$ if and only if $E$ admits a section (and therefore, $E \cong E_{m-1}\oplus\varepsilon^1$ for some rank $m-1$ vector bundle $E_{m-1}$). The obstructions to lifting $f$ lie in $H^k(X; \pi_{k-1}(F))$ where $F$ is the homotopy fiber of the map $BO(m-1) \to BO(m)$. A lift exists if and only if all these obstructions vanish.

In general, if $H$ is a subgroup of $G$, then the homotopy fiber of $BH \to BG$, induced by the inclusion $H\hookrightarrow G$, is $G/H$. So in the above case, $F = O(m)/O(m-1) = S^{m-1}$. So we see that the primary obstruction to lifting $f$ occurs when $k = m$. If $n = \dim X < m$, then the primary obstruction, and all other obstructions, vanish for dimension reasons, so such a lift exists.

That is, if $\operatorname{rank} E > \dim X$, then $E \cong E_{m-1}\oplus\varepsilon^1$ for some rank $m - 1$ vector bundle $E_{m-1}$. If $\operatorname{rank}E_{m-1} > \dim X$, then by the same argument, $E_{m-1} \cong E_{m-2}\oplus\varepsilon^1$ for some rank $m - 2$ vector bundle $E_{m-2}$, so $E \cong E_{m-2}\oplus\varepsilon^2$. We can continue this argument until $\operatorname{rank}E_{m-i} = \dim X$; note, this occurs when $i = m - n$. So we see that $E \cong E_n\oplus\varepsilon^{m-n}$ for some rank $n$ vector bundle $E_n$.

Note, if we tried to go one step further, there would be an obstruction to finding a section of $E_n$ which would lie in $H^n(X; \pi_{n-1}(S^{n-1})) = H^n(X; \mathbb{Z})$. This obstruction is the Euler class of $E_n$ (provided $E$, and hence $E_n$, is orientable).


Another way to proceed is to instead consider the inclusion $O(n) \to O(m)$ and the induced map $BO(n) \to BO(m)$. The map $f : X \to BO(m)$ lifts to $\tilde{f} : X \to BO(n)$ if and only if $E$ admits $m - n$ linearly independent sections (and therefore $E \cong E_n\oplus\varepsilon^{m-n}$ for some rank $n$ vector bundle $E_n$). The homotopy fiber of the map $BO(n) \to BO(m)$ is $O(m)/O(n)$ which is the Stiefel manifold $V_{m-n}(\mathbb{R}^m)$. In general, $V_p(\mathbb{R}^q)$ is $(q - p - 1)$-connected, so $V_{m-n}(\mathbb{R}^m)$ is $(n-1)$-connected. Therefore the primary obstruction to lifting $f$ lies in $H^{n+1}(X; \pi_n(V_{m-n}(\mathbb{R}^m)))$ which vanishes for dimension reasons, as do all the other obstructions.

Note, if we tried the same thing with $BO(n-1) \to BO(m)$, we would run into an obstruction in $H^n(X; \pi_{n-1}(V_{m-n+1}(\mathbb{R}^m)))$. In general,

$\pi_{q-p}(V_p(\mathbb{R}^q)) = \begin{cases} \mathbb{Z} & q - p\ \text{is even or}\ p = 1,\\ \mathbb{Z}_2 &\text{otherwise}. \end{cases}$

As $m = \operatorname{rank} E > \dim X = n$, we have $m - n + 1 > 1$. So we see that if $n$ is even, the primary (and only) obstruction lies in $H^n(X; \mathbb{Z}_2)$, while if $n$ is odd, it lies in $H^n(X; \mathbb{Z})$. In the first case, the obstruction is $w_n(E)$ (see page 143 of Characteristic Classes by Milnor and Stasheff). In the latter case, the obstruction is $W_n(E) = \beta w_{n-1}(E)$ (see section $38.8$ of Topology of Fibre Bundles by Steenrod).

In the case where $n$ is odd, the obstruction $W_n(E) = W_n(E_n)$ coincides with the Euler class of $E_n$, provided $E_n$ is orientable (see Theorem $12.5$ of Characteristic Classes). When $n$ is even however, the two obstructions do not coincide, but the vanishing of the Euler class of $E_n$ does imply the vanishing of $w_n(E_n) = w_n(E)$ (see property $9.5$ in Characteristic Classes). What this tells us is that if $E_n$ is rank $n$ vector bundle over an $n$-dimensional CW complex, it may be possible to find $l + 1$ linearly independent sections of $E_n\oplus\varepsilon^l$ even if $E_n$ does not admit a section. An easy example is $TS^{2n} \to S^{2n}$; it does not admit a section as it has non-zero Euler class, but $TS^{2n}\oplus\varepsilon^l \cong \varepsilon^{2n + l}$ so it admits $l + 1$ linearly independent sections (in fact, it admits $2n + l$ such sections).