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Corollary 9 in these notes by Ralph Cohen has grabbed my attention.

I do not undestand how to show that if we have a rank $k$ bundle which is stably isomorphic to the stable normal bundle then there is a virtual normal bundle of rank $k$.

This seems to boil down to proving the following thing:

Let $M^m$ a smooth manifold and suppose we are given $f:M^m\to BO(k)$. Let $n>k$ and let $g: M^m\to BO(n)$ be such that

$$TM\oplus g^*(EO(n)) \simeq \varepsilon^{n+m}.$$

In other words, $g$ is the classifying map for a virtual normal bundle of rank $n$ ($\varepsilon$ is the trivial bundle).

Suppose that $g$ is homotopic to $i\circ f$ where $i:BO(k)\to BO(n)$ is the obvious inclusion. Then we would like to show that

$$TM\oplus f^*(EO(k)) \simeq \varepsilon^{k+m}.$$

However the only thing I can conclude from the homotopy $g\simeq i\circ f$ is that

$$TM\oplus f^*(EO(k))\oplus \varepsilon^{n-k} \simeq \varepsilon^{n+m},$$

which is not enough in general.

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  • $\begingroup$ What's a "virtual normal bundle"? (it seems from your description of g that it's synonymous with "stable normal bundle"?) $\endgroup$
    – kiran
    Jun 22 at 15:51
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    $\begingroup$ @kiran Thank you for your comment, I borrowed this def. from Cohen's notes. In general given an immersion $f:M\to N$, a virtual normal bundle for $f$ is a bundle $\nu$ such that $TM\oplus \nu \simeq f^*N$. In the above setting we are dealing with immersions into Euclidean spaces so we can say that $\nu$ is virtual normal bundle of rank $k$ if $TM\oplus \nu\simeq \varepsilon^{n+k}$, this implies by Hirsch-Smale theory that $M$ has an immersion into $\mathbb R^{n+k}$ with normal bundle isomorphic to $\nu$. If $k$ is large, $\nu$ is isomorphic to the stable normal bundle, for small $k$ idk. $\endgroup$ Jun 22 at 16:00

1 Answer 1

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This follows from obstruction theory; also see this answer.

If $E \to X$ is a rank $r$ real vector bundle over a CW complex $X$, then the obstructions to finding a nowhere-zero section lie in $H^i(X; \pi_{i-1}(S^{r-1}))$. In particular, if $r > \dim X$, then such a section exists so $E\cong E_0\oplus\varepsilon^1$ where $E_0$ has rank $r - 1$. However, $E_0$ may not be unique. For example, the bundle $\varepsilon^{n+1} \to S^n$ has rank $r = n+1 > n$ and decomposes as $\varepsilon^n\oplus\varepsilon^1$ and $TS^n\oplus\varepsilon^1$.

The obstructions to the uniqueness of a nowhere-zero section up to scale lie in $H^i(X; \pi_i(S^{r-1}))$. In particular, if $r > \dim X + 1$, then $E$ admits a unique nowhere-zero section up to scale, so $E \cong E_0\oplus\varepsilon^1$ where $E_0$ has rank $r - 1$ and is unique up to isomorphism. Note, in the example above, $r = \dim X + 1$. If the rank of $E_0$, namely $r - 1$, is still larger than $\dim X + 1$, then we can apply the same argument to split off a trivial line bundle with a unique complement. It follows that if $V\oplus\varepsilon^p \cong W\oplus\varepsilon^p$ and $\operatorname{rank} V = \operatorname{rank} W > \dim X$, then $V \cong W$. In particular, if $\operatorname{rank} V > \dim X$ is stably trivial, then it is in fact trivial.

Since $TM\oplus f^*(EO(k))$ is stably trivial and $TM\oplus f^*(EO(k))$ has rank $m + k > m = \dim M$, we see that $TM\oplus f^*(EO(k)) \cong \varepsilon^{m+k}$.

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