# Stable normal bundle and immersions

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.

• What's a "virtual normal bundle"? (it seems from your description of g that it's synonymous with "stable normal bundle"?) Jun 22 at 15:51
• @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. Jun 22 at 16:00

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