# Extending a monomorphism of bundles, Lemma 7.3, Atiyah, Shapiro

This is from Lemma 7.2, pg17

Let $$E,F$$ be vector bundles on $$X$$ and $$f:E \rightarrow F$$ a monomorphism on $$Y$$. Then if $$\dim F > \dim E+ \dim X$$, $$f$$ can be extended to a monomorphismon on $$X$$ and any two such extensions are homotopic rel $$Y$$.

I completely do not follow the supplied proof.

It would be great if someone could provide references (or proof ) for each of the claims.

Consider the fibre bundle $$Mon(E,F)$$ where the fibers are the monomorphisms of $$E_x \rightarrow F_x$$.

1. This is homeomoprhic to $$GL(n)/GL(n-m)$$
2. This space is $$n-m-1$$ connected.
3. cross sections can be extended and are all homeomorphic if $$\dim X \le \dim F- \dim E -1$$

1) Let $$V,W$$ be $$\mathbb{K}$$-vector spaces with $$\dim(V)\leq\dim (W)$$. In particular there always there exists a linear monomorphism $$\varphi:V\hookrightarrow W$$, a fact you can verify easily by choosing bases for $$V$$ and $$W$$ (a vector space always admits a basis). Now if $$A\in Gl(W)$$ then $$A\varphi:V\hookrightarrow W$$ is also monic, and this observation gives us an action

$$Gl(W)\times Mon(V,W)\rightarrow Mon(V,W),\qquad (A,\varphi)\mapsto A \varphi.$$

It's not difficult to check that this action is transitive, again by choosing bases. Moreover all the spaces and maps are suitably nice, so if we fix a basepoint monomorphism $$\varphi_0\in Mon(V,W)$$, then we get an induced homeomorphism

$$Gl(W)/Stab(\varphi_0)\cong Mon(V,W),$$

where $$Stab(\varphi_0)\leq Gl(W)$$ is the stabiliser subgroup of $$\varphi_0$$ with respect to the above action.

Now use the fixed $$\varphi_0$$ to identify $$V$$ as a subspace of $$W$$ and choose a complement $$V^\perp\leq W$$ to get a direct sum decomposition $$V\oplus V^\perp$$. Vector subspaces always admit complements - just choose bases. A slightly fancier way to generate a complement (as the notation suggets) is to fix an arbitrary inner product on $$W$$ (choose a basis) and let $$V^\perp$$ be the orthogonal complement with respect to this inner product.

Now there is a subgroup inclusion $$Gl(V^\perp)\hookrightarrow Gl(W)$$ which sends $$B$$ to $$id_V\oplus B$$, and it is clear that

$$Stab(\varphi_0)=Stab(V)\cong Gl(V^\perp)$$

with respect to this inclusion. Putting everything together we have a homeomorphism

$$Gl(W)/Gl(V^\perp)\cong Gl(V\oplus V^\perp)/Gl(V^\perp)\cong Mon(V,W).$$

To see things most clearly fix bases for $$V$$, $$W$$ (sigh) so that $$V\cong \mathbb{K}^n$$ and $$W\cong\mathbb{K}^{n+m}$$ and take $$\varphi_0$$ to be the inclusion of $$\mathbb{K}^n$$ in $$\mathbb{K}^{n+m}$$ as the first $$n$$ non-zero coordinates. In this case our previous homeomorphism is just

$$Gl(\mathbb{K}^{n+m})/Gl(\mathbb{K}^m)\cong Mon(\mathbb{K}^n,\mathbb{K}^{n+m}).$$

Now this is all unparametrised, but the same construction can be carried out fibrewise. With your notation we have $$Mon(E,F)_x=Mon(E_x,F_x)$$ for $$x\in X$$, and so

$$Mon(E,F)\cong \bigcup_{x\in X}Mon(E_x,F_x).$$

The fibre over a fixed basepoint $$x_0\in X$$ is just $$Mon(E,F)_{x_0}=Mon(E_{x_0},F_{x_0})$$, and since $$E_{x_0}\cong\mathbb{K}^n$$ and $$F_{x_0}\cong\mathbb{K}^{n+m}$$ for some $$n,m\in\mathbb{N}_0$$ by assumption we have

$$Mon(E,F)_{x_0}\cong Gl(\mathbb{K}^{n+m})/Gl(\mathbb{K}^m)$$

as above.

2) What we've actually shown above is that we have a fibration sequence

$$Gl(\mathbb{K}^n)\rightarrow Gl(\mathbb{K}^{n+m})\rightarrow Mon(\mathbb{K}^n,\mathbb{K}^{n+m}).$$

You can get this by running through the standard theorems. $$Gl(\mathbb{K}^{n+m})$$ is a Lie group, $$Mon(\mathbb{K}^n,\mathbb{K}^{n+m})$$ is an open submanifold of $$Mat_{n\times(n+m)}(\mathbb{K})$$ and the action is smooth. Moreover $$Gl(\mathbb{K}^n)$$ is closed in $$Gl(\mathbb{K}^{n+m})$$, so the projection onto the oribt space is the just projection of a Lie group onto its quotient by a closed subgroup (in particular it admits local sections and is thus a fibration).

Now take the fibration sequence above and set $$m=1$$. Then $$Mon(\mathbb{K}^n,\mathbb{K}^{n+1})$$ is just the unit sphere

$$S(\mathbb{K}^{n+1})=\{x\in\mathbb{K}^{n+1}\mid |x|^2=1\},$$

and you can see this directly by thinking of points of $$S(\mathbb{K}^{n+1})$$ as $$1$$-dimensional subspaces of $$\mathbb{K}^{n+1}$$. Thus we have a fibration sequence

$$Gl(\mathbb{K}^n)\rightarrow Gl(\mathbb{K}^{n+1})\rightarrow S(\mathbb{K}^{n+1})$$

and in particular a long exact sequence of homotopy groups. We have

$$S(\mathbb{K}^{n+1})=\{x\in\mathbb{K}^{n+1}\mid |x|^2=1\}=\begin{cases}S^n&\mathbb{K}=\mathbb{R}\\ S^{2n+1}&\mathbb{K}=\mathbb{C}\\ S^{4n+3}&\mathbb{K}=\mathbb{H}\end{cases}$$

so the map $$Gl(\mathbb{K}^n)\rightarrow Gl(\mathbb{K}^n)$$ is as connected as the indicated sphere is. In the real case this map is $$(n-1)$$-connected. On the other hand, the map$$Gl(\mathbb{K}^{n+1})\rightarrow Gl(\mathbb{K}^{n+1})$$ is $$(n+1)$$-connected, and therefore the composite inclusion $$Gl(\mathbb{K}^n)\hookrightarrow Gl(\mathbb{K}^{n+2})$$ is as connected as the fist inclusion, so is $$(n-1)$$-connected.

Iterating this gives us that $$Gl(\mathbb{K}^n)\hookrightarrow Gl(\mathbb{K}^{n+m})$$ is $$(n-1)$$-connected, so from our fibration sequence the first non-trivial homotopy group of $$Mon(\mathbb{K}^n,\mathbb{K}^{n+m})$$ occurs in degree $$n$$, so the space is $$(n-1)$$-connected. This matches up with your statement when you recall that they have written $$n$$ where I have written $$n+m$$, and $$n-m$$ where I have written $$n$$.

3) Your final query is just an exercise in obstruction theory using what we now now about the connectivity of $$Mon(\mathbb{K}^n,\mathbb{K}^{n+m})$$. If $$E\rightarrow B$$ is a suitable fibration with $$(n-1)$$-connected fibre, and you are given a map $$f:X\rightarrow B$$, then there are a chain of obstructions to lifting to map into $$E$$ which covers $$f$$, and these lie in the group $$H^{k+1}(X;\pi_kF)$$, the first living in $$H^{n+1}(X;\pi_nF)$$. If all these obstructions vanish then a lift can be found. Clearly if $$\dim(X), then all the obstructions live in trivial groups, so vanish showing that a lift exists.

I'm afraid that here is not the best place to explain much further details, but a good place to strat reading about obstruction theory is in Hatcher's book Algebraic Toplogy on pg. 415. Davis and Kirk also do a very good treatement in their book (whose title escapes me right now). Classic references (although a little more difficult) are Steenrod, Whitehead and Spanier.

• The book title is "Lecture notes in algebraic topology". I agree this is one of the best places to first learn obstruction theory. – user98602 Feb 1 at 14:11