I am trying to understand the proof of lemma 7.2 in the paper Clifford modules by Atiyah, Bott and Shapiro (Topology, Vol. 3, Supp. 1, 3-38). However I encountered a confusing statement:

Right at the beginning they state ($E$ and $F$ are vector bundles on $X$):

Consider the fibre bundle $Mon(E,F)$ on $X$ whose fibre at $x\in X$ is the space of all monomorphisms $E_x \rightarrow F_x$. This fibre is homeomorphic to $GL(n)/GL(n-m)$ where $n = dim(F), m = dim(E)$, and so it is $(n-m-1)$-connected.

I think that the claim in sentence 2 is wrong and the word homeomorphic should be replaced with homotopic? By dimensional counting I would find that $GL(n)/GL(n-m)$ would have dimension $n^2 - (n-m)^2 = m (2n - m)$, while the space of monomorphisms has dimension $nm$.

The conclusion is not changed and this might just be a typo. Is this a correct observation? If not, what did I get wrong?


Let $V\cong \mathbb{R}^{m}, W\cong \mathbb{R}^{n}$. The claim is the space of monomorphisms $$ Hom_{\textrm{monomormphism}}(V, W)\cong GL(n,\mathbb{R})/Gr(n-m,n) $$ To see why this true it suffice to "add" $n-m$ additional vectors to $V$ such that $V\oplus M\cong W$. The choices for all isomorphisms is precisely $GL(n, \mathbb{R})$. Since we have freedom of choosing any basis we like for $M$, the basis choices we need to mod out for a fixed isomorphism are precisely $Gr(n-m,n)$ as the map has been fixed already. So I believe the authors did made a mistake.

  • $\begingroup$ While I understand your explanation of the quotient formula, I have difficulty to reconcile it with dimensional counting: In the case $n=4$, $m=2$, we would find that a monomorphism is specified by $2\times 4 = 8$ parameters. The "Atiyah-Bott-Shapiro claim" would imply $4^2 - (4-2)^2 = 12$ parameters. In my understanding monomorphisms parametrize the space of m-dimensional subspaces in an n-dimensional subspace together with a choice of basis, i.e. I expect its dimension to be $dim(Gr(n,m)) + dim(GL(m)) = m (n-m) + m^2 = n m$. $\endgroup$ – S.K. Jul 22 '17 at 19:59
  • $\begingroup$ Should have been $dim(Gr(m,n))$ instead $dim(Gr(n,m))$ in the previous comment. $\endgroup$ – S.K. Jul 22 '17 at 20:09
  • $\begingroup$ @S.K.: Updated. $\endgroup$ – Bombyx mori Jul 22 '17 at 23:46

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