Given a Hausdorff space $X$, the set of all vector bundles of finite dimensions $Vect(X)$ (up to isomorphism) with respect to the direct sum is a monoid. We can use the group completion construction to turn $Vect(X)$ into a group, generally denoted by $K(X)$. So now elements in $K(X)$ look like $E-F$, where $E,F$ are vector bundles over $X$. Furthermore, by using the fact that $X$ is Hausdorff, any element in $K(X)$ can be written as $E-\epsilon^n$, where $\epsilon^n$ is the $n$-dimensional trivial bundle over $X$.

Question: I can't imagine how the expression $E-F$ would carry any geometric meaning. How should we think of $E-F$, the 'difference' between vector bundles?

What I am trying to get to is that writing '-3 apples' doesn't make much sense. However in the context of money (debts), negative numbers do turn out to be useful and provides a new way for us to think of negative numbers.

So are there any ways we can look at $E-F$ in which we can assign some kind of meaning to $E-F$?

For example we can look at elements in a cyclic group as some kind of a rotation of some object. I can 'visualise' $Vect(X)$, but I don't know how I should look at $K(X)$. The elements $E-F$ are just symbols to me at the moment.

  • 3
    $\begingroup$ You can think of it as a chain complex of vector bundles with $E$ in degree $0$ and $F$ in degree $1$. $\endgroup$ Apr 30, 2013 at 5:51
  • 1
    $\begingroup$ A quote I heard once from Paul Baum of Baum-Connes might help a bit: "They are just elements in this group we make." Strictly speaking, the Grothendieck construction actually erases information from the Whitney semigroup of vector bundles, by descending to stable equivalence classes of bundles. This seems to be OK in most settings in noncommutative topology. But there are situations where one wants to keep the semigroup. $\endgroup$
    – Michael
    Jun 16, 2013 at 13:38


You must log in to answer this question.

Browse other questions tagged .