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I saw this question: What does "formal" mean? But I don't think it answers my question.

In Hatcher's book "Vector Bundles and K-Theory", he defines $K^0(X)$ to be the group of formal differences of vector bundles over the space $X$, with the equivalence relation that $E_1-E_1'=E_2-E_2'$ iff $E_1\oplus E_2\cong_s E_2\oplus E_1'$, where $\cong_s$ means they are stably isomorphic. By defining it like this, it can easily be turned into an abelian group.

Are these differences meaningless and form part of a trick to get the group to work, or do they actually hold some sort of topological significance?

George
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2 Answers2

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Well, I'm not sure what you mean by "meaningless", but these "differences" are just symbols. You could consider $E_1-E_1'$ as just a notation for the ordered pair $(E_1,E_1')$. The group $K^0(X)$ is then defined as the set of equivalence classes of ordered pairs of vector bundles on $X$ under the equivalence relation $(E_1,E_1')\sim(E_2,E_2')$ iff $E_1\oplus E_2'\cong_s E_2\oplus E_1'$. It then makes sense to write the equivalence class of an ordered pair $(E_1,E_1')$ as $E_1-E_1'$ because this operation behaves formally like subtraction. (And in fact, it is the universal way to turn the monoid of vector bundles up to isomorphism under $\oplus$ into a group.)

It is perhaps instructive to observe that if you define an equivalence relation on $\mathbb{N}^2$ by $(a,b)\sim (c,d)$ iff $a+d=c+b$, then the equivalence classes can naturally be identified with the integers, sending $(a,b)$ to the integer $a-b$. Here we're doing the same sort of thing to abstractly allow ourselves to subtract vector bundles, by representing the result of such a subtraction as an equivalence class of pairs of vector bundles.

Eric Wofsey
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You should think of the $K$ theory classes as stable equivalence classes of vector bundles. Two classes are equal, i.e. $[E]=[F]$ or equivalently $[E]-[F]=[0]$ if and only if the bundles $E$ and $F$ are stably equivalent. That means that there is some $r$ such that the bundles $E\oplus \mathbb{R}^r$ and $F\oplus \mathbb{R}^r$ are isomorphic as ordinary vector bundles. Thus by constructing a group out of vector bundles you are forgetting information.

Thomas Rot
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