K-Theory Equivalence Classes

Question

Let $M$ be a finite dimensional compact manifold and $(Vect(M),\oplus)$ be the abelian monoid of complex vector bundles on $M$. I just read that it is possible to construct an equivalence relation on $Vect(M)^2$ by $$(E_1,F_1)\sim (E_2,F_2) \iff \exists N\in \mathbb N \, s.t. \, E_1\oplus F_2\oplus \mathbb C^N\cong E_2\oplus F_1\oplus \mathbb C^N$$ In the book i'm reading (Salomon's notes on SW theory), it says that "the additional summand $\mathbb C^N$ is needed to obtain an equivalence relation". It seems to me that the only problem of not having it could be the transitivity property: is it not necessarily true that if $E_1\oplus F_2\cong E_2\oplus F_1$ and $E_2\oplus F_3\cong E_3\oplus F_2$, then $E_1\oplus F_3\cong E_3\oplus F_1$? I'm not able to prove the transitivity property also for the adjusted relation above. Can someone help me with that?

Answer 1

The conceptual way to define the equvalence relation is that $$ (E_1,F_1)\sim (E_2,F_2) \Leftrightarrow \exists G\in Vect(M) s.t. E_1\oplus F_2\oplus G\cong E_2\oplus F_1\oplus G. $$ In this formulation the transitivity is easier to prove. This relation is the same as the one you used since it is well known that for $G\in Vect(M)$, there is $\tilde G\in Vect(M)$ such that $G\oplus\tilde G\cong \mathbb C^N$ for some $N\in\mathbb N$.

Edit (in view of the comment by @MichaelAlbanese): Without "stabilization" as expressed by the definition of the equivalence relation, transitivty is lost. Indeed start with arbitrary bundles $E$, $\tilde E$ and $F$ and then put $E_1:=E$, $F_1=0$, $E_2=E\oplus F$, $F_2=F$, $E_3=\tilde E$ and $F_3=0$. Then $E_1\oplus F_2\cong E_2\oplus F_1$ is satisfied automatically, while $E_2\oplus F_3\cong E_3\oplus F_2$ exactly means that $E\oplus F\cong \tilde E\oplus F$. If transitivity would hold then this would imply that $E\cong\tilde E$. But it is well known that vector budnels can be stably isomorphic without being isomorphic.The standard example (for real vector bundles) is $TS^2$ which is non-trivial by the hairy ball theorem, but its sum with the normal bundle to $S^2$ (which is easily seen to be trivial) is $T\mathbb R^3|_{S^2}$, which is also trivial. Hence $TS^2\oplus\mathbb R\cong\mathbb R^3\cong\mathbb R^2\oplus\mathbb R$ but $TS^2$ is not isomorphic to $\mathbb R^2$.