most recent 30 from http://mathoverflow.net 2013-05-18T20:49:59Z http://mathoverflow.net/feeds/question/108923 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108923/euler-characteristics-and-the-difference-bundle-construction Kofi 2012-10-05T14:48:18Z 2012-10-08T22:10:06Z

<p>I am reading on K theory in Lawson and Michelson (Spin Geometry). One has the "exact sequence spaces" $L(X,Y)$ and there is the theorem that there is a unique equivalence of functors $\chi$ between $L$ and $K$ (the Euler characteristic) such that in the case $Y = \emptyset$ $$ \chi([V_0, \dots, V_n]) = \sum_{k=0}^n (-1)^k [V_k] \in K(X, \emptyset).$$ To give a more explicit characterization of the map in the case $n=1$, they make quite a complicated construction by defining a space $Z$ by gluying two copies of $X$ together along $Y$, defining bundles on $Z$ and using the isomorphism $K(Z, X_1) \cong K(X, Y)$. I compared to the original paper of Atiyah, Bott and Shapiro, and they make the same construction. However, I feel that one does not need all those surgical methods.</p> <p>Instead, to define $\chi([V_0, V_1])$, look at the exact sequence $$0 \rightarrow K(X, Y) \stackrel{i}{\rightarrow} K(X) \stackrel{j}{\rightarrow} K(Y) \rightarrow 0$$ and notice that the element $[V_0] - [V_1] \in K(X)$ is in the kernel of $j$ since $V_0$ and $V_1$ are isomorphic when restricted to $Y$. Now define the Euler characteristic as the preimage of $[V_0] - [V_1]$ under $j$, which is well-defined by exactness.</p> <p>Did I miss something or are the constructions that I referenced needlessly complicated? </p>

http://mathoverflow.net/questions/108923/euler-characteristics-and-the-difference-bundle-construction/109188#109188 Johannes Ebert 2012-10-08T22:10:06Z 2012-10-08T22:10:06Z

<p>You missed that the sequence is not exact at $K(X,Y)$ (neither is it exact at $K(Y)$, but that does not matter here). There is an ambiguity coming from $K^{-1} (Y)$, i.e. automorphisms of bundles. If $Y$ is a point, your construction works, and the purpose of the arguments in Atiyah-Bott-Shapiro is to extend it to $Y \neq \ast$.</p>