K-Thy mapping cone in pullback diagram - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T23:06:57Z http://mathoverflow.net/feeds/question/98078 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/98078/k-thy-mapping-cone-in-pullback-diagram K-Thy mapping cone in pullback diagram santker heboln 2012-05-27T01:00:29Z 2012-05-27T01:00:29Z <p>Let $X, Y$ be connected, compact Hausdorff spaces and $\mathcal{H}_1, \mathcal{H}_2$ Hilbert spaces with $A \subset \mathcal{L}(\mathcal{H}_1), B \subset \mathcal{L}(\mathcal{H}_2)$ $\ast$-subalgebras. Set $\mathcal{K}_1 := \mathcal{K}(\mathcal{H}_1)$ for the compact operators and $\mathcal{K}_2 := \mathcal{K}(\mathcal{H}_2)$.</p> <p>Assume that we have the canonical isomorphisms $A / \mathcal{K}_1 \cong C(X), B / \mathcal{K}_2 \cong C(Y)$ (via the quotient maps).</p> <p>Let $\Sigma$ be a $C^{\ast}$-algebra pullback: $\Sigma := \{(F, G) \in C(X, B) \oplus C(Y, A) : q_Y(F) = q_X(G)\}$</p> <p>with the maps $$q_Y \colon C(X, B) \to C(X \times Y), q_X \colon C(Y, A) \to C(X \times Y)$$</p> <p>which are the pointwise quotient maps.</p> <p>We can write down the Mayer-Vietoris theorem in K-theory and then we have the middle terms $K_i(C(X, A)) \oplus K_i(C(Y, B))$ in this sequence.</p> <p>I would prefer $K^i(C(X)) \oplus K^i(C(Y)) \cong K^i(X) \oplus K^i(Y)$, i.e. a six-term exact sequence in K-theory no longer depending on $A, B$ and nontrivial examples of such situations.</p> <p>I have the following observations (question at the end):</p> <p>The mapping cone of $q_Y$ is defined as follows $$C q_Y := \{(F, f) \in C(X, B) \oplus C C(X \times Y) : f(0) = q_Y(F)\} \$$</p> <p>for $q_X$ $$C q_X := \{(G, g) \in C(Y, A) \oplus C C(X \times Y) : g(0) = q_X(G)\}$$</p> <p>Note that with the short exact sequence $$0 \to C(X, \mathcal{K}_2) \to C q_Y \to C C(X \times Y) \to 0$$</p> <p>we have the isomorphism $K_i(C q_Y) \cong K_i(\ker q_Y) = K^i(X)$ and similarly for $q_X$.</p> <p>Now define the (special) cone $$C_{\Sigma} := \{(F, G, f) \in \Sigma \oplus C C(X \times Y) : f(0) = q_Y(F) = q_X(G)\}$$ </p> <p>We have an injective homomorphism $\kappa \colon C_{\Sigma} \to C q_X \oplus C q_Y, (F, G, f) \mapsto (F, f) \oplus (G, f)$. </p> <p>Consider the short exact sequence $$0 \to SC(X \times Y) \to C_{\Sigma} \to \Sigma \to 0$$</p> <p>Where $j \colon SC(X \times Y) \to C_{\Sigma}, f \mapsto (0, 0, f)$ and $\varphi \colon C_{\Sigma} \to \Sigma, (F, G, f) \mapsto (F, G)$.</p> <p>Now we can calculate the six-term exact sequence in K-theory. What I would like to have are isomorphisms in K-theory $K_i(C_{\Sigma}) \cong K_i(C q_X) \oplus K_i(C q_Y) \cong K^i(X) \oplus K^i(Y)$.</p> <p>We would obtain a Mayer-Vietoris sequence which no longer depends on $A, B$.</p> <p>But these cones are not homotopy equivalent. I would just like to know a concrete counterexample that such an isomorphism in K-theory does not hold (given $X, Y, A, B$ etc.).</p>