Kasparov's Dirac element and the index map - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T17:33:31Z http://mathoverflow.net/feeds/question/110563 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110563/kasparovs-dirac-element-and-the-index-map Kasparov's Dirac element and the index map Zhaoting Wei 2012-10-24T17:32:53Z 2012-10-25T13:59:29Z <p>In Kasparov's 1988 paper <a href="http://www.springerlink.com/content/g53222057r7k8868" rel="nofollow">Equivariant KK-theory and the Novikov conjecture</a> section 4 he defined the Dirac element for a (non-spin) $G$- Riemanian manifold $X$ as an element in the $K$-homology $K^0_G(C_{\tau}(X))=KK^G(C_{\tau}(X),\mathbb{C})$, where $C_{\tau}(X)$ is the algebra of continuous sections vanishing at infinity of the complex Clifford bundle on $X$.</p> <p>In more details, let $\mathcal{H}=L^2(\bigwedge^*(X))$ be the Hilbert space of $L^2$-forms on $X$. $C_{\tau}(X)$ acts on $\mathcal{H}=L^2(\bigwedge ^ * (X))$ (from the left) by the usually Clifford multiplication. Let $d$ be the exterior derivation and $d^*$ be its adjoint using the Riemanian structure on $X$. $\mathcal{D}:=d+d ^ *$. Of course $\mathcal{D}$ is not bounded but there exits an operator $\mathcal{F}=\mathcal{D}(1+\mathcal{D}^2)^{-1/2}\in L(\mathcal{H})$. The pair $(\mathcal{H}, \mathcal{F})$ gives the Dirac element $[d_X]\in K^0_G(C_{\tau}(X))=KK^G(C_{\tau}(X),\mathbb{C})$.</p> <p>If $X$ has a $G$-equivariant $\text{spin}^{c}$ structure, there exists a vector bundle $S$ on $X$ such that $C_{\tau}(X)=\text{End}(S)$ hence $C_{\tau}(X)$ is Morita equivalent to $C_0(X)$, the algebra of continuous functions on $X$ vanishing at infinity. Let $H$ be the Hilbert space of the $L^2$ sections of $S$ and $D$ denote the Dirac operator on $S$ and $F:=D(1+D^2)^{-1/2}\in L(H)$. $(H,F)$ gives an element in $K^{\dim X}_G(C_0(X))=KK^G_{\dim X}(C_0(X),\mathbb{C})$. It can be proved that under the Morita equivalence $C_{\tau}(X)\sim C_0(X)$, $(H,F)$ represents the same element as $(\mathcal{H}, \mathcal{F})$ in the last paragraph (The main reason is that $d+d^*$ and $D$ have the same principal symbol). This justifies the name "Dirac element" for $[d_X]=(\mathcal{H}, \mathcal{F})$ in the last paragraph.</p> <p>We also have the descent homomorphism $$\phi: KK^G(C_0(X),\mathbb{C})\rightarrow KK(C^ * _r (G; C_0(X)), C^ * _r (G))$$ hence $\phi([d_X]) \in KK(C^ * _r (G; C_0(X)), C^ * _r (G))$.</p> <p>Now if we defined the equivariant K-theory $K^ G _n (X)$ to be $K_0(C^ * _r (G; C_0(X\times\mathbb{R}^ n)))$ where $C^ * _r (G; C_0(X\times \mathbb{R}^ n))$ is the reduced cross product $C^∗$-algebra. It can be verified that when $G$ is compact, this definition coincide with the original equivariant K-theory given by Atiyah and Segal. We now use the element $\phi([d_X]) \in KK( C ^ * _ r(G;C_0(X)),C^*_ r(G))$ in the last paragraph and through the right multiplication in Kasparov product it gives a map<br> $$K^G_{i+\dim X}(X) \rightarrow K^ G_i(pt)$$ In fact more generally it give a map $K^G_{i+\dim X}(X\times Y)\rightarrow K^G_i(Y)$ for any space $Y$ and $i=0,1$ when $X$ is $\text{spin}^{c}$ . </p> <p>On the other hand, notice that when $U &lt; G$ is the maximal compact subgroup of $G$, if $X=G/U$ and $Y= pt$, we can define the index map $$K^G_{i+\dim G/U}(G/U)\rightarrow K^G_i(pt)$$ in the statement of Connes-Kasparov conjecture, whether $X=G/U$ is $\text{spin}^{c}$ or not, see Penington and Plymen's 1983 paper <a href="http://www.sciencedirect.com/science/article/pii/0022123683900356" rel="nofollow">The Dirac operator and the principal series for complex semisimple Lie groups</a>. </p> <p>Let me say more words about their construction and the relation to ours (thanks for the comments of Alain Valette). By definition $K^G_0(G/U)=K_0(C^*_r(G;C_0(G/U)))$ and $C^*_r(G;C_0(G/U))$ is Strong Morita equivalent to $C^*_r(U)$. Since $U$ is compact, $K_0(C^*_r(U))=R(U)$ the representation ring of $U$ hence $K^G_0(G/U)=K_0(C^*_r(U))=R(U)$ and similarly $K^G_1(G/U)=K_1(C^*_r(U))=0$. In Penington-Plymen paper the index map is defined to be $$R(U)\rightarrow K _ {\dim G/U}(C^*_r(G))$$ using the index map of the Dirac operator when $G/U$ is $\text{spin}^{c}$. When $G/U$ is not $\text{spin}^{c}$, they use the double covering of $G$ and then do the similar construction. </p> <p>My question is: if $X$ is not $\text{spin}^{c}$, can we also use the Dirac element $[d_X]$ to give a index map $K^G_{i+\dim X}(X\times Y)\rightarrow K^G_i(Y)$? </p> http://mathoverflow.net/questions/110563/kasparovs-dirac-element-and-the-index-map/110660#110660 Answer by Alain Valette for Kasparov's Dirac element and the index map Alain Valette 2012-10-25T13:59:29Z 2012-10-25T13:59:29Z <p>I'm not quite sure I understood your question completely, but let me try something. Since the Clifford algebra of $\mathbb{R}^n$ is unital, you have an inclusion $i:C_0(X)\rightarrow C_\tau(X)$ (sections which are pointwise multiples of the unit), which you may use to pull-back $[d_X]$, so you have $i^*[d_X]\in KK^0_G(C_0(X),\mathbb{C})$; then you may apply amplification $\tau_Y: KK^0_G(C_0(X),\mathbb{C})\rightarrow KK^0_G(C_0(X\times Y),C_0(Y))$, then the descent map $\phi$, you get an element in $KK_0(C_0(X\times Y)\rtimes G,C_0(Y)\rtimes G)$, and right Kasparov product with this element gives you a map $K^*_G(X\times Y)\rightarrow K^*_G(Y)$ - with no shift of dimension however.</p>