In Kasparov's 1988 paper Equivariant KK-theory and the Novikov conjecture 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 $X$.

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 $[d_X]\in K^0_G(C_{\tau}(X))=KK^G(C_{\tau}(X),\mathbb{C})$.

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.

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 $\phi([d_X]) \in KK(C^ * _r (G; C_0(X)), C^ * _r (G))$.

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
$$ K^G_{i+\dim X}(X) \rightarrow K^ G_i(pt) $$ 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 $\text{spin}^{c}$ .

On the other hand, notice that when $ U < 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 The Dirac operator and the principal series for complex semisimple Lie groups.

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.

My question is: if $X$ is not $\text{spin}^{c}$, can we also use the Dirac element $[d_X]$ to give a index map $K^G_{i+\dim X}(X\times Y)\rightarrow K^G_i(Y)$?

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    $\begingroup$ First, I think that you should be more specific about the properties you expect of an index map. Second, I checked the Penington-Plymen paper and there is no mention of the index map you indicate; their index map goes from the representation ring $R(U)$ to the K-theory $K_*(C^*_r(G))$ of the reduced $C^*$-algebra of $G$. $\endgroup$ Commented Oct 24, 2012 at 23:05
  • $\begingroup$ @Alain: Thank you for your comments! Sure I should be more clear about what I want. The second point you made may be because of the non-standard terminology I used. In fact for any Lie group $G$ I define $K^G_0(X)=K(C_r^*(G; C_0(X)))$ where $C_r^*(G; C_0(X))$ is the reduced cross product $C^*$-algebra. Hence the $K_*(C^*_r(G))$ is exactly $K^G_*(pt)$ in my question. As for $R(U)$, since $U$ is compact, $R(U)=K_0(C^*_r(U))$. We also notice that $C^*_r(U)$ is strong Morita equivalent to $C^*_r(G; C_0(G/U))$, hence $R(U)=K^G_0(G/U)$ in this context. The index map I am looking for is the same as $\endgroup$ Commented Oct 25, 2012 at 3:29
  • $\begingroup$ @Alain: (Continued) The index map I am looking for is the same as that in Penington-Plymen paper. I apologize for the confusing I've made and I have made some modifications in the question body to define equivariant K-theory more clearly. Nevertheless it is really appreciated that you pointed out the problem. $\endgroup$ Commented Oct 25, 2012 at 3:34
  • $\begingroup$ @Zhaoting: Thanks a lot for the clarification. I think however that you should write $K^n_G(X)$ rather than $K^G_n(X)$, because it is K-theory, i.e. a theory contravariant in $X$. $\endgroup$ Commented Oct 25, 2012 at 7:35

1 Answer 1


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.


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