I have the following, probably very elementary question: Let $Cl^{p,q}$ be the Clifford algebra on generators $e_i$, $i=1, \ldots, p+q$ with $e_i e_j = -e_j e_i$ and $e_{i}^{2}=-1$ for $i=1,\ldots,p$, $e_{i}^{2} = 1 $ for $p+1 \leq i \leq p+q$.

Let $X$ be a compact space. Karoubi (in his book ''K-Theory'', section III.4) defines $K^{p,q} (X)$ as the abelian group, manufactured by the following recipe:

Consider triples $(E,\eta_0,\eta_1)$, where $E \to X$ is a finite-dimensional vector bundle with a $Cl^{p,q}$-module structure, and $\eta_i$ is a $Cl^{p,q}$-antilinear involution (i.e., it anticommutes with the generators) on $E$. Take the free abelian group generated by the isomorphism classes of these things, and divide out the following equivalence relations:

$(E,\eta_0,\eta_1) + (F,\zeta_0,\zeta_1)=(E \oplus F, \eta_0 \oplus \zeta_0, \eta_1 \oplus \zeta_1)$; $(E, \eta_0, \eta_1)=0$ if $\eta_0$ is homotopic to $\eta_1$ (as an $Cl^{p,q}$-antilinear involution).

There are several reformulations of this equivalence relation possible (loc. cit.). One of Karoubi's main results is that the group $K^{p,q}(X)$ is isomorphic to $KO^{p-q}(X)$.

On the other hand, we have real Kasparov theory. Karoubi's $K^{p,q}(X)$ is isomorphic to $KK (Cl^{p,q};C(X)) \cong KK (R;Cl^{q,p}\otimes C(X))$.

Recall that an element in $KK (Cl^{p,q};C(X))$ is represented by a Kasparov module, i.e. a triple $(H,\phi,F)$; here $H$ is a $Z/2$-graded real Hilbert bundle on $X$. $F$ is a family of Fredholm operators on $H$, which are odd (i.e., if $\iota$ defines the grading of $H$, then $F \iota = - \iota F$). $\phi$ is a graded $\ast$-homomorphism of $C^{\ast}$-algebras $Cl^{p,q}\to B (H)$ into the bounded operators on $H$. Moreover, the operators $F-F^{\ast}$, $[F,\phi(a)]$, $(F^2-1)$ are compact for all $a \in Cl^{p,q}$ (the bracket is the graded commutator; for the experts this is because $Cl^{p,q}$ is unital).

Here is my innocent question: how can I write down an isomorphism $K^{p,q}(X) \to KK (Cl^{p,q};C(X))$ \emph{explicitly}? I assume this is easy and a matter of pure linear algebra (therefore the tag), but, as often in this area, the published literature does not delve into concrete details of this sort. I am not interested in an abstract existence proof, since I know where to find it in the literature.

EDIT: Karoubi develops another model for $K^{p,q}(X)$, let me call it $F^{p,q}(X)$. This is the abelian group, generated by pairs $(H,F)$, where $H$ is a Hilbert space with $C^{p,q+1}$-action, and $F$ is a map from $X$ to Fredholm operators on $H$, such that for all $x \in X$, $F$ is $Cl^{p,q+1}$-antilinear and selfadjoint. The equivalence relation is given by homotopy and direct sum, and invertible operators are equivalent to $0$.

It is easy to map $F^{p,q}(X)$ into Kasparov theory: If $F$ is such a family, by a spetral deformation, one can achive that $F^2 -1 $ is compact. Consider the last generator $e_{p+q+1}$ as a $Z/2$-grading. Let $\phi: Cl^{p,q} \to B(H)$ be the map given by the Clifford action. Then $(H,\phi,F)$ represents the desired element in $KK$-theory. The isomorphism $F^{p,q} \cong K^{p,q}$ is implicit; using Kuipers theorem and the long exact sequence in $K$-theory.

Thus I can reformulate my question: How do I write down explicitly the isomorphism $K^{p,q}(X) \to F^{p,q}(X)$ (in this direction, not the other one).

  • $\begingroup$ @Johannes I've looked at Karoubi's book and I'm not sure whether you've misunderstand his definition of $K^{p,q}$. In his Definition II 2.13 and Definition III 4.11, $K^{p,q}(X)$ should be constructed by triples $(E,F, \eta)$ instead of $(E, eta_0, \eta_1)$. Hopefully this helps. $\endgroup$ May 21, 2013 at 2:23
  • $\begingroup$ @Zhaoting: Theorem III.4.22, loc. cit, states that the definition I gave in my question agrees with Def III.4.11. This is not the source of my confusion. $\endgroup$ May 21, 2013 at 8:58
  • $\begingroup$ If I'm not mistaken, $Cl^{1,0}$ (with the appropriate grading data) is a generator for $K^{1,0}(point)$, and there is a natural Clifford map $c: \mathbb{R} \to Cl^{1,0}$. $c$ is an unbounded multiplier on $Cl^{1,0}$, but we can compress it to the bounded operator $c(1 + c^2)^{-1}$ using the functional calculus. This in turn determines a Kasparov cycle. Using similar remarks for $K^{0,1}(point)$ and products, this should extend to a map on $K^{p,q}(X)$ (the product in the unbounded model for KK probably helps here). This is obviously all very sketchy, but it seems like the right idea. $\endgroup$ May 21, 2013 at 13:05
  • $\begingroup$ @Paul: ''unbounded multiplier''? We are on a finite-dimensional space, and all $C^{\ast}$-algebras in sight are unital. Moreover, $c^2 =-1$, whence $(1+c^2 )=0$. Also, if it is well-defined at all, the element you define is zero, since you have an invertible operator. $\endgroup$ May 21, 2013 at 13:42
  • $\begingroup$ Sorry, I really confused myself this morning; let me try again. I'll write $V = \mathbb{R}^{p+q}$ and $Cl(V) = Cl^{p,q}$. The map $c$ I am interested in is the embedding $c \colon V \to Cl(V)$ (which satisfies $c(v)^2 = -||v||^2 1$). It is an unbounded multiplier of the C*-algebra $\mathcal{C}(V) = C_0(V, Cl(V))$ and hence $c(1+c^2)^{-1}$ defines a class in $KK(\mathbb{R},\mathcal{C}(V))$. If I'm not mistaken, this KK-group is isomorphic to $KK(\mathbb{R},Cl^{p,q})$ by Bott periodicity. Actually, this should work over any base space (not just a point). $\endgroup$ May 21, 2013 at 16:31

2 Answers 2


About 15 years ago I used Karoubi's description of $K$-theory to solve prove some cutting and pasting formula for the index families of elliptic problems. To do so I needed rephrase Karoubi's theory into something more flexible and more computable. Some of the interpretations I found might be relevant to your question. I will briefly describe one such interpretation referring for proofs and many more details to the original source, my old paper Generalized symplectic geometries and the index of families of elliptic problems, Mem. A.M.S., vol. 128, no.609, 1997. I will refer to this as the old paper. $\newcommand{\bsH}{\mathscr{H}}$

To simplify the presentation let me define a $C^{p,q}$-module to be a Hilbert space $H$ equipped with a $8$_morphism of $C^*$_algebras $\phi: C^{P,q}\to B(H)$. A graded $C^{p,q}$-module can then be identified witha $C^{p,q+1}$-module. $\newcommand{\eF}{\mathscr{F}}$

Suppose that $H$ is a $C^{p,q+1}$-module. Define $\eF^{p,q}$ (or $\eF^{p,q}(H)$ to be the space of closed, densely defined, Fredholm, selfadjoint operators $T: H\to H$ that super-commute with the $C^{p,q}$-structure, i.e.,

$$ Te_k+ e_k T=0,\;\;\forall k=1,\dotsc, p+q. $$

The space $\eF^{p,q}$ carries a natural topology defined by the metric

$$ d(T_1,T_2)= \Vert \Psi(T_1-\Psi(T_2)\Vert,\;\;\Psi(\lambda=\lambda(1+\lambda^2)^{\frac{1}{2}}. $$

Denote by $\newcommand{\eBF}{\mathscr{BF}}$ $\eBF^{p,q}$ the subspace of $\eF^{p,q}$ consisting of bounded operators. Then one can show (see here) that the inclusion $\eBF^{p,q}\hookrightarrow \eF^{p,q}$ is a homotopy equivalence and that $\eBF^{p,q}$ is a classifying space for Karoubi's $KO^{p,q}$, which for simplicity I will denote by $K^{p,q}$.

Thus, to a compact $CW$-complex and a continuous map $T: X\to\eF^{p,q}$ one can associate an element $(E,\eta_0,\eta_1)\in K^{p,q}(X)$. To effectively describe this correspondence

$$ (X\stackrel{T}{\to}\eF^{p,q}) \to (E,\eta_0,\eta_1), $$

one needs a new, symplectic description of Karoubi's $K$-theory, and it is through the symplectic prism that I got to see Kasparov's KK lurking in the background.

Let $T\in \eF^{p,q}(H)$. Recall that, by construction $H$ is a $C^{p,q+1}$-mpodule. The direct sum $\hat{H}=H\oplus H $ has a richer structure of $C^{p+1,q+1}$-module (see section 5.2 in the old paper)

The graph $\Gamma_T$ of $T$ is a closed subspace of $\hat{H}$. Denote by $R_T:\hat{H}\to\hat{H}$ the orthogonal reflection in $\Gamma_T$. Observe that the subspace $H\oplus 0\in \hat{H}$ can be identified with $\Gamma_0$, the graph of the trivial linear map. We set

$$\Gamma_\infty=0\oplus H\subset \hat{H}. $$

Then $R_T^2=1$, and the condition $T\in\eF^{p,q}$ is equivalent with the following requirements.

  1. $R_T$ supercommutes tith the $C^{p+1,q+1}$-structure on $\hat{H}$.
  2. The pair of subspaces $(\Gamma_0,\Gamma_T)$ is a Fredholm pair.
  3. The subspace $\Gamma_T$ does not intersect $\Gamma_\infty$.

$\newcommand{\eFL}{\mathscr{FL}}$. We denote by $\eFL^{p+1,q+1}(\hat{H})$ the set of closed subspaces $L$ of $\hat{H}$ such that the reflection $R_L$ in $L$ satisfies the conditions $1$ and $2$ above. (In the old paper I called these spaces generalized lagrangian spaces of type $(p+1,q+1)$. We have thus produced an inclusion

$$\eF^{p,q}(H)\to \eFL^{p+1,q+1}(\hat{H}). $$

One can show two things. First, the space $\eFL^{p+1,q+1}$ classifies $K^{p+1,q+1}$ and second, the above inclusion is a homotopy equivalence. (The proof uses Bott periodicity.) One can use a process of symplectic reduction to canonically associate to continuous family $L: X\to \eFL^{p+1,q+1}(X)$ an element

$$ (E,\eta_0,\eta_1)\in K^{p+1,q+1}(X)\cong K^{p,q}(X). $$

Observe that the elements of $\eFL^{p+1,q+1}$ can be identified with selfajoint operators $R:\hat{H}\to \hat{H}$ such that $R^2=1$ and super-commute with the $C^{p+1,q+1}$-structure and they satisfy condition 2. This almost looks like a Kasparov element.

To actually get a Kasparov element consider a smooth, odd, nondecreasing function $\newcommand{\bR}{\mathbb{R}}$

$$\beta : \bR\to \bR,\;\;\beta (t)=\pm 1\;\;\mbox{if $\pm t>1$}. $$

For $\newcommand{\ve}{{\varepsilon}}$ $\ve>0$ we set $\beta_\ve(t)=\beta(t/\ve)$.

Then for any $T\in\eF^{p,q}$, $\beta_\ve (F)$ defines a Kasparov element for $\ve>0$ sufficiently small.

Remark. I will describe below an explicit map

$$K^{p,q}(X)\to [X, \eF^{p+1,q+1}]. $$

This is more or less what you need since $K^{p,q}(X)\cong K^{p+1,q+1}(X)$ and more generally, $K^{p_0,q_0}(X)\cong K^{p_1,q_1}(X)$ if $p_0-q_0\equiv p_1-q_1\bmod 8$.

The symplectic point of view is crucial since the above map is inspired by Floer's work on symplectic (Floer) homology. To justify the symplectic terminology let me discuss a simple example.

Let us look at a $C^{1,0}$-module. This is a real Hilbert space equipped with a orthogonal operator $J: H\to H$ such that $J^2=-1$, $J^*=-J$. In the finite dimensional case think $H=\bR^n\oplus \bR^n$, $J(x,y)=-(y,x)$.) A subspace $L\subset H$ is called Lagrangian if $JL=L^\perp$.

In the case of $\bR^n\oplus \bR^n= T^*\bR^n$ equipped with the canonical symplectic structure we obtain in this fashion the classical notion of Lagrangian subspace. If $P_L$ denotes the orthogonal projection onto $L$ and $R_L=2P_L-1$ denotes the orthogonal reflection in the subspace $L$, then $L$ is Lagragian iff $R_L$ anticommutes with $J$,

$$R_LJ+JR_L=0. $$

In algebraic terms $R_L$ defines a $\newcommand{\bZ}{\mathbb{Z}}$ $\bZ/2$-grading of the $C^{1,0}$-module $H$. However the symplectic point of view has more flexibility because it leads to certain operations which algebraically do not seem natural. (I'm thinking here of the process symplectic reduction.)

In general given a $C^{p,q}$-module $H$, we define a $(p,q)$-Lagrangian in $H$ to be a subspace $L\subset H$ such that $R_L$ supercommutes with the $C^{p,q}$-structure, i.e., $R_L$ is a $\bZ/2$-grading of the $C^{p,q}$-module $H$. Denote by $\DeclareMathOperator{\Lag}{Lag}$ $\Lag^{p,q}(H)$ the space of $(p,q)$-lagragians in $H$. Observe that and element in $K^{p,q}(X)$ is defined by a continuous map

$$ X\to {\Lag}^{p,q}(H)\times {\Lag}^{p,q}(H), \;\;x\mapsto (L_0(x), L_1(x))$$

where $H$ is a finite dimensional $C^{p,q}$-module. Thus an element in $K^{p,q}$ is a pair of continuous families of lagrangian subspaces in a $C^{p,q}$-module. Moreover, it suffices consider only the case when one of the families $L_0(x)$ is constant.

Fix a finite dimensional $C^{p',q'}$-module $H$, $p'=p+1$, $q'=q+1$. Denote by $J$ the operator on $H$ defined by the multiplication by $e_{p+1}$ so that $J^*=-J$, $J^2 =-1$.

To a pair of lagrangians $L_0, L_1\in \Lag^{p',q'}(H)$ we associate an operator $T_{L_0,L_1}\in\eF^{p,q}(\bsH)$ where

$$\bsH=L^2(0,1, H), $$

and $T_{L_0,L_1}$ is the closed, Fredholm selfadjoint unbounded operator on $\bsH$ with domain

$$ D(T_{L_0,L_1})=\bigl\lbrace u\in L^{1,2}(0,1; H);\; u(0)\in L_0,\;\;u(1)\in L_1\bigr\rbrace, $$

such that

$$ T_{L_0,L_1} u(s)=J\frac{du}{ds},\;\;s\in (0,1). $$

Above, $L^{1,2}$ denotes the Sobolev spaces of functions with first order derivative in $L^2$.

The motivation for the operator $T_{L_0,L_1}$ comes from symplectic Floer homology. In his papers on the (Floer) homology of a pair of lagrangian submanifolds A. Floer investigated the operators $T_{L_0,L_1}$ in the case $p=1, q=0$ and the indices of one-parameter families of such operators. Note that only the domain of $T_{L_0,L_1}$ depends on $L_0,L_1$. The action of $_{L_0,L_1}$ is independent of the lagrangians $L_0,L_1$>

To an element $\alpha\in K^{p',q'}(X)$ represented by a continuous map

$$ X\ni x \to (L_0(x), L_1(x))\in {\Lag}^{p',q'}(H)\times {\Lag}^{p',q'}(H) $$

we can associate an element $T_\alpha\in [X,\eF^{p,q}]$ given by the continuous map

$$ X\ni x\mapsto T_{L_0(x), L_1(x)}\in\eF^{p,q}(\bsH). $$

As mentioned before, $\eF^{p,q}(\bsH)$ classifies $K^{p,q}$ and thus the map $T_\alpha$ defines an element ${\rm ind}\; T_\alpha\in K^{p,q}(X)$. In that old paper I proved that ${\rm ind}\; T_\alpha \in K^{p,q}(X)$ coincides with $\alpha\in K^{p',q'}(X)$ via the canonical isomorphism $K^{p,q}(X)\to K^{p',q'}(X)$. The proof uses crucially the process of symplectic reduction. For details see Thm. 5.5 in the old paper.

The familly $ T_{\alpha}$ can be given a Kasparov descriptition as explained above.

  • $\begingroup$ This is not what I would call an explicit map. $\endgroup$ May 21, 2013 at 10:26
  • $\begingroup$ Mapping from the space $\mathcal{F}^{p,q}$ to $KK$-theory is much simpler. First of all, it should be the space of $Cl^{p,q+1}$-antilinear Fredholms. When you consider the last basis vector as a grading, you get a graded Hilbert space, with a graded $Cl^{p,q}$-action. The operator $F$ anticommutes with that action, hence it defines a Kasparov module for $KK(Cl^{p,q};R)$. $\endgroup$ May 21, 2013 at 10:35
  • $\begingroup$ Alternatively, I could rephrase my question: How do I make $K^{p,q}(X) \to [X; \mathcal{F}^{p,q}]$ explicit (in this direction; otherwise it is useless in the context where this question arose) $\endgroup$ May 21, 2013 at 10:36
  • $\begingroup$ In the old paper I describe an explicit map $K^{p,q}(X)\to [X,\mathscr{FL}^{p,q}]$. The symplectic reduction is a sort of inverse of this map. Because of space I will give the details as an update to my post. $\endgroup$ May 21, 2013 at 11:25
  • $\begingroup$ Still, it feels too complicated. What I imagined is that to a Karoubi triple $(E, \eta_0,\eta_1)$, one can associate a Kasparov element $(H,F)$, where $H$ is the space of sections in a \emph{finite-dimensional} graded $Cl^{q,p}$-vector bundle and $F$ some operator. Or is this too naive and there exists an obstruction against this? $\endgroup$ May 21, 2013 at 16:01

To get such an isomorphism one needs to require Karoubi's triples $(E,\eta_0,\eta_1)$ to consist of Euclidean vector bundles, and gradings of such (as well as the $Cl^{p,q}$-module on $E$). Given such a triple, one obtains a Kasparov $Cl^{p,q},C_0(X)$-module $(H,\varphi,F)$ as follows. $H$ is the completion of $C_c(X,E)$ with respect to the obvious $C_0(X)$-inner-product, $\varphi: Cl^{p,q}\longrightarrow \mathcal{L}(H)\cong C_b(X,E)$ is induced by the $Cl^{p,q}$-module $\rho:Cl^{p,q}\longrightarrow \mathcal{L}(E)$, and $F:=F_{\frac{1}{2}(\eta_0+\eta_1^*)}$; where for $f\in \mathcal{L}(E)$, $F_f(\phi)(x):=f(\phi(x)), \phi\in C_c(X,E)$. The converse is an application of Serre-Swan Theorem (and then the operator $F$ in the Kasparov module, viewed as a linear endomorphism of a hermitian vector bundle $E$, provides the gradings $\eta_0=F$ and $\eta_1=F^*$ ) ...

  • $\begingroup$ This cannot possibly be correct, because there is an important piece of structure missing: there is no $Z/2$-grading on your $H$. This is absolutely essential. Here is the reason: if $X$ is compact and if $(E,\phi)$ is a \emph{finite-dimensional} $Cl^{p,q}-C(X)$-bimodule, then $\mathcal{K}(E)=\mathcal{L}(E)$. Thus EACH graded operator $F$ yields a Kasparov module $(E,\phi,F)$, and they all represent the same element in $KK$. If one wants to represent a KK-class by a finite-dimensional Hilbert module, all the information is contained in the grading. -1; sorry about that. $\endgroup$ May 21, 2013 at 16:50
  • $\begingroup$ @Johannes: Sorry, you are right, that was a very stupid mistake. In fact, the most convenient picture of $K^{p,q}(X)$ to use here is the Fredholm one given in Karoubi's paper Algèbres de Clifford et operateurs de Fredholm, CRAS, t. 267, p. 305 (1968) (Section II): $K^{p,q}(X)$ is generated by pairs $(\mathcal{E},F)$ where $\mathcal{E}\longrightarrow X$ is a graded Hilbert bundle equipped with a $Cl^{p,q}$-module structure, and $F\in C(X,Fred_\mathcal{E})$ is a Fredholm section of degree $1$ anticommuting with the generators of $Cl^{p,q}$. Such a pair is but a $KK$-class. $\endgroup$
    – Mkouboi
    May 22, 2013 at 14:03
  • $\begingroup$ ... in fact $F$ is self-adjoint! $\endgroup$
    – Mkouboi
    May 22, 2013 at 14:05
  • $\begingroup$ Yes, that is a convenient picture, but not the one I wish to consider (I have a concrete application in mind). $\endgroup$ May 22, 2013 at 20:10

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