Context: Let $A$ be an ungraded (not necessarily unital) $C^*$ algebra. $\mathcal{K}$ space of compact bounded operators on an infinite separable graded Hilbert space $H=H_0 \oplus H_1$. Consider the space $$ A \otimes \mathcal{K} $$ Let us suppose there is a unique norm.

Edit: I replaced a large part of text which can be seen in history. For streamlining the post.

Claim 1' If we begin with a graded homomoprhism, $\mathcal{S} \rightarrow A \otimes \mathcal{K}$, then the unitary $u$ we obtain this way (via the Cayley transform) has the property that $\alpha(u)=u^*$.

Claim 2: For any unital graded $C^*$ algebra $B$ containing $A \otimes \mathcal{K}$, consider the grading element, $$ \epsilon = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} $$ which grades $\mathcal{K}$. Any skew unitary $u$ is equal to $$p_\epsilon = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} $$ modulo $A \otimes \mathcal{K}$, i.e. $p_\phi-p_\epsilon \in A \otimes \mathcal{K}$.

May someone elaborate the details? These are from page 43, proof of Prop 3.17 , Higson's notes.

Questions regarding Aweygan's reply

So $p_\phi - p_\epsilon \in A \otimes \mathcal{K}$, impies $[p_\phi]-[p_\epsilon]$ in fact may be regarded as an element $$K_0(A) = \ker [ K_0(A_+) \rightarrow K_0(\Bbb C) ] $$

Then how do we know $[p_\phi]-[p_\epsilon] = [p']-[q']$ the original element we were given? But then judging from the computations given by Aweygan, it seems that we have to prove, we let $u(0)=a$.

$$ \begin{pmatrix} 1+p'a/2 & 0 \\ 0 & -q'a/2 \end{pmatrix} - \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} p' & 0 \\ 0 & 0 \end{pmatrix} - \begin{pmatrix} q' & 0 \\ 0 & 0 \end{pmatrix} $$ are equivalent in $G(V(A_+))$ the group completion of the commutative monoid of projections over $A_+$.

More: So if $j:S^1 \hookrightarrow \Bbb C$ is inclusion, its decomposition $j=u+1$, where $u \in C_0(\Bbb R)$, can be computed $(j-1) \circ c$, where $c: \Bbb R \rightarrow S^1 $is Cayley tramsform. This gives $u(0)=-2$, - which I now substitute for $a$. It is still unclear if these represents the same $k$ theory element (which I have made a separate post).

  • $\begingroup$ For claim 1, it looks like you might be misenterpreting the data. I suggest rereading this material. Otherwise, I'll post an answer once I've thoroughly understood claim 2. $\endgroup$ – Aweygan Mar 7 at 17:31

For claim 1, the information copied here is not quite what they have stated in the book.

Suppose $A$ is a graded unital $C^*$-algebra, with the grading given by a $*$-automorphism $\alpha:A\to A$. A unitary $u\in A$ is called a skew-unitary if $\alpha(u)=u^*$.
If the grading is internal, i.e., there is some self-adjoint unitary $\varepsilon\in A$ such that $\alpha(x)=\varepsilon x\varepsilon$ for all $x\in A$, then the map from skew unitaries to projections given by $u\mapsto \frac12(1+u\varepsilon)$ is a bijection.

With this information, it should be clear why $\frac12(1+u\varepsilon)$ is a projection if $\varepsilon$ is a self-adjoint unitary and $\varepsilon u\varepsilon=u^*$.

For claim 2, the authors aren't claiming that any skew unitary is equivalent $p_\epsilon$, only a very special one. In this section, $\phi:\mathcal S\to A\otimes\mathcal K$ is a graded $*$-homomorphism. Via the Cayley transform, $\phi$ induces a unital $*$-homomorphism $\tilde\phi$ from $C(S^1)$ to the unitization of $A\otimes\mathcal K$. The unitary in question is then $\tilde\phi(z)$, where $z:S^1\to\mathbb C$ is the inclusion map.

Using the grading on $\mathcal K$, we can consider the algebra $B$ in question as the algebra of all $2\times 2$-matrices with entries in $\widetilde{A\otimes\mathcal K}$ (the unitization of $A\otimes\mathcal K$), graded by diagonal matrices (even part) and off-diagonal matrices (odd part). Then to say that $b=(b_{ij})\in B$ lies in $A\otimes\mathcal K$ precisely means that the scalar part of each entry $b_{ij}$ is zero.

From the graded $*$-homomorphism $\phi:C_0(\mathbb R)\to A\otimes\mathcal K$, we obtain a unital $*$-homomorphism $\tilde\phi:\widetilde{C_0(\mathbb R)}\to B$ (by mapping units to units, and everything else by $\phi$). Note that $\widetilde{C_0(\mathbb R)}=C(S^1)$ is generated by a single unitary $u$. Then $u=1+f$ for some $f\in C_0(\mathbb R)$, and
$$\tilde\phi(u)=\begin{pmatrix}1+v_{11}&v_{12}\\v_{21}&1+v_{22} \end{pmatrix}$$ where $\phi(f)=(v_{ij})\in A\otimes\mathcal K$. Then we have $$p_\phi=\frac12(1+\tilde\phi(u)\epsilon=\frac12\left(\begin{pmatrix}1&0\\0&1\end{pmatrix}+\begin{pmatrix}1+v_{11}&v_{12}\\v_{21}&1+v_{22} \end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\right)=\begin{pmatrix}1+\frac{v_{11}}{2}&\frac{-v_{12}}{2}\\\frac{v_{21}}{2}&\frac{-v_{22}}{2}\end{pmatrix},$$ so that $$p_\phi-p_\epsilon=\begin{pmatrix}\frac{v_{11}}{2}&\frac{-v_{12}}{2}\\\frac{v_{21}}{2}&\frac{-v_{22}}{2}\end{pmatrix}\in A\otimes\mathcal K.$$

Further Questions:

  1. Note that $A\otimes\mathcal K$ is isomorphic to $M_2(A\otimes\mathcal K)$, by decomposing the Hilbert space $H$ that $\mathcal K$ acts on into a direct sum $H=H_0\oplus H_1$ (This is also how the grading on $\mathcal K$ is defined). So when I say an element $(b_{ij})$ of $B$ lies in $A\otimes\mathcal K$ when the scalar part of each $b_{ij}$ is zero, I really mean that $(b_{ij})$ lies in $M_2(A\otimes\mathcal K)$.

  2. As I said above, $A\otimes\mathcal K$ is graded so that it looks like $M_2(A\otimes\mathcal K)$. Thus the homomorphism $\phi:\mathcal S\to A\otimes\mathcal K$ looks like a homomorphism $\mathcal S\to M_2(A\otimes\mathcal K)$.

  3. How I have $B$ defined, a typical element of $B$ looks like a $2\times 2$ matrix $(b_{ij})=(a_{ij}+\lambda_{ij})$, where $a_{ij}\in A\otimes\mathcal K$ and $\lambda_{ij}\in\mathbb C$. The embedding $A\otimes\mathcal K\to B$ is just $(a_{ij})\mapsto(a_{ij}+0)$, with scalar part $0$.

  • $\begingroup$ The grading of $\mathcal S$ is by even and odd functions. You can consider $B$ to be the unitization of $M_2(A\otimes\mathcal K)$, graded by diagonal matrices (even part) and off-diagonal matrices (odd part). They don't claim that $u-p_\varepsilon\in A\otimes\mathcal K$, but $p_\phi-p_\varepsilon$ is, where $p_\varepsilon=\frac12(1+u\varepsilon)$. I'm still trying to understand this myself, and I will edit my answer once I see this. $\endgroup$ – Aweygan Mar 17 at 14:33
  • $\begingroup$ I think I have it. I'll write up what I have; let me know if you have any questions. $\endgroup$ – Aweygan Mar 19 at 17:42
  • $\begingroup$ The unitary in question is the function $z:S^1\to\mathbb C$ given by inclusion: $z(e^{i\theta})=e^{i\theta}$. When we say that $C(S^1)$ is generated by $z$, we mean that the $*$-subalgebra of all polynomials in $z$ and $z^*$ is dense in $C(S^1)$ (this follows from the Stone-Weierstrass theorem). $\endgroup$ – Aweygan Mar 21 at 17:03
  • $\begingroup$ I edited my answer a few days ago. Was this not what you were looking for? $\endgroup$ – Aweygan Mar 25 at 22:35
  • $\begingroup$ Hi Aweygan, i'm really sorry for the pester, may you explain of my argument for showing that this is indeed an inverse on the right path? In fact I don't see how this is an inverse. PS. I 'd like to give more points for all the effort you've put into reply my dumb questions (even though the points probably won't matter for you :P ) $\endgroup$ – CL. Apr 14 at 22:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.