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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).

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  • $\begingroup$ 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
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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 $0$.

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  • $\begingroup$ 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

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