# Special elements in the $C^*$ algebra $A \otimes \mathcal{K}$.

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.

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).

• 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. – 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 $$0$$.

• 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. – Aweygan Mar 17 at 14:33
• I think I have it. I'll write up what I have; let me know if you have any questions. – Aweygan Mar 19 at 17:42
• 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). – Aweygan Mar 21 at 17:03
• I edited my answer a few days ago. Was this not what you were looking for? – Aweygan Mar 25 at 22:35
• 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 ) – CL. Apr 14 at 22:03