# Meeting 4: Shintaro Nishikawa

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#### Remark: Comparison fails in C*-algebra case

The conclusion of Theorem 3.2 (mutually subequivalent projections are Murray-von Neumann equivalent) is special to von Neumann algebra setting. In a C*-algebra, it is possible to have two inequivalent projections each of which is subequivalent to the other. Here are three examples of this phenomenon:

1. The Cuntz algebra $\mathcal{O}(3)$ is generated by three orthogonal isometries whose range projections $e_i = s_is_i^*$ are an orthgonal decomposition of the identity: $e_1+e_2+e_3=1$. It is a fact that the $K_0$ group of $\mathcal{O}(3)$ is cyclic of order two and generated by the class of $1$. Of course, each $e_i$ is manifestly Murray-von Neumann equivalent to $1$, so $[e_1]=[e_2]=[e_3]=[1] \in K_0(\mathcal{O}(3))$. Let $p = e_1+e_2$. On the one hand, $p$ is a subprojection of $1$. On the other hand, $1 \sim_\text{MvN} e_1 \leq p$, so that $1$ is subequivalent to $p$. However, it is impossible for $p$ and and $1$ to be Murray-von Neumann equivalent, because $[p] = [e_1]+[e_2]=[1]+[1] \neq [1]$ in $K_0(\mathcal{O}(3))$.
2. Let $\mathcal{T}$ be the Toeplitz algebra generated by the unilateral shift operator $S$ on $\ell^2(\mathbb{N})$ which satisfies $S^*S = 1$, $SS^*=1-e_0$, with $e_0$ the rank-1 projection onto the first coordinate. Define $A$ to be the "double" of $\mathcal{T}$, putting $A = \{ (T_1,T_2) \in \mathcal{T} \oplus \mathcal{T} : T_1=T_2 \text{ mod compacts} \}$. In $A$, each of $1= (1,1)$ and $p =(1,1-e_0)$ is subequivalent to the other. Indeed, $p \leq 1$ and, in the other direction, $1 \sim_\text{MvN} (1-e_0,1-e_0) \leq p$. However, there cannot be any $W=(W_1,W_2) \in A$ with $W^*W=1$ and $WW^*=p$. This would entail that $W_1$ is a unitary, having index $0$, while $W_1$ is an isometry, having index $1$. This is impossible, because Fredholm operators which are equal mod compacts have the same index.
3. We can consider the universal C*-algebra $C^*(s,p)$ generated by a projection $p$ and an isometry $s$ satisfying $ss^* p = ss^*$. Evidently $p$ and $1$ are mutually subequivalent. They cannot, however, be Murray-von Neumann equivalent, or else we could map this universal example onto either of the examples above.

#### Kaplansky Density Theorem

Notation: Given a subset $S$ of C*-algebra, we use the following notations

• $S_\text{s.a.}$ for the self-adjoint part of $S$
• $S_+$ for the positive part of $S$
• $S_1$ for the intersection of $S$ with the closed unit ball.
• $U(S)$ for the unitary part of $S$.
We also mix these notations, so that, for example, $S_{+,1}$ denotes $\{x \in S: x \geq 0, \|x\| \leq 1\}$.

Theorem 4.1 (Kaplansky Density):Let $\mathscr{A}$ be a $*$-algebra of bounded operators on some Hilbert space $H$. Let $A$ and $M$ be, respectively, the C*-algebra and von Neumann algebra generated by $\mathscr{A}$. Then, the following are s.o.t. dense containments (which is the same as saying they are w.o.t. dense containments).

1. $\mathscr{A}_\text{s.a.} \subseteq M_\text{s.a.}$
2. $\mathscr{A}_{\text{s.a.},1} \subseteq M_{\text{s.a.},1}$
3. $\mathscr{A}_1 \subseteq M_1$.
4. $\mathscr{A}_{+,1} \subseteq M_{+,1}$
5. $U(\mathscr{A}) \subseteq U(M)$, if $\mathscr{A}$ is a C*-algebra with unit.

Lemma: Let $\mathscr{A}$ be a dense $*$-algebra in a C*-algebra $A$. Then the following containments are (norm) dense.

1. $\mathscr{A}_\text{s.a.} \subseteq A_\text{s.a.}$
2. $\mathscr{A}_{\text{s.a.},1} \subseteq A_{\text{s.a.},1}$
3. $\mathscr{A}_1 \subseteq A_1$.
4. $\mathscr{A}_{+,1} \subseteq A_{+,1}$

Proof:

1. Take $x \in A_\text{s.a.}$. Find $x_n \in \mathscr{A}$ norm-converging to $x$. Then, $y_n = (1/2)(x_n+x_n^*) \in \mathscr{A}_\text{s.a.}$ also norm-converge to $x$.
2. Take $x \in A_{\text{s.a.},1}$. Find $x_n \in \mathscr{A}_\text{s.a.}$ norm-converging to $x$. Then, $y_n = x_n/\|x_n\| \in \mathscr{A}_{\text{s.a.},1}$ also norm-converge to $x$.
3. Take $x \in A_1$. Find $x_n \in \mathscr{A}$ norm-converging to $x$. Then $y_n = x_n/\|x_n\| \in \mathscr{A}_1$ also norm-converge to $x$.
4. Take $x \in A_{+,1}$. Find $x_n \in \mathscr{A}_1$ norm-converging to $x^{1/2}$. Thenm $y_n = x_n^*x_n \in \mathscr{A}_{+,1}$ also norm-converge to $x$.
$\square$

NOTES NOT COMPLETE!!

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