Showing invertibility of bounded operators.

Solution 1:

Since $A$ acts irreducible on $H$ by (b), we have $$ A'=\{R\in L(H)\mid RX=XR\text{ for all }X\in A\}=\mathbb C I $$ by Schur's lemma. Direct calculations using (3)-(5) show that $S^\ast S,T^\ast T\in A'$. Combined with (2) this implies $S^\ast S=tI$, $T^\ast T=(1-t)I$ for some $t\in[0,1]$. If $t=0$ or $t=1$, then $A$ would be commutative, but all irreducible representations of commutative $C^\ast$-algebras are $1$-dimensional, in contradiction to (a). Thus $t\in (0,1)$.

Moreover, $SS^\ast=S^\ast S=t I$ for $t\neq 0$ implies $\ker S=\{0\}$ and $\operatorname{ran}S=H$. Hence $S$ is invertible, and the same holds for $T$. Furthermore, the operators $U=t^{-1/2}S$ and $V=(1-t)^{-1/2}T$ are surjective isometries, hence unitaries, and anticommute because $S$ and $T$ do.

I don't see how to prove (c) at the moment. I'll come back to that later or maybe someone else can help out.

Solution 2:

Too long for a comment, but building on MaoWao's answer it is not hard to show that $\dim H=2$. The spirit of the argument is to notice that necessarily $$ S=\sqrt t\,\begin{bmatrix} 1&0\\0&-1\end{bmatrix},\qquad T=\tfrac{\sqrt{1-t}}2\begin{bmatrix} 0&1\\1&0\end{bmatrix} $$ (properly, some unitary conjugation of those). I expect that the argument can be simpler, I didn't spend time on it and just went with what came naturally first.

With all the properties established by MaoWao, we are now given unitaries $U$ and $V$ such that $UV=-VU$ and the C$^*$-algebra $A$ they generate is irreducible.

From $UV=-VU$ we get that $U^2V=VU^2$. From Flugede-Putnam we also get $U^2V^*=V^*U^2$. So $U^2\in A'=\mathbb CI$. This implies that after multiplying by an appropriate scalar we may assume that $U^2=I$ and the spectrum of $U$ is $\{1,-1\}$ (it cannot be just $\{1\}$ because $U\ne I$, and multiplying $U$ by a scalar will still generate the same algebra and satisfy the same relations). Similarly, we can assume $\sigma(V)=\{1,-1\}$.

We can now write $U=2P-I$, $V=2Q-I$ for certain projections $P$ and $Q$ (these are the respective spectral projections corresponding to $1$, so $U=P-(I-P)$, $V=Q-(I-Q)$). The equality $UV=-VU$ now reads $$ 4PQ-2Q-2P+I=-(4QP-2P-2Q+I). $$ This we may write as $$\tag1 4PQ+4QP-4Q-4P+2I=0. $$ If we multiply by $I-P$ on the left and right, $$ -4(I-P)Q(I-P)+2(I-P)=0. $$ That is, $$(I-P)Q(I-P)=\tfrac12\,(I-P).$$ Conjugating $(1)$ with $P$ we get $$ PQP=\tfrac12\,P. $$

If we take $W=\sqrt2\,PQ(I-P)$, it is easy to check from the above formulas (or variations of them using the symmetry of the roles) that $W^*W=I-P$, $WW^*=P$. Thus $$\tag2 P\simeq I-P. $$

The relation $(2)$ allow us to assume below that $PH$ and $(I-P)H$ are the same space.

Now we express everything as $2\times2$ matrices in terms of the decomposition $H=PH\oplus (I-P)H$. So $$ P=\begin{bmatrix} 1&0\\0&0\end{bmatrix},\qquad Q=\begin{bmatrix} 1/2&T\\ T^*&1/2\end{bmatrix}. $$ From $Q^2=Q$ we get that $T=\operatorname{Re} T$, and $T^2=\tfrac14\,I$. So $T$ is selfadjoint and $\sigma(T)=\{1/2,-1/2\}$. This means that there exists a projection $R$ with $T=R-\tfrac12\,I$. The element $$ \begin{bmatrix} R&0\\0&R\end{bmatrix} $$ commutes with both $P$ and $Q$, and so it is in $A'$. This means that $R=0$ or $R=I$, as it is a projection. Thus $T=\pm\tfrac12$.

It is now trivial to check via matrix computations that $A'$ (that is the set of operators that commute with both $P$ and $Q$) consists of all operators of the form $$ \begin{bmatrix} Z&0\\0&Z\end{bmatrix}. $$ As $A'=\mathbb C I$, such $Z$ has to be a scalar multiple of the identity. This is only possible if $P$ has rank 1. Thus $I-P$ has rank 1 by $(2)$, and $\dim H=2$.