Dimension of an operator module

Question

Let $H$ be an infinite dimensional Hilbert space. Consider unital $C^*$-sub-algebra of $\mathcal{A}\subset\mathcal{B}(H)$ such that there exist a family of isometries $\{S_n:n\in N\}$ with pairwise orthogonal images. It is easy to see that $\mathcal{A}$ is infinite dimensional.

My question: Is there exist finite dimensional $\mathcal{A}$-module $Y\subset\mathcal{B}(H)$ with action defined by composition of operators.

Answer 1

If $\cal A$ contains infinitely many injective operators $S_n$ with pairwise orthogonal images, then for any nonzero $T \in B(H)$ the operators $S_n T$ must be linearly independent. So the only finite-dimensional left $\cal A$-module in $B(H)$ is $\{0\}$.

Answer 2

I see a little generalization of idea proposed by Robert Israel.

Let $\mathcal{A}$ be unital $C^*$-algebra such that there exist two elements $s_1$, $s_2$ with property $s_k^*s_l=\delta_k^l 1_{\mathcal{A}}$ (this is anlogues of isometries with pairwise orthgonal images). Consider elements $S_n=s_1^{n-1}s_2$. It is easy to check that $$ S_k^*S_l=\delta_k^l 1_{\mathcal{A}} \tag{1} $$ i.e. we built a family of isometries with pairwise orthogonal images.

Let $Y$ be unital left $\mathcal{A}$-module and $y\in Y\setminus\{0\}$. Consider elements $\{S_n\cdot y:n\in\mathbb{N}\}$. Assume that $\sum_{k=1}^N \lambda_k S_k\cdot y=0$. Multiplying this equality by $S_n^*$ where $n\in\{1,\ldots,N\}$ with usage of (1) we obtain $\lambda_n y=0$. Since $y\neq 0$, then $\lambda_n=0$ for all $n\in\{1,\ldots,N\}$. This means that $\{S_n\cdot y:n\in\mathbb{N}\}$ linearly independent and $\operatorname{dim}Y=+\infty$.