The C* algebra generated by $T$ has no projection?

Question

Let $T$ is a self-adjoint operator on a separable infinite dimensional Hilbert space $H$ such that $\|T^2- T\| < 1/4$ .

Prove that there is no (non trivial) projection in $C^*(T)$ the C* algebra generated by $T$ (and just $T$ without $1$)

Answer 1

Put aside a trivial case, when the $C^*$-algebra generated by $T$ and $I$ is one dimensional, or equivalently the set $\sigma(T)$ consists of a single point $\lambda$ (which satisfies $\lambda^2-\lambda<{1\over 4}).$

Assume that $\sigma(T)=\{\lambda_1,\lambda_2,\ldots, \lambda_n\},$ $n\ge 2.$ Then $$T=\sum_{k=1}^n \lambda_kP_{\lambda_k},$$ where $P_{\lambda_k}$ denotes the orthogonal projection on ${\rm ker}(T-\lambda_k I).$ Every projection $P_{\lambda_k}$ belongs to the $C^*$-algebra generated by $T$ and $I.$ As $\|T^2-T\|<{1\over 4},$ every eigenvalue should satisfy $\lambda_k^2-\lambda_k<{1\over 4}.$

The statement is not true for infinite dimensional spaces. Consider the Hilbert space $L^2((0,{1\over 4}),dx)$ and the operator $$(Tf)(x)=xf(x).$$ Then $\sigma(T)=\left[0,{1\over 4}\right ],$ therefore $$\sigma(T^2-T)=\left [-{3\over 16},0\right ], \quad \|T^2-T\|={3\over 16}\qquad (*).$$ The $C^*$-algebra generated by $T$ and $I$ is isometrically isomorphic to $C[0,{1\over 4}],$ hence it does not contain any nontrivial projection.
In particular the subalgebra generated just by $T$ has no projections. By the way, this subalgebra is isometrically isomorphic to $C_0\left [0,{1\over 4}\right ],$ the functions which vanish at $x=0.$

In general nontrivial projections exist if and only if the spectrum of $T$ is a not connected subset of $\mathbb{R}.$ For example consider the operator $T$ analogous to the one above, but defined on the space $L^2((0,{1\over 4})\cup ({3\over 4},1),dx).$ Then $\sigma(T)=\left[0,{1\over 4}\right ]\cup \left[{3\over 4},{1}\right ]$ and $(*)$ holds. The $C^*$-algebra generated by $T$ and $I$ is isometrically isomorphic to $C([0,{1\over 4}]\cup [{3\over 4},1]),$ hence it contains nontrivial projections. The subalgebra generated by $T$ is isometrically isomorphic to $C_0([0,{1\over 4}]\cup [{3\over 4},1]),$ therefore it also contains a nontrivial projection.

Remark. There is nothing special about the constant ${1\over 4}.$ One may consider any positive constant $a>0.$