Spectral value of reduced von Neumann algebra

Question

Let $M$ a von Neumann algebra and $x$ positive and $a$ a projector, $N_x$ is the von Neumann algebra generated by $x$.

Do you know if $min Spec_{aN_xa}(axa)\in Spec_M(x)$ please?

[Attempt added from comments:]

I thought that $a(x - \lambda.1)a$ is non invertible so $x - \lambda.1$ is non invertible with $\lambda = \min \sigma_{a N_x a} (axa)$.

Answer 1

First I'll address the attempt. It is not the case that if $aya$ fails to be invertible ($a$ a projection, $y$ in $M$) then $y$ must fail to be invertible. Note that "invertible" has at least two potentially different meanings. If the projection $a$ is not $\mathbf{1}$, then no element of the algebra $a M a$ is invertible in $M$. The algebra $a M a$ is a nonunital subalgebra of $M$, although it is a unital algebra with a different unit (namely $a$). So results that compare spectrum in an algebra with the spectrum in a subalgebra that contains the same unit, which you may have seen, and which may have motivated the original question(?) do not apply directly. Note also that there is nothing in the attempt that uses the fact that your chosen element of $\operatorname{spec}_{a N_x a} (axa)$ is on the boundary of that spectrum (i.e. that it is the least element, and not just some element). When a hypothesis does not seem to be used in an argument, this is a sign that maybe something stronger is true, or maybe something is wrong with the argument.

It may help to think about one concrete example.

Let $M$ be the algebra of $2 \times 2$ complex matrices (usual involution and usual addition and multiplication), let $x = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$, and let $a = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$.

Note that:

• $x$ is positive and $\operatorname{spec}_M(x) = \{0, 2\}$.
• $a N_x a = \{\lambda a: \lambda \in \mathbb{C}\}$ (whether by $N_x$ you mean the unital subalgebra of $M$ generated by $\mathbf{1}$ and $x$, or the potentially nonunital subalgebra of $M$ generated by $x$ alone),
• $axa = a$ is the multiplicative identity of $a N_x a$ and $\operatorname{spec}_{a N_xa}(axa) = \{1\}$.

So in this example $\operatorname{spec}_{aN_xa}(axa) = \{1\}$ is not a subset of $\operatorname{spec}_M(x) = \{0,2\}$.