Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $(\mathcal{A},*)$ be a $*$-algebra, we have the following observation:

Let $\|\cdot\|_1$ and $\|\cdot\|_2$ be two norms on $\mathcal{A}$ such that the involution is an isometry with respect to both norms, and that $\|\cdot\|_1$ and $\|\cdot\|_2$ are equivalent on $\mathcal{A}_{sa}$. Then the two norms are equivalent on $\mathcal{A}$.

That is, if we have two $*$-norms ($\|a\|=\|a^*\|$) that are equivalent on $\mathcal{A}_{sa}$, then they are equivalent on the whole algebra. The crucial part of the proof of the observation is that $a\in\mathcal{A}$ can be expressed as linear combination of a fixed number ($\le 2$) of self-adjoint elements ($\frac{a+a^*}{2}$, $i\frac{a-a^*}{2}$), which are themselves convex combination of $a$ and $a^*$ (both have norm $\|a\|$).

I wonder how far we can generalize this observation. There seem to be three directions, i) from $\mathcal{A}_{sa}$ to other spanning set (projections?); ii) from norms to other subadditive functions; and the one I am most interested in now iii) from $\|\cdot\|$-topology to other locally convex topologies.

Thus my question: is the topology of a $*$-algebra determined by the topology of self-adjoint elements?

share|improve this question

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.