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Has anything been written on the following question:

Let $(\mathfrak{A}, \phi)$ consist of a C*-algebra $\mathfrak{A}$ equipped with a tracial state $\phi$. Let $\pi : \mathfrak{A} \twoheadrightarrow \mathfrak{A}^\mathrm{ab}$ be the quotient map of $\mathfrak{A}$ onto its abelianization $\mathfrak{A}^\mathrm{ab}$. Under what conditions does there exist a state $\hat{\phi}$ on $\mathfrak{A}^\mathrm{ab}$ such that $\phi = \hat{\phi} \circ \pi$?

The obvious answer is that you need $\ker \phi \supseteq \left( \left\{ xy - yx : x, y \in \mathfrak{A} \right\} \right)$, where $(S)$ means the smallest closed two-sided ideal containing $S$. This means that $\phi$ being tracial is a necessary condition. But it doesn't look like it should be a sufficient condition, since $\ker \phi$ isn't generally an ideal.

I apologize for the perhaps overly broad question, but I was surprised how hard it was to find anyone considering the question of when a state on a C*-algebra can descend to a state on a quotient.

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