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Recall Kaplansky's conjecture which states that every algebra homomorphism from the Banach algebra C(X) (where X is a compact Hausdorff topological space) into any other Banach algebra, is necessarily continuous. The conjecture is equivalent to the statement that every algebra norm on C(X) is equivalent to the usual uniform norm.

By work of Dales and Esterle, CH refutes the conjecture. On the other hand Solovay (building on work of H. Woodin) proved the consistency of Kaplansky's conjecture with ZFC.

As far as I know, the models constructed for the consistency of Kaplansky's conjecture also satisfy the Martin's axiom (MA). So my question is:

Question. Is Kaplansky's conjecture + the negation of MA consistent with ZFC?

Is this question open?

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    $\begingroup$ It was open in the early 90s, at least. Part of the difficulty is that $\mathsf{PFA}$, a strengthenings of $\mathsf{MA}$, implies Kaplansky's conjecture (as shown by Todorcevic). $\endgroup$ – Andrés E. Caicedo Sep 10 '13 at 6:11
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    $\begingroup$ Since MA is fragile under forcing (e.g. adding a Cohen real destroys it), perhaps one strategy would be to analyze Kaplansky's conjecture after adding a Cohen real. $\endgroup$ – Joel David Hamkins Sep 10 '13 at 10:38
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The answer is yes.

Recall that if $S$ is a Suslin tree, then PFA($S$) denotes the forcing axiom for the class of all proper posets which preserve $S$. PFA($S$) is consistent with ZFC relative to a supercompact cardinal, like PFA, and shares many of the consequences of PFA, but a model of PFA($S$) naturally does not satisfy MA (since $S$ is Suslin).

Claim: PFA($S$) implies Kaplansky's conjecture.

To prove this it suffices to show that each of the forcings used in Todorcevic's PFA proof preserve the Suslin tree $S$. I won't go through all of the details of the PFA proof; let me just point to the appropriate references.

Woodin proved (see Dales-Woodin, chapter 3) that if Kaplansky's conjecture fails, then there is a nonprincipal ultrafilter $U$ and an order-embedding

$$ (\langle \mathrm{id} \rangle/U, <_U) \to (\omega^\omega, <^*) $$

where $\langle \mathrm{id} \rangle$ is the set of all $f\in\omega^\omega$ with $\lim f(n) - 1 = \lim n - f(n) = \infty$, and $<^*$ is eventual dominance. Assume PFA($S$) and suppose such an embedding exists.

MA implies that for any nonprincipal ultrafilter $U$, there is an order-embedding $(2^{\omega_1}, <_{lex}) \to (\omega^\omega/U, <_U)$. ($<_{lex}$ is the lexicographical ordering.) Todorcevic's proof of this (see Partition Problems in Topology, Theorem 7.7) uses a poset that is not only ccc but Knaster, i.e. every uncountable sequence of conditions has an uncountable subsequence which are pairwise compatible. It's well known that such posets preserve Suslin trees; so PFA($S$) also gives us such an embedding. It's not hard to show that this embedding can be modified to map into $\langle\mathrm{id}\rangle$, so we get an embedding

$$ (2^{\omega_1}, <_{lex})\to (\omega^\omega, <^*) $$

Now Todorcevic shows (PPIT Theorem 8.8) that there can't be such an embedding in a model of PFA. The poset he uses is

$$ \mathrm{Col}(\omega_1, 2^\omega) * \mathbb{P}_{\mathcal{A},\mathcal{B}} $$

where $\mathrm{Col}(\omega_1, 2^\omega)$ is the usual countably-closed collapse of $2^\omega$ to $\omega_1$, $(\mathcal{A},\mathcal{B})$ is an $(\omega_1,\omega_1)$-gap found in the extension, and $\mathbb{P}_{\mathcal{A},\mathcal{B}}$ is the ccc poset which freezes that gap. It's again well-known that countably-closed forcings preserve Suslin trees, so we just need to show that the same holds for $\mathbb{P}_{\mathcal{A},\mathcal{B}}$, i.e. after forcing with $\mathbb{P}_{\mathcal{A},\mathcal{B}}$, $S$ is still ccc. For this it suffices to show that after forcing with $S$, $\mathbb{P}_{\mathcal{A},\mathcal{B}}$ is still ccc. But $\mathbb{P}_{\mathcal{A},\mathcal{B}}$ is ccc whenever $(\mathcal{A},\mathcal{B})$ is truly a gap, i.e. it's not split by some real $a\subseteq\omega$, and since $S$ adds no reals it follows that $\mathbb{P}_{\mathcal{A},\mathcal{B}}$ is still ccc in the extension by $S$. This completes the proof.

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