Let $\mathscr{C}$ be the Coleman-Mazur-Buzzard eigencurve of some fixed tame level $N$. Are there any known examples of a singular point $x\in \mathscr{C}$ which lies in a unique irreducible component of $\mathscr{C}$? Certainly the eigencurve can be singular at points where irreducible components of different flavors (e.g., Eisenstein vs. cuspidal, CM vs. non-CM, distinct inertial types at some prime dividing $N$) cross each other.
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asked May 26, 2013 at 15:41
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1$\begingroup$ Dear David, take $N=11, p=3$. The ordinary family has degree two over the Iwasawa algebra. Let $\Delta$ be the discriminant. Specializing in weight $2$, the ordinary Hecke algebra is $(a,b) \in \mathbf{Z}_3 \times \mathbf{Z}_3$ with $a \equiv b \mod 3$, so $\Delta$ specializes to $9$ times a unit. In weight four, one can check that the ordinary Hecke algebra is $\mathbf{Z}_3[\sqrt{\pm 3}]$, so $\Delta$ specializes to $3$. From this, one sees that $\Delta$ is neither constant nor a square. This implies that there is only one component, and (Weierstrass preparation) that $\Delta$ has a root. $\endgroup$– Sausage RollCommented May 26, 2013 at 19:56
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$\begingroup$ Dear Sausage, Couldn't this situation correspond to something like $\mathbb T = \mathbb Z_p[[k]][X^2 - (15 - 3k)]?$ (Here I am changing variables in $\Lambda = \matbb Z[[T]]$ from $T$ to $k$ in some region of weight space, just to make things easier.) This ring is also just $\mathbb Z_p[[X]],$ with the map to weight space being given by $k = (15 - X^2)/3$, so the ordinary family itself is not singular, it is just that there is ramification in the map to weight space. Regards, Matt $\endgroup$– EmertonCommented May 27, 2013 at 15:44
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$\begingroup$ P.S. Probably I should be a little careful if I was working integrally, since there is a division by $3$ in the formula relating $k$ and $X$, but I don't think this will cause trouble on the rigid analytic fibres, which is what the OP is asking about. $\endgroup$– EmertonCommented May 27, 2013 at 15:46
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1$\begingroup$ Dear David, After having SR sign off on my comments, I think it's possible that the answer to your question is "no", that such singular points are not known. They are not known to me at least, and while I've certainly had conversations about these issues with various experts over the years, unfortunately I don't remember now what I was told in them. If Joel sees this question here, than you might get an answer, since he's thought a lot about these sorts of questions. If not, perhaps you could ask him directly. Cheers, Matt $\endgroup$– EmertonCommented May 27, 2013 at 19:54
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1$\begingroup$ Dear David, that's not quite a coincidence; I first heard of that example from Glenn Stevens in 2003 who saw it originally from Rob. Around that time, I computed by hand equations for this Hida family (using the trace formula and Weierstrass preparation) modulo small powers of $(3,T)$, enough to show that $\Delta$ had a unique root $t_0$ in $\Lambda$, but that $\Delta'(t_0) \ne 0$, which shows in this case that the family is smooth. I wondered then about the same question you are asking now, but I'm afraid to say that I don't have any profound thoughts on the matter. $\endgroup$– Sausage RollCommented May 27, 2013 at 23:22
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