Number of minima in a ribbon disk? I am asking this question mainly with the hope of finding a reference to (presumably well-trodden) topic.  Let $K$ be a ribbon knot and define $I(K)$ to be the minimum over number of minima of all ribbon disks of $K$ ("I" for invariant - I'm guessing that it has a name and a literature).
Is there a family of ribbon knots that show that $I(K)$ can be arbitrarily large?  In other words, how do I get a hold on lower bounds for $I(K)$ - hopefully without assuming slice = ribbon.   
 A: A handle decomposition of a ribbon disk gives a band presentation of the knot.  If I understand how this goes correctly, the band presentation is an unlink along with attached bands between them such that


*

*performing embedded arc surgery along these bands (which carry framing information for the surgery) gives the knot, and

*if the unknots are thought of as vertices and the bands as edges, the graph forms a tree (since loops in the graph would correspond to a maxima, which don't exist).
The number of components in the unlink is the number of minima, and so your $I(K)$ is the minimal number of such components over all band presentations of the ribbon knot $K$.
For example, here is a band presentation of a square knot:

By the way, there appear to be some moves that can go between any two band presentations for the same ribbon disk of a ribbon knot:

(See at least Figure 3 of https://arxiv.org/abs/1804.09169 or the paper they cite: 
Swenton, Frank J., On a calculus for 2-knots and surfaces in 4-space, J. Knot Theory Ramifications 10, No. 8, 1133-1141 (2001). ZBL1001.57044.).  I wonder what additional moves you might need to go between any two ribbon disks.
Anyway, there is a nice way to generate ribbon knots with fundamental groups that surject onto a group finitely generated by the conjugates of some element, as you are probably aware:
Johnson, Dennis, Homomorphs of knot groups, Proc. Am. Math. Soc. 78, 135-138 (1980). ZBL0435.57003.
Edit/warning: I tried fixing what follows, but I'm not sure it's salvageable.  I had thought that if you looked at Johnson's construction carefully, you could get the result that each $I(K)$ was an upper bound for $\operatorname{rank}(\pi_1(S^3-K))$ when $K$ is a ribbon knot.  However, I failed to consider that there might be a band presentation with fewer minima by knotting up the bands, which is a source of additional rank!
Each generator is assigned an unknot, and each band sum introduces a single additional relation depending on which unknots the band passes through. If we start with some knot $K$, then there is a ribbon knot $K'$ such that there is a surjection $\pi_1(S^3-K')\to \pi_1(S^3-K)$, which implies that $I(K')\geq \operatorname{rank}(\pi_1(S^3-K))$, where the rank is the minimal number of generators over all presentations of the finitely generated group $\pi_1(S^3-K)$.
The following paper implies that there are knot groups of arbitrary rank:
Weidmann, Richard, On the rank of amalgamated products and product knot groups, Math. Ann. 312, No. 4, 761-771 (1998). ZBL0926.20019.
In particular, the $n$-fold connect sum of any nontrivial knot $K$ has $\operatorname{rank}(\pi_1(S^3-\mathop{\#}_nK))\geq n+1$.  Thus, by following Johnson's construction one gets an infinite family of ribbon knots $K_1,K_2,\dots$ with $I(K_n)\geq n$.
(If there were a way to modify Johnson's construction to generate prime knots, then one could get an infinite family of prime ribbon knots with arbitrary rank.)
(This answer used to claim that $I(K)\geq \operatorname{rank}\pi_1(S^3-K)$ for a ribbon knot, but now I'm not so sure about that.)
A: Since the Johnson paper only gives a quotient of the knot group, it is not clear why $I(\#_n K)\geq n$, and in particular, it is not clear if $I$ can be arbitrarily large. Am I missing something obvious?
