The question

Is there a prime for which the bound will be tight?

Does it give sharp bounds? Yes: the quartic at $p=89$, from $10$ to $4$.
Does it always improve the reduction bound? No: at $p=31$ the quartic already has the sharp bound $4$.
Does more precision always make the bound sharp? No: the quintic at $p=29$ stays at $3$, although its Picard number is $1$.
Can base change lose sharpness? Yes: the real multiplication example at $83$ would give $16$ over $\mathbf{Q}$, but at least $17$ using only the squared Frobenius over $\mathbf{Q}(\sqrt{2})$.

Is there always a sharp prime over $\mathbf{Q}$?

SOURCE: O:L1462-1480

What it does. Ends the lecture, and the break falls here.

PANEL FLAG: the panel holds the unrestricted question to be settled, since unconditional algorithms to compute the geometric Neron-Severi group of a K3 surface over a number field exist (Poonen-Testa-van Luijk, Theorem 8.38, https://math.mit.edu/~poonen/papers/compute_ns.pdf). As the panel reads it, the question needs scoping to this particular obstruction method and to a stated base field, and the preceding real-multiplication example is not a failure of the obstruction method. The wording is the author's own, so it is unchanged pending the author's decision.

EDITORIAL: the deck's blockquote wording is not in the source frame, so the frame's own question is now on the slide; the frame's question is already scoped to "a prime for which the bound will be tight", which is the scoping the panel asked for. The third visible line, "Not a claim that the method always works, and not a claim that it cannot be made to.", is likewise deck-written, not on the frame.

Spoken: Is there a way to get a sharp upper bound on the Picard number of a K3 surface?

Spoken: Open, with the RM case above as the known obstacle. Hopeful over $\mathbf{Q}$.

TRANSCRIBED FROM: frobenious-dist/nyc-jnts.tex:1462-1480 (first item)