Is it not true that violations of Gram's law implies transitions of both $\arg\left(\zeta\left(\frac12+i\,T\right)\right)$ and $\Im\left(\log\zeta\left(\frac12+i\,T\right)\right)$ in the neighborhood of Gram points where Gram's law is violated?

$\log \zeta(s)$ has a branch analytic on any simply connected domain not containing a zero/pole. When removing the horizontal 'lines' $(-\infty,1]$ and $(i\Im(\rho)-\infty,i\Im(\rho)+1]$ I get a simply connected domain where $\log \zeta(s)$ is analytic. This function has some jumps on vertical lines on $\Re(s)\le 1$.

@StevenClark It is a bit frustrating that you don't reply when I am asking if those few complex analysis prerequisites are clear to you. To understand what mathematica is doing you should show us a ComplexPlot (modulus/phase) of $\log \zeta(s)$ on $\Re(s)\in (-1,3),\Im(s)\in (-50,50)$

My argument is based on what I've read about Gram points and Gram's law and the equivalence of $\Im\left(\log\zeta\left(\frac12+i\,T\right)\right)$ and $\arg\left(\zeta\left(\frac12+i\,T\right)\right)$ as well as the Mathematica results. I don't believe Mathematica is doing anything different in terms of analytic continuation. When I run a simplify operation on $\Im(\log\zeta(s))$ Mathematica gives the result $\arg(\zeta(s))$, so it's possible when Mathematica is asked to evaluate $\Im(\log\zeta(s))$ it actually ends up evaluating $\arg(\zeta(s))$.

$\Im \log$ and $arg$ are the same thing, this is not the point, for a function with zeros and poles there are many ways to define $\Im \log F(s)$, adding or substracting $2 k\pi$ differently at every $s$. Gram points don't have anything to do in my answer.

I'll take a look at generating some density plots. I can also generate 3D and contour plots if needed. I can generate separate plots for abs value, real component, imaginary component, and arg if you want but I'm not sure all of these are really needed.

It is as if you didn't care of my answer.

Sorry, I was busy today. When I saw your last comment yesterday indicating "This function has some jumps on vertical lines on R(s)≤1" I thought perhaps you were investigating and beginning to come around to my position.

The following link implies $\Re(\log(\zeta(s)))=\log(|\zeta (s)|)$: wolframalpha.com/input/?i=Re%5BLog%5BZeta%5Bs%5D%5D%5D. I evaluated a density plot of $\Re(\log(\zeta(s)))-\log(|\zeta (s)|)$ and this does indeed seem to be the case. Doesn't this along with $\Im(\log\zeta(s))=\arg(\zeta(s))$ tell you everything you need to know about Mathematica's analytic continuation of $\log\zeta(s)$?

See my edit ${}{}$

Thanks. I'll study your answer and dig through some of my books to see if I can find more information on the standard definition of $\arg\zeta(s)$.

I noticed Mathematica is truncating the plot range of the 3D plot of $\Im(\log\zeta(s)))$ that I illustrated in my question above so I defined a standard version of the $\arg(s)$ function (where $-\pi<\arg(s)<\pi$) as $\arg_{\pi}(s)=-i \log\left(\frac{s}{|s| }\right)$. I then compared $\arg_{\pi}(\zeta(s))$ to the Mathematica implementation of $\arg(\zeta(s))$. Contour and density plots seem to illustrate a difference but I can't seem to confirm this difference with normal plots.

Mathematica's implementation of $\arg(\zeta(s))$ has a lot of branches for $\Re(s)<\frac{1}{2}$ and $\Im(s)\ne \rho$. Just to clarify, are you saying there are no branches in the standard version of $\arg(\zeta(s))$ for $\Re(s)<1/2$ where $\Im(s)$ does not correspond to the imaginary part of a non-trivial zeta zero?

In a comment on my question above @Conrad indicated "analytic continuity on the segment $2 \to 2+it \to 1/2+it$" which seemed inconsistent with your answer.

It is not, this is what I am doing in my answer. We start with $\log \zeta(s)$ analytic on $\Re(s) > 1$ then we continue analytically along horizontal lines.

Conrad's comment seems to imply to me the analytic continuity is only valid for $Re(s)\ge\frac{1}{2}$, whereas your answer seems to imply to me the analytic continuity extends to the left of the critical line.

No. You really need to learn about this topic: analyticity / analytic continuation (not continuity). $\log F(s)$ can be continued analytically along a curve iff $F$ is analytic with no zeros/poles on that curve. This is because $\log F(s)$ is the primitive of $F'(s)/F(s)$.