Re: [AUCTeX-devel] glitch in displayed math formatting (\[ ... \])

[Uwe Brauer]

   > Hi Uwe,
   > 2016-09-11 13:51 GMT+02:00 Uwe Brauer <address@hidden>:

   > What do you mean by "does not work"?  What do you expect?  What's the
   > behavior you see?


Take the following example using equations

\begin{equation}
\label{eq:6.3}
 \left\{
 \begin{array}{rrr}
  u^{\nu}\nabla_{\nu}\epsilon + (\epsilon+p) \nabla_{\nu}u^{\nu}    &= 0 \\
  (\epsilon+p)   P_{\alpha\beta}u^{\nu}\nabla_{\nu}{u^\beta} + P^\nu{}_\alpha   
 
\nabla_\nu p     &=    0 \\ u^\nu\nabla_\nu s & = 0
 \end{array}\right..
\end{equation}

Rifilling leads to

\begin{equation}
\label{eq:6.3}
 \left\{
   \begin{array}{rrr}
     u^{\nu}\nabla_{\nu}\epsilon + (\epsilon+p) \nabla_{\nu}u^{\nu}    &= 0 \\
     (\epsilon+p)   P_{\alpha\beta}u^{\nu}\nabla_{\nu}{u^\beta} + 
P^\nu{}_\alpha    
     \nabla_\nu p     &=    0 \\ u^\nu\nabla_\nu s & = 0
   \end{array}\right..
\end{equation}


Now I swore this did not work for \[\] last time I checked that is
\[
\label{eq:6.3}
 \left\{
 \begin{array}{rrr}
  u^{\nu}\nabla_{\nu}\epsilon + (\epsilon+p) \nabla_{\nu}u^{\nu}    &= 0 \\
  (\epsilon+p)   P_{\alpha\beta}u^{\nu}\nabla_{\nu}{u^\beta} + P^\nu{}_\alpha   
 
\nabla_\nu p     &=    0 \\ u^\nu\nabla_\nu s & = 0
 \end{array}\right..
\]

However I just checked and indeed it seems to work, that is filling
produce the desired behavior. 
\[
\label{eq:6.3}
 \left\{
   \begin{array}{rrr}
     u^{\nu}\nabla_{\nu}\epsilon + (\epsilon+p) \nabla_{\nu}u^{\nu}    &= 0 \\
     (\epsilon+p)   P_{\alpha\beta}u^{\nu}\nabla_{\nu}{u^\beta} + 
P^\nu{}_\alpha    
     \nabla_\nu p     &=    0 \\ u^\nu\nabla_\nu s & = 0
   \end{array}\right..
\]

I will watch this however and once I see something odd I report back.