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

   > Hi Uwe,
> 2016-09-11 13:51 GMT+02:00 Uwe Brauer <o...@mat.ucm.es>:

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

Take the following example using equations

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

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

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.

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