# Size of brackets containing subscripts with descender

When placing a_j in \left(...\right), the dynamic parentheses scale to bigl(...bigr).

How can I tell LaTeX to treat subscript j (and J, f, g, p, q, y) as if it had no descender? I'm using LyX which is the reason for all the \left(...\right) stuff. Just using (...) makes formulas much harder to edit.

I know that I could use \left(a_{\smash[b]{j}}\right)_{\smash[b]{j}}, but I don't want to modify every equation.

Note: Computer Modern Math and Latin Modern Math don't have that problem for whatever reason. In any other math fonts I tried (e.g. Euler, Kurier, or Libertine) it does exist.

I've found \left[ and \right] resizing differently for very similar formulae. Don't know how to fix it. and Subscripts after a macro for \left( and \right) parentheses which solve the problem partially.

• using \left\right here is wrong anyway you get unwanted horizontal space as well as the wrong size delimiter, just use (a_i) – David Carlisle Jul 21 '18 at 15:49
• If you can't avoid \left and \right with LyX, then it's time to leave it alone and type in the code yourself. – egreg Jul 21 '18 at 20:57

To avoid running into autosizing issues, while still being able to place the outer index well below the inner index, write

(a_i{)}_i


and

(a_j{)}_j


Encasing the ) symbol in curly braces serves to change its math “status” from math-close to math-ord which, in turn, influences how the outer subscripts are placed.

A full MWE (note the relative positions of the outer subscripts):

\documentclass{article}
\usepackage{palatino,eulervm} % to mimic the "look" of the OP's screenshot
\begin{document}
$(a_i{)}_i$ vs.\ $(a_i)_i$

\medskip
$(a_j{)}_j$ vs.\ $(a_j)_j$
\end{document}

• This is simply wrong, sorry. – egreg Jul 21 '18 at 20:57
• The placement of the subscript doesn't depend on the type of the atom. In the second case, the subscript is to a subformula: try $(a_i\mathord)_j$ and see. – egreg Jul 21 '18 at 21:06
• @egreg: It appears to matter whether one writes \mathord) or \mathord{)}. I just posted an MWE to illustrate my earlier, somewhat terse answer. – Mico Jul 21 '18 at 21:10
• Yes, of course: \mathord) doesn't create a subformula, \mathord{)} does. Which is exactly the point. – egreg Jul 21 '18 at 21:18
• @egreg - (Just got back from the TUG conference banquet dinner.) I continue to believe that the main point of my answer is about the difference in placement of the subscript i, depending on whether one writes $)_i$ or ${)}_i$. If the latter expression entails "creating a subformula", I have no quarrel with that. – Mico Jul 22 '18 at 1:30