In This question, the following appears:

$$[0,2] \times \left[0,4-\sqrt{3}\right] \times \left[0,4-\sqrt{3}\right] \times \mathop{\Large\times}_{i=1}^{k-3}[0,2]$$

Thinking that that last $\times$ should be somewhat comparable in size to $\sum$ or $\prod$ or $\bigoplus$ or $\bigcup,$ I tried changing it to this: $$ [0,2] \times \left[0,4-\sqrt{3}\right] \times \left[0,4-\sqrt{3}\right] \times \mathop{\Huge\times}_{i=1}^{k-3}[0,2] $$ The subscript $i=1$ definitely seems too high. Contrast these: $$ \begin{array}{c} \displaystyle \sum_{i=1}^{k-3} [0,2] \qquad\qquad \mathop{\Huge\times}_{i=1}^{k-3} [0,2] \qquad \qquad \sum_{i=1}^{k-3} [0,2] \\ \hline \end{array} $$ Can this be corrected?

`\varprod`

is available in MathJax. (At least in the version currently available on MO and SE.) Neither is`\operatornamewithlimits`

. $\endgroup$`\raisebox`

work here? … After experimentation, now edited out, no. Well, if it did, then`\raisebox{-experimentally determined amount}`

might be the easiest solution. But it doesn't.`\bigtimes`

also doesn't work. $\endgroup$bug reportor anissuewith MathJax. In real LaTeX,`\Huge`

cannot be used inside a math formula in the first place (so this is a convenience feature of MathJax), but if you replace it with the correct`\hbox{\Huge$\times$}`

, it will behave exactly as seen here: even if wrapped in`\mathop`

, it will be set too high, unless you use`\vcenter`

to correct it. $\endgroup$`\varprod`

pointed out by @MartinSleziak (which is also not a bug report, but may be something that MathJax developers are willing to act on—by "at least for bug reports" I didn't mean "this is a bug report", but rather that, while they weren't necessarily here for this, they might act on it anyway). $\endgroup$