# Why use the control sequences \bigl, \biggl, \bigr or \biggr, as I can always use \big or \bigg?

If my manuscript file has

$$\biggl({\partial^2\over \partial x^2}+ {\partial^2 \over \partial y^2}\biggr)\bigl|\varphi(x+iy)\bigr|^2 = 0$$


or

$$\bigg({\partial^2\over \partial x^2}+ {\partial^2 \over \partial y^2}\bigg)\big|\varphi(x+iy)\big|^2 = 0$$


the result will be the same :

• to post code snippets, use 4 leading spaces, not . Commented May 30, 2011 at 19:26
• @Raphink Didn't follow you. Could you give more details ? Commented May 30, 2011 at 19:30
• @Alan: Oh sorry, I thought that was an attempt at enforcing a code higlighting by using two $ instead of a single one. My bad ;-) Commented May 30, 2011 at 19:34 • See also: difference between \big[ and \bigl[ Commented May 30, 2011 at 20:07 ## 3 Answers \big (and its friends) create an Ord(inary) atom. \bigl and \bigr (and their friends) create Open and Close atoms. Spacing between atoms can vary. For example, TeX inserts a thin space between an Op (big operator) atom and an Ord atom, while it inserts no space between an Op atom and an Open atom. Compare (Op and then Ord): $\sum\big($  with (Op and then Open): $\sum\bigl($  You can see the a list of the 13 kinds of atoms in page 158 of the TeXbook. The table in page 170 shows the spacing between pairs of adjacent atoms. EDIT: In the example give by bellochio, there's no difference in spacing due to the kind of atoms involved in both expressions; the only part which could be considered problematic is the initial part: \biggl({\partial^2\over \partial x^2}  versus \bigg({\partial^2\over \partial x^2}  However, using \showlists for the first expressions produces (only the relevant parts are shown): ### display math mode entered at line 1 \mathopen .\hbox(14.5001+9.50012)x7.36115 ..\mathon ..\hbox(14.5001+9.50012)x7.36115 ...\hbox(0.39998+23.60025)x7.36115, shifted -14.10013 [] ...\vbox(14.5+0.0)x0.0 ...\hbox(0.0+0.0)x0.0, shifted -2.5 ..\mathoff \mathord .\fraction, thickness = default .\\mathord .\.\fam1 @ .\^\fam0 2 ./\mathord ./.\fam1 @ ./\mathord ./.\fam1 x ./^\fam0 2  so the first atoms are of type Open, and Ord. For the second expression \showlists produces: ### display math mode entered at line 1 \mathord .\hbox(14.5001+9.50012)x7.36115 ..\mathon ..\hbox(14.5001+9.50012)x7.36115 ...\hbox(0.39998+23.60025)x7.36115, shifted -14.10013 [] ...\vbox(14.5+0.0)x0.0 ...\hbox(0.0+0.0)x0.0, shifted -2.5 ..\mathoff \mathord .\fraction, thickness = default .\\mathord .\.\fam1 @ .\^\fam0 2 ./\mathord ./.\fam1 @ ./\mathord ./.\fam1 x ./^\fam0 2  so the first atoms are of type Ord, and Ord. According to the table in page 170, TeX inserts no space between an Open and an Ord atom and also between two Ord atoms and that's why there's no difference in the spacing. • From the table on page 170 we have : space between an Ord atom and an Inner atom = (1). Space between an Open atom and an Inner atom = 0. How come the spaces after the first parentheses were equal in my example, in the two cases ? Commented May 30, 2011 at 21:30 • @bellochio: I've updated my answer given an explanation. Commented May 30, 2011 at 23:19 • I'm giving you the credit for the answer, but there is still something I didn't understand : according to your explanation, the distance between the first parenthesis an the first partial derivative should be 0. But you can see clearly that there's at least one pixel between those two atoms.How that could be explained ? Commented May 31, 2011 at 0:30 • @bellochio: I did a test boxing the symbols with a frame with \fboxsep set to 0pt and the space belong to the symbols themselves. Commented May 31, 2011 at 19:24 • It is strange, though, that page 158 of the TeXBook refers to "an inner atom like '$1\over2'" when fractions appear to be treated as ordinary atoms for spacing purposes. – MSC Commented Sep 4, 2014 at 15:27 As has already been pointed out in the other answers, using l ("ell") immediately following \Bigg, \bigg, \Big, and \Big informs TeX that the subsequent "fence symbol" -- (, [, \{, etc -- is to be given math-class "Math Open" rather than "Math Ordinary". The consequences of this difference in the math class status of the fence symbol ("Open" vs "Ordinary) are particularly striking if the first math atom that follows the fence symbol is an arithmetic operator such as + (plus), - (minus), or \times: • If the fence was assigned class "Math Open", i.e., if \Biggl, \biggl, etc was used, TeX will -- correctly -- not insert a bit of extra space between the fence symbol and the -/+/\times symbol, resulting in the symbol (correctly!) be typeset as a unary operator. • Absent the l ("ell") qualifier, the fence is assigned class "Math Ordinary", and TeX will incorrectly interpret the -/+/\times symbols as binary operators (class "Math Bin") and thus also insert a bit more whitespace between these symbols and the next math atom. The first few examples illustrate the resulting differences in spacing, both between the opening fence and the arithmetic operators and between the arithmetic operators and what follows. The final two examples show that this issue isn't limited to opening-fence/unary operator cases. Indeed, it also occurs for (a) opening-fence/mathop cases and (b) ending-unary-operator/closing fence cases. \documentclass{article} \usepackage{amsmath,booktabs} \begin{document} \begin{align*} \Biggl[-2x - 4y\Biggr]& \quad\text{\emph{with} l'': } \begin{array}{l} \text{\textbullet\ tight spacing after opening fence;}\\ \text{\textbullet\ first -$'' symbol treated as unary operator} \end{array}\\ \Bigg[ -2x - 4y\Bigg]& \quad\text{without l'': } \begin{array}{l} \text{\textbullet\ loose spacing after opening fence;}\\ \text{\textbullet\ first $-'' symbol treated as binary operator} \end{array}\\ \addlinespace \biggl(+3u + 7v\biggr)&\\ \bigg( +3u + 7v\bigg) &\\ \addlinespace \Bigl\vert \div 2x \times 4y\Bigr\vert &\\ \Big\vert \div 2x \times 4y\Big\vert &\\ \addlinespace \bigl\{ \times 3u \div 7v\bigr\} &\\ \big\{ \times 3u \div 7v\big\} &\\ \addlinespace \lvert \sin\theta \rvert &\\ \vert \sin\theta \vert &\\ \addlinespace \lvert a- \rvert &\\ \vert a- \vert & \text{without r'': -'' not treated as a unary operator} \end{align*} \end{document}  • Is this happening even if I group the -,+ with the number that follows? Commented Jun 30, 2014 at 9:22 • @percusse -- If one encases a group of math atoms -- a math "molecule"? -- in curly braces, the resulting math atom will be of type "Math Ord" unless one explicitly assigns a different math type. Grouping the + and - symbols with the subsequent material will thus succeed in avoiding getting too much whitespace between the opening fence and the following math atom. Speaking for myself, I do find it more natural to assign math-open status explicitly to an opening fence than to have to remember to encase the interior material in curly braces. – Mico Commented Jun 30, 2014 at 11:07 • @Andrew - Usage of either \bigr) or "just" \big) might matter for two reasons. First, if the last item before the fence symbol were a symbol that's supposed to be treated as a "trailing unary operator" (aside: is there such a thing outside of C programming, e.g., a++?!), it would be important to write ... a+ \bigr)$. Second, the spacing between the closing fence and the next item in the math list (assuming there is one) may also be affected by the fence having type math-close (i.e., if it was created via \bigr)) or math-ord (which will be the case if it was created via \big). – Mico Commented Jun 11, 2015 at 19:02 • @Andrew - See also Stefan Kottwitz's answer, in which he contrasts $\bigl[ \times \bigr]$ against $\big[ \times \big]$. – Mico Commented Jun 11, 2015 at 19:23 • Thanks Mico. I was just wondering since you didn't mention the right-hand variants. Very helpful! – user30471 Commented Jun 11, 2015 at 19:40 In that case, yes, perhaps, but try $\log\bigl($vs.$\log\big(\$


they are not the same, because \big( is AFAIR not of type math open

there are also examples regarding the type of the right fence (though I've forgotten about it at the moment)

• You mean the difference is the space between the 'g' and the '(' ? Why this difference didn't show up in my example ? Commented May 30, 2011 at 19:40