The follwing LaTex lines don't print out what I hope to see:

calculate $\rho_i^{t} = \| \mathcal{w}_{i}^{t} - \mathcal{w}_{i}^{t-1} \|_2$

what I get instead of the two w is that:

see here

Can anyone help me please?

Thank you!


1 Answer 1


That's because the font does not have lowercase characters. This is also documented in the TeXbook (in plain TeX you get the effect of \mathcal using \cal).

For example, ‘$\cal A$’ produces ‘𝒜‘ and ‘$\cal Z$‘ produces ‘𝒵’. But beware: This works only with the letters A to Z; you'll get weird results if you apply \cal to lowercase or Greek letters.

As you can see from the symbols table below, there are other characters encoded in the slots where lowercase usually is (w maps to slot 119).

enter image description here

That being said, some Unicode math fonts have a lowercase script alphabet, e.g. XITS Math. The Unicode version of Latin Modern unfortunately does not have a lowercase script alphabet.

\setmathfont{XITS Math}
$\rho_i^{t} = \| \mathscr{w}_{i}^{t} - \mathscr{w}_{i}^{t-1} \|_2$

enter image description here

  • Wonderful, thank you. I used now \mathit{w}. Looks also pretty satisfactorily. Commented Apr 24, 2018 at 6:42
  • 3
    @FelixLaumann Instead of \mathit{w} you can also just use plain w. The font should be the same. The only difference is that \mathit enables character kerning (compare \mathit{ff} and ff), i.e. it is to be used for actual words. Commented Apr 24, 2018 at 6:46
  • 3
    @HenriMenke or rather \mathit is wrong, it uses the text italic font (so multi-letter identifiers work) w on its own will use math italic (which typically has wider sidebearings, and potentially different character shapes) Commented Apr 24, 2018 at 7:08

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