When writing an integral, it seems like something should be done to separate the "d", as in \int f(x) dx
, so as not to confuse it with a variable. I've seen it left as-is, bolded, and straightened. Even among those options there are several ways to accomplish each task; e.g., I could do a \mathrm
or a \operatorname
. What is the preferred method of dealing with the "d"?
Math-Mode – Should I Use \mathrm for the ‘d’ in My Integrals
best practicesmath-mode
Related Solutions
\textrm
(and the corresponding switch \rmfamily
) changes an aspect of the font – in this case the family. If possible other aspects – like the series (bold or not) or the shape (italic) – are preserved. \text...
-commands can be nested and their "aspect changes" are combined:
\documentclass{article}
\usepackage[T1]{fontenc}
\begin{document}
\sffamily\bfseries\itshape abc
\textrm{abc}\textrm{\textup{abc}}
\end{document}
\mathrm
and the other math alphabets don't change only an aspect of a font. They switch to specific fonts. They make it possible to print symbols in special fonts. It is quite possible to setup things so that e.g. \mathrm
use times, \mathbf
is a bold palatino, \mathit
a slanted Arial etc. You can also confuse everyone by defining \mathbf
to be light itshape palatino and \mathit
a bold Arial. You can define your own special math fonts e.g. \mathvector
or \mathmatrix
or \mathfancy
. As all these commands switch to dedicated font they can't be nested. The last command (the inner one) always wins:
\documentclass{article}
\usepackage[T1]{fontenc}
\begin{document}
$\mathit{A}
\mathbf{A}
\mathit{\mathbf{A}}
\mathbf{\mathit{A}}$
\end{document}
As a general rule increasing the linespacing is bad. (Of course this assumes that the default spacing has been set appropriate to the current font). If the inline math doesn't fit within the specified line spacing consider changing that before changing the global line spacing.
For the examples given
\documentclass{article}
\usepackage{amsmath}
\DeclareMathOperator{\Hom}{Hom}
\usepackage{mathpazo}
\showoutput
\showboxdepth5
\begin{document}
\noindent
\parbox{.4\textwidth}{
xx xx xx xx xx xx xx\\
%$\left(\frac{2\pi i}{r}\right)$\\
$(\frac{2\pi i}{r})$\\
xx xx xx xx xx xx xx\\
$\Hom^{(c,d)}$\\
xx xx xx xx xx xx xx
}
\parbox{.4\textwidth}{
xx xx xx xx xx xx xx\\
xx xx xx xx xx xx xx\\
xx xx xx xx xx xx xx\\
xx xx xx xx xx xx xx\\
xx xx xx xx xx xx xx}
\end{document}
using Computer Modern they both fit into the exiting space. Using mathpazo
the \Hom
example fits but the \left \right
construction is too big. Using \left
\right
in inline math is a bit suspect anyway and if you just use normal size brackets it fits within the specified baseline.
In general you should set your settings for the space that you consider acceptable and then if the math doesn't fit in the space, use display math or a variant notation or something (sometimes you can just use \smash
to hide the hight even though it could then overprint a descender on the line above, if you know there is no descender then...) Here for example is your passage unchanged on the original baselineskip set by the font package, but with superscripts made smaller. \lineskip
glue is not used, only \baselineskip
glue as shown in the log, so confirming that baseline spacing has been preserved. There are alternatives such as playing with the font dimen parameters that control script positions. But I ended up not using that, so commented out.
\documentclass[12pt]{article}
\usepackage{amsmath}
\DeclareMathOperator{\Hilb}{Hilb}
\DeclareMathOperator{\Hom}{Hom}
\newcommand{\C}{\mathbb{C}}
\showoutput
\showboxdepth4
\usepackage{mathpazo}
\DeclareMathSizes{12pt}{12}{6.5}{5}
\begin{document}
\sbox0{$aaa$}
\typeout{
13 \the\fontdimen13\textfont2^^J
14 \the\fontdimen14\textfont2^^J
15 \the\fontdimen15\textfont2^^J
16 \the\fontdimen16\textfont2^^J
17 \the\fontdimen17\textfont2^^J
18 \the\fontdimen18\textfont2^^J
19 \the\fontdimen19\textfont2^^J
19 \the\fontdimen8\textfont3^^J
}
%
% \fontdimen13\textfont2=\dimexpr(\fontdimen13\textfont2)/10\relax
% \fontdimen14\textfont2=\dimexpr(\fontdimen14\textfont2)/10\relax
% \fontdimen15\textfont2=\dimexpr(\fontdimen15\textfont2)/10\relax
% \fontdimen16\textfont2=\dimexpr(\fontdimen16\textfont2)/10\relax
% \fontdimen17\textfont2=\dimexpr(\fontdimen17\textfont2)/10\relax
% \fontdimen18\textfont2=\dimexpr(\fontdimen18\textfont2)/10\relax
% \fontdimen19\textfont2=\dimexpr(\fontdimen19\textfont2)/10\relax
% \fontdimen8\textfont3=\dimexpr(\fontdimen8\textfont3)/10\relax
The case $\Hilb^G_G$ where $v=G$ is the regular representation of $G$ has been studied
in particular depth, and is frequently referred to as GHilb. Since the ideal sheaf of
a generic point on $[\C^2/G]$ corresponds to the regular representation of $G$, the
Hilbert-Chow morphism $\Hilb^G_G\to\C^2/G$ is an isomorphism away from $0$. Since
$\Hilb^G_G$ is smooth, this is a resolution of singularities. In fact, it turns out
that it is the minimal resolution of $\C^2/G$, and has been much studied.
The action of $(\C^*)^2$ on $\C^2$ induces an action on $\Hilb_n(\C^2)$, and since if
$G\subset (\C^*)^2$ then in particular $G$ and $(\C^*)$ commute, we also have a
$(\C^*)^2$ action on $\Hilb_G(\C^2)$. The fixed points of this action on
$\Hilb_n(\C^2)$ are precisely the monomial ideas. Since $G\subset (\C^*)^2$, we see
that the monomial ideals are also the fixed points of the action on $\Hilb_n(\C^2)^G$.
The space of $\Hom^{(c,d)}_\C(\mathcal{I},R/\mathcal{I})$ of weight $(c,d)$ vector
space maps corresponds to the cells in $\lambda$ and above $P_\lambda(c,d)$, while the
space $\Hom^{(c,d)}_R (\mathcal{I},R/\mathcal{I})$ of weight $(c,d)$ maps of $R$-
modules corresponds to the bounded regions below $P_\lambda$ and above
$P_\lambda(c,d)$.
\end{document}
Best Answer