Vertical alignment of the `aligned` environment in table cells

Question

I think most people will agree that the vertical alignment in the table cells below is less than ideal--especially if I want to remove the \cmidrule to be more in line with accepted "best practices" for tables. What I'd like is the following: For every cell in the lower two rows of the table, the content should be flush with the top of the cell. (Possible exception: I'm not entirely convinced this is right for the leftmost column.) The headings E1|C, etc. should remain vertically centered.

Important note: ideally, all the columns except the leftmost column should have their width determined automatically. I would prefer not to have to specify the width of a parbox or similar (although if you can solve it using these, that is still better than I have been able to do so far).

I'm sure the solution to this is contained somewhere in the labyrinth of related questions, but...well...the related questions are a labyrinth, and I was unable to find what I was looking for.

Here's the code for this example:

\documentclass{article}
\usepackage{amsmath}
\usepackage{booktabs,array}

\newcommand*{\sheaf}[1]{\mathcal{#1}}

\begin{document}
\begin{tabular}{@{}>{\raggedright}m{10.5em} l l l l@{}}
\toprule
Rational curve: \\$[t,s] \mapsto$ & $\sheaf E_{1} |_C$ & $\sheaf E_{3} |_C$ & $\sheaf E_{4} |_C$ & $\sheaf E_{5} |_C$ 
\\ \midrule 
$[-s t^{8}+2s^{8} t+2s^{9},\linebreak[0]-s^{4} t^{5},\linebreak[0]s^{5} t^{4}-s^{9},\linebreak[0]2t^{9}-s^{9}]$ 
& $\begin{aligned}&\sheaf O(-3)^{3}\end{aligned}$ 
& $\begin{aligned}&\sheaf O(-2)^{8}\\ &\oplus \sheaf O(-1)^{11}\end{aligned}$ 
& $\begin{aligned}&\sheaf O(-2)^{3}\\ &\oplus \sheaf O(-1)^{30}\\ &\oplus \sheaf O\end{aligned}$ 
& $\begin{aligned}&\sheaf O(-1)^{45}\\ &\oplus \sheaf O^{10}\end{aligned}$ 
%
\\ 
\cmidrule(l){2-5}
%
$[s t^{8}-s^{2} t^{7},\linebreak[0]2t^{9}-s^{8} t,\linebreak[0]-s^{9},\linebreak[0]t^{9}]$ 
& $\begin{aligned}&\sheaf O(-6)\\ &\oplus \sheaf O(-2)\\ &\oplus \sheaf O(-1)\end{aligned}$ 
& $\begin{aligned}&\sheaf O(-4)\\ &\oplus \sheaf O(-3)\\ &\oplus \sheaf O(-2)^{4}\\ &\oplus \sheaf O(-1)^{12}\\ &\oplus \sheaf O\end{aligned}$ 
& $\begin{aligned}&\sheaf O(-3)\\ &\oplus \sheaf O(-2)^{4}\\ &\oplus \sheaf O(-1)^{25}\\ &\oplus \sheaf O^{4}\end{aligned}$ 
& $\begin{aligned}&\sheaf O(-2)^{3}\\ &\oplus \sheaf O(-1)^{39}\\ &\oplus \sheaf O^{13}\end{aligned}$ 
\\ \bottomrule
\end{tabular}

\end{document}

Answer 1

You can use t for the optional argument of aligned to get top alignment; I also changed the first column to be of type p{...}:

\documentclass{article}
\usepackage{amsmath}
\usepackage{booktabs,array}

\newcommand*{\sheaf}[1]{\mathcal{#1}}

\begin{document}
\noindent\begin{tabular}{@{}>{\raggedright}p{10.5em} l l l l@{}}
\toprule
Rational curve: \\$[t,s] \mapsto$ & $\sheaf E_{1} |_C$ & $\sheaf E_{3} |_C$ & $\sheaf E_{4} |_C$ & $\sheaf E_{5} |_C$ 
\\ \midrule 
$[-s t^{8}+2s^{8} t+2s^{9},\linebreak[0]-s^{4} t^{5},\linebreak[0]s^{5} t^{4}-s^{9},\linebreak[0]2t^{9}-s^{9}]$ 
& $\begin{aligned}[t]&\sheaf O(-3)^{3}\end{aligned}$ 
& $\begin{aligned}[t]&\sheaf O(-2)^{8}\\ &\oplus \sheaf O(-1)^{11}\end{aligned}$ 
& $\begin{aligned}[t]&\sheaf O(-2)^{3}\\ &\oplus \sheaf O(-1)^{30}\\ &\oplus \sheaf O\end{aligned}$ 
& $\begin{aligned}[t]&\sheaf O(-1)^{45}\\ &\oplus \sheaf O^{10}\end{aligned}$ 
%
\\ 
\cmidrule(l){2-5}
%
$[s t^{8}-s^{2} t^{7},\linebreak[0]2t^{9}-s^{8} t,\linebreak[0]-s^{9},\linebreak[0]t^{9}]$ 
& $\begin{aligned}[t]&\sheaf O(-6)\\ &\oplus \sheaf O(-2)\\ &\oplus \sheaf O(-1)\end{aligned}$ 
& $\begin{aligned}[t]&\sheaf O(-4)\\ &\oplus \sheaf O(-3)\\ &\oplus \sheaf O(-2)^{4}\\ &\oplus \sheaf O(-1)^{12}\\ &\oplus \sheaf O\end{aligned}$ 
& $\begin{aligned}[t]&\sheaf O(-3)\\ &\oplus \sheaf O(-2)^{4}\\ &\oplus \sheaf O(-1)^{25}\\ &\oplus \sheaf O^{4}\end{aligned}$ 
& $\begin{aligned}[t]&\sheaf O(-2)^{3}\\ &\oplus \sheaf O(-1)^{39}\\ &\oplus \sheaf O^{13}\end{aligned}$ 
\\ \bottomrule
\end{tabular}

\end{document}

Using a \parbox for the first entry of the first row, gives vertically centered entries for the row:

\documentclass{article}
\usepackage{amsmath}
\usepackage{booktabs,array}

\newcommand*{\sheaf}[1]{\mathcal{#1}}

\begin{document}
\noindent\begin{tabular}{@{}>{\raggedright}p{10.5em} l l l l@{}}
\toprule
\parbox{10.5em}{Rational curve: \\$[t,s] \mapsto$} & $\sheaf E_{1} |_C$ & $\sheaf E_{3} |_C$ & $\sheaf E_{4} |_C$ & $\sheaf E_{5} |_C$ 
\\ \midrule 
$[-s t^{8}+2s^{8} t+2s^{9},\linebreak[0]-s^{4} t^{5},\linebreak[0]s^{5} t^{4}-s^{9},\linebreak[0]2t^{9}-s^{9}]$ 
& $\begin{aligned}[t]&\sheaf O(-3)^{3}\end{aligned}$ 
& $\begin{aligned}[t]&\sheaf O(-2)^{8}\\ &\oplus \sheaf O(-1)^{11}\end{aligned}$ 
& $\begin{aligned}[t]&\sheaf O(-2)^{3}\\ &\oplus \sheaf O(-1)^{30}\\ &\oplus \sheaf O\end{aligned}$ 
& $\begin{aligned}[t]&\sheaf O(-1)^{45}\\ &\oplus \sheaf O^{10}\end{aligned}$ 
%
\\ 
\cmidrule(l){2-5}
%
$[s t^{8}-s^{2} t^{7},\linebreak[0]2t^{9}-s^{8} t,\linebreak[0]-s^{9},\linebreak[0]t^{9}]$ 
& $\begin{aligned}[t]&\sheaf O(-6)\\ &\oplus \sheaf O(-2)\\ &\oplus \sheaf O(-1)\end{aligned}$ 
& $\begin{aligned}[t]&\sheaf O(-4)\\ &\oplus \sheaf O(-3)\\ &\oplus \sheaf O(-2)^{4}\\ &\oplus \sheaf O(-1)^{12}\\ &\oplus \sheaf O\end{aligned}$ 
& $\begin{aligned}[t]&\sheaf O(-3)\\ &\oplus \sheaf O(-2)^{4}\\ &\oplus \sheaf O(-1)^{25}\\ &\oplus \sheaf O^{4}\end{aligned}$ 
& $\begin{aligned}[t]&\sheaf O(-2)^{3}\\ &\oplus \sheaf O(-1)^{39}\\ &\oplus \sheaf O^{13}\end{aligned}$ 
\\ \bottomrule
\end{tabular}

\end{document}

Personally, I would opt for the first option (everything top aligned).

Perhaps you could also use aligned environments on the first column too to get a little more spacing between the consecutive lines?

\documentclass{article}
\usepackage{amsmath}
\usepackage{booktabs,array}

\newcommand*{\sheaf}[1]{\mathcal{#1}}

\begin{document}
\noindent\begin{tabular}{@{}>{\raggedright}p{10.5em} l l l l@{}}
\toprule
Rational curve: \\$[t,s] \mapsto$ & $\sheaf E_{1} |_C$ & $\sheaf E_{3} |_C$ & $\sheaf E_{4} |_C$ & $\sheaf E_{5} |_C$ 
\\ \midrule 
$\begin{aligned}[t]
&[-s t^{8}+2s^{8} t+2s^{9},\\[-0.65ex]
&-s^{4} t^{5}, s^{5} t^{4}-s^{9}, \\[-0.65ex]
&2t^{9}-s^{9}]
\end{aligned}$ 
& $\begin{aligned}[t]&\sheaf O(-3)^{3}\end{aligned}$ 
& $\begin{aligned}[t]&\sheaf O(-2)^{8}\\ &\oplus \sheaf O(-1)^{11}\end{aligned}$ 
& $\begin{aligned}[t]&\sheaf O(-2)^{3}\\ &\oplus \sheaf O(-1)^{30}\\ &\oplus \sheaf O\end{aligned}$ 
& $\begin{aligned}[t]&\sheaf O(-1)^{45}\\ &\oplus \sheaf O^{10}\end{aligned}$ 
%
\\ 
\cmidrule(l){2-5}
%
$\begin{aligned}[t]
&[s t^{8}-s^{2} t^{7},2t^{9}-s^{8} t, \\[-0.65ex]
& -s^{9}, t^{9}]
\end{aligned}$ 
& $\begin{aligned}[t]&\sheaf O(-6)\\ &\oplus \sheaf O(-2)\\ &\oplus \sheaf O(-1)\end{aligned}$ 
& $\begin{aligned}[t]&\sheaf O(-4)\\ &\oplus \sheaf O(-3)\\ &\oplus \sheaf O(-2)^{4}\\ &\oplus \sheaf O(-1)^{12}\\ &\oplus \sheaf O\end{aligned}$ 
& $\begin{aligned}[t]&\sheaf O(-3)\\ &\oplus \sheaf O(-2)^{4}\\ &\oplus \sheaf O(-1)^{25}\\ &\oplus \sheaf O^{4}\end{aligned}$ 
& $\begin{aligned}[t]&\sheaf O(-2)^{3}\\ &\oplus \sheaf O(-1)^{39}\\ &\oplus \sheaf O^{13}\end{aligned}$ 
\\ \bottomrule
\end{tabular}

\end{document}

Using the tabularx package, you can avoid having to guess the right width for the first column:

\documentclass{article}
\usepackage{amsmath}
\usepackage{booktabs,array,tabularx}

\newcommand*{\sheaf}[1]{\mathcal{#1}}

\begin{document}
\noindent\begin{tabularx}{\textwidth}{@{}>{\raggedright}X l l l l@{}}
\toprule
Rational curve: \\$[t,s] \mapsto$ & $\sheaf E_{1} |_C$ & $\sheaf E_{3} |_C$ & $\sheaf E_{4} |_C$ & $\sheaf E_{5} |_C$ 
\\ \midrule 
$[-s t^{8}+2s^{8} t+2s^{9},\linebreak[0]-s^{4} t^{5},\linebreak[0]s^{5} t^{4}-s^{9},\linebreak[0]2t^{9}-s^{9}]$ 
& $\begin{aligned}[t]&\sheaf O(-3)^{3}\end{aligned}$ 
& $\begin{aligned}[t]&\sheaf O(-2)^{8}\\ &\oplus \sheaf O(-1)^{11}\end{aligned}$ 
& $\begin{aligned}[t]&\sheaf O(-2)^{3}\\ &\oplus \sheaf O(-1)^{30}\\ &\oplus \sheaf O\end{aligned}$ 
& $\begin{aligned}[t]&\sheaf O(-1)^{45}\\ &\oplus \sheaf O^{10}\end{aligned}$ 
%
\\ 
\cmidrule(l){2-5}
%
$[s t^{8}-s^{2} t^{7},\linebreak[0]2t^{9}-s^{8} t,\linebreak[0]-s^{9},\linebreak[0]t^{9}]$ 
& $\begin{aligned}[t]&\sheaf O(-6)\\ &\oplus \sheaf O(-2)\\ &\oplus \sheaf O(-1)\end{aligned}$ 
& $\begin{aligned}[t]&\sheaf O(-4)\\ &\oplus \sheaf O(-3)\\ &\oplus \sheaf O(-2)^{4}\\ &\oplus \sheaf O(-1)^{12}\\ &\oplus \sheaf O\end{aligned}$ 
& $\begin{aligned}[t]&\sheaf O(-3)\\ &\oplus \sheaf O(-2)^{4}\\ &\oplus \sheaf O(-1)^{25}\\ &\oplus \sheaf O^{4}\end{aligned}$ 
& $\begin{aligned}[t]&\sheaf O(-2)^{3}\\ &\oplus \sheaf O(-1)^{39}\\ &\oplus \sheaf O^{13}\end{aligned}$ 
\\ \bottomrule
\end{tabularx}

\end{document}