Why does `sort <(ls -l)` work but `sort < (ls -l)` fail?

Because that's how it's meant to be.

<(...) in bash is the syntax for process substitution. It's copied from the same operator in ksh.

<, (, ), |, &, ; are special lexical tokens in bash that are used to form special operators in different combinations. <, <(, <<, <&... each have their role. < is for redirection. <file, < file would redirect input from a file. <'(file)' would redirect input from a file called (file), but <(file) is a different operator that is not a redirection operator.

< (file) would be < followed by (file). In that context, in bash, (file) is not valid. (...) can be valid as a single token in some contexts like:

(sub shell)
func () {
  ...
}
var=(foo bar)

But not in

sort < (cmd)

In the fish shell, it's different. In fish, (...) is for command substitution (the equivalent of $(...) in bash). And < is for input redirection like in Bourne-like shells.

So in fish:

sort <(echo file)

would be the same as:

sort < (echo file)

That is:

sort < file

But that's something completely different from bash's process substitution.

In the yash shell, another POSIX shell, <(...) is not for process substitution but for process redirection

In there,

sort <(ls -l)

Short for:

sort 0<(ls -l)

is a redirection operator. It's more or less equivalent to:

ls -l | sort

While in bash, the <(ls -l) is expanded to the path of a pipe, so it's more like:

ls -l | sort /dev/fd/0

In zsh, (...) is overloaded as a globbing operator ((*.txt|*.png) would expand to txt and png files) and as glob qualifier (*(/) for instance expands to directory files).

In zsh, in:

sort < (ls -l)

That (ls -l) would be treated as a glob qualifier. The l glob qualifier is to match on number of links and expects a number after l (as in ls -ld ./*(l2) would list the files with 2 links), so that's why you get a zsh: number expected error there.

sort < (w) would have given a zsh: no matches found: (w) error instead as (w) matches the files with empty name that are writeable.

sort < (w|cat) would have sorted the content of the w and/or cat files in the current directory...

short answer

you can't use cmd1 < cmd2. (where cmdX are commands)

long answer

• sort <(ls -l) will create a temporary file with the result of ls -l and "give" it to sort. it is equivalent to :

  ls -l > u
  sort u
  

• sort < (ls -l) will

  1. fork a new shell, with ls -l
  2. try to use redirect operator < to give output to sort.

unfortunatly, you must use pipe ( | ) and its variant to redirect output of a shell/command to another.

correct syntax would have been

ls -l | sort