Mysterious horizontal space when using conditional macro

Question

In order to deal with both burdensome expressions $K$-vector space and $K$-vector space $E$ at once, and this in both text and math mode alike, I defined a fine working shortcut taking two arguments, the second of them optional, as follows:

\makeatletter
\newcommand{\vs}[2]{%
    \relax\ifmmode
        \ifx\relax#2\relax
            #1\text{-vector space}%
        \else
            #1\text{-vector space }#2%
        \fi
    \else
        \ifx\relax#2\relax
            $#1$-vector space%
        \else
            $#1$-vector space $#2$%
        \fi
    \fi}
\makeatother

Thanks to the nice macro \ifemptyarg created by egreg in another post, I could redefine \vs in a thriftier way:

\documentclass{article}

\makeatletter
\def\ifemptyarg#1{%
    \if\relax\detokenize{#1}\relax    % H. Oberdiek
        \expandafter\@firstoftwo%
    \else
        \expandafter\@secondoftwo%
    \fi}
\makeatother

\newcommand{\vs}[2]{%
    \relax\ifmmode
        \ifemptyarg{#2}
            {#1\text{-vector space}}
            {#1\text{-vector space }#2}
    \else
    \ifemptyarg{#2}
        {$#1$-vector space}
        {$#1$-vector space $#2$}
    \fi}


\begin{document}

For any field $K$, a consequence of the Axiom of Choice
is that there is a basis for every \vs{K}{E}, but less known
is the fact that if there is a basis for every \vs{K}{},
and this for any $K$, then the Axiom of Choice holds!

\end{document}

But mysteriously, with this version of \vs the result of both [...] \vs{K}{E}, and [...] \vs{K}{}, is the same as if instead I was respectively writing [...] $K$-vector space\ , and [...] $K$-vector space $E$\ ,, namely, with an undesired extra space after each comma. Why?

Answer 1

You have several spaces inserted by newlines...

\newcommand{\vs}[2]{%
    \relax\ifmmode
        \ifemptyarg{#2}%
            {#1\text{-vector space}}%
            {#1\text{-vector space }#2}%
    \else
    \ifemptyarg{#2}%
        {$#1$-vector space}%
        {$#1$-vector space $#2$}%
\fi}

works. Notice that the "%" after the \ifemptyarg{#2} is not necessary, but well...