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The notation ${}^t g$ for the transpose of a linear transformation is, in my view, quite unusual: otherwise (at least in many areas of math), one almost never sees subscripts or superscripts appearing on the left.

Although $g^t$ strikes me as a more natural choice of notation, I have noticed that the notation ${}^t g$ seems to be the "highbrow" choice, especially common whenever linear transformations are being emphasized rather than matrices.

Is there any reason for this other than historical accident?

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    $\begingroup$ I don't know about 'highbrow', but putting it on the left keeps it from being confused with exponentiation. $\endgroup$
    – Robert Bryant
    Commented Oct 18, 2014 at 23:17
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    $\begingroup$ I guess to highlight that transposition itself is a linear operation, so it makes "pedantic" sense to have it appear before (though I find it grotesque, rather than "highbrow")... $\endgroup$
    – Suvrit
    Commented Oct 18, 2014 at 23:35
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    $\begingroup$ I think the left t is the Bourbaki choice. $\endgroup$
    – Gerald Edgar
    Commented Oct 18, 2014 at 23:41
  • $\begingroup$ @GeraldEdgar: I see, that's where it comes from! $\endgroup$
    – Suvrit
    Commented Oct 19, 2014 at 2:03
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    $\begingroup$ I always regarded the placement of it to the left as a reminder that the transpose reverses the order of multiplication. Differential geometers write coordinates as $x^i$, so I never thought that it would be confused with an exponent when it's in the upper right (since the reader ought to know what the context is). $\endgroup$
    – KConrad
    Commented Oct 19, 2014 at 2:21

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$$^t(g^n)=(^tg)^n$$ for $n=-1$ (inverse), $n=2$ (squaring), ... so we can just write $$^tg^n$$

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    $\begingroup$ In numerical linear algebra, where transposes normally go on the right, the notation $A^{-T} := (A^{-1})^T = (A^T)^{-1}$ is quite common, and on rare occasions I have seen $A^{2T}$. So there is an analogously suggestive solution even putting transposes on the right. $\endgroup$
    – Federico Poloni
    Commented Oct 19, 2014 at 7:17
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    $\begingroup$ (Incidentally, unfortunately one cannot extend these operations to a fully-fledged "transposition algebra" with $T^2=1$, since $A^{1+T}\neq A^{T+1}$, i.e., $A^TA\neq AA^T$.) $\endgroup$
    – Federico Poloni
    Commented Oct 19, 2014 at 7:43

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