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What are the intuitive and historical reasons for choosing the word "exterior" for the concept of an exterior derivative of a form?

The reasoning I've heard about it is the following: let p(t) be a continuous parametric curve, then if you fix t_0, the tangent line to the curve p(t) at t_0 lies "exterior" of the curve p(t), since it is an approximation of p(t) itself.

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2 Answers 2

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I) The term exterior multiplication ("äussere Multiplication") is due to Grassmann, who introduced the term in his book (written in 1844)

Die Wissenschaft der extensiven Grösse oder die Ausdehnungslehre, eine neue Mathematische Disciplin"

As you can check in the table of contents of the book (on page 276), paragraphs §§34,35 are called Grundgesetze der äusseren Multiplication (Basic laws of exterior multiplication). Here is the scan of this book by Google .

II) The terminology exterior differential ("différentielle extérieure") was introduced in the 1930's by articles of Elie Cartan, inspired by Grassmann.

Here is a secondary reference from an Analysis course by Chatterji and another by Chern and Chevalley, in their analysis of Elie Cartan's mathematical contributions (cf. in particular pages 229 and 230 )

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    $\begingroup$ I'm going to go out on a limb here and say that the reason why $\wedge$ is called exterior multiplication is because the wedge product of two one-forms, say, is not a one-form and hence it lies "outside" the space of one-forms. In fact, whenever I've taught differential forms to undergraduates and introduce the wedge product for the first time, I always have to remark that the wedge product $dx^i \wedge dx^j$ is just that and not something we have already seen in class. $\endgroup$ Aug 31, 2010 at 22:24
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    $\begingroup$ For a comparison, the automorphisms f(x) = gxg^{-1} on a group or the derivations D(x) = [x,y] on a Lie algebra are called "inner" since they are defined in terms of an element inside the algebraic structure itself. $\endgroup$
    – KConrad
    Sep 1, 2010 at 3:42
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I always presumed it is due to its relationship with the exterior product $\wedge$. (The latter's name seems natural as a complement to the interior or inner product $\cdot$; one is anti-symmetric while the other is symmetric, etc.)

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    $\begingroup$ This only begs the question of why the inner product is called "inner". $\endgroup$
    – Deane Yang
    Aug 31, 2010 at 19:59
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    $\begingroup$ This doesn't beg the question, it raises the question. Begging the question is something else. Not to be pedantic or anything. $\endgroup$
    – MTS
    Aug 31, 2010 at 20:59
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    $\begingroup$ MTS, you're right. Thanks for the correction. Details here: en.wikipedia.org/wiki/Begging_the_question $\endgroup$
    – Deane Yang
    Aug 31, 2010 at 21:48
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    $\begingroup$ MTS - Be pedantic, if you like, but how does this very slightly altered form of Deane's comment strike you "Nate's answer merely begs the question, since we don't know why the inner product got its name." $\endgroup$ Aug 31, 2010 at 21:53
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    $\begingroup$ Kevin: at sites.google.com/site/grassmannalgebra/thegrassmannalgebrabook is a modern rewriting of Grassmann's book and the table of contents mentions an exterior, interior, and regressive product. $\endgroup$
    – KConrad
    Sep 1, 2010 at 3:39

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