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Does anyone know what's the rationale for using the adjectives inner and outer for certain algebraic products?

Also, I've seen the term exterior algebra. Does the exterior here have anything to do with the outer of outer product? If so, is there an interior algebra corresponding to inner products?

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Have you read the "inner product" entry at jeff560.tripod.com/i.html –  TTS Jun 21 '13 at 14:03
    
Rationales aside, in my impression, the term "exterior product" appeared sooner than "outer product", but both were already in use in the 80's, where "exterior product" was more popular in mathematics literature while "outer product" was more popular in engineering literature. –  user1551 Jun 21 '13 at 14:08
    
    
@TTS: +++ for that link; this page will be a great resource for those few enlightened expositors of mathematics who have caught on the fact that sometimes mathematics nomenclature introduces a fair bit of "cognitive interference" that silently hinders some students' understanding. –  kjo Jun 21 '13 at 19:35

4 Answers 4

Take the direct product of two groups. In the inner version, you are given a group $G$, and two subgroups $H, K$, such that $H \cap K = \{ 1 \}$, and $H, K$ commute elementwise. Then the subgroup $\langle H, K \rangle$ is called the inner direct product of $H$ and $K$, as everything takes place inside $G$.

Alternatively, if you are only given two groups $H$ and $K$, you may build a group $P$ on the set $H \times K$, with the multiplication rule $(h_1, k_1) (h_2, k_2) = (h_1 h_2, k_1 k_2)$. This is called the outer direct product, as you may say that you go out of $H$ and $K$ to build $P$.

Now $H' = \{ (h, 1) : h \in H\}$ is a subgroup of $P$ isomorphic to $H$, and $K' = \{ (1, k) : k \in K\}$ is a subgroup of $P$ isomorphic to $K$. One checks that $H' \cap K' = \{ 1 \}$, that $H', K'$ commute elementwise, and that the inner product of $H', K'$ inside $P$ is isomorphic to the outer direct product of $H, K$.

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+1 I agree that these two names are logical. But what about "inner product" for a pos. def. symmetric bilinear form on a real vector space? –  user29743 Jun 21 '13 at 14:31

I have no idea what the real reason is. But elements of exterior algebra quite often represents sub-spaces of vector space and exterior product of two elements represents union of these two sub-spaces and inner product represents their intersection.

Hence the names:

inner - biggest subspace inside

exterior - smallest subspace outside

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If you think of column vectors as matrices you can't multiply them directly, because they are the wrong shapes. You have to take the transpose of one of them. If you take the inner product $$\left( \begin{array}{c} x_1\\y_1\\z_1\end{array}\right)\cdot\left( \begin{array}{c} x_2\\y_2\\z_2\end{array}\right) = \left( \begin{array}{c} x_1\\y_1\\z_1\end{array}\right)^T\left( \begin{array}{c} x_2\\y_2\\z_2\end{array}\right)$$ The $T$ goes on the inside, if you take the outer product $$\left( \begin{array}{c} x_1\\y_1\\z_1\end{array}\right)\otimes\left( \begin{array}{c} x_2\\y_2\\z_2\end{array}\right) = \left( \begin{array}{c} x_1\\y_1\\z_1\end{array}\right)\left( \begin{array}{c} x_2\\y_2\\z_2\end{array}\right)^T$$ the $T$ goes on the outside.

The idea has been abstracted from column vectors to other areas, but the name stuck.

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Do you have a reference? Especially since your "exterior product" is different from what I would call an exterior product -- namely, the wedge product. –  M Turgeon Jun 21 '13 at 14:05
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I just remember it from a linear algebra course when I was an undergraduate. Remember that inner and outer products are not the same thing as interior and exterior products. –  Tim Jun 21 '13 at 14:20
    
Right, my mistake. –  M Turgeon Jun 21 '13 at 16:14
    
Nice. So "inner" and "outer" just tell you where the "T" goes. But the explanation only works for people who like to write vectors as columns. In my business, people typically write vectors as rows. Maybe we're unusual. Dunno. –  bubba Jun 22 '13 at 1:31

The exterior product has a direct relation with the outer product in dimension $3$. Consider the fundamental relations:

\begin{aligned} i \times i &= 0\\ j \times j &= 0\\ k \times k &= 0\\ i \times j &= k\\ j \times k &= i\\ k \times i &= j \end{aligned} Now, look at the relations for $1$-forms:

\begin{aligned} dx \wedge dx &= 0\\ dy \wedge dy &= 0\\ dz \wedge dz &= 0 \end{aligned}

Now, associate $dx, dy, dz$ with $i, j, k$, and associate $dx \wedge dy, dy \wedge dz, dz \wedge dx$ also with $i, j, k$. That's just duality.

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What does it have to do with the question the OP asked? –  M Turgeon Jun 21 '13 at 16:08

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