# Urge/reason for inventing interior product ( Grassmann algebra )

Hello everyone,

I wanted to lecture on Grassmann and his works , and I have been reading the collected works of Grassmann " Die Lineale Ausdehnungslehre ". There Grassmann introduced something called " Interior product " ( Left and Right interior products ) . So I was completely stuck up there, the Bourbaki papers don't speak on the reason or urge in creating such manipulations, but its understood by someone who really mastered them.

Can anyone suggest me a good definition of the left and right interior products and explain the purpose of introducing them along with the intuition ?

I would be really honored to hear that.

Thank you.

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Contrary to your claims, Bourbaki does explain the geometric meaning of interior products in a way that is accessible for somebody unfamiliar with the subject. See sections III.11.12 and III.11.13 in Algebra (3rd edition). In fact, Bourbaki is the only book known to me that explains the geometric meaning of interior products in exterior algebras. –  Dmitri Pavlov Jul 23 '12 at 10:31
@DmitriPavlov : I have edited my question. Thanks. –  Shanmukha_Srinivasan Jul 23 '12 at 16:45
I still very strongly disagree with your characterization of Bourbaki's book. The paragraphs I referred to are very easy to read and the geometric meaning behind them is crystal clear and stated very explicitly in the text. –  Dmitri Pavlov Jul 23 '12 at 17:17
@DmitriPavlov : I am not looking about the geometric meaning, but the need to introduce them. –  Shanmukha_Srinivasan Jul 23 '12 at 17:27
Well, the geometric meaning also immediately explains why we should introduce them. Orthogonal complements constantly pop up in all branches of mathematics. –  Dmitri Pavlov Jul 23 '12 at 19:52

Here is an elementary motivation for interior products .

Suppose $V$ is an $n$-dimensional vector space .
To every non-zero vector $v\in V$ you can associate the complex $$0\to V\to...\to \Lambda ^kV\to \Lambda ^{k+1}V\to...\to \Lambda ^nV\to 0 \\quad (\star)$$ where the linear map $ext_k(v):\Lambda ^kV\to \Lambda ^{k+1}V$ is the map $\omega \mapsto v\wedge \omega$.
But how do you prove that it is exact? Answer: with interior products!

Choose a linear form $f\in V^*$ such that $f(v)=1$ and introduce the interior product map $int_k(f):\Lambda ^kV\to \Lambda ^{k-1}V$ which on a decomposable vector reduces to $$int_k(f)(v_1\wedge...\wedge v_k)=\sum_{j=1}^k (-1)^{j+1}f(v_j)v_1\wedge...\wedge \widehat {v_j}\wedge v_k$$ It is then easy to show the relation $$int_{k+1}(f)\circ ext_k(v)+ext_{k-1}(v)\circ int_k(f)=Id_k: \Lambda ^k(V)\xrightarrow {=} \Lambda ^k(V)$$ which immediately implies that the complex $(\star)$ is exact at the $k$-th slot since $$ext_k(v)(\omega)=0\implies \omega = ext_{k-1}(v)[int_{k-1}(f)(\omega)]$$ Differential geometers make essentially the same calculation in the context of the De Rham complex of differential forms: see Exercise 4 of Chapter 7 (Volume I) in Spivak's wonderful A Comprehensive Introduction to Differential Geometry

NB The above is the toy version of the theory of the Koszul complex.
It can always be souped up to any desired degree of unintelligibility (in my case not much souping up is necessary) .

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I think that the relation should be $\mathrm{int}_{k+1}(f) \circ \mathrm{ext}_{k}(v) + \mathrm{ext}_{k-1}(v) \circ \mathrm{int}_{k}(f) = f(v) \mathrm{id}$. Note the similarity to mathoverflow.net/questions/68378/clifford-algebra-non-zero These are the operators used to define the representation of the Clifford algebra of $V \oplus V^*$ (with symmetric bilinear form coming from the dual pairing) on the exterior algebra $\Lambda(V)$. –  MTS Jul 23 '12 at 21:22
This becomes an utmostly natural thing to do if one factorizes the complex $\Lambda V$ as the tensor product $\Lambda(v)\oplus\Lambda V'$ with $(v)$ the subspace spanned by $v$ and $V'$ a complement of $(v)$ in $V$. Standard general nonsense implies one need only show one of the two factors is exact, and dealing with $\Lambda(v)$ is immediate. Translating that standard general nonsense back to the complex $\Lambda V$ gives interior products. –  Mariano Suárez-Alvarez Jul 23 '12 at 22:21
ALthough I agree with all of this, isn't $int_k(v): \Lambda^{k+1}V^* \rightarrow \Lambda^kV^*$ just the naturally defined transpose of $ext_k(v): \Lambda^kV \rightarrow \Lambda^{k+1}V$? –  Deane Yang Jul 23 '12 at 22:24
Your nota bene is quite on the mark: the Kozszul complex is one of those very simple things than can be complicated into being as inintelligible as one can imagine :-) –  Mariano Suárez-Alvarez Jul 23 '12 at 23:58
Dear Shanmukha, I'm afraid I don't know: I haven't read Grassmann's original work. I only know that his contemporaries were very puzzled by his apparently unmotivated research, a common phenomenon with very original thinkers. –  Georges Elencwajg Jul 24 '12 at 8:42

Grassmann's original motivations came from mechanics and geometry. Here is what he wrote in the foreword to his 1844 book "Die Lineale Ausdehnungslehre, ein neuer Zweig der Mathematik (Linear Extension Theory, a new branch of mathematics)":

"While I was pursuing the concept of product in geometry as it had been established by my father, I concluded that not only rectangles but also parallelograms in general may be regarded as products of an adjacent pair of their sides, provided one again interprets the product, not as the product of their lengths, but as that of the two displacements with their directions taken into account. When I combined this concept of the product with that previously established for the sum, the most striking harmony resulted; thus whether I multiplied the sum (in the sense just given) of two displacements by a third displacement lying in the same plane, or the individual terms by the same displacement and added the products with due regard for their positive and negative values, the same result obtained, and must always obtain."

More can be found at http://www-history.mcs.st-and.ac.uk/Extras/Grassmann_1844.html

There is a very accessible account in the talk "Grassmann, Geometry and mechanics" by John Browne: