Urge/reason for inventing interior product ( Grassmann algebra ) - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T06:38:21Z http://mathoverflow.net/feeds/question/102917 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/102917/urge-reason-for-inventing-interior-product-grassmann-algebra Urge/reason for inventing interior product ( Grassmann algebra ) Shanmukha_Srinivasan 2012-07-23T08:42:41Z 2012-07-24T14:45:39Z <p>Hello everyone, </p> <p>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. </p> <p>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 ? </p> <p>I would be really honored to hear that. </p> <p>Thank you.</p> http://mathoverflow.net/questions/102917/urge-reason-for-inventing-interior-product-grassmann-algebra/102959#102959 Answer by Georges Elencwajg for Urge/reason for inventing interior product ( Grassmann algebra ) Georges Elencwajg 2012-07-23T20:30:41Z 2012-07-23T21:42:24Z <p>Here is an elementary motivation for interior products . </p> <p>Suppose $V$ is an $n$-dimensional vector space .<br> 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 $\omega \mapsto v\wedge \omega$.<br> But how do you prove that it is exact? Answer: with interior products! </p> <p>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 href="http://www.mathpop.com/bookhtms/dg1.htm" rel="nofollow"><em>A Comprehensive Introduction to Differential Geometry</em></a></p> <p><strong>NB</strong> The above is the toy version of the theory of the Koszul complex.<br> It can always be souped up to any desired degree of unintelligibility (in my case not much souping up is necessary) .</p> http://mathoverflow.net/questions/102917/urge-reason-for-inventing-interior-product-grassmann-algebra/103007#103007 Answer by Margaret Friedland for Urge/reason for inventing interior product ( Grassmann algebra ) Margaret Friedland 2012-07-24T14:45:39Z 2012-07-24T14:45:39Z <p>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)":</p> <p>"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."</p> <p>More can be found at <a href="http://www-history.mcs.st-and.ac.uk/Extras/Grassmann_1844.html" rel="nofollow">http://www-history.mcs.st-and.ac.uk/Extras/Grassmann_1844.html</a></p> <p>There is a very accessible account in the talk "Grassmann, Geometry and mechanics" by John Browne:</p> <p><a href="https://sites.google.com/site/grassmannalgebra/hermanngrassmann" rel="nofollow">https://sites.google.com/site/grassmannalgebra/hermanngrassmann</a></p>