Geometrical meaning of Grassmann Algebra - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T16:16:54Z http://mathoverflow.net/feeds/question/22247 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/22247/geometrical-meaning-of-grassmann-algebra Geometrical meaning of Grassmann Algebra Neil 2010-04-22T20:01:02Z 2012-09-04T07:52:20Z <p>I don't understand wedge product and Grassmann algebra. However, I heard that these concepts are obvious when you understand the geometrical intuition behind them. Can you give this geometrical meaning or name a book where it is explained?</p> http://mathoverflow.net/questions/22247/geometrical-meaning-of-grassmann-algebra/22254#22254 Answer by Dmitri Pavlov for Geometrical meaning of Grassmann Algebra Dmitri Pavlov 2010-04-22T20:53:24Z 2010-05-07T08:30:05Z <p>For a brief explanation of the geometric meaning of exterior product, interior product of a k-form and l-vector, Hodge dual etc. see my answer here: <a href="http://mathoverflow.net/questions/4648/when-to-pick-a-basis/4900#4900" rel="nofollow">http://mathoverflow.net/questions/4648/when-to-pick-a-basis/4900#4900</a></p> <p>The best reference for this stuff is Bourbaki, Algebra, Chapter 3.</p> http://mathoverflow.net/questions/22247/geometrical-meaning-of-grassmann-algebra/22256#22256 Answer by Changwei Zhou for Geometrical meaning of Grassmann Algebra Changwei Zhou 2010-04-22T20:57:59Z 2010-05-15T01:16:25Z <p>I think a good introductory book is Federer's book "Geometric measure theory", I remember the first chapter is on Grassmann algebra. Another more accessible book is "The Road to Reality" by Roger Penrose, you can check the chapter on Grassmann algebra and Clifford algebra. </p> http://mathoverflow.net/questions/22247/geometrical-meaning-of-grassmann-algebra/22275#22275 Answer by Miguel for Geometrical meaning of Grassmann Algebra Miguel 2010-04-22T23:00:29Z 2010-04-22T23:00:29Z <p>You could start by considering the vector product in 3 dimensions...</p> http://mathoverflow.net/questions/22247/geometrical-meaning-of-grassmann-algebra/22277#22277 Answer by babubba for Geometrical meaning of Grassmann Algebra babubba 2010-04-22T23:07:08Z 2010-04-22T23:07:08Z <p>I guess Bott-Tu is one book to check out. But probably my all-time favourite is Morita's Geometry of Differential Forms.</p> <p><a href="http://books.google.com/books?id=5N33Of2RzjsC&amp;printsec=frontcover&amp;dq=morita+geometry+of+differential+forms&amp;cd=1#v=onepage&amp;q&amp;f=false" rel="nofollow">http://books.google.com/books?id=5N33Of2RzjsC&amp;printsec=frontcover&amp;dq=morita+geometry+of+differential+forms&amp;cd=1#v=onepage&amp;q&amp;f=false</a></p> <p>Oh, of course there's always John M. Lee's Smooth Manifolds book, which has a lucid and thorough explanation of every single concept he introduces.</p> http://mathoverflow.net/questions/22247/geometrical-meaning-of-grassmann-algebra/22278#22278 Answer by Jacques Carette for Geometrical meaning of Grassmann Algebra Jacques Carette 2010-04-22T23:22:22Z 2010-04-22T23:22:22Z <p>You could consider going back to the source! There is a good English translation of Grassmann's original work, which is all rooted in his geometric intuition for what is now called multilinear algebra and Grassmann algebras. Of course, you'll also have to suffer through a lot of metaphysical and theological mumbo-jumbo to get at the mathematics. But the mathematics is brilliant indeed. I have long thought that his examples were always more convincing to me than any of the modern texts -- although the modern texts have much clearer mathematical definitions! I really wish that modern texts were written with the mathematical clarity and rigor of 'now', but with the detours into motivation and intuition best seen in the classics (i.e. mathematical papers from the early 1700s to the early 1920s).</p> http://mathoverflow.net/questions/22247/geometrical-meaning-of-grassmann-algebra/22279#22279 Answer by Deane Yang for Geometrical meaning of Grassmann Algebra Deane Yang 2010-04-22T23:48:24Z 2010-04-22T23:48:24Z <p>Most books just introduce the formalism of exterior algebra and dive directly into differential forms, without really explaining the geometric interpretation of just the exterior algebra of a vector space. My recollection is that the book by Harold Edwards,</p> <p><a href="http://www.amazon.com/Advanced-Calculus-Differential-Forms-Approach/dp/0817637079" rel="nofollow">http://www.amazon.com/Advanced-Calculus-Differential-Forms-Approach/dp/0817637079</a></p> <p>explains everything very nicely.</p> http://mathoverflow.net/questions/22247/geometrical-meaning-of-grassmann-algebra/22657#22657 Answer by John for Geometrical meaning of Grassmann Algebra John 2010-04-27T00:20:28Z 2010-04-27T00:20:28Z <p>You could look at</p> <p><a href="http://sites.google.com/site/grassmannalgebra" rel="nofollow">http://sites.google.com/site/grassmannalgebra</a></p> <p>There is a free pdf book draft which discusses Grassmann algebra from a geometric point of view, and without too much mathematical terminology.</p> http://mathoverflow.net/questions/22247/geometrical-meaning-of-grassmann-algebra/22665#22665 Answer by Allen Knutson for Geometrical meaning of Grassmann Algebra Allen Knutson 2010-04-27T01:29:50Z 2010-04-27T01:29:50Z <p>I'm going to vote for Guillemin and Pollack's chapter of "Differential Topology".</p> <p>Basically, a k-form should be something that you can integrate over k-dimensional submanifolds. And it shouldn't matter how you parametrize them. That means that there should be determinants baked in to the definition, since those measure how the volume changes when you change coordinates. The determinant of a matrix negates when you switch two rows, so a k-form should be antisymmetric this same way. That's pretty much it.</p> http://mathoverflow.net/questions/22247/geometrical-meaning-of-grassmann-algebra/24012#24012 Answer by Arvid for Geometrical meaning of Grassmann Algebra Arvid 2010-05-09T10:32:10Z 2010-05-09T10:32:10Z <p>I can recommend reading the book "Geometric Algebra For Computer Science, An Object Oriented Approach to Geometry". It covers not only covers the geometrical meaning of Grassmann Algebra, but even better, Clifford Algebra. </p> <p>About your question:</p> <p><em>"To extend the representational capabilities of linear algebra, Chapter 2: Spanning Oriented Subspaces introduces the outer product. The outer product of two vectors is algebraically a 2-blade. In its most simple geometric interpretation such a 2-blade represents the 2-dimensional subspace through the origin, spanned by the vectors, as shown in Figure 2.3 (left). An outer product of three vectors is a 3-blade, see Figure 2.3 (right). A 3-blade represents the volume spanned by the three vectors. Such extended geometrical entities are now basic elements of algebraic computation. We use the blades of a geometric algebra to algebraically represent all geometrical primitives. The scalars in a vector space are represented as 0-blades, the vectors by 1-blades, and the oriented area elements are 2-blades. In Part II, we will give enriched interpretations to these blades. For example, in the homogeneous model 2-blades are used to represent lines and 3-blades represent planes; in the conformal model 3-blades represent circles, and so on."</em></p> <p>-- <a href="http://www.geometricalgebra.net/tour.html" rel="nofollow">http://www.geometricalgebra.net/tour.html</a></p> http://mathoverflow.net/questions/22247/geometrical-meaning-of-grassmann-algebra/70688#70688 Answer by Norm Cimon for Geometrical meaning of Grassmann Algebra Norm Cimon 2011-07-18T23:36:02Z 2011-09-27T05:50:10Z <p>The best synopsis, one developed by a real teacher, is from David Hestenes who not coincidentally developed a relatively complete curriculum built around it, and an elegant symbolism. The outer product is simply the multi-dimensional extension of the notion of a directed line segment to include directed planes and directed volume elements. William Clifford incorporated this into the surprisingly straightforward notion of a geometric algebra, something that's been a dream of mathematicians for a very long time. In one of the truly unfortunate events in mathematical history, Clifford died of tuberculosis at the age of 34, derailing his ideas for almost one hundred years, and giving us the bizarre pantheon of often incompatible systems we currently use.</p> <p>What does geometric algebra have for elements? They are called multi-vectors and they have scalar, vector, bi-vector (a directed plane), and - in general - multi-vector components. What Clifford defined was the product of multi-vectors as the sum of the inner (dot) product and Grassman's outer product. Geometric algebra is built up from this basic operation, and the rules that guide it, the key one being that the outer-product is anti-symmetric: reverse the order of the elements in the outer-product and the sign changes. That one simple rule turns out to be the key to unlocking one of the greatest advances in the history of mathematics and physics, an algebra for manipulating objects in space.</p> <p>Just as a complex numbers "tag" real and imaginary parts, objects in geometric algebra are "tagged" by the basis elements which are extended to include, not just one dimensional basis elements, but multi-dimensional basis elements as well - those directed planes and volume elements mentioned above. It's an extremely elegant and very seductive extension of linear algebra to obtain an algebra that unifies the hodgepodge of systems currently used including differential geometry, matrix algebra, vector algebra and tensors. I'm not making this up, it really does this.</p> <p>Geometric algebra also manipulates geometric objects in space <i>without having to resort to coordinates</i>. It provides what computer scientists might call a "wrapper" for complex numbers, vectors, rotors, spinors, and the physical concepts derived from those objects. In the initial paper I read, the third page was devoted to showing, almost off-handedly, that <i>geometric algebra has an almost trivial isomorphism that is in every way equivalent to the complex numbers</i>. But there's more. Difficult concepts in physics emerge naturally, almost casually, through the manipulation of geometric space using the algebra. It's greatest benefit may be this: It allows "specialists" to actually talk to each other. Who woulda thunk it!</p> <p>Here are some links: <a href="http://geocalc.clas.asu.edu/pdf-preAdobe8/UnifiedLang.pdf" rel="nofollow">Hestenes, D. - A Unified Language for Mathematics and Physics</a> <a href="http://www.mrao.cam.ac.uk/~clifford/publications/abstracts/imag_numbs.html" rel="nofollow">Gull, S; Lasenby, A; Doran, C - Imaginary Numbers Are Not Real - The Geometric Algebra of Spacetime</a> <a href="http://geocalc.clas.asu.edu/html/Oersted-ReformingTheLanguage.html" rel="nofollow">Hestenes, D - Reforming the Mathematical Language of Physics</a> <a href="http://www.mrao.cam.ac.uk/~clifford/publications/abstracts/dll_millen.html" rel="nofollow">Lasenby, J; Lasenby A; Doran, C - A Unified Language for Physics and Engineering in the 21st Century</a></p> <p>It can't be emphasized too strongly that geometric algebra is not just another technique. It is, instead, an all encompassing framework.</p> <p>The best mathematical introduction I've found, to date, is from <a href="http://faculty.luther.edu/~macdonal/index.html#GA&amp;GC" rel="nofollow">Alan Macdonald of Luther College in Iowa</a>. His freely available paper, <a href="http://faculty.luther.edu/~macdonal/GA&amp;GC.pdf" rel="nofollow">A Survey of Geometric Algebra and Geometric Calculus</a>, is exceptional, but you have to be prepared to spend time with it. It is rigorous and comprehensive, and every page is a new adventure as you learn the operations of geometric algebra.</p> <p>Macdonald also has a recently (2009) published <a href="http://www.amazon.com/Linear-Geometric-Algebra-Alan-Macdonald/dp/1453854932/ref=sr_1_1?ie=UTF8&amp;qid=1317102508&amp;sr=8-1" rel="nofollow">book</a>, the first undergraduate text to cover both linear algebra and geometric algebra.</p> http://mathoverflow.net/questions/22247/geometrical-meaning-of-grassmann-algebra/106308#106308 Answer by Jonathan Manton for Geometrical meaning of Grassmann Algebra Jonathan Manton 2012-09-04T07:52:20Z 2012-09-04T07:52:20Z <p>The key ingredient, in my mind, is to realise that the Grassmann algebra of a $d$-dimensional vector space $V$ is concerned primarily with $d$-dimensional volumes of parallelotopes and that lower-dimensional parallelotopes are merely building blocks for $d$-dimensional parallelotopes. This is explained at length here: <a href="http://jmanton.wordpress.com/2012/09/03/introduction-to-the-grassmann-algebra-and-exterior-products/" rel="nofollow">http://jmanton.wordpress.com/2012/09/03/introduction-to-the-grassmann-algebra-and-exterior-products/</a></p> <p>All volumes are relative volumes. We act as if we do not know what the underlying metric on $V$ is and we only want to make statements such as "this parallelotope is twice as big as that parallelotope" if it is true with respect to all metrics, not just a single metric.</p> <p>Since the volume of a $d$-dimensional cube equals the $d$-fold product of its side length, it is not unreasonable to hope that the (signed) volume of an (oriented) parallelotope is some sort of product of its side lengths. In fact, the axioms of a "product" of two things essentially agree with the axioms of a bilinear function, and the volume of a parallelotope is indeed given by a multi-linear function of its sides, leading to the standard definition of the exterior algebra in terms of (alternating) multi-linear maps. Regardless, thinking of volume as a product of lengths gives some intuition as to why the wedge <em>product</em> is used to define parallelotopes.</p> <p>The notation $v_1 \wedge \cdots \wedge v_i$ should be understood to refer to the parallelotope made from the vectors $v_1,\cdots,v_i \in V$. If $i &lt; d = \dim V$ then the "volume" of the parallelotope $v_1 \wedge \cdots \wedge v_i$ is always zero; keep in mind the key point that the Grassmann algebra on $V$ is <em>a priori</em> concerned with $d$-dimensional volume. Lower-dimensional parallelotopes are merely building blocks for top-dimensional parallelotopes. For example, we say $v_1 \wedge \cdots \wedge v_i = w_1 \wedge \cdots \wedge w_i$ if and only if, for all $u_1,\cdots,u_{d-i}$, it is true that $v_1 \wedge \cdots \wedge v_i \wedge u_1 \wedge \cdots \wedge u_{d-i} = w_1 \wedge \cdots \wedge w_i \wedge u_1 \wedge \cdots \wedge u_{d-i}$ where the latter means the (signed) volumes of the two $d$-dimensional parallelotopes are equal (with respect to every possible metric).</p> <p>The classical results now follow from this. For example, $v_1 \wedge \cdots \wedge v_i = \lambda w_1 \wedge \cdots \wedge w_i$ for some $\lambda$ if and only if, either both sides are zero because they are degenerate parallelotopes, or $\operatorname{span}{v_1,\cdots,v_i} = \operatorname{span}{w_1,\cdots,w_i}$. It is <em>a posteriori</em> acceptable to interpret $v_1 \wedge \cdots \wedge v_i = \lambda w_1 \wedge \cdots \wedge w_i$ as meaning the $i$-dimensional volume of the parallelotope $v_1 \wedge \cdots \wedge v_i$ is $\lambda$ times the $i$-dimensional volume of the parallelotope $w_1 \wedge \cdots \wedge w_i$, but the underlying reason is that they behave the same way when used as building blocks.</p> <p>The importance of thinking in terms of top-dimensional parallelotopes is that it is otherwise difficult to explain why $v_3 = v_1 + v_2$ does not imply that the length of $v_1$ plus the length of $v_2$ equals the length of $v_3$. In the Grassmann algebra, vectors and lower-dimensional parallelotopes do not have an independent life of their own but are primarily building blocks for top-dimensional parallelotopes. Vector addition in a Grassmann algebra relates to addition of top-dimensional volume, not to lower-dimensional volumes.</p>