Why is the exterior algebra so ubiquitous? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T15:51:57Z http://mathoverflow.net/feeds/question/1684 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/1684/why-is-the-exterior-algebra-so-ubiquitous Why is the exterior algebra so ubiquitous? Qiaochu Yuan 2009-10-21T16:53:22Z 2011-06-18T15:24:09Z <p>The exterior algebra of a vector space V seems to appear all over the place, such as in</p> <ul> <li>the definition of the cross product and determinant,</li> <li>the description of the Grassmannian as a variety,</li> <li>the description of irreducible representations of GL(V),</li> <li>the definition of differential forms in differential geometry,</li> <li>the description of fermions in supersymmetry.</li> </ul> <p>What unifying principle lies behind these appearances of the exterior algebra? (I should mention that what I'm really interested in here is the geometric meaning of the Gessel-Viennot lemma and, by association, of the principle of inclusion-exclusion.)</p> http://mathoverflow.net/questions/1684/why-is-the-exterior-algebra-so-ubiquitous/1696#1696 Answer by Greg Muller for Why is the exterior algebra so ubiquitous? Greg Muller 2009-10-21T17:31:02Z 2009-10-21T17:31:02Z <p>For me, the exterior algebra is the free polynomial algebra in anti-commutative variables. Of course, this begs the question, why do anti-commutative variables come up so much?</p> <p>As a homological algebraist, the reason for this that jumps out at me is that the boundary map d in a complex is an anticommuting operator, which can be seen in the Koszul sign rule for commuting boundary maps across each other. Of course, this doesn't really explain all the instances of anti-commuting variables.</p> http://mathoverflow.net/questions/1684/why-is-the-exterior-algebra-so-ubiquitous/1699#1699 Answer by Andrea Ferretti for Why is the exterior algebra so ubiquitous? Andrea Ferretti 2009-10-21T18:02:08Z 2009-10-21T18:02:08Z <p>I will only answer for the link between determinants, differential forms and the Grassmannian.</p> <p>The fact is that determinant, up to a sign, represents the volume of a parallelepipedal having n assigned vectors as sides. The sign is determined by orientation of this solid.</p> <p>Indeed the axioms for the determinant can be translated geometrically: for instance the fact that the determinant vanishes when two columns are equal corresponds to the fact that a solid lieing in a hyperplane has 0 volume.</p> <p>Now take a a linear map f expressed by a matrix A: the image of the unit cube is the solid generated by the columns of A; so f stretchs volumes by a factor |det(A)|, by the previous remark.</p> <p>This is the infinitesimal expression of the usual formula for the change of variables in the integral, and it is the reason why the Jacobian determinant appears there. It is just the infinitesimal factor by which you multiply volumes. I hope this gives a rough explanation why the determinant appears in this formula.</p> <p>Now to differential forms. Assume you want to integrate a quantity on a manifold, say a function. You may want to try to integrate it in local coordinates, but the result will depend on the coordinates chosen. So in order to get something well-defined you need a quantity whose local expression changes by the inverse factor (ok, I'm neglecting orientation here). This is exactly a n-form, whose local expression changes by the determinant of the Jacobian of the inverse change of coodinates.</p> <p>This vague discussion should so far give an idea why differential form of maximal degree are apt to be integrated on oriented manifolds. Now choose a manifold M. You can integrate k-forms on M on k-subvarieties of M, so differential forms of any degree appear as dual elements of subvarieties of the corresponding dimension. Pushing this correspondence a bit explain why the complex of differential form gives the cohomology of M. But this is a topological invariant, so it has plenty of other constructions.</p> <p>So we get an analytic tool (differential forms) which describes part of the topology of M; something which of course is worthy studying. Feeew!! If you got this far, you can understand what kind of link I see between determinants and differential forms.</p> <p>As a particular case, this also give an explanation of the link with Grasmmannian: to a given subspace A you just associate the (constant) differential forms dual to it, up to multiples; this allows you to think of point of the Grassmannian as a point in a projective space, giving (more or less) the usual Plucker embedding. I mean: dual elements to general subvarieties are noncostant differential forms, but if you just restrict to subspaces you can just use costant differential forms.</p> <p>I don't have an intuitive explanation of the link with irreducible representations of GL and I don't know Fermions, so I can't help you there.</p> http://mathoverflow.net/questions/1684/why-is-the-exterior-algebra-so-ubiquitous/1705#1705 Answer by Ben Webster for Why is the exterior algebra so ubiquitous? Ben Webster 2009-10-21T18:47:17Z 2009-10-26T18:34:22Z <p>Just to use a buzzword that Greg didn't, the exterior algebra is the symmetric algebra of a purely odd supervector space. So, it isn't "better than a symmetric algebra," it is a symmetric algebra.</p> <p>The reason this happens is that super vector spaces aren't just Z/2 graded vector spaces, they also have a slightly different tensor category structure (the flip map on the tensor product of two odd vector spaces is -1 times the usual flip map, and the usual flip map for all other pure vector spaces). If you look at all the formulas from homological algebra, for things like how to take the tensor product of two complexes, they always have a bunch of weird signs showing up; these always can be though of as coming from the fact that you should take the tensor product on graded vector spaces inherited from super vector spaces, not the boring one.</p> <p>Of course, this just raises the question of why supervector spaces show up so much. Greg had about as good an answer as I could give for that.</p> http://mathoverflow.net/questions/1684/why-is-the-exterior-algebra-so-ubiquitous/1770#1770 Answer by S. Carnahan for Why is the exterior algebra so ubiquitous? S. Carnahan 2009-10-22T01:00:34Z 2011-06-17T18:46:17Z <p>One good reason for the ubiquity of the exterior algebra construction is that it has nice basic properties (which if made precise will uniquely define it):</p> <ol> <li>It is a functor from vector spaces to (strictly) supercommutative algebras.</li> <li>Direct sums are taken to tensor products of algebras.</li> <li>It plays well with base change and descent, i.e., one can make exterior algebra bundles by gluing.</li> <li>It takes a line to an odd line plus an even line (used to great effect <a href="http://mathoverflow.net/questions/68004/natural-pairings-between-exterior-powers-of-a-vector-space-and-its-dual/68033#68033" rel="nofollow">in Torsten's answer here</a> - note the last paragraph comparing exterior with symmetric cases).</li> </ol> <p>Bonus properties in the finite rank case include:</p> <ol> <li>There is a multiplicative determinant subfunctor to invertible objects (read: graded lines).</li> <li>It yields a Hopf algebra.</li> <li>You have a perfect "Hodge dual" pairing valued in the determinant (as pointed out by Marc Nieper-Wißkirchen).</li> </ol> <p>Regarding the ways the exterior algebra is useful when the symmetric algebra is not, I think all of the applications you listed revolve around the finite rank properties - in particular, the distinguished nature of the determinant as a canonical one-dimensional tensor. The only one-dimensional symmetric tensor is the trivial one, which carries no information. Any attempt to make things like volumes, cup products, or Hodge stars requires an orientation, which can be viewed as a determinant (see e.g., <a href="http://mathoverflow.net/questions/47596/orientations-for-zero-dimensional-manifolds/47597#47597" rel="nofollow">my answer here</a>).</p> <p>As <a href="http://en.wikipedia.org/wiki/Exterior_algebra" rel="nofollow">Wikipedia</a> mentions, exterior algebras satisfy a universal property: for any linear map from a vector space $V$ to an associative algebra $A$ landing in the square zero subspace, there is a unique algebra homomorphism from $\bigwedge V$ to $A$ making a certain triangular diagram commute. This yields a description of the exterior algebra functor as a left adjoint to a forgetful functor from strictly supercommutative algebras. If 2 is invertible, then it is equivalent to the parity-shifted symmetric algebra functor, but in general, it represents a genuinely different functor. In particular, the fact that determinants exist in characteristic 2 is an indication that the exterior algebra is more important than the shifted symmetric algebra.</p> <p>I should emphasize that supersymmetry is different from the existence of fermions. In short, fermions are just odd fields that transform a certain way under ordinary spacetime symmetries, but supersymmetry is the specification of additional odd symmetries of spacetime. This is substantially more exotic: theories containing fermionic particles (like electrons) can exist without supersymmetry, and in fact existed happily for about 50 years before supersymmetry was hatched.</p> http://mathoverflow.net/questions/1684/why-is-the-exterior-algebra-so-ubiquitous/1859#1859 Answer by David Bar Moshe for Why is the exterior algebra so ubiquitous? David Bar Moshe 2009-10-22T10:02:43Z 2009-10-22T10:02:43Z <p>In my opinion, the unifying object of all the cases you mentioned is the Dirac-operator:</p> <ol> <li><p>The Dirac operator acts on the exterior (possibly twisted) algebra of differential forms.</p></li> <li><p>Some cases in representation theory (where GL(V) is a special case) can be formulated using the Kostant-Dirac operator.</p></li> <li><p>Fermions satisfy the Dirac equation.</p></li> <li><p>The (infinite dimensional) Grassmannian in second quantization can be constructed from the (one-particle) Dirac spectrum (above and below thw Dirac sea).</p></li> <li><p>Finally, the determinant, can seen as the Jacobian of a linear transformation of Grassmann variables which are "classical" (in the Berezin's sense) counterparts of fermions.</p></li> </ol> http://mathoverflow.net/questions/1684/why-is-the-exterior-algebra-so-ubiquitous/1870#1870 Answer by pasquale zito for Why is the exterior algebra so ubiquitous? pasquale zito 2009-10-22T13:25:07Z 2009-10-22T13:25:07Z <p>Somehow, I can't resist (re)formulating an answer in the following minimalistic and tautological way: the exterior algebra appears each time you consider the tensor algebra generated by a vector space (e.g., you want do define a notion of volume for n-parallelepipeds spanned by n-tuples of vectors) and you want to quotient w.r.t. the ideal generated by products of the same vector (you want flat parallelepipeds to have zero volume).</p> <p>It's an "I won't repeat myself" statement which, in its peremptory simplicity, is likely to appear in the early evolutionary stages of many mathematical ideas. </p> http://mathoverflow.net/questions/1684/why-is-the-exterior-algebra-so-ubiquitous/6728#6728 Answer by Marc Nieper-Wißkirchen for Why is the exterior algebra so ubiquitous? Marc Nieper-Wißkirchen 2009-11-24T20:18:03Z 2009-11-24T20:18:03Z <p>If the exterior algebra was just the symmetric algebra (up to a degree shift), it wouldn't be a useful notion by itself.</p> <p>But to me it seems it isn't: Consider an ordinary vector space <em>V</em> over a field <em>k</em> of characteristic 2. For simplicity lets assume that <em>V</em> is one-dimensional with generator $x$. As $x \wedge x = 0$, the Graßmann algebra of <em>V</em> is <em>k</em> in weight 0 and 1, and trivial in all other weights.</p> <p>Now change the parity of the vector space <em>V</em>, i.e. consider <em>x</em> to be of odd degree. Let us call the resulting odd vector space $W$. What is the symmetric algebra over $W$? Neglecting the parity, it is simply the polynomial algebra <em>k</em> in one variable, which is <em>k</em> in all positive weights. The discrepancy comes from the fact that in characteristic 2, the symmetric algebra does not see the difference between even and odd elements.</p> <p>As to the ubiquity of the exterior algebra: Given an operator <em>A</em> acting on any space, it is quite natural to ask whether $A \circ A = 0$ (e.g. any would-be differential in homological algebra). Whenever I have a vector space of operators that has this property, the exterior algebra shows up. And such a situation looks quite common.</p> <p>Furthermore, the exterior algebra enjoys a property the symmetric algebra (of the shifted space) does not have: If V is free of rank n, the natural pairing $\Lambda^p V \otimes \Lambda^{n - p} V \to \Lambda^n V$ is a perfect pairing.</p> http://mathoverflow.net/questions/1684/why-is-the-exterior-algebra-so-ubiquitous/36817#36817 Answer by Gilbert Bernstein for Why is the exterior algebra so ubiquitous? Gilbert Bernstein 2010-08-26T22:17:47Z 2010-10-11T17:24:42Z <p>Here is an answer which would make sense to an elementary school student—if they understood what you were asking. It's Grassmann's original argument for considering anticommutativity. I don't have a reference handy, but I'm pretty sure it shows up in the intro material to one of the Ausdehnungslehre, or perhaps a summary essay.</p> <p>Grassmann's goal was to find a way to "arithmetize" geometry. So, let's do that very naively. Suppose you have a line segment AB and another collinear line segment BC:</p> <p>A---------B----------C</p> <p>Then, through visual inspection, AB + BC = AC. However, now suppose C lies in the middle instead of B:</p> <p>A---------C----------B</p> <p>Writing down the obvious equation from this arrangement we get AC + CB = AB</p> <p>If we solve the resulting system of two equations, we realize that BC = -CB</p> <p>Anticommutativity!</p> <hr> <p>An additional thought for Qiaochu, have you looked at Klein and Rota's book <em>Introduction to Geometric Probability</em>? There are some interesting analogies there between combinatorial structures and geometry that might give you some thoughts. In particular, they link inclusion-exclusion and the Euler characteristic as the unique 0-dimensional invariant valuations in the combinatorial and geometric settings respectively.</p> http://mathoverflow.net/questions/1684/why-is-the-exterior-algebra-so-ubiquitous/68134#68134 Answer by Greg Graviton for Why is the exterior algebra so ubiquitous? Greg Graviton 2011-06-18T10:03:56Z 2011-06-18T11:41:52Z <p>As Gilbert said, Grassmann's original intention was to arithmetrize geometry, and I think the exterior product captures this quite nicely.</p> <p>The intuition is that for any vector space $V$, the exterior produt $\Lambda^k V$ corresponds to the <em>$k$-dimensional subspaces of $V$</em>. In other words, the amazing insight of Grassmann is that subspaces can be captured by an algebraic product, at least to some extend. Namely, consider two sets of vectors $v_1,\dots,v_k$ and $w_1,\dots,w_k$. These sets span the same $k$-dimensional subspace in $V$ if and only if</p> <p>$v_1 \wedge \dots \wedge v_k = \lambda \cdot w_1 \wedge \dots \wedge w_k \neq 0$</p> <p>in $\Lambda^k V$. Put differently, a pure wedge product of vectors can be identified with their linear span (if it has full dimension).</p> <p>This view perfectly explains the first four points of the question:</p> <ul> <li>The extra factor in the wedge product can be interpreted as a measure of volume.</li> <li>Reformulation of the intuition above.</li> <li>Elements of $GL(V)$ also permute the higher-dimensional subspaces of $V$.</li> <li>Differential forms = infinitesimal subspaces + volumes.</li> </ul> <p>The fermions don't fit in, their only relation to the above seems to be their antisymmetry.</p> http://mathoverflow.net/questions/1684/why-is-the-exterior-algebra-so-ubiquitous/68136#68136 Answer by Roland Bacher for Why is the exterior algebra so ubiquitous? Roland Bacher 2011-06-18T10:38:01Z 2011-06-18T10:38:01Z <p>A point which is perhaps worthwile of explicite mention is that the symmetric group (where we consider only the subgroup of all elements with finite support if $E$ is infinite) of a set $E$ has two one-dimensional representations. The trivial representation is of course related to the symmetric algebra and the signature representation corresponds to the exterior algebra. </p> http://mathoverflow.net/questions/1684/why-is-the-exterior-algebra-so-ubiquitous/68153#68153 Answer by Orr Shalit for Why is the exterior algebra so ubiquitous? Orr Shalit 2011-06-18T15:24:09Z 2011-06-18T15:24:09Z <p>I think that what unifies some of the different examples of when the exterior algebra occurs is that it is the structure that transforms the action of a commutative ring on a module (or, more concretely, the action of several commuting linear operators on a vector space) into a chain complex. The ubiquitous structure is really <em>commutative</em> rings and modules over commutative rings. The exterior algebra actually encodes commutativity, in a sense. (One might then ask why is it that mathematicians love so much commutative rings, commuting operators, etc.)</p> <p>When tensoring a commutative action with the graded algebra one gets the Koszul complex, and the anti-commutative nature of the graded algebra is precisely what, when coupled with the commutative action of the ring, makes possible the definition of a dirac operator $d$ with $d^2=0$.</p> <p>I got these ideas after thinking about Joseph Taylor's paper, "A joint spectrum for several commuting operators", J. Funct. Anal. , 1970.</p>