Nice proofs of the Poincaré–Birkhoff–Witt theorem - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T22:03:16Z http://mathoverflow.net/feeds/question/87402 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/87402/nice-proofs-of-the-poincarebirkhoffwitt-theorem Nice proofs of the Poincaré–Birkhoff–Witt theorem AH 2012-02-03T05:41:51Z 2013-02-06T16:19:02Z <p>Let $\mathfrak{g}$ be a finite-dimensional Lie algebra with an ordered basis $x_1 &lt; x_2 &lt; ... &lt; x_n$.</p> <p>We define the universal enveloping algebra $U(\mathfrak{g})$ of $\mathfrak{g}$ to be the free noncommutative algebra $k\langle x_1,...,x_n\rangle$ modulo the relations $(x_ix_j - x_jx_i = [x_i,x_j])$.</p> <p>The Poincaré–Birkhoff–Witt theorem states that $U(\mathfrak{g})$ has a basis consisting of lexicographically ordered monomials i.e. monomials of the form $x_1^{e_1}x_2^{e_2}...x_n^{e_n}$. Checking that this basis spans $U(\mathfrak{g})$ is trivial, so the work lies in showing that these monomials are linearly independent.</p> <p>One standard proof of PBW is to construct a $\mathfrak{g}$-action on the commutative polynomial ring $k[y_1,...,y_n]$ by setting $x_1^{e_1}x_2^{e_2}...x_n^{e_n}\cdot 1 = y_1^{e_1}y_2^{e_2}...y_n^{e_n}$ and verify algebraically that this gives rise to a well-defined representation of $\mathfrak{g}$. Details can be found in Dixmier's book on enveloping algebras.</p> <p>What other proofs of PBW are there out there?</p> <p>Are there nice reformulations of the above proof from a different perspective, such as one that emphasizes the universal property of $U(\mathfrak{g})$?</p> <p>However, I would be especially interested in learning about proofs which are not just repackaged versions of the same algebraic manipulations used in the above proof (for example, geometric proofs which appeal to some property of $U(\mathfrak{g})$ as differential operators, etc.). If we allow ourselves more tools than just plain algebra, what other proofs of PBW can we get?</p> http://mathoverflow.net/questions/87402/nice-proofs-of-the-poincarebirkhoffwitt-theorem/87414#87414 Answer by darij grinberg for Nice proofs of the Poincaré–Birkhoff–Witt theorem darij grinberg 2012-02-03T09:48:40Z 2013-02-05T14:23:22Z <p>The nicest one I have ever seen uses a mix of universal algebra and combinatorial algebra, and was given by <a href="http://www.sciencedirect.com/science/article/pii/0021869369900866" rel="nofollow">P. J. Higgins in <em>Baer Invariants and the Birkhoff-Witt Theorem</em>, Journal of Algebra 11, pp. 469-482 (1969)</a> (free PDF linked).</p> <p>Then there is the purely computational one which works over any $\mathbb Q$-algebra as base "field" and was given in the book by Deligne-Morgan. See <a href="http://mathoverflow.net/questions/66683" rel="nofollow">http://mathoverflow.net/questions/66683</a> for details.</p> <p>Emanuela Petracci gave in her <a href="http://www.iecn.u-nancy.fr/~petracci/tesi.pdf" rel="nofollow">thesis</a> another computational proof, which uses the language of bialgebras to make the manipulations manageable.</p> <p>There is also Cohn's <a href="http://jlms.oxfordjournals.org/content/s1-38/1/197.full.pdf" rel="nofollow"><em>A remark on the Poincaré-Birkhoff-Witt Theorem</em></a>, J. London Math. Soc. (1963) s1-38(1): 197-203. It also has a <a href="http://mathoverflow.net/questions/61954" rel="nofollow">discussion topic on MO</a>.</p> <p>If $\mathfrak g$ is the Lie algebra of a Lie group over $\mathbb R$, then you can indeed prove PBW using geometry: see, e. g., Proposition 1.9 in PDF 1 of Chapter 2 of <a href="http://ocw.mit.edu/courses/mathematics/18-755-introduction-to-lie-groups-fall-2004/lecture-notes/" rel="nofollow">Helgason's Lie Groups lecture notes</a>. However, I don't think it is realistic to use this as a general proof for PBW; Lie's Third Theorem seems to be hard and require PBW itself.</p> <p>Poincaré might have proven PBW himself (at least over a field of characteristic $0$), but I <a href="http://mathoverflow.net/questions/74923" rel="nofollow">don't understand his proof</a> (at least in a modern translation, which might itself be erroneous).</p> <p>I hate to say but the only of the above references that I found easily readable is Higgins's paper...</p> http://mathoverflow.net/questions/87402/nice-proofs-of-the-poincarebirkhoffwitt-theorem/88173#88173 Answer by n m for Nice proofs of the Poincaré–Birkhoff–Witt theorem n m 2012-02-11T03:07:47Z 2012-02-11T03:07:47Z <p>Another proof is given here; Positelsky, L, Functional Analysis and Its Applications, 1993, 27:3, 197–204 :</p> <p>I quote: </p> <p>''The classical PBW theorem attains its natural place in this context as a particular case of the fact that every Kozsul CDG-algebra corresponds to a QLS-algebra; here a QLS=quadratic linear scalar algebra is roughly an algebra defined by (generators and) non-homogenious relations of degree 2. </p> <p><a href="http://www.mathnet.ru/php/archive.phtml?wshow=paper&amp;jrnid=faa&amp;paperid=712&amp;option_lang=rus" rel="nofollow">text in russian</a>, <a href="http://www.springerlink.com/content/g05531513220774r/" rel="nofollow">text in English</a> </p> http://mathoverflow.net/questions/87402/nice-proofs-of-the-poincarebirkhoffwitt-theorem/88177#88177 Answer by Jim Conant for Nice proofs of the Poincaré–Birkhoff–Witt theorem Jim Conant 2012-02-11T04:11:30Z 2012-02-11T04:11:30Z <p>I'm partial to Dylan Thurston's proof in his <a href="http://arxiv.org/abs/math.QA/0006083" rel="nofollow">Ph.D thesis.</a> He proves the Duflo isomorphism (which is stronger than PBW) in a graphical/knot-theoretic context. The paper is a real pleasure to read.</p> http://mathoverflow.net/questions/87402/nice-proofs-of-the-poincarebirkhoffwitt-theorem/120871#120871 Answer by David Speyer for Nice proofs of the Poincaré–Birkhoff–Witt theorem David Speyer 2013-02-05T15:55:06Z 2013-02-05T16:57:43Z <p>$\def\fg{\mathfrak{g}}$This is an idea I've had for a bit which I both like and dislike: It comes from trying to take the very nice proof which holds when $\fg$ is the Lie algebra of a Lie group over $\mathbb{R}$ and push it to become purely algebraic. As darij writes, this nice proof can be found well explained in <a href="http://ocw.mit.edu/courses/mathematics/18-755-introduction-to-lie-groups-fall-2004/lecture-notes/chap2_topic1to7.pdf" rel="nofollow">Helgason's notes</a> (Prop 1.9). Of course, some may argue that I've destroyed the niceness by trying to remove the geometry.</p> <p>Let our ground field $k$ have characteristics $0$. Let $k[[\fg^{\ast}]]$ be the completion of $\mathrm{Sym}(\fg^{\ast})$ at the origin, so elements of $k[[\fg^{\ast}]]$ can be thought of roughly as germs of functions on $\fg$ near $0$. For $g \in \fg$, we have the natural derivation $\delta(g)$ on $k[[\fg^{\ast}]]$, given by extending the linear pairing $\fg \times \fg^{\ast} \to k$ to a continuous derivation. </p> <p><p>For $g \in \fg$, define a derivation $\partial(g)$ by $$\partial(g) = \sum_{n=0}^{\infty} \frac{B_n}{n!} \delta\left( (ad \ u)^n(g) \right).$$ Here $u$ is the formal variable which one can think of as "valued in (a formal nbhd of the identity in) $\fg$" and $t/(1-e^{-t}) = \sum (B_n/n!) t^n$.</p> <p>It should be possible to prove that $\partial(g)$ is a Lie algebra action directly. If so, one then finishes exactly as in Helgason's notes, seeing that the PBW basis of $U(\fg)$ pairs independently against $k[[\fg^{\ast}]]$.</p> http://mathoverflow.net/questions/87402/nice-proofs-of-the-poincarebirkhoffwitt-theorem/120930#120930 Answer by David Jordan for Nice proofs of the Poincaré–Birkhoff–Witt theorem David Jordan 2013-02-06T02:24:10Z 2013-02-06T16:19:02Z <p>While the OP asked for a more geometric rather than algebraic proof, I would like to point out Bergman's very nice paper,</p> <p>The diamond lemma for ring theory, Advances in Mathematics 29 (1978) 178-218.</p> <p>Here's a link to the paper (thanks to Darij for pointing this out!):</p> <p><a href="http://www.sciencedirect.com/science/article/pii/0001870878900105" rel="nofollow">http://www.sciencedirect.com/science/article/pii/0001870878900105</a></p> <p>Also, there is this blog post by David Speyer which summarizes it, in the context of other "diamond lemma" results:</p> <p><a href="http://sbseminar.wordpress.com/2009/11/20/the-diamond-lemma/" rel="nofollow">http://sbseminar.wordpress.com/2009/11/20/the-diamond-lemma/</a></p> <p>Essentially, for algebras whose defining relations fit into a framework of "find a certain type of monomial, and replace it with a linear cominbation of simpler ones", for some suitable notion of "simple", there are obvious obstructions to linear independence of reduced expressions: one may have a monomial (AB)C=A(BC), such that both AB and BC can be reduced. In that case, the ambiguity is called resolvable if upon further simplifying to reduced expressions, the two different expressions yield the same result. Of course if they didn't, then it would mean there was a linear relation amongst reduced monomials.</p> <p>Bergman's result says that if you can resolve such simple overlap ambiguities, featuring just one overlap (and also something called inclusion ambiguity - which are rarer, and can always be weeded out anyways, as he explains), then the PBW monomials indeed form a basis. In the case of Lie algebras, checking this reduces to nothing more than the Jacobi identity, as he shows.</p> <p>I really like the approach via Diamond Lemma as it is very elementary to prove, and the conditions of the Lemma are easy to verify in practice. It is general enough to apply to very many "almost" commutative algebras, such as arise in quantum groups and related areas.</p>