Basis-free definition of Casimir element? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T16:12:24Z http://mathoverflow.net/feeds/question/52587 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/52587/basis-free-definition-of-casimir-element Basis-free definition of Casimir element? Tim Campion 2011-01-20T06:53:22Z 2011-07-16T17:49:39Z <p>Let $V$ be a finite-dimensional vector space and let $\mathfrak g \subset \mathfrak{gl}(V)$ be a representation of a semisimple Lie algebra on $V$. Let $e_1, \dots, e_n$ be a basis for $V$. Let $e_1', \dots, e_n'$ be the dual basis of ${e_i}$ under the Killing form $B_V(X,Y) = \mathrm{trace}(X \circ Y)$. The <em>Casimir element</em> of the universal enveloping algebra $U(\mathfrak g)$ (or $\mathrm{Sym}^2(V)$, if you prefer) is defined by</p> <p>$c_V = \sum_i e_i \cdot e_i'$</p> <p>One then proves that this definition is independent of the choice of basis.</p> <p>My question is: is there a good definition of $c_V$ that can be stated without referring to a basis?</p> <p>A perfunctory glance at Bourbaki, Humphreys, Fulton and Harris, and Wikipedia turned up none but the above definition.</p> <p>I'm mostly interested in representations of ordinary finite-dimensional Lie algebras (over $\mathbb{C}$, even), but if the issue is more naturally addressed in the context of some more general sort of algebra, then I'd be interested to learn about it, even though I don't really know much about, say, Hopf algebras.</p> <p><strong>EDIT: Qiaochu's answer below looks like just what I'm looking for. I wrote this clarification and will leave it up in case anyone' interested.</strong> But Qiaochu also improves on most of what I say here.</p> <p>Peter McNamara gives a sensible definition below which I ought to have been aware of: the Casimir is the unique element of degree 2 in the center of the universal enveloping algebra, up to some normalization. I suppose this definition will probably work with some modification in cases where $\mathfrak g$ is not the adjoint rep. This is a nice definition: it sheds some light on what makes the Casimir tick, and points our attention to the center of the unversal enveloping algebra as an object of interest. <strong>Where is a good reference to read about the center of a universal enveloping algebra systematically?</strong></p> <p>I'm not entirely satisfied with this definition, though, and I'd like to illustrate why with a pair of examples: coordinate-free definitions of the determinant, and of the trace.</p> <p>The determinant on an $n$-dimensional vector space $V$ may be defined (up to normalization) as the <em>unique element of $\Lambda^{n}(V)$</em> (the highest-dimension grading of the alternating algebra). This definition is highly analogous to Peter McNamara's definition of the Casimir. In either case, we define <em>widget</em> by finding an algebra $A$ of which <em>widget</em> is naturally an element, and by careful analysis of $A$, we show that <em>widget</em> is uniquely characterized by the features of $A$. One of the benefits of this sort of definition is that it points our attention towards the algebra $A$ as a worthy object of study. On the other hand, one might wish to avoid using $A$ altogether.</p> <p>In contrast, here is a definition of the trace on a finite-dimensional vector space $V$. Consider the isomorphism $\mathrm{End}(V) \cong \mathrm{End}(V)^{*}$ given by viewing $\mathrm{End}(V)$ as $V \otimes V^{*}$, swapping the factors, and using the canonical isomorphism $V \cong V^{**}$. The trace is the image of the identity under this map. It might be said that the tensor algebra is used in an auxiliary manner, but there is (I think) a difference: only the most general, "hands-off" properties of tensor products are used. The heaviest lifting comes from the isomorphism $V^{**} \cong V$ (in order to show that it is actually an isomorphism, one must keep track of dimensions of a finite-dimensional vector space under dualization). So the only hard work comes on the original object of interest: the vector space $V$ itself. In contrast, the definition of the determinant required some detailed understanding of the alternating algebra, which was not originally in question.</p> <p>There might be a logical way to state this: something along the lines of whether the definition can be given in an extremely weak fragment of logic (on the order of Horn logic?). It would have to be a fragment which doesn't easily allow definitions. More likely there's no real difference between the definition of the trace and the determinant beyond the former being "easier" than the latter. Nevertheless, if there were such an "easy" definition of the Casimir, it ought to be enlightening. </p> http://mathoverflow.net/questions/52587/basis-free-definition-of-casimir-element/52594#52594 Answer by Qiaochu Yuan for Basis-free definition of Casimir element? Qiaochu Yuan 2011-01-20T09:07:44Z 2011-01-20T09:25:33Z <p>The Casimir element is dual to the Killing form. (I think. I am somewhat uncertain about this because nobody has ever said this to me, even though it seems like the right thing to say, and frankly I don't know why Lie algebra textbooks don't just say this.) That is, the nondegeneracy of the Killing form is equivalent to its providing an isomorphism $\mathfrak{g} \to \mathfrak{g}^{\ast}$, and writing this isomorphism as a tensor exhibits it as an element of $\mathfrak{g} \otimes \mathfrak{g}$ - precisely the element $\sum e_i \otimes e_i'$ where $e_i$ is a basis. This embeds into $U(\mathfrak{g}) \otimes U(\mathfrak{g})$ and the multiplication map into $U(\mathfrak{g})$ gives the Casimir element.</p> <p>(Oh good, I see that this is the same definition Akhil gives in the blog post darij linked to above. That makes me feel better.) </p> <p>Also, you're using the "wrong" definition of the determinant. The exterior powers are all functors, and they take linear maps $T : V \to V$ to linear maps $T : \Lambda^n V \to \Lambda^n V$. Since $\Lambda^n V$ is one-dimensional when $n = \dim V$, the linear maps from $\Lambda^n V$ to itself are <em>canonically</em> isomorphic to the base field $k$. There is also a slightly more transparent definition of the trace: $\text{End}(V)$ is canonically isomorphic to $V^{\ast} \otimes V$, and then one composes with the dual pairing $V^{\ast} \times V \to k$.</p> <p>It seems to me the notion you're looking for is what general notion of <a href="http://en.wikipedia.org/wiki/Monoidal_category" rel="nofollow">monoidal category</a> supports the definition you're looking at. Traces can be defined in monoidal categories with duals, although I'm not sure what the natural setting for determinants is.</p> http://mathoverflow.net/questions/52587/basis-free-definition-of-casimir-element/52605#52605 Answer by John Klein for Basis-free definition of Casimir element? John Klein 2011-01-20T12:01:06Z 2011-01-20T13:56:22Z <p>Concerning your first question: it's really an issue about finite dimensional vector spaces.</p> <p>(1). If $V$ is a finite dimensional vector space over a field $k$, then one has an evaluation map</p> <p>$$e: V^\ast \otimes V \to k$$</p> <p>defined by $f\otimes v \mapsto f(v)$. Then $$e \in (V^{\ast} \otimes V)^\ast .$$ There is a canonical isomorphism $V \otimes V^\ast\to (V^\ast \otimes V)^\ast$ given by $$(v,f) \mapsto ((g,w) \mapsto g(v)f(w)) .$$ Using this isomorphism, we can regard $e$ corresponds to element $e'\in V\otimes V^\ast$. This is the Casimir element.</p> <p>Remark: $e'$ is characterized by the following property: it is the unique vector in $V\otimes V^{\ast}$ such that for all vector spaces $W$, the map $$\hom(V,W) \to W\otimes V^\ast$$ given by $f\mapsto (f\otimes 1_{V^\ast})(e')$ is an isomorphism.</p> <p>(2). More generally, if one is given a perfect pairing: $$e : U \otimes V \to k$$ (meaning that the adjoint $U \to V^\ast$ is an isomorphism), one gets by the same procedure an element $e'\in V\otimes U$.</p> <p>(3). $V = \frak g$ is a semi-simple Lie algebra, then the killing form $V\otimes V \to k$ is perfect. So we get by (2) the (Casimir) element $e' \in V\otimes V$.</p> http://mathoverflow.net/questions/52587/basis-free-definition-of-casimir-element/52646#52646 Answer by Emerton for Basis-free definition of Casimir element? Emerton 2011-01-20T16:18:00Z 2011-01-20T16:59:16Z <p>This is an answer to the question about the general structure of the centre of the enveloping algebra:</p> <p>The centre of the enveloping algebra (in particular when $\mathfrak g$ is reductive) is one of the basic objects in the study of infinite-dimensional representations of Lie algebras and Lie groups. It was first described in general (for reductive $\mathfrak g$) by Harish-Chandra, as far as I know, who proved the following: let $\mathfrak g$ be a reductive Lie algebra (over $\mathbb C$, say) and let $\mathfrak h$ be a Cartan subalgebra. Let $W$ be the Weyl group, which acts on $\mathfrak h$ via the adjoint action. </p> <p>Then there is an isomorphism $Z(\mathfrak g) \cong Sym(\mathfrak h)^W.$ (Here $Z(\mathfrak g)$ denotes the centre of the enveloping algebra $U(\mathfrak g)$, and $Sym(\mathfrak h)$ denotes the symmetric algebra of $\mathfrak h,$ which is the same as the enveloping algebra $U(\mathfrak h)$, since $\mathfrak h$ is abelian.)</p> <p>E.g. if $\mathfrak g = {\mathfrak sl}_2$, then a Cartan subalgebra is one-dimensional, therefore $Sym(\mathfrak h)^W$ turns out to be a polynomial ring in one generator, where this generator can be taken to be the Casimir.</p> <p>If $\mathfrak g$ is semi-simple of rank $l$, then the centre turns out to be a polynomial ring in $l$ generators, one of which is the Casimir, and the others of which can be taken to be the so-called higher Casimirs. (Their degrees are the so-called exponents of $\mathfrak g$, or perhaps the exponents shifted by one; I'm unsure about the standard normalization. [Added: As Mike Skirvin confirms in a comment below, they are the exponents shifted by $1$.])</p> <p>To learn more you can google Harish-Chandra isomorphism, or look in one of the many representation-theory texts that are out there. (I like Knapp's book <em>Representation Theory of Semisimple Groups: An Overview Based on Examples</em>, but there are lots of choices.)</p> http://mathoverflow.net/questions/52587/basis-free-definition-of-casimir-element/70513#70513 Answer by paul garrett for Basis-free definition of Casimir element? paul garrett 2011-07-16T17:49:39Z 2011-07-16T17:49:39Z <p>I can't help but add: consider the chain of $G$-equivariant natural maps, using the identification of the Lie algebra $g$ with its dual via the non-degenerate $G$-equivariant Killing form: $$End(g) \approx g\otimes g^* \approx g\otimes g \subset \bigotimes{}^\bullet g \rightarrow Ug$$ The identity endomorphism of $g$ commutes with the (Adjoint) action of $G$, so its image on the far end does, as well, and is Casimir. The often-seen expressions in coordinates (chosen depending on the circumstance), derive from the expression for the identity map in $g\otimes g$.</p> <p>(I also can't help but add that trying to understand where these quadratic expressions "live", it was a great relief at some point to understand that such computations take place in the enveloping algebra... so, even if it is possible to talk about Casimir without the enveloping algebra, it may be easier to explain the context <em>with</em> the enveloping algebra.)</p>