# Basis-free definition of Casimir element?

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 Casimir element of the universal enveloping algebra $U(\mathfrak g)$ (or $\mathrm{Sym}^2(V)$, if you prefer) is defined by

$c_V = \sum_i e_i \cdot e_i'$

One then proves that this definition is independent of the choice of basis.

My question is: is there a good definition of $c_V$ that can be stated without referring to a basis?

A perfunctory glance at Bourbaki, Humphreys, Fulton and Harris, and Wikipedia turned up none but the above definition.

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.

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. But Qiaochu also improves on most of what I say here.

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. Where is a good reference to read about the center of a universal enveloping algebra systematically?

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.

The determinant on an $n$-dimensional vector space $V$ may be defined (up to normalization) as the unique element of $\Lambda^{n}(V)$ (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 widget by finding an algebra $A$ of which widget is naturally an element, and by careful analysis of $A$, we show that widget 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.

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.

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.

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By definition, Killing form is trace(XY) where one is considering the adjoint rep of g. A dual basis only exists when bilinear form is non-degenerate, which means we have to assume g is semisimple. If g is simple, can characterise Casimir up to scalar as unique degree 2 element of U(g), then can normalise by looking at the scalar which it acts on an irrep. So if g simple, it certainly can be done, but question is whether this is a useful defn. I know there is a school of thought that says try to avoid taking a basis if at all possible, but in this case, picking a basis gives a good defn. –  Peter McNamara Jan 20 '11 at 7:23
Try go in such direction - the Casimir element is the only element of the center $Z(\mathfrak{g})$ of degree 2. –  Melania Jan 20 '11 at 7:31
@ Peter McNamara: good point, I forgot semisimple and have edited accordingly. Is there another name for the bilinear form I've defined above when $V$ is not the adjoint rep? Also edited to hopefully clarify in light of your answer. @ Melania: I guess you mean center of the enveloping algebra? –  Tim Campion Jan 20 '11 at 8:07
For semisimple Lie algebras: amathew.wordpress.com/2010/01/31/… gives an invariant definition. –  darij grinberg Jan 20 '11 at 9:03
Two comments: 1) It's good to appreciate that there is an abstract characterization of the Casimir element in the enveloping algebra setting, but to use it in representation theory you have to make a specific choice of Lie algebra basis. Historically this type of concrete construction naturally came first and doesn't even require the abstract enveloping algebra. 2) The Casimir isn't uniquely characterized as a central element of degree 2 in the enveloping algebra, since such an element is only determined up to a scalar. –  Jim Humphreys Jan 20 '11 at 13:50

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.

(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.)

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 canonically 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 $V^{\ast} \times V \to k$.

It seems to me the notion you're looking for is what general notion of monoidal category 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.

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Aha! That is very, very nice. I see you're right about the determinant. I think the definition I gave was a shortcut we used in my first linear algebra class. I now recall being briefly dissatisfied with it at some point later, and now I understand why. Thanks. Now I see that the real place finite-dimensionality enters in defining the trace is in the initial endomorphism $\mathrm{End}(V) \cong V \otimes V^*$. –  Tim Campion Jan 20 '11 at 10:09
At least for the trace and determinant, at some point I ought to read the nlab article on Schur functors ncatlab.org/nlab/show/Schur+functor I suppose the distinction I want to make is not essentially logical, but rather categorical: it's about whether a definition can be expressed in the internal logic of a some sort of category... –  Tim Campion Jan 20 '11 at 10:12
It's worth mentioning that you can recover an infinite-dimensional form of $\text{End}(V) \simeq V^{\ast} \otimes V$ (I think there are good reasons to put the dual first) if you work with Hilbert spaces and Hilbert-Schmidt operators between them. –  Qiaochu Yuan Jan 20 '11 at 12:27
The determinant is also just the trace of $\Lambda^n(T)$, (although this does not render the homomorphism property as obvious). Actually, all the coefficients of the characteristic polynomial $\chi(x)= \sum^n_{k=0}\Lambda^{n−k}(−T) x^k$ including the trace itself are of this form. By the way, is the intrinsic definition of trace useful for anything besides, say, concluding the "dimension" is an invariant? Something about modules over rings, for instance? –  Phil Isett Sep 8 '11 at 2:33
@Phil: it generalizes in various interesting ways. See, for example, Ponto and Shulman: math.ucsd.edu/~mshulman/papers/traces_sym.pdf –  Qiaochu Yuan Sep 8 '11 at 3:56

This is an answer to the question about the general structure of the centre of the enveloping algebra:

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.

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.)

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.

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 $1$.])

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 Representation Theory of Semisimple Groups: An Overview Based on Examples, but there are lots of choices.)

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This obviously isn't very important, but the degrees of the generators of $Sym(\mathfrak{h})^W$ are the exponents shifted by +1. For example, when $\mathfrak{g} = \mathfrak{sl}_n,$ the degree of the generators are $2, \ldots, n,$ but the exponents are $1, \ldots, n-1.$ While this may just seem like a silly difference in terminology, there are several other ways to define exponents which gives credence to the shift. –  Mike Skirvin Jan 20 '11 at 16:55
Dear Mike, Thanks for this! Best wishes, Matt –  Emerton Jan 20 '11 at 16:58
Some of the related history of Casimir elements was discussed in another recent MO post mathoverflow.net/questions/41150 –  Jim Humphreys Jan 20 '11 at 18:21
Thank you, Emerton: this was exactly what I was hoping for. I see now that I should have been able to find this from the "Casimir invariant" Wikipedia article, which links to "Harish-Chandra homomorphism", as well as from the mathoverflow question that Professor Humphreys links to! –  Tim Campion Jan 20 '11 at 19:14
Let me also mention that the eigenvalues of the elements in the center of U(g) on finite-dimensional irreps distinguishes irreps, so that's one basic reason you should care about the center of U(g). This is a somewhat rare property for an associative algebra to have (for example, it is possesed by C[G] where G is a finite group). This is discussed, for example, in Kirillov's book. –  Qiaochu Yuan Jan 20 '11 at 20:40

Concerning your first question: it's really an issue about finite dimensional vector spaces.

(1). If $V$ is a finite dimensional vector space over a field $k$, then one has an evaluation map

$$e: V^\ast \otimes V \to k$$

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.

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.

(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 $e'\in V\otimes U$.

(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 $e' \in V\otimes V$.

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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 $g\otimes g$.

(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 with the enveloping algebra.)

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