Basis-Free Definition of Casimir Element – Representation Theory

lie-algebrasrt.representation-theory

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.

Best Answer

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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