Talk:Casimir element

comments

need to disambiguate rank. there is currently no page for the rank of a Lie algebra/Lie group. does this deserve its own page? or is it better explained in the Lie algebra and Lie group articles?Lethe 02:04, 9 Mar 2004 (UTC)

Formulation

Added the precise formulation for the case of semisimple Lie algebras, but unsure what modification is required in other possible cases - Most sources (Knapp, Helgason, & Jacobson) appear to only consider the semisimple case. Dmaher 01:09, 8 July 2006 (UTC)

Cleanup & Expert tags

I have added "cleanup" and "expert" tags to the article for the following reasons. Currently (16:24, 11 July 2006 (UTC)) what the article discusses, and the introduction defines, is just a quadratic (second-order) Casimir operator. This is just one of many possible Casimir operators, although perhaps the most common and best known (especially in physical applications). General Lie algebras can possess other, higher-dimensional Casimir operators. A more general definition should be given by someone who is an expert on this topic. 131.111.8.98 16:24, 11 July 2006 (UTC)

Well, one thing is definitely missing: the article completely sidesteps the issue where do these operators actually "live" — it should be the (center of) the universal enveloping algebra, although of course, they induce an operator in any its representation. As for "higher Casimir operators", this terminology is rarely used in representation theory, at least within mathematics. I can think of only one major book, Zhelobenko's "Compact Lie groups and their representations", AMS Translations, vol 40, 1973, which employs this terminology. I know that the convention is different in physics literature, where the term "Casimir operator" may refer to more general elements of the center of the universal enveloping algebra. In that case, though, they are called "Higher Casimir elements" or something similar. I think the problem partly arises from the following incorrect sentence in the universal enveloping algebra article:

Under this representation, the elements of U(L) invariant under the action of L (i.e. such that any element of L acting on them gives zero) are called invariant elements. They are generated by the Casimir invariants.

Arcfrk 11:32, 10 March 2007 (UTC)

The usage "higher", while perhaps rare in mathematics, is common in physics, and is used to distinguish the quadratic case from more complex invariants, e.g. on supersymmetric manifolds, etc.

Anyway, I'm removing the "needs expert attention" tags; by the criteria given, 98% of all math articles on WP need "expert attention".

FWIW, my PhD thesis had the words "Casimir operator" in its subtitle, and I would be rather hard-pressed to give a general definition that corresponds to the Lie-algebra definition. In my case, it was built from a Dirac operator, integrated over a volume. It was a cocycle (i.e. vanishing derivative), and so corresponded to a conserved current, by Noether's theorem. The "conserved charge" of this current was the Casimir invariant. I still don't know how to turn this into a general definition. linas 13:55, 15 May 2007 (UTC)

Requested move

The following discussion is an archived discussion of the proposal. Please do not modify it. Subsequent comments should be made in a new section on the talk page. No further edits should be made to this section.

The result of the proposal was moved. --BDD (talk) 17:53, 15 May 2013 (UTC)

Casimir invariant → Casimir element – Because a Casimir element doesn't have to be an operator. Also, the latter name is much commoner, at least in the math community Taku (talk) 01:19, 8 May 2013 (UTC).

The above discussion is preserved as an archive of the proposal. Please do not modify it. Subsequent comments should be made in a new section on this talk page. No further edits should be made to this section.

example: so(3)

If I multiply L_x²+L_y²+L_z² I always get -2 times identity. Don't know if I do something wrong, just wanted to tell you. 88.66.141.244 (talk) 18:45, 15 August 2013 (UTC)

Maybe an i is missing (e.g. define them as i*L_x etc.)? 88.66.141.244 (talk) 14:29, 17 August 2013 (UTC)

There are definitely errors in this section. Multiplication of the $L_{\{x,y,z\}}$ with the imaginary unit i will not help as this is effectely a change of basis in the complexification of so(3). A change of bases does not affect the Killing form and thus also does not affect the definition of the Casimir element.

A correct calculation of the Casimir element is obtained as follows: If we use $L_x,L_y, L_z$ as the ordered basis of so(3), then the matrices of the adjoint representation of $L_x,L_y,L_z$ are the matrices $L_x,L_y,L_z$ themselves. From this, we compute the Killing form to be minus twice the Kronecker delta, i.e. $B(L_a,L_b) = -2δ_{a,b}$ for $a,b\in \{x,y,z\}$. Therefore, the dual basis element of $L_a$ with respect to the killing form is $-(1/2)L_a$ for $a\in \{x,y,z\}$. Hence, the Casimir Element is $\Omega = (-1/2) ( L_x^2 + L_y^2 + L_z^2)$.

Using an explicit description of the irreducible matrix representations on $so(3)$ which I omit here, I obtain that on an irreducible representation of dimension $2\ell+1$, the Casimir element $\Omega$ acts as $\frac{\ell (\ell + 1)}{2}$.

The author of this section denotes an element $L^2$ as the casimir element, which is called $\Omega$ in the denition. I suspect that $L^2$ might have some physical relevance, since we have $L^2 := L_x^2 + L_y^2 + L_z^2 = -2\Omega$.

99.241.86.114

99.241.86.114 (talk · contribs · deleted contribs · logs · filter log · block user · block log), a.k.a. latex to html, is a well-known person (see Special:WhatLinksHere/user talk:99.241.86.114), to whose main agenda I’m largely indifferent. But now I bring to evaluation one aspect of his/her activity yet unknown to me.

Good version	After 99.241.86.114 [1]
<math>n</math>-dimensional (WP:HYPHEN)	''n''−dimensional the minus sign
G-invariant (same as above)	''G''–invariant the en dash character
<math>(2\ell+1)</math>-dimensional	(2ℓ + 1)–dimensional. en dash, unprotected spaces

There will be no trouble if someone will attempt to improve the typography, even using 99.241.86.114’s drafts, but nothing (but an edit war against me) will come from an attempt to reinstate his/her “contribution” without fixes.

Regards, Incnis Mrsi (talk) 09:10, 22 January 2014 (UTC)

Are you aware that most readers won't notice the difference between -, − and other horizontal dashes (and consequently don't care), but that inline Latex often (always in case you use PNG) looks like shit? YohanN7 (talk) 05:06, 23 January 2014 (UTC)

I said: I’m indifferent to 99.241.86.114’s main agenda, but not to the collateral damage. If I will change English words from Latin script to Cyrillic, then you reverted me, right? Or may I change the spelling to my non-standard (but easily legible) variant? There is an expressed consensus about punctuation exactly as there is a consensus that English Wikipedia uses English, that it is written in Latin letters and with a standard orthography. The minus sign is at all a symbol, not a kind of other horizontal dash as you contemptuously said. A user who persistently edits against the consensus may end with his/her contribs ruined. Especially if such contribs consist mostly of various tweaks of doubtful value: there is no consensus about the math formatting. Incnis Mrsi (talk) 05:27, 23 January 2014 (UTC)

I'm just amazed that you want to go through all the trouble of reverting to a more-than-year-old version, reinstating several subsequent edits, instead of just replacing the eye-hurting horizontal dash with your favorite horizontal dash, and in the process make 99.241.86.114 look like a very suspect individual. YohanN7 (talk) 06:32, 23 January 2014 (UTC)

Is there any way to improve the way in-line Latex looks on my computer? What is PNG? The image file format?

178.38.88.134 (talk) 21:49, 27 February 2015 (UTC)

Please explain

(1) What is ρ(Xⁱ)? I don't know what it means to apply the representation to the dual vector {{Xⁱ}}.

(2) What is the representation-theoretic meaning, or characterization, of $\ell$ in the expression $\ell (\ell +1)$?

178.38.88.134 (talk) 21:49, 27 February 2015 (UTC)

Only for (1). $X^{i}$ is not a dual vector (linear functional) but just dual to $X_{i}$; i.e., $B(X_{i},X^{j})=\delta _{ij}$. -- Taku (talk) 22:57, 10 October 2015 (UTC)

$\ell$ is spin (physical terminology), whereas $2\ell$ is the highest weight of the corresponding irreducible representation of ${\mathfrak {so}}(3)$; of course, as the article mentions, $2\ell +1$ is the dimension of this representation. Arcfrk (talk) 08:04, 13 October 2015 (UTC)