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June 28[edit]

Commutators and "ad"[edit]

On the page commutator, what does "ad" stand for? It says that ad(x)(y) = [x,y]. --HappyCamper 04:44, 28 June 2007 (UTC)[reply]

The article says that's a useful alternative definition, with the particular bonus that ad(x) is then something that can act like an operator - ad(x)(y) = [x,y], (ad(x))^3(y) = [x, [x, [x, y]]] etc. Confusing Manifestation 05:49, 28 June 2007 (UTC)[reply]

Here "ad" is short for "adjoint"; for any Lie algebra g, you get a map from g to the Lie algebra of endomorphism gl(g) = End(g) by mapping the element X of g to the endomorphism "send Y to [X,Y]". This is called the adjoint representation of the Lie algebra g. Tesseran 07:37, 28 June 2007 (UTC)[reply]

I've linked Adjoint representation in the article not sure if this is the most appropriate link as Adjoint has several related terms. --Salix alba (talk) 07:46, 28 June 2007 (UTC)[reply]

(edit conflict) We need to be careful with capitalization, because we have both

\operatorname {Ad} \colon G\to \operatorname {Aut} (T_{e}G)\,\!

and

\operatorname {ad} \colon T_{e}G\to \operatorname {End} (T_{e}G).\,\!

Fulton & Harris (Representation Theory, ISBN 978-0-387-97495-8, pp. 106–107) define "ad" using the differential of "Ad", then define the commutator using "ad".

Caution: The term "adjoint" is used in several contexts, and the meaning varies (though there is usually a family resemblance). For example, we have adjoint matrix, a different meaning for adjoint matrix, adjoint operator, and adjoint functors; none of which are the same as the adjoint representation idea under discussion.

Does this help, or merely pile on confusion? :-) --KSmrq^T 08:05, 28 June 2007 (UTC)[reply]

I don't know that F&H's choices of notation are relevant -- I'm familiar with Chapter 8 of F&H, and have always found their approach extremely opaque. Regardless of how they choose to endow the tangent space to a Lie group with the structure of a Lie algebra, the adjoint representation of a Lie algebra is inherent, and a priori has nothing to do with Lie groups at all. (There are certainly other ways to give the tangent space a bracket; my personal favorite is to identify the tangent space with the space of left-invariant vector fields, and then the bracket of vector fields gives you your Lie algebra structure. [Not that this is news to you, KSmrq.]) I believe that "ad" for the adjoint representation of a Lie algebra is standard; I haven't seen enough notation introduced for the Lie group concept to say what's accepted there. Tesseran 09:29, 28 June 2007 (UTC)[reply]

Our own adjoint representation article uses the Ad/ad notation, and cites Fulton & Harris; therefore I feel it helpful to point out the distinction. It would be horribly confusing to overlook it, especially since I link to that article. Likewise, I feel that it would be unkind to force HappyCamper to stumble through all the different meanings of "adjoint". In other words, my post is intended to provide orientation and an explanatory link.

While we're disambiguating, perhaps we should caution that brackets and commutators come in different flavors as well. I know, "A foolish consistency is the hobgoblin of little minds, …" (Emerson, my emphasis); but sometimes I wish mathematicians could try for a bit more. All the variations make it tough on us encyclopedia editors (not to mention the generations of mathematics students)! --KSmrq^T 11:49, 28 June 2007 (UTC)[reply]

OK, so I guess it was a bit of wishful thinking that "adjoint" here has any analogies with any of the other adjoints. --HappyCamper 16:37, 28 June 2007 (UTC)[reply]

Sorry, KSmrq; I thought the link was to Adjoint representation of a Lie algebra (which redirects to Adjoint endomorphism). That page treats things from an inherent perspective (until the end). Thanks for your clarifying remark, and sorry if I seemed antagonistic. Tesseran 01:45, 29 June 2007 (UTC)[reply]

impact of wind speed hitting a door at full force[edit]

An OHS query in context but a mathematical one. If the wind speed hitting a door rose from 40kph to 65kph how much pressure would it add to a door?

Leedale1 06:42, 28 June 2007 (UTC)[reply]

Let's start from first principles. The likely factors in determining the pressure are the density of the air and the wind speed. Use dimensional analysis to find a function of density and velocity that has the same dimensions as pressure. Assuming constant density, you now know that presure is proportional to some power of the wind speed. So if the wind speed has increased by a factor of 65/40 = 1.625, you can now find the relative effect on pressure. You can't find the absolute pressure without knowing (a) the density of the air and (b) any numerical constants in the pressure/density/wind speed relation. Gandalf61 12:28, 28 June 2007 (UTC)[reply]

Oh, are you sure setting up a formula with dimensional analysis like that will yield a correct result? Mabye it will, often it does, but I don't know why that has to be the case here. —Bromskloss 12:53, 28 June 2007 (UTC)[reply]

Yes, dimensional analysis gives the correct result, modulo a constant factor. The actual formula is in a Wikpedia article with initials d.... p... but I thought that Leedale1 might learn more from doing a bit on dimensional analysis than by just blindly putting numbers into a formula. Gandalf61 13:41, 28 June 2007 (UTC)[reply]

I get your reference. (But aren't we forgetting the s… p…? If at all the question is well defined.) Anyway, in other situations, couldn't dimensional analysis mislead us? Time dilation in special relativity is

t'=t/{\sqrt {1-v^{2}/c^{2}}}

, but that's not apparent from the dimensions of the variables involved. In general, it seems to me like we run into trouble when there are dimensionless quantities involved or when the equation doesn't consist exclusively of products of powers of the quantities. —Bromskloss 18:54, 28 June 2007 (UTC)[reply]

Associated Legendre functions[edit]

I feel that the article associated Legendre function can be improved somewhat, see here. To that end one needs to prove that $P_{l}^{(|m|)}\propto P_{l}^{(-|m|)},\quad |m|=0,1,\ldots ,l$ . I believe that it is sufficient to show that both functions are solutions of the same associated Legendre equation, although this is a second order differential equation and has two solutions. Concrete questions: (i) Is it indeed sufficient to show that positive and negative m functions are solutions for the proportionality to be true? (ii) How do we show that $P_{l}^{(m)}$ , given in its form with the l+m-th derivative of $(x^{2}-1)^{l}$ , is a solution?--P.wormer 09:04, 28 June 2007 (UTC)[reply]