Wikipedia:Reference desk/Archives/Mathematics/2011 June 4

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June 4

Minor confusion on "if G has faithful complex irreducible representation then Z(G) is cyclic"

Hello all,

I am confronted with the following proof:

Claim: If G has faithful complex irreducible representation then the centre Z(G) is cyclic.

Proof: Let $\rho :G\to GL(V)$ be a faithful irreducible complex representation, and $z\in Z(G)$: so $zg=gz\,\forall \,v\in V$. This is a G-endomorphism on V and so is multiplication by some scalar $\tau _{z}$ say, by Schur's lemma. Then the map $Z(G)\to \mathbb {C} ^{\times },\,z\to \tau _{z}$ is a representation of Z and is faithful as $\rho$ is faithful. Thus, Z(G) is isomorphic to a finite subgroup of $\mathbb {C} ^{\times }$ and so is cyclic.

Now my question is this: where do we actually use the fact that we're working with the centre? I'm sure it's in there somewhere, but I can't spot why this only works with elements commuting with everything in G, and why we can't just apply the argument to the whole of G. Could anyone explain? Thanks very much. Mathmos6 (talk) 12:31, 4 June 2011 (UTC)

Schur's lemma only applies for z in the center. More specifically, you're using the fact that the image of z is in the center of End(V).--RDBury (talk) 21:30, 4 June 2011 (UTC)

Rotation of conic sections in polar form

Hi Reference Desk,

I have been given a conic section in polar form and I was asked to find the angle of rotation and then sketch the graph.

r = 1 / (1 - 2 cos(t) + 2 sin(t))

I rewrote r into

r = 1 / (1 - e cos(t + t'))

where e is the eccentricity

I solved for e and t' and this gives me

r = 1 / (1 - 2 cos(t + pi/4))

Does this mean the conic section is rotated anticlockwise by pi/4 or clockwise by pi/4? Does the sign in front of the cos or sin matter?

Are there any general rules for rotation of conic sections in the polar form? I have read a lot articles on rotation of conic sections but they were all written in terms of the standard form and not the polar form.

Thanks a lot! — Preceding unsigned comment added by 169.232.101.13 (talk) 22:39, 4 June 2011 (UTC)

You probably meant r = 1 / (1 - 2 √2 cos(t + pi/4)) so the eccentricity is 2 √2. In general, replacing t by t+α in a polar equation rotates the graph clockwise by α, just as replacing x by x+a in a Cartesian equation shifts left by a.--RDBury (talk) 05:49, 5 June 2011 (UTC)

Problem with Lagrange multipliers

Question:

Find the maximum and minimum values of f(x,y) = x^2 + y^2 subject to the constraint g(x,y) = x^6 + y^6 = 1

This is how I went about solving it:

grad f = t grad g (where t is a constant) 2x = 6t x^5 2y = 6t y^5

1 = 3 t x^4 --> (1/3t)^(1/4)=x 1 = 3 t y^4 --> (1/3t)^(1/4)=y

I substituted x and y into g(x,y)to determine t

(1/3t)^(3/2) + (1/3t)^(3/2) = 1 (1/3t)^(3/2) = 1/2 1/3t = (1/4)^(1/3)

Substituting 1/(3t) back into x and y

x=(1/4)^(1/12) y=(1/4)^(1/12)

To determine whether this is a maximum or a minimum, I need to use the second derivative test

fx=2x fxx=2 fxy=0

fy=2y fyy=2 fyx=0

The Hessian matrix D = 4 > 0, fxx = 2 > 0. This usually means that the point is a minimum, but the problem is that we did not even substitute x = (1/4)^(1/12) into the determinant of the Hessian matrix.

How should this problem be done? — Preceding unsigned comment added by 169.232.101.13 (talk) 22:52, 4 June 2011 (UTC)

First, you're missing a few solutions. What Lagrange really says is that the vectors (2x,2y) and (6x^5,6y^5) are linearly dependent. So solve by setting the determinant (2x)(6y^5)-(2y)(6x^5) equal to 0. This gives x=0, y=0 and x=±y, and putting these back into the original equation gives 8 solutions (0,±1), (±1,0), (±2^−1/6,±2^−1/6). Second, while there probably is a second derivative test for Lagrange multipliers somewhere, I can't find it the books I have. The Hessian only applies for unconstrained problems. In any case, the second derivative test only tells you if a point is a local maximum or minimum and the problem is asking for global. So just evaluate the function at the critical points; the ones where the values is highest are the maxima and the ones where the values is lowest are the maxima. The constraint curve is bounded so you don't have to worry about what happens at infinity.

Another way to solve this is by setting x=cos^1/3t, y=sin^1/3t. This reduces the problem is one dimension and you can use simpler methods on it. (The algebra might get more complicated though.)--RDBury (talk) 06:25, 5 June 2011 (UTC)

Question about 'a convenient way to type symbols?'

I searched on google but can't find good result, so I came here for help. The only way I type symbols like 'sets relation symbol' 'predicate logic symbol' 'very strange operators' is opening the unicode list then copy it to my sheet. Is there some trick to make this operation faster? (Also I found this 'uncommon symbols' on Wikipedia are showed as PNG pictures.)Nilman (talk) 23:00, 4 June 2011 (UTC)

The Edit tools, which appear below the edit window, have various sets including 'Math and logic'. Select that from the popup menu then click on a symbol to insert it as you type. If a symbol you use a lot doesn't appear there you can request that it's added.--JohnBlackburne^words_deeds 23:14, 4 June 2011 (UTC)

More generally there are two ways of getting symbols that don't appear on your keyboard: (1) Use something like <math> mode that allows you to use special descriptive names such as \alpha for the Greek letter α, or (2) find an input method that does what you need. (The Edit tools are an input method supplied by the Wikipedia editor.) Looie496 (talk) 23:47, 4 June 2011 (UTC)

Thank you.Nilman (talk) 11:35, 5 June 2011 (UTC)

You can also use a program like AutoHotkey to set up keyboard shortcuts to specific symbols. This has the advantage of working in (almost) any program. -Elmer Clark (talk) 15:35, 7 June 2011 (UTC)