Wikipedia:Reference desk/Archives/Mathematics/2013 February 16

= February 16 =

Quadratic roots
Is is possible to have two equal complex roots of a quadratic equation, where a,b,c are real numbers is a quadratic equation? — Preceding unsigned comment added by 117.227.197.26 (talk) 13:06, 16 February 2013 (UTC)


 * No - only complex conjugate, if they are complex. Using the standard form of the quadratic equation:
 * $$ax^2 + bx + c = 0 $$
 * where as usual a, b, c are real numbers, think about the quadratic formula solution:
 * $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
 * so for negative discriminant,
 * $$ b^2-4ac < 0$$
 * the roots are complex conjugate. M&and;Ŝc2ħεИτlk 13:25, 16 February 2013 (UTC)


 * Altenative method - if the roots are equal and both equal to k then
 * $$ax^2+bx+c=a(x-k)^2$$
 * $$\Rightarrow ax^2+bx+c=ax^2-2akx+ak^2$$
 * $$\Rightarrow b=-2ak$$
 * $$\Rightarrow k=-\frac{b}{2a}$$
 * and so k is real since a and b are real. Gandalf61 (talk) 14:10, 16 February 2013 (UTC)

Notation for representation theory of the Lorentz group
The notations D(Λ) = "representation of the Lorentz group" and (m, n) = "finite dimensional irreducible representations" seem clear enough, however the notation (say) "D(1/2, 0)" (i.e. including the superscript) is confusing...


 * My burning questions are...


 * is D(2m) = (m, 0) for half-integer m, or equivalently D(m) = (m/2, 0) for integer m? Or...
 * is D(m) = (m, 0) for integer or half-integer m?

Just alternative notations/conventions?...


 * In this paper by Gábor Zsolt Tóth (see appendix A4); letting m and n be integers:
 * D(m) = (m/2, 0) &oplus; (0, m/2),
 * $\overline{D}$(m) = (m/2, m/2),
 * D(m, n) = (m/2, n/2) &oplus; (n/2, m/2),
 * what does the last expression have for equivalent notation:
 * D(m, n) = D(?) &oplus; D(?) ?
 * while...


 * The WP article takes m, n to be half-integers in (m, n), so does that translate to
 * "D(2m, 2n) = (m, n) &oplus; (n, m)" ?
 * and this has what notation:
 * "D(m, n) = D(?) &oplus; D(?) ?


 * In all... is the statement
 * D(m, n) = (m/2, n/2) &oplus; (n/2, m/2) = "(2m  +  1)(2n  +  1)-dimensional irreducible representations of D(Λ)"
 * true?


 * Are any of the differing conventions, where to put the 1/2 factor, or just the choice of what is integer and half-integer, is "the" standard?

Thanks in advance for any and all replies. Best, M&and;Ŝc2ħεИτlk 13:23, 16 February 2013 (UTC)


 * Don't get lost in the notation. The finite dimensional real irreducible representations of the spin group of the Lorentz group all have one of the following forms (for n and m integers):
 * $$(n/2,m/2)\oplus (m/2,n/2), (n/2,n/2).$$
 * These are irreducible as real representations, which means you cannot decompose them further without breaking invariance under complex conjugation. However those of the first kind do decompose as a sum of two complex conjugate represnetations.  The source you cite calls the first kind of representation $$D^{(n,m)}$$ and the second kind $$\widetilde{D}^{(n)}$$.  The fact that these are irreducible implies that there is no decomposition of $$D^{(n,m)}$$ into a direct sum of two real representations D's.   Sławomir Biały  (talk) 14:40, 16 February 2013 (UTC)
 * It might be added that the above exhaustive list applies to O+(1;3) which includes parity inversions (the orthochronous Lorentz group). In physics SO+(1;3) (the proper orthochronous Lorentz group) is often of interest because parity is not a symmetry in every theory (see e.g. weak interactions). For SO+(1;3) the irreducible representations are of the form
 * $$(m/2,n/2)$$
 * for n and m integers.YohanN7 (talk) 16:18, 16 February 2013 (UTC)


 * Excellent explanations of the facts - if that's all there is to it (the notation and it's meaning), that's fine. Thanks (again)! M&and;Ŝc2ħεИτlk 23:43, 16 February 2013 (UTC)