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

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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∧Ŝc²ħεИ_τlk 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

See also: Talk:Representation theory of the Lorentz group § Feedback

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) ⊕ (0, m/2),

D^(m) = (m/2, m/2),

D^{(m, n)} = (m/2, n/2) ⊕ (n/2, m/2),

what does the last expression have for equivalent notation:

D^{(m, n)} = D^(?) ⊕ 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) ⊕ (n, m)" ?

and this has what notation:

"D^{(m, n)} = D^(?) ⊕ D^(?) ?

• In all... is the statement

D^{(m, n)} = (m/2, n/2) ⊕ (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∧Ŝc²ħεИ_τlk 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∧Ŝc²ħεИ_τlk 23:43, 16 February 2013 (UTC)