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Centre of the ring of quaternions

  • Thread starter Wingeer
  • Start date
  • #1
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Homework Statement


What is the centre of the ring of the quaternions defined by:
[tex]\mathbf{H}=\{ \begin{pmatrix}
a & b \\
-\bar{b} & \bar{a} \end{pmatrix} | a,b \in \mathbf{C} \}[/tex]?

Homework Equations



The definition of the centre of a ring:
The centre Z of a ring R is defined by [tex]Z(R)=\{A | AX=XA, \forall X \in R\}[/tex]

The Attempt at a Solution


I figured that multiples of the 2x2 identity matrix must be in the centre.
Also if we denote an element of H by:
[tex]\begin{pmatrix} x & y \\
-\bar{y} & \bar{x} \end{pmatrix}[/tex]
where [tex]x=x_1 + ix_2[/tex] and similarly for a,b and y that:
1. [tex]b\bar{y}=\bar{b}y[/tex]
2. [tex]y(a-\bar{a})=b(x-\bar{x})[/tex]
3. [tex]\bar{b}(x-\bar{x})=\bar{y}(a-\bar{a})[/tex]

Then for instance we get from the first equation that:
[tex]b_2x_1=a_1y_2[/tex]
But I am not sure whether this approach really is any useful at all. Some hints would be greatly appreciated.
 

Answers and Replies

  • #2
76
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Anyone?
I actually have another question about the quaternions. I am asked to show that:
[tex]\mathbf{H'} = \{ a+bi+cj+dk | a,b,c,d \in \mathbf{R} \}[/tex]
with: i^2=j^2=k^2=-1, ij=k=-ji, ik=-j=-ki and jk=i=-kj.

is isomorphic as rings to the quaternions defined in the previous post.
I started by noticing that (where x,y are complex numbers):
[tex]\begin{pmatrix}
x & y \\
-\bar{x} & \bar{y} \end{pmatrix} = \begin{pmatrix}
a+bi & c+di \\
c-di & a-bi \end{pmatrix} = \begin{pmatrix}
a & 0 \\
0 & a \end{pmatrix} + \begin{pmatrix}
bi & 0 \\
0 & bi \end{pmatrix} + \begin{pmatrix}
0 & c \\
-c & 0 \end{pmatrix} + \begin{pmatrix}
0 & di \\
di & 0 \end{pmatrix}[/tex]

And so we see that every element in H is a linear combination of these matrices which all are linearly independent as well. This means we have found a basis for H.
So if we define a function f:H -> H' by:
[tex]1= f \left( \begin{pmatrix}
1 & 0 \\
0 & 1 \end{pmatrix} \right)[/tex]
[tex]i=f \left( \begin{pmatrix}
i & 0 \\
0 & i \end{pmatrix}\right)[/tex]
[tex]j=f \left( \begin{pmatrix}
0 & 1 \\
-1 & 0 \end{pmatrix}\right)[/tex]
[tex]k=f \left( \begin{pmatrix}
0 & i \\
i & 0 \end{pmatrix}\right)[/tex]

We see that obviously f is both surjective and injective as these are the only values f are defined for. Therefore f is an bijection and H and H' are isomorphic. Do I have to mix ring homomorphisms in this? Or?
 
  • #3
Dick
Science Advisor
Homework Helper
26,258
618
You've have the right general ideas there. Let's call your matrices M1, Mi, Mj and Mk. By the way, I think you've Mi wrong, check it again. You need to define f for all matrices. But that's easy just define it to be the linear map H->H' defined by your mapping of the basis elements. That you have a bijection between H and H' isn't really in question, because you've mapped the basis for a four dimensional real vector space into the basis of another one. So it's a bijection. Now you have to worry whether it's a ring homomorphism. Ring addition is not a problem just because f is linear. It's multiplication you have to check. If f(xy)=f(x)f(y). For example is f(Mi*Mj)=f(Mi)f(Mj)?
 
Last edited:
  • #4
Deveno
Science Advisor
906
6
note that it is sufficient to check the 16 possible products of M1,Mi,Mj,Mk because of linearity and the distributive laws.

your formula for Mi is indeed wrong, as the lower right coordinate is not the complex conjugate of the upper left coordinate, so that matrix isn't even in H.

*****

with regard to your first problem, note that

[a 0]
[0 a] is not in Z(H) unless a is real, because:

[x+iy .0..][0 i]....[.0.. -y+ix]
[.0.. x-iy][i 0] = [b+ia ...0..]

whereas:

[0 i][x+iy .0.]....[...0.. y+ix]
[i 0][.0.. x-iy] = [-y+ix .0..], these two matrices aren't equal unless y = 0.

remember that an element of Z(H) has to commute with ALL of H, so if you find just ONE element of H a certain matrix doesn't commute with, that matrix cannot be in the center. so i suggest you find which matrices commute with your matrices Mi, Mj and Mk.
 

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