# Centre of the real Hamilton Quaternions H

1. Jan 20, 2012

### Math Amateur

Can anyone help me with the following exercise from Dummit and Foote?

============================================================

Describe the centre of the real Hamilton Quaternions H.

Prove that {a + bi | a,b R} is a subring of H which is a field but is not contained in the centre of H.

============================================================

Regarding the problem of describing the centre - thoughts so far are as follows:

Let h = a + bi + cj + dk

Then investigate conditions for i and h to commute!

i $\star$ h = i $\star$ ( a + bi + cj + dk)
= ai + b$i^2$ + cij + dik
= ai - b + ck - dj

h $\star$ i = ( a + bi + cj + dk) $\star$ i
= ai + b$i^2$ + cji + dki
= ai - b -ck + dj

Thus i and h commute only if c = d = 0

Proceeding similarly we find that

j and h commute only if b = d = 0

and

k and h commute only if b = c = 0

Thus it seems as if I am being driven to the conclusion that the only Hamilton Quaternions that commute with every element of the ring of Hamilton Quaternions are elements of the form

a + 0i + 0j + 0k

But I am unsure of how to formally and validly argue from the facts established above to conclude this!

Can anyone help or at least confirm that I am on the right track!

2. Jan 20, 2012

### micromass

That is correct. What exactly is your problem?

3. Jan 20, 2012

### Deveno

you've proven that only quaternions of the form:

a + bi commute with i (this makes sense if you think about it, as neither j nor k commute with i).

well if q = a + bi + cj + dk is in Z(H), it has to commute with i, since i is a quaternion. so that alone narrows down the possibilities right there.

what you've done with j and k is fine, although you could also show that if a + bi commutes with j, then b = 0, and likewise for k.

this means that only real quaternions (b = c = d = 0) can possibly be in the center. it's not hard to show that all real quaternions are indeed central, which settles the matter.

EDIT: a little reflection should convince you that the center of the quaternions has to be a field. the only possibilities for the dimension (as a vector space over R) of this field is either 1 or 2 (if it was 2, it would have to be an isomorph of the complex numbers). the fact that i doesn't commute with j or k, effectively kills this possibility. it actually makes more sense to think of i,j and k being "3 identical copies of √-1", because there's no real way to tell them apart from one another. this is why H is sometimes viewed as "scalars+vectors", the "pure quaternion (non-real)" part, acts very much like a vector in R3, which was actually Hamilton's original goal-to find an "algebra" for 3-vectors.

the fact that Z(Q8), the center of the group of quaternion units, is equal to {-1,1} should reassure you that your conclusion is correct.

Last edited: Jan 20, 2012
4. Jan 20, 2012

### Math Amateur

Great!

Thanks so much for the help!