Centre of the real Hamilton Quaternions H

[Math Amateur]

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

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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.

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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!

 

[micromass]

That is correct. What exactly is your problem?

 

[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.

 

[Math Amateur]

Great!

Thanks so much for the help!

 

[blue_raver22]

Your approach is on the right track. To prove that the centre of the real Hamilton Quaternions is {a + 0i + 0j + 0k | a R}, we can use the fact that for any element h = a + bi + cj + dk in H, the condition for h to commute with all elements of H is that it must commute with the generators i, j, and k. As you have shown, this leads to the conclusion that h must have only one non-zero component, which is the real part a. Therefore, the centre of H is indeed {a + 0i + 0j + 0k | a R}.

To prove that {a + bi | a,b R} is a subring of H, we need to show that it is closed under addition, multiplication, and additive and multiplicative identities, and that it is a non-empty subset of H. All of these conditions can be easily verified, so we can conclude that {a + bi | a,b R} is a subring of H.

To prove that this subring is a field, we need to show that every non-zero element in this subring has a multiplicative inverse. However, for any non-zero element a + bi in this subring, we can see that its multiplicative inverse is 1/(a^2 + b^2)(a - bi). This inverse is also an element of the subring, so we have shown that every non-zero element in the subring has a multiplicative inverse, making it a field.

Finally, to show that this subring is not contained in the centre of H, we can simply choose any element that does not commute with all elements in H. For example, let's take h = 1 + i, which is an element of the subring but does not commute with j or k. This shows that the subring is not contained in the centre of H.

Overall, your approach and thoughts were correct. It is important to note that the subring is not contained in the centre because it contains elements that do not commute with all elements in H, as shown by the example of h = 1 + i.