Centre of the real Hamilton Quaternions H

In summary, the centre of the real Hamilton Quaternions is a field, but is not contained in the centre of H.
  • #1
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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 [itex] \star [/itex] h = i [itex] \star [/itex] ( a + bi + cj + dk)
= ai + b[itex]i^2[/itex] + cij + dik
= ai - b + ck - dj

h [itex] \star [/itex] i = ( a + bi + cj + dk) [itex] \star [/itex] i
= ai + b[itex]i^2[/itex] + 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!
 
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  • #2
That is correct. What exactly is your problem?
 
  • #3
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.
 
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  • #4
Great!

Thanks so much for the help!
 
  • #5


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.
 

FAQ: Centre of the real Hamilton Quaternions H

What is the Centre of the Real Hamilton Quaternions H?

The Centre of the Real Hamilton Quaternions H, also known as the center of the quaternion algebra H, is the set of elements that commute with all other elements in the algebra. In other words, it is the set of elements that satisfy the equation ab = ba for all a and b in H.

Why is the Centre of the Real Hamilton Quaternions H important?

The Centre of the Real Hamilton Quaternions H plays a crucial role in understanding the algebraic structure of H. It allows for the simplification of calculations and helps in identifying important subalgebras within H.

How is the Centre of the Real Hamilton Quaternions H related to the real numbers?

The Centre of the Real Hamilton Quaternions H contains the real numbers as a subalgebra. In fact, the real numbers are the only elements in H that commute with all other elements, making them the center of H.

Can the Centre of the Real Hamilton Quaternions H be extended to other quaternion algebras?

Yes, the concept of a center can be extended to other quaternion algebras. However, the center of a specific algebra may not necessarily be the same as the center of another algebra.

How is the Centre of the Real Hamilton Quaternions H related to the identity element?

The identity element, or 1, is always a member of the Centre of the Real Hamilton Quaternions H. This is because it commutes with all other elements in H, making it an element of the center. In fact, the center of H always contains all scalar multiples of the identity element.

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