I have to calculate the center of the real quaternions, $\mathbb{H}$.

So, I assumed two real quaternions, $q_n=a_n+b_ni+c_nj+d_nk$ and computed their products. I assume since we are dealing with rings, that to check was to check their commutative product under multiplication. So i'm looking at $q_1q_2=q_2q_1$. When I do this, I find that clearly the constant terms are identical, so it is clear that the subset $\mathbb{R}$ is in the center. So, perhaps then that $\mathbb{C}\le\mathbb{H}$. However i ended up, after direct calculation with the following system;

$$c_1d_2=c_2d_1$$ $$b_1d_2=b_2d_1$$ $$b_1c_2=b_2c_1$$

So the determination is then found by solving this system. Intuitively, I felt that this lead to $0$'s everywhere and thus the center of $\mathbb{H}$, $Z(\mathbb{H})=\mathbb{R}$. I then checked online for some confirmation and indeed it seemed to validate my result. However, the proof method used is something I haven't seen. It was pretty straight forward and understandable, but again, I've never seen it. It goes like this;

Suppose $b_1,c_1,$ and $d_1$ are arbitrary real coefficients and $b_2, c_2,$ and $d_2$ are fixed. Considering the first equation, assume that $d_1=1$ (since it is arbitrary, it's value can be any real...). This leads to $$c_1=\frac{c_2}{d_2}$$ And that this is a contradiction, since $c_1$ is no longer arbitrary (it depends on $c_2$ and $d_2$)

I really like this proof method, although it is unfamiliar to me. I said earlier that for my own understanding, it seemed intuitively obvious, but that is obviously not proof:

1) What are some other proof methods for solving this system other than the method of contradiction used below? I was struggling with this and I feel I sholnd't be.

2) What other proofs can be found in elementary undergraduate courses that use this method of "assume arbitrary stuff", and "fix some other stuff" and get a contradiction? I found this method very clean and fun, but have never seen it used (as far as I know) in any elementary undergraduate courses thus far...

  • $\begingroup$ It doesn't appear to me that you have proven that $Z(\mathbb H)=\mathbb R$; it may help to actually write down the complete proof you have in mind. $\endgroup$ – pre-kidney Jan 1 '16 at 6:44

I am not sure where the contradiction lies exactly in your proof by contradiction. But here is another method.

An element $x\in \mathbb H$ belongs to the center if and only if $[x,y]=0$ for all $y\in \mathbb H$, where $[x,y]=xy-yx$ denotes the commutator of two elements.

We see immediately that $[x,1]=0$, whereas if $x=a+bi+cj+dk$ we have $$ [x,i]=-2ck+2dj. $$ Thus $[x,i]=0$ if and only if $c=d=0$. Similarly $[x,j]=0$ if and only if $b=d=0$. Thus the only elements $x$ which commute with both $i$ and $j$ are $x\in \mathbb R$; in particular, it follows that $Z(\mathbb H)\subset \mathbb R$. Since it is clear that $\mathbb R\subset Z(\mathbb H)$, the result follows.

Idea behind the proof: There are three special copies of the complex numbers sitting inside $\mathbb H$: the subspaces $$ \mathbb C_i=\mathbb R[i],\qquad \mathbb C_j=\mathbb R[j],\qquad \mathbb C_k=\mathbb R[k]. $$ Over $\mathbb H$, all of these subspaces are their own centers: $Z_{\mathbb H}(\mathbb C_i)=\mathbb C_i$ and so forth. Since $$\mathbb H=\mathbb C_i+ \mathbb C_j+ \mathbb C_k,$$ it follows that $Z(\mathbb H)=Z(\mathbb C_i)\cap Z(\mathbb C_j)\cap Z(\mathbb C_k)=\mathbb R$.

  • $\begingroup$ The proof i alluded to is here. Thank you for your solution here. $\endgroup$ – Iceman Jan 1 '16 at 14:48
  • $\begingroup$ Thanks for the helpful feedback $\endgroup$ – pre-kidney Jan 2 '16 at 1:03

If $a+bi+cj+dk$ is in center, then it should commute with generators $$i,j,k,\mbox{ and reals}.$$ For example, see what do we get for $(a+bi+cj+dk).i=i.(a+bi+cj+dk)$?

  • $\begingroup$ such products are easier to compute and compare, than computation of arbitrary two quaternions $\endgroup$ – p Groups Jan 1 '16 at 6:43
  • $\begingroup$ It's a good idea $\endgroup$ – Learnmore Jan 1 '16 at 7:39

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