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I saw that if $a+bi\in \mathbb C$, it's also an element of the split quaternions ($\mathbb P$), since $a+bi=a+bi+0j+0k$. Does this mean $\mathbb C\subset\mathbb P$? If so, does it follow that all Cayley-Dickenson Constructions are a subset of the split version of twice the dimensions?

I remember from set theory that if $(x \in A)\rightarrow(x\in B)$ then $A\subset B$, but I don't know if it applies here, since $A$ and $B$ might have different properties.

William C.
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    If you take all split quaternions of the form $a + bi$, then that set does look like exactly like the complex numbers, since the same multiplication and addition formulas hold. – Joppy Feb 20 '20 at 07:51

2 Answers2

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It's cleaner to say that there's a natural embedding $\Bbb C\hookrightarrow \Bbb P$ of the ring (field) of complex numbers to the ring of split quaternions, making $\Bbb C$ isomorphic to a subring of $\Bbb P$.

Note that this is the general pattern in any construction step of numbers: $$\Bbb N\ \hookrightarrow\ \Bbb Z\ \hookrightarrow\ \Bbb Q\ \hookrightarrow\ \Bbb R\ \hookrightarrow\ \Bbb C$$ In particular, this last step can be introduced by defining $\Bbb C:=\Bbb R[x]/(x^2+1)$, and then we can identify elements of $\Bbb R$ with the cosets of constant polynomials, thus defining an injective homomorphism $\Bbb R\to\Bbb C$, but we have no mere inclusion $\Bbb R\subseteq\Bbb C$.

Berci
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  • While the construction of $\mathbb C$ from polynomials is a common one, I would not say it is generally introduced this way. I think the far more common introduction is as pairs of real numbers with explicit formulas for addition and multiplication. – celtschk Feb 20 '20 at 12:18
  • You are right. In the latter case we only have identifications (real $x$ with pair $(x,0)$) as well. – Berci Feb 20 '20 at 16:28
  • But in split-quaternions calculating $\sqrt{j+k}$ requires introduction of the second complex unity. It would be $-\frac{(-1)^{3/4} j+(-1)^{3/4} k+(-1-i_0)}{2^{3/4}}$ where $i_0$ is not a split-quaternion. – Anixx Nov 05 '22 at 08:23
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Split-quaternions can be represented as arbitrary real $2\times2$ matrices $\left( \begin{array}{cc} a & b \\ c & d \\ \end{array} \right)$. On the other hand, complex numbers can be represented by $2\times2$ matrices of the following form: $\left( \begin{array}{cc} a & b \\ -b & a \\ \end{array} \right)$, so they are isomorphic to a subset of split-quaternions.

Anixx
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