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To construct $\Bbb C$, we consider $\Bbb R^2$ endowed with the operations: $$\begin{align} (a,b) + (c,d) &:= (a+c, b+d) \\ (a,b) \cdot (c,d) &:= (ac - bd, ad+bc)\end{align} $$ then write $(0,1_{\Bbb R}) = i$, go on writing $(a,b)$ as $a+ib$, etc. Maybe the question is silly and has a trivial explanation, but nevertheless, I'll ask: Has anyone ever tried to repeat the procedure with $\Bbb C^2$, "nesting imaginary units"? More specifically, I mean, define on $\Bbb C^2$ the operations: $$\begin{align} (z_1,z_2) + (w_1,w_2) &:= (z_1 + w_1, z_2+w_2) \\ (z_1,z_2) \cdot (w_1,w_2) &:= (z_1w_1 - z_2w_2, z_1w_2+z_2w_1)\end{align}$$ then write $(0,1_{\Bbb C}) = j$, where $j$ is another "imaginary unit" such that $j^2 = -1_{\Bbb C}$, write $(z_1,z_2) = z_1 + jz_2$ and so on? (I'm just using this subscript $\Bbb C$ for emphasis). It seems to me that we would get $i^2 = j^2$, but that wouldn't mean necessarily that $i = j$, would it? Also, would be this related in any way to the quaternions?

This, way, if we can make $\Bbb C^2$ a field, we could make any $\Bbb R^{2n}, n \in \Bbb Z$ a field, by repeating the process, no? I don't know if any problem would appear if I expanded everything in terms of the real and imaginary parts of every component. We would get some mixed terms $ij$ and $ji$. Surely I could adventure myself in the calculations, but if someone already thought of this, and it ended up being meaningless, I won't keep hitting my head on the wall. Thank you for the attention, and I hope I managed to get my idea through.

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You're essentially adjoining a new square root of $-1$ to $\Bbb C$, namely $(0,1)$. In effect you are constructing the ring $\Bbb C[x]/(x^2+1)$ (where $x$ represents this new sqrt of $-1$). By the Chinese Remainder Theorem (the one used in abstract algebra), this is isomorphic to

$$\frac{\Bbb C[x]}{((x+i)(x-i))}\cong\frac{\Bbb C[x]}{(x+i)}\times\frac{\Bbb C[x]}{(x-i)}\cong\Bbb C\times\Bbb C .$$

This is a direct product of $\Bbb C$ with itself. It's certainly not a field; it has many zero divisors (nonzero things that multiply into zero: e.g. $(a,0)(0,b)=(0,0)$ in $\Bbb C\times\Bbb C$) and four idempotents (things that satisfiy $u^2=u$). If you adjoin yet another one, you get

$$\frac{(\Bbb C\times\Bbb C)[x]}{(x^2+1)}\cong\frac{\Bbb C[x]}{(x^2+1)}\times\frac{\Bbb C[x]}{(x^2+1)}\cong\Bbb C\times\Bbb C\times\Bbb C\times\Bbb C.$$

The first isomorphism is $(a,b)+(c,d)x+(x^2+1)\mapsto (a+cx+(x^2+1),b+dx+(x^2+1))$.

Notice this new ring again has many zero divisors, so it's not a field or a division ring (like the quaternions $\Bbb H$ are), and it's also commutative (unlike $\Bbb H$) and four-dimensional over $\Bbb C$ (as opposed to $\Bbb H$, which is four-dimensional over $\Bbb R$) hence eight-dimensional over $\Bbb R$.

Frobenius theorem states that the only real division algebras over $\Bbb R$ are $\Bbb R,\Bbb C$ and $\Bbb H$, so you will not be able to get any higher-dimensional ones no matter how clever your construction.

If you alter the $2$-tuple construction of $\Bbb C$ out of $\Bbb R$ with conjugation, you get the Cayley-Dickson construction: if you apply this construction to $\Bbb C$, you get $\Bbb H$ (no longer commutative), and if you apply it to $\Bbb H$ you get the octonions $\Bbb O$ (no longer associative), and then the sedenions (no longer "alternative") and so on. Another generalization of $\Bbb R,\Bbb C,\Bbb H$ are Clifford algebras.

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  • $\begingroup$ Great answer. Thank you very much. I've got a lot to read about, now. +1 $\endgroup$ – Ivo Terek Jun 22 '14 at 4:07
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Close. To get the complexes, use real matrices $$ \left( \begin{array}{rr} a & b \\ -b & a \end{array} \right) . $$

To get the quaternions, use complex matrices $$ \left( \begin{array}{rr} \alpha & \beta \\ -\bar{\beta} & \bar{\alpha} \end{array} \right) . $$

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Yes, Hamilton and many others considered trying to do something of the sort; as it turns out we can't create an algebra structure in three dimensions, but we can do this in four. You can read about Hamilton's discovery of it here. This gives what are called quaternions: we have $i,j,k$ so that $i^2=j^2=k^2$; this gives $ij=k$ among other identities. However we lose commutativity: $ij=-ji$, for example, so this is no longer a field. They are, however, a division algebra, and an example of a skew field due to anticommutativity.

In eight dimensions there is a further construction of the octonions but we further lose associativity. We cannot do this in higher dimensions.

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  • $\begingroup$ I had saw something like this, about the dimensions once, some time ago, but I hadn't thought it would be related to what I asked, right now. I'll read about it, thank you! $\endgroup$ – Ivo Terek Jun 22 '14 at 4:09

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