# Could the set composed of real quaternion be represented by a Hilbert space?

Assume we have "real" quaternions: $$Q = a+bi+cj+dk$$ where $$a,b,c,d$$ are real numbers. The dot product between any two real quaternions is an inner product, and we can define the length of a quaternion $$Q$$ as $$|Q| = \sqrt{}$$.

I am still very new to the analysis part. I wonder if we can represent the set of real quaternions as Hilbert space then?

Thank you.

• When you say 'real quaternions' do you mean the 'usual' quaternions, the ones of the form $a+bi+cj+dk$? Commented Nov 8, 2019 at 22:03
• Yes, the usual quaternions with $a,b,c,d$ being real numbers. I have also edited my question. Sorry for the confusion.
– Ali
Commented Nov 9, 2019 at 18:13
• A Hilbert space is just a complete inner product space. The definition doesn't "know" about the multiplicative structure. It might as well be $\mathbb{R}^4$. So if you just slap on the usual Euclidean inner product onto $Q$ it becomes a (real) Hilbert space. Commented Nov 9, 2019 at 18:21

The key challenge is choosing a field for $$\Bbb H$$ to be a space over. (Bear in mind $$\Bbb H$$, unlike fields, isn't commutative.) One way to do this is to consider $$\Bbb H$$ a $$2$$-dimensional Hilbert space over $$\Bbb C$$ with basis $$1,\,j$$, so $$a+bi+cj+dk=(a+bi)1+(c+di)j$$ is a unique decomposition. Then $$\langle a+bi+cj+dk,\,e+fi+gj+hk\rangle:=(a-bi)(e+fi)+(c-di)(g+hi)$$is a suitable inner product. In particular$$\langle a+bi+cj+dk,\,a+bi+cj+dk\rangle=(a-bi)(a+bi)+(c-di)(c+di)=a^2+b^2+c^2+d^2,$$as expected. What we can't do, however, is use the ordinary quaternion multiplication $$q_1^\ast q_2$$, in which $$q_1$$ is conjugated, as an inner product, because IPs have to live in the field, in this case $$\Bbb C$$.