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 $|Q| = \sqrt{<Q,Q>}$. 
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.

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 $\Bbb C$.
